ch 3 -earthquake and buildings

S E C T I O N 3 E A R T H Q U A K E S A N D B U I L D I N G S

3. Earthquakes and Buildings

3.1 Earthquake or seismic analysis and design 3.1.1 Introduction Earthquakes can cause local soil failure, surface ruptures, structural damage and human deaths. The most significant earthquake effects on buildings or their structural components result from the seismic waves which propagate outward in all directions from the earthquake focus. These different types o f waves can cause significant ground movements up to several hundred miles from the source. The movements depend upon the intensity, sequence, duration and the frequency content o f the earthquake-induced ground motions. For design purposes ground motion is described by the history o f hypothesized ground acceleration and is commonly expressed in terms o f the response spectrum derived from that history. When records are unavailable or insufficient, smoothed response spectra are devised for design purposes to characterize the ground motion. In principle, the designers describe the ground motion in terms o f two perpendicular horizontal components and a vertical component for the entire base o f the structure.

When the history o f ground shaking at a particular site, or the response spectrum derived from this history is known, a building's theoretical response can be calculated by various methods; these are described later. Researchers (373-524) have carried out thorough assessments o f structures under earthquakes.

3.1.2 The world earthquake countries Table 3.1 gives the list o f countries with lethal earthquakes with average return periods. The statistics recording casualties due to earthquakes include a wide range o f events such as fires, tsunamis generated by offshore events, rock falls, landslides and other hazards triggered by earthquakes. Most large-scale disasters given by Table 3.2 due to earthquakes result from the collapse o f buildings. The number o f significant earthquakes can also be viewed in Fig. 3.1.

1 4 8 B A N G A S H

Table 3.1. The world's earthquake countries

Earthquake No. of lethal Average Earthquake No. of lethal Average

ranking earthquakes return ranking earthquakes return

1900-1996 period 1900-1996 period

(years) (years)

1 CHINA 151 0.6

2 JAPAN 78 1

3 ITALY 4 3 2

4 IRAN 8 2 1

5 TURKEY 1 1 0 1

6 C I S ( U S S R ) 4 6 2

7 PERU 53 2

8 PAKISTAN 4 8 5

9 CHILE 3 4 3

1 0 INDONESIA 4 8 2

1 1 GUATEMALA 1 4 7

1 2 INDIA 2 1 4

1 3 NICARAGUA 3 3 1

1 4 MOROCCO 2 4 6

1 5 NEPAL 4 2 3

1 6 PHILIPPINES 2 4 4

1 7 MEXICO 4 4 2

1 8 TAIWAN 4 9 2

1 9 ECUADOR 2 6 4

2 0 ARGENTINA 1 2 8

2 1 ALGERIA 2 0 5

2 2 Y E M E N 3 3 1

2 3 EL SALVADOR 1 3 7

2 4 ROMANIA 4 2 3

2 5 COSTA RICA 7 1 3

2 6 YUGOSLAVIA 1 8 5

2 7 COLOMBIA 2 9 3

2 8 AFGHANISTAN 1 1 8

2 9 U S A 3 2 3

3 0 GREECE 4 3 2

3 1 JAMAICA 2 50

3 2 BURMA 1 0 9

33 VENEZUELA 1 6 6

3 4 ALBANIA 1 4 7

3 5 PAPUA N E W

GUINEA 9 1 0

3 6 BULGARIA 6 1 5

3 7 JORDAN 2 50

38 L IBYA 2 50

3 9 N E W ZEALAND 6 1 5

4 0 UGANDA 1 1 0 0 +

4 1 LEBANON 1 1 0 0 +

4 2 PORTUGAL 4 2 3

4 3 PUERTO RICO 1 1 0 0 +

4 4 DOMINICAN

REPUBLIC 1 1 0 0 +

4 5 ZAIRE 3 3 0

4 6 ETHIOPIA 3 3 0

4 7 SOLOMON ISLANDS 2 5 0

4 8 CANADA 3 3 0

4 9 BANGLADESH 5 1 8

5 0 PANAMA 2 5 0

5 1 FRANCE 2 50

5 2 CYPRUS 2 5 0

53 SOUTH AFRICA 6 1 5

5 4 MONGOLIA 1 1 0 0 +

55 EGYPT 3 3 0

56 GHANA 1 1 0 0 +

57 TUNISIA 1 1 0 0 +

58 AUSTRALIA 2 5 0

59 MALAWI 1 1 0 0 +

6 0 CUBA 2 5 0

6 1 HAITI 2 5 0

6 2 SPAIN 4 2 3

63 BOLIVIA 4 2 3

6 4 POLAND 2 5 0

6 5 TANZANIA 2 5 0

6 6 HONDURAS 3 3 0

6 7 CZECHOSLOVAKIA 1 1 0 0 +

68 HUNGARY 2 5 0

6 9 BELGIUM 1 1 0 0 +

7 0 ICELAND 1 1 0 0 +

7 1 GERMANY 1 1 0 0 +

7 2 ISRAEL 2 -73 BURUNDI 1 -7 4 IRAQ 1 -7 5 SUDAN 1 -

7 6 SYRIA 1 -

FIGURES BASED ON THE MARTIN CENTRE DATABASE OF DESTRUCTIVE EARTHQUAKES 1 9 0 0 - 1 9 9 0 .

E A R T H Q U A K E S A N D B U I L D I N G S 1 4 9

Table 3.2. Locations of large-scale earthquakes

Earthquake Magnitude

1976 TANGSHAN CHINA 1.1 1920 KANSU CHINA 8.5

1923 KANTO JAPAN 8.3

1970 ANCASH PERU 7.8

1908 MESSINA ITALY 7.5

1927 TSINGHAI CHINA 8.0

1990 MANJIL IRAN 7.3

1939 ERZINCAN TURKEY 8.0

1 9 1 5 AVEZZANO ITALY 7.5

1939 CHILIAN CHILE 8.3

1935 QUETTA PAKISTAN 7.5

1988 ARMENIA U S S R 6.9

1976 GUATEMALA GUATEMALA 7.5

1974 CHINA CHINA 6.8

1948 ASHKHABAD U S S R 7.3

1905 KANGRA INDIA 8.6

1978 TABAS IRAN 7.7 1 9 1 7 INDONESIA INDONESIA N A

1962 BUYIN ZHARA IRAN 7.3

1968 DASHT-I BIYAZ IRAN 7.3

1960 AGADIR MOROCCO 5.9

3.1.3 The intensity scales The genealogy of the intensity scales is described in a step-by-step manner in Fig. 3.2. The most well-known intensity scales, Modified Mercalli (Richter) and Medvedev Sponheuer Karnik (MSK), are fully described in Tables 3.3 and 3.4. These tables are needed to assess the damage to structures such as buildings and their components. A guide to earthquake magnitude is given in Table 3.5.

3.2 Structural dynamics and response spectra 3.2.1 General introduction to basic dynamics A number of textbooks and research papers have been written on the subject of structural dynamics in relation to earthquakes. Various mathematical models exist. In such models the number of independent coordinates specify the position or configuration at any time. They are known as degrees of freedom. The structures can be idealized in some cases from a single-degree-of-freedom to multiple-degrees-of-freedom, depending upon the nature of the problem. Sections 3.2.2 to 3.2.5 summarize these cases. They are included so that the reader becomes familiarized with the concepts of structural dynamics associated with seismic activities.

1 5 0 B A N G A S H

Number of Significant Earthquakes (M>5.0) Recorded in each 5° x 5° Area, 1900 - 1990

Source: British Geological Survey Worldwide Earthquake Database

Note: 5° x 5° cell are not the same area worldwide, but smaller toward the poles. Accurate calculations of seismicity per unit area should take this into account.

Number of Earthquakes in 5 ' x 5* Area

1 1 - 1 0

2 1 1 - 5 0

3 5 1 - 1 0 0

4 1 0 1 - 5 0 0

5 501 - 1 0 0 0

6 1 001 - 1 5 0 0

Major Cities, Population 1990

Q >20m

O " l5 -20m

O 1 0 - 1 5 m

O 5 - 1 0 m

o 2 - 5 m

Figure 3.1. Seismicity of the world (courtesy of Cartographic Data Atlas version 1.02, British Geological Survey Worldwide Earthquake Database)

A n effective earthquake-resisting design for buildings in an area subject to seismic activity depends on the characteristics o f earthquakes, namely frequency content, intensity and duration. The type o f construction, soil conditions and a number o f other complex factors concerning dynamic modelling, load-structure-soil interactions, etc. are additional areas which need to be investigated. For primary and secondary torsional design provisions, dynamic eccentricity and associated dynamic amplification o f static eccentricity are employed in the majority o f building codes which are described later on in this section. The dynamic amplification phenomenon cannot give a complete assessment o f the effects o f torsional coupling over the full ranges o f the ratio o f torsional to lateral frequencies, the structural damping and the frequency content.


Historical Evolution of Seismic Intensity Scales

1783 Domenico Pignatoro grades seismic shocks for Italian earthquakes: "Slight...to...Violent" 1828 Egen uses grades of perceived shaking for geolographical mapping of a single event Scale 1 -6

18th-19th Century Personal Intensity Scales used by their authors as a shorthand for their own investigations e.g. Robert Mallet 1858 and 1862

1874

1878

1883

1883

Michele Stefano De Rossi

Francois Forel

-10 point scale e.g-

10 point scale

"8: ...Very Strong Shock. Fall of chimneys and cracks in buildings..

Rossi - Forel Intensity Scale (R-F) h-10 point scale e.g. "8: ...Fall of Chimneys, cracks in the walls of buildings..."

Giuseppe Mercalli 10 point scale to describe higher intensities

1880s to 1915 1888

1900

1904

1912

1917

1936

1931

1956

levels 4 and 5 of R-F scales combined and level 10 divided into two

"8: ...Partial ruin of some houses and frequent and considerable cracks in others..."

Attempts to define Absolute Intensity Scales bases on Acceleration E.S. Holden First 'Absolute Scale of Earthquake Intensity' based on acceleration (irregular values)

for Californian earthquakes Prof. Omori Absolute Intensity Scale for Japan: Seven Grades, based on shaking table studies

r Cancani Acceleration values added to Mercalli Scale, regular exponential values for 1 -10, plus two additional acceleration values for possible higher levels, 11 and 12

Plus a number of others, listed in Freeman (1930)

Mercalli Cancani Seiberg (MCS) 12 point scales

h- Two points added by -J Cancani and descriptions for them added by Seiberg

"8: ...Even though solidly constructed, houses of European architecture generallysuffer heavy damage by gaping fissures in the walls..."

MCS Scale adopted by International Seismological Association Richter's Instrumental Measurment of Magnitude supercodes intensity for comparing size of different earthquakes. Intensity takes Roman Numerals (I - XII), to distinguish from Magnitude Scale

Wood and Newmann Modified Mercalli (MM)

Charles Richter Modified Mercalli (MM-1956)

1930s-1970s

1964

1964

1976

1980

1990

— 12 point scale — For use in United States

and for modern building types

— 12 point scale — [Masonry used as indicator|

of intensity. 4 grades of masonry proposed

"VIII: ...Damage considerable to ordinary substantial building, partial collapse..."

Regional Intensity Scales

Different Scales used in different areas of the world: Europe: MCS (1912) USA: MM (1931) Japan: JMA (Based on 7 point Omori Scale, 1900) USSR: Soviet Scale (1931) 12 point scale similar to MCS China: Chinese Scale (1956) 12 point scale similar to Soviet Scale and MM

"VIII: ...Damage or partial collapse to Masonry C (Ordinary workmanship and mortar) Some damage to masonry B (Good workmanship andmortar, reinforced)..

Medvedev Sponhuer Karnik (MSK) 12 point scale

To standardise intensity assessment internationally]

and provide damage functions for vulnerability

assessment

"VIII: ...Structure type B (ordinary brick buildings) Many (about 50 %) damage degree 3 (heavy damage, large and deep cracks in walls) and single (about 5%) damage 4 (partial collapse)..."

MSK' International Intensity Scale' Offically Adopted at Unesco Intergovernmental Conference on Seismology 1 Adopts modifications suggested by Working Groups, including reservations about the existance of intensity levels XI and XII MSK Revisions 1976 (MSK-76)

1980 (MSK-81) Further working group revisions, published as MSK -1981

^ I ^ S l d S ? ^ t o S S m t 0 U P d 3 t e P r o b l e m s b e i n 9 addressed: inclusion of new building types, revision of scale tor wider application d a m a g e d j s t r i b u t i o n S ) n o n-iinearity between levels Viand VII.

Figure 3.2. The 'genealogy' of seismic intensity scales

1 5 2 B A N G A S H

Table 3.3. Modified Mercalli intensity scale

I . NOT FELT EXCEPT BY A VERY FEW UNDER EXCEPTIONALLY FAVOURABLE CIRCUMSTANCES.

I I . FELT BY PERSONS AT REST, ON UPPER FLOORS, OR FAVOURABLY PLACED.

I I I . FELT INDOORS; HANGING OBJECTS SWING; VIBRATION SIMILAR TO PASSING OF LIGHT TRUCKS; DURATION

MAY BE ESTIMATED; MAY NOT BE RECOGNIZED AS AN EARTHQUAKE.

I V . HANGING OBJECTS SWING; VIBRATION SIMILAR TO PASSING OF HEAVY TRUCKS, OR SENSATION OF A JOLT

SIMILAR TO A HEAVY BALL STRIKING THE WALLS; STANDING MOTOR CARS ROCK; WINDOWS, DISHES, AND

DOORS RATTLE; GLASSES CLINK AND CROCKERY CLASHES; I N THE UPPER RANGE OF I V WOODEN WALLS

AND FRAMES CREAK.

V . FELT OUTDOORS; DIRECTION MAY BE ESTIMATED; SLEEPERS WAKENED, LIQUIDS DISTURBED, SOME

SPILLED; SMALL UNSTABLE OBJECTS DISPLACED OR UPSET; DOORS SWING, CLOSE, OR OPEN; SHUTTERS AND

PICTURES MOVE; PENDULUM CLOCKS STOP, START, OR CHANGE RATE.

V I . FELT BY ALL; MANY FRIGHTENED AND RUN OUTDOORS; WALKING UNSTEADY; WINDOWS, DISHES AND

GLASSWARE BROKEN; KNICK-KNACKS, BOOKS, ETC., FALL FROM SHELVES AND PICTURES FROM WALLS;

FURNITURE MOVED OR OVERTURNED; WEAK PLASTER AND MASONRY D * CRACKED; SMALL BELLS RING

(CHURCH OR SCHOOL); TREES AND BUSHES SHAKEN (VISIBLY, OR HEARD TO RUSTLE).

V I I . DIFFICULT TO STAND; NOTICED BY DRIVERS OF MOTOR CARS; HANGING OBJECTS QUIVER; FURNITURE

BROKEN; DAMAGE TO MASONRY D , INCLUDING CRACKS; WEAK CHIMNEYS BROKEN AT ROOF LINE; FALL OF

PLASTER, LOOSE BRICKS, STONES, TILES, CORNICES (ALSO UNBRACED PARAPETS AND ARCHITECTURAL

ORNAMENTS); SOME CRACKS I N MASONRY C * ; WAVES ON PONDS; WATER TURBID WITH M U D ; SMALL

SLIDES AND CAVING I N ALONG SAND OR GRAVEL BANKS; LARGE BELLS RING; CONCRETE IRRIGATION DITCHES

DAMAGED.

V I I I . STEERING OF MOTOR CARS AFFECTED; DAMAGE TO MASONRY C OR PARTIAL COLLAPSE; SOME DAMAGE TO

MASONRY B * ; NONE TO MASONRY A * ; FALL OF STUCCO AND SOME MASONRY WALLS; TWISTING AND FALL

OF CWMNEYS, FACTORY STACKS, MONUMENTS, TOWERS AND ELEVATED TANKS; FRAME HOUSES MOVED

ON FOUNDATIONS IF NOT BOLTED DOWN; LOOSE PANEL WALLS THROWN OUT; DECAYED PILING BROKEN OFF;

BRANCHES BROKEN FROM TREES; CHANGES I N FLOW OR TEMPERATURE OF SPRINGS AND WELLS; CRACKS I N

WET GROUND AND ON STEEP SLOPES.

I X . GENERAL PANIC; MASONRY D DESTROYED; MASONRY C HEAVILY DAMAGED, SOMETIMES WITH

COMPLETE COLLAPSE; MASONRY B SERIOUSLY DAMAGED; GENERAL DAMAGE TO FOUNDATIONS; FRAME

STRUCTURES IF NOT BOLTED SHIFTED OFF FOUNDATIONS; FRAMES RACKED; SERIOUS DAMAGE TO

RESERVOIRS; UNDERGROUND PIPES BROKEN; CONSPICUOUS CRACKS I N GROUND; I N ALLUVIATED AREAS

SAND AND M U D EJECTED, EARTHQUAKE FOUNTAINS AND SAND CRATERS APPEAR.

X . MOST MASONRY AND FRAME STRUCTURES DESTROYED WITH THEIR FOUNDATIONS; SOME WELL-BUILT

WOODEN STRUCTURES AND BRIDGES DESTROYED; SERIOUS DAMAGE TO DAMS, DYKES AND

EMBANKMENTS; LARGE LANDSLIDES; WATER THROWN ON BANKS OF CANALS, RIVERS, LAKES, ETC.; SAND

AND M U D SHIFTED HORIZONTALLY ON BEACHES AND FLAT LAND; RAILS BEND SLIGHTLY

X I . RAILS BENT GREATLY; UNDERGROUND PIPELINES COMPLETELY OUT OF SERVICE.

X I I . DAMAGE NEARLY TOTAL; LARGE ROCK MASSES DISPLACED; LINES OF SIGHT AND LEVEL DISTORTED; OBJECTS

THROWN INTO THE AIR.

MASONRY A : GOOD WORKMANSHIP, MORTAR, AND DESIGN; REINFORCED, ESPECIALLY LATERALLY, AND BOUND

TOGETHER BY USING STEEL, CONCRETE, ETC., DESIGNED TO RESIST LATERAL FORCES.

MASONRY B : GOOD WORKMANSHIP AND MORTAR; REINFORCED, BUT NOT DESIGNED I N DETAIL TO RESIST

LATERAL FORCES.

MASONRY C : ORDINARY WORKMANSHIP AND MORTAR; NO EXTREME WEAKNESSES LIKE FAILING TO TIE I N AT

CORNERS, BUT NEITHER REINFORCED NOR DESIGNED AGAINST HORIZONTAL FORCES.

MASONRY D : WEAK MATERIALS, SUCH AS ADOBE; POOR MORTAR; LOW STANDARDS OF WORKMANSHIP; WEAK

HORIZONTALLY.


Table 3.4 (pages 153-155). Seismic intensity scale MSK-81 Medvedev-Sponheuer-Karnik 1981 Revision

Intensity degree I: Not noticeable

T H E INTENSITY OF THE VIBRATION IS BELOW THE LIMIT OF SENSIBILITY; THE TREMOR IS DETECTED AND RECORDED BY SEISMOGRAPHS ONLY.

Intensity degree II: Scarcely noticeable (very slight)

VIBRATION IS FELT ONLY BY INDIVIDUAL PEOPLE AT REST I N HOUSES, ESPECIALLY ON UPPER FLOORS OF BUILDINGS.

Intensity degree III: Weak

T H E EARTHQUAKE IS FELT INDOORS BY A FEW PEOPLE, OUTDOORS ONLY I N FAVOURABLE CIRCUMSTANCES. T H E

VIBRATION IS WEAK. ATTENTIVE OBSERVERS NOTICE A SLIGHT SWINGING OF HANGING OBJECTS, SOMEWHAT MORE

HEAVILY ON UPPER FLOORS.

Intensity degree IV: Largely observed

T H E EARTHQUAKE IS FELT INDOORS BY MANY PEOPLE, OUTDOORS BY A FEW. HERE AND THERE PEOPLE AWAKE,

BUT NO-ONE IS FRIGHTENED. T H E VIBRATION IS MODERATE. WINDOWS, DOORS AND DISHES RATTLE. FLOORS AND

WALLS CREAK. FURNITURE BEGINS TO SHAKE. HANGING OBJECTS SWING SLIGHTLY. LIQUIDS I N OPEN VESSELS ARE

SLIGHTLY DISTURBED. I N STANDING MOTOR CARS THE SHOCK IS NOTICEABLE.

Intensity degree V: Strong

Effects on people and surroundings: T H E EARTHQUAKE IS FELT INDOORS BY MOST, OUTDOORS BY MANY.

M A N Y SLEEPING PEOPLE WAKE. A FEW RUN OUTDOORS. ANIMALS BECOME UNEASY. BUILDINGS TREMBLE

THROUGHOUT. HANGING OBJECTS SWING CONSIDERABLY. PICTURES SWING OUT OF PLACE. OCCASIONALLY

PENDULUM CLOCKS STOP. UNSTABLE OBJECTS MAY BE OVERTURNED OR SHIFTED. OPEN DOORS AND WINDOWS

ARE THRUST OPEN AND SLAM BACK AGAIN. LIQUIDS SPILL I N SMALL AMOUNTS FROM WELL-FILLED OPEN

CONTAINERS. T H E VIBRATION IS STRONG, RESEMBLING SOMETIMES THE FALL OF A HEAVY OBJECT I N THE BUILDING.

Effects on structures: DAMAGE OF GRADE 1 I N A FEW BUILDINGS OF TYPE A IS POSSIBLE.

Effects on nature: SOMETIMES CHANGE I N FLOW OF SPRINGS.

Intensity degree VI: Slight damage

Effects on people and surroundings: FELT BY MOST INDOORS AND OUTDOORS. M A N Y PEOPLE I N BUILDINGS

ARE FRIGHTENED AND RUN OUTDOORS. A FEW PERSONS LOSE THEIR BALANCE. DOMESTIC ANIMALS RUN OUT OF

THEIR STALLS. I N A FEW INSTANCES DISHES AND GLASSWARE MAY BREAK. BOOKS FALL DOWN. HEAVY FURNITURE

MAY POSSIBLY MOVE AND SMALL STEEPLE BELLS MAY RING.

Effects on structures: DAMAGE OF GRADE 1 IS SUSTAINED I N SINGLE BUILDINGS OF TYPE B AND I N MANY OF

TYPE A . DAMAGE I N A FEW BUILDINGS OF TYPE A IS OF GRADE 2 .

Effects on nature: I N A FEW CASES CRACKS UP TO WIDTHS OF 1 C M ARE POSSIBLE I N WET GROUND; I N

MOUNTAINS OCCASIONAL LAND-SLIPS; CHANGE I N FLOW OF SPRINGS AND I N LEVEL OF WELL-WATER ARE OBSERVED.

Intensity degree VII: Damage to buildings

Effects on people and surroundings: MOST PEOPLE ARE FRIGHTENED AND RUN OUTDOORS. M A N Y FIND IT

DIFFICULT TO STAND. T H E VIBRATION IS NOTICED BY PERSONS DRIVING MOTOR CARS. LARGE BELLS RING.

Effects on structures: I N MANY BUILDINGS OF TYPE C DAMAGE OF GRADE 1 IS CAUSED; I N MANY BUILDINGS

OF TYPE B DAMAGE IS OF GRADE 2 . M A N Y BUILDINGS OF TYPE A SUFFER DAMAGE OF GRADE 3, A FEW OF GRADE

4. I N SINGLE INSTANCES LANDSLIPS OF ROADWAY ON STEEP SLOPES; LOCAL CRACKS I N ROADS AND STONE WALLS.

Effects on nature: WAVES ARE FORMED ON WATER, AND WATER IS MADE TURBID BY M U D STIRRED UP.

WATER LEVELS I N WELLS CHANGE, AND THE FLOW OF SPRINGS CHANGES. I N A FEW CASES DRY SPRINGS HAVE THEIR

FLOW RESTORED AND EXISTING SPRINGS STOP FLOWING. I N ISOLATED INSTANCES PARTS OF SANDY OR GRAVELLY

BANKS SLIP OFF.

1 5 4 B A N G A S H

Table 3.4. (Continued)

Intensity degree VIII: Destruction of buildings Effects on people and surroundings: GENERAL FRIGHT; A FEW PEOPLE SHOW PANIC, ALSO PERSONS DRIVING

MOTOR CARS ARE DISTURBED. HERE AND THERE BRANCHES OF TREES BREAK OFF. EVEN HEAVY FURNITURE MOVES

AND PARTLY OVERTURNS. HANGING LAMPS ARE I N PART DAMAGED.

Effects on structures: M A N Y BUILDINGS OF TYPE C SUFFER DAMAGE OF GRADE 2 , AND A FEW OF GRADE 3.

M A N Y BUILDINGS OF TYPE B SUFFER DAMAGE OF GRADE 3, AND A FEW OF GRADE 4 . M A N Y BUILDINGS OF TYPE A

SUFFER DAMAGE OF GRADE 4 , AND A FEW OF GRADE 5. MEMORIALS AND MONUMENTS MOVE AND TWIST.

TOMBSTONES OVERTURN. STONE WALLS COLLAPSE.

Effects on nature: SMALL LANDSLIPS ON HOLLOWS AND ON BANKED ROADS ON STEEP SLOPES; CRACKS I N

GROUND UP TO WIDTHS OF SEVERAL CENTIMETRES. N E W RESERVOIRS COME INTO EXISTENCE. SOMETIMES DRY

WELLS REFILL AND EXISTING WELLS BECOME DRY. I N MANY CASES CHANGE I N FLOW AND LEVEL OF WATER OR WELLS.

Intensity degree IX: General damage to buildings Effects on people and surroundings: GENERAL PANIC; CONSIDERABLE DAMAGE TO FURNITURE. ANIMALS

RUN TO AND FRO I N CONFUSION AND CRY.

Effects on structures: M A N Y BUILDINGS OF TYPE C SUFFER DAMAGE OF GRADE 3, AND A FEW OF GRADE 4 .

M A N Y BUILDINGS OF TYPE B SHOW DAMAGE OF GRADE 4 , A FEW OF GRADE 5. M A N Y BUILDINGS OF TYPE A

SUFFER DAMAGE OF GRADE 5. MONUMENTS AND COLUMNS FALL. RESERVOIRS MAY SHOW HEAVY DAMAGE.

I N INDIVIDUAL CASES RAILWAY LINES ARE BENT AND ROADWAYS DAMAGED.

Effects on nature: O N FLAT LAND OVERFLOW OF WATER, SAND AND M U D IS OFTEN OBSERVED. GROUND CRACKS

TO WIDTHS OF UP TO 1 0 C M , I N SLOPES AND RIVER BANKS MORE THAN 1 0 CM; FURTHERMORE A LARGE NUMBER OF

SLIGHT CRACKS I N GROUND; FALLS OF ROCK, MANY LANDSLIDES AND EARTH FLOWS; LARGE WAVES ON WATER.

Intensity degree X: General destruction of buildings Effects on structures: M A N Y BUILDINGS OF TYPE C SUFFER DAMAGE OF GRADE 4 , A FEW OF GRADE 5. M A N Y

BUILDINGS OF TYPE B SHOW DAMAGE OF GRADE 5, MOST OF TYPE A COLLAPSE. D A M S , DYKES AND BRIDGES

MAY SHOW SEVERE TO CRITICAL DAMAGE. RAILWAY LINES ARE BENT SLIGHTLY. ROAD PAVEMENT AND ASPHALT

SHOW WAVES.

Effects on nature: I N GROUND, CRACKS UP TO WIDTHS OF SEVERAL DECIMETRES, SOMETIMES UP TO 1 METRE.

BROAD FISSURES OCCUR PARALLEL TO WATER COURSES. LOOSE GROUND SLIDES FROM STEEP SLOPES. CONSIDERABLE

LANDSLIDES ARE POSSIBLE FROM RIVER BANKS AND STEEP COAST. I N COASTAL AREAS DISPLACEMENT OF SAND AND

M U D ; WATER FROM CANALS, LAKES, RIVER, ETC. THROWN ON LAND. N E W LAKES OCCUR.

Intensity degree XI: Catastrophe Effects on structures: DESTRUCTION OF MOST AND COLLAPSE OF MANY BUILDINGS OF TYPE C . EVEN WELL-

BUILT BRIDGES AND DAMS MAY BE DESTROYED AND RAILWAY LINES LARGELY BENT, THRUSTED OR BUCKLED;

HIGHWAYS BECOME UNUSABLE; UNDERGROUND PIPES DESTROYED.

Effects on nature: GROUND FRACTURED CONSIDERABLY BY BROAD CRACKS AND FISSURES, AS WELL AS BY

MOVEMENT I N HORIZONTAL AND VERTICAL DIRECTIONS; NUMEROUS LANDSLIDES AND FALLS OF ROCK. T H E

INTENSITY OF THE EARTHQUAKE REQUIRES TO BE INVESTIGATED ESPECIALLY.

Intensity degree XII: Landscape changes Effects on structures: PRACTICALLY ALL STRUCTURES ABOVE AND BELOW GROUND ARE HEAVILY DAMAGED OR

DESTROYED.

Effects on nature: T H E SURFACE OF THE GROUND IS RADICALLY CHANGED. CONSIDERABLE GROUND CRACKS

WITH EXTENSIVE VERTICAL AND HORIZONTAL MOVEMENT ARE OBSERVED. FALLS OF ROCK AND SLUMPING OF RIVER

BANKS OVER WIDE AREAS; LAKES ARE DAMMED; WATERFALLS APPEAR, AND RIVERS ARE DEFLECTED. T H E INTENSITY

OF THE EARTHQUAKE REQUIRES TO BE INVESTIGATED ESPECIALLY.


Table 3.4. (Continued)

Type of structures (building not anti-seismic) A BUILDINGS OF FIELDSTONE, RURAL STRUCTURES, ADOBE HOUSES, CLAY HOUSES.

B ORDINARY BRICK BUILDINGS, LARGE BLOCK CONSTRUCTION, HALF-TIMBERED STRUCTURES, STRUCTURES OF

HEWN BLOCKS OF STONE.

C PRECAST CONCRETE SKELETON CONSTRUCTION, PRECAST LARGE PANEL CONSTRUCTION, WELL-BUILT WOODEN

STRUCTURES.

Classification of damage to buildings GRADE 1 : SLIGHT DAMAGE: FINE CRACKS I N PLASTER; FALL OF SMALL PIECES OF PLASTER.

GRADE 2 : MODERATE DAMAGE: SMALL CRACKS I N WALLS; FALL OF FAIRLY LARGE PIECES OF PLASTER; PANTILES

SLIP OFF; CRACKS I N CHIMNEYS; PARTS OF CHIMNEYS FALL DOWN.

GRADE 3: HEAVY DAMAGE: LARGE AND DEEP CRACKS I N WALLS; FALL OF CHIMNEYS.

GRADE 4 : DESTRUCTION: GAPS I N WALLS; PARTS OF BUILDINGS MAY COLLAPSE; SEPARATE PARTS OF THE

BUILDINGS LOSE THEIR COHESION; INNER WALLS AND FILLED-IN WALLS OF THE FRAME COLLAPSE.

GRADE 5: TOTAL DAMAGE: TOTAL COLLAPSE OF BUILDINGS.

For the design o f buildings three approaches are specified and are given below:

{a) Seismic coefficient method Various codes recommend this method. Flexibility or stiffness methods are employed. Specified shear distribution as cubic parabola in various building storeys is recommended. This approach is adopted for buildings no more than 40 m high.

(b) Modal analysis Modal analysis using average acceleration spectra and a multiplying factor is adopted for buildings within the range o f 40 to 90 m height. The response spectra are given in this section. Direct integration method is recommended as a method o f solving various differential equations.

(c) Detailed dynamic analysis Here the building structure is dynamically analysed using actual earthquake accelerogram and time-wise integration o f the dynamic response for tall buildings o f more than 90 m. The response spectra can be elastic or inelastic. Finite element and other methods associated with static and dynamic condensation techniques are employed to solve for large building structures located in seismic zones.

A greater detail o f such methods is given in Part B o f the book, Methods o f Analysis, and in the Appendix.

Design for drift and lateral stability are important and should be addressed in the early stages o f the building design development. The concept o f lateral stability and its relationship to drift and the P — A effect has been codified. In an

1 5 6 B A N G A S H

Table 3.5. A guide to earthquake magnitude

Magnitudes less than 4.5

MAGNITUDE 4.5 REPRESENTS AN ENERGY RELEASE OF ABOUT 1 0 8 KILOJOULES AND IS THE EQUIVALENT OF ABOUT

1 0 TONS OF T N T BEING EXPLODED UNDERGROUND. BELOW ABOUT MAGNITUDE 4 .5 , IT IS EXTREMELY RARE FOR

AN EARTHQUAKE TO CAUSE DAMAGE, ALTHOUGH IT MAY BE QUITE WIDELY FELT. EARTHQUAKES OF MAGNITUDE 3

AND MAGNITUDE 2 BECOME INCREASINGLY DIFFICULT FOR SEISMOGRAPHS TO DETECT UNLESS THEY ARE CLOSE

TO THE EVENT. A SHALLOW EARTHQUAKE OF MAGNITUDE 4.5 CAN GENERALLY BE FELT FOR 5 0 TO 100 K M FROM

THE EPICENTRE.

Magnitudes 4.5 to 5.5 - local earthquakes

MAGNITUDE 5.5 REPRESENTS AN ENERGY RELEASE OF AROUND 1 0 9 KILOJOULES AND IS THE EQUIVALENT OF ABOUT

1000 TONS OF T N T BEING EXPLODED UNDERGROUND. EARTHQUAKES OF MAGNITUDE 5.0 TO 5.5 MAY CAUSE

DAMAGE IF THEY ARE SHALLOW AND IF THEY CAUSE SIGNIFICANT INTENSITY OF GROUND SHAKING I N AREAS OF

WEAKER BUILDINGS. EARTHQUAKES UP TO MAGNITUDES OF ABOUT 5.5 CAN OCCUR ALMOST ANYWHERE I N THE

WORLD - THIS IS THE LEVEL OF ENERGY RELEASE THAT IS POSSIBLE I N NORMAL NON-TECTONIC GEOLOGICAL

PROCESSES SUCH AS WEATHERING AND LAND FORMATION. A N EARTHQUAKE OF MAGNITUDE 5.5 MAY WELL BE FELT

100 TO 2 0 0 K M AWAY.

Magnitudes 6.0 to 7.0 - large magnitude events

MAGNITUDE 6 REPRESENTS AN ENERGY RELEASE OF THE ORDER OF 1 0 1 0 KILOJOULES AND IS THE EQUIVALENT OF

EXPLODING ABOUT 6 0 0 0 TONS OF T N T UNDERGROUND. A MAGNITUDE 6.3 IS GENERALLY TAKEN AS BEING

ABOUT EQUIVALENT TO AN ATOMIC B O M B BEING EXPLODED UNDERGROUND. A MAGNITUDE 7.0 REPRESENTS AN

ENERGY RELEASE OF 1 0 1 2 KILOJOULES. LARGE MAGNITUDE EARTHQUAKES, OF MAGNITUDE 6.0 AND ABOVE ARE

MUCH LARGER ENERGY RELEASES ASSOCIATED WITH TECTONIC PROCESSES. I F THEY OCCUR CLOSE TO THE SURFACE

THEY MAY CAUSE INTENSITIES AT THEIR CENTRE OF V I I I , I X OR EVEN X , CAUSING VERY HEAVY DAMAGE OR

DESTRUCTION IF THERE ARE TOWNS OR VILLAGES CLOSE TO THEIR EPICENTRE. SOME OF THESE LARGE MAGNITUDE

EARTHQUAKES, HOWEVER, ARE ASSOCIATED WITH TECTONIC PROCESSES AT DEPTH AND MAY BE RELATIVELY

HARMLESS TO PEOPLE ON THE EARTH'S SURFACE. THERE ARE ABOUT 2 0 0 LARGE MAGNITUDE EVENTS SOMEWHERE

I N THE WORLD EACH DECADE. A MAGNITUDE 7.0 EARTHQUAKE AT SHALLOW DEPTH MAY BE FELT AT DISTANCES

5 0 0 K M OR MORE FROM ITS EPICENTRE.

Magnitudes 7.0 to 8.9 - great earthquakes

A MAGNITUDE 8 EARTHQUAKE RELEASES AROUND 1 0 1 3 KILOJOULES OF ENERGY, EQUIVALENT TO MORE THAN 4 0 0

ATOMIC BOMBS BEING EXPLODED UNDERGROUND, OR ALMOST AS MUCH AS A HYDROGEN BOMB. T H E LARGEST

EARTHQUAKE YET RECORDED, MAGNITUDE 8.9 , RELEASED 1 0 1 4 KILOJOULES OF ENERGY. GREAT EARTHQUAKES ARE

THE MASSIVE ENERGY RELEASES CAUSED BY LONG LENGTHS OF LINEAR FAULTS RUPTURING I N ONE BREAK. I F THEY

OCCUR AT SHALLOW DEPTHS THEY CAUSE SLIGHTLY STRONGER EPICENTRAL INTENSITIES THAN LARGE MAGNITUDE

EARTHQUAKES BUT THEIR GREAT DESTRUCTIVE POTENTIAL IS DUE TO THE VERY LARGE AREAS THAT ARE AFFECTED BY

STRONG INTENSITIES.

earthquake area the relative displacement o f buildings is measured by an overall drift ratio or index. This ratio is either o f the following:

(i) the ratio o f the relative displacement o f a particular floor to the storey height at that level;

(ii) relative displacement between two adjacent floors.


The drift index is the drift divided by the storey height. Stability analysis sections and the Appendix provide details for drift, lateral

stability and P — A . The limitations o f these are given by various codes briefly described in Sections 3.2.6.1 to 3.2.6 .14.

3.2.2 Single-degree-of-freedom systems - undamped free vibrations 3.2.2.1 The mathematical models A mass M is suspended by a spring having a linear constant K (the force necessary to cause unit change o f length, also called spring stiffness; units: lbf/in or N/mm). Horizontal constraints are assumed, so that m can only be displaced vertically (Fig. 3.3):

W - KASTAT = 0

The coordinate x defines the position o f mass M at any time; positive 6 is taken downwards, with the same sign convention for forces, accelerations, etc.

3.2.2.2 Setting up the differential equation (a) By Newton's second law of motion The magnitude o f the acceleration o f a particle is proportional to the resultant force acting upon it and has the same direction and sense as this force:

— = x (velocity) at

^-^ = x, (acceleration)

W = KA. STAT ( 3 . 1 )

(Fig. 3 . 4 ) .

d ^ = -K(ASTAT + 6)+W (3.2)

Unstrained position

K A 'STAT.

Equilibrium position M M

Figure 3.3.

Free Body Diagram - T W = Mg

The mass and its equilibrium position

158 B A N G A S H

>K Equilibrium position

/////

Equilibrium position M Dynamic position

K

A

M

Dynamic Free Body

Complete Free Body * W = Mg

Figure 3.4. Displacement of a mass from the equilibrium position and released

or

x + Kx = 0 (3.3)

(b) By energy method For a conservative system, the total energy o f the system (potential and kinetic energy) is unchanged at all times.

KE + PE = constant

or

= (KE + PE)=0 dt

KE = \Mx2

•0 f 0 [ W - K(ASTAT + x)} dx = - Kxdx =

x Jx

PE =

d (Mx1 Kx2

Kx1

(3.4)

(3.5)

(3.6)

dt 0 or Mx + Kx = 0

The motion defined is said to be harmonic, because o f its sinusoidal form. The motion is repeated, with the time for one cycle being defined by the value ut equals 2ir. The period T or time for one cycle is:

nn 2 7 T T = —

u

(3.7)

The reciprocal o f T is called frequency f (in cycles per unit time):

f = —

The term u> is called circular frequency, and it is measured in radians per second:

M

Kq W V A S T A X

^ = 32.2ft/s 2 or 386in/s 2 or 9.81 m/s 2

A S T A T = static deflection o f the system

(3.8)

(3.9)


Remember to use consistent units when calculating u or / . The last o f the expressions (3.9) seems to be the most convenient in structural applications.

The velocity x and the acceleration x are expressed by time derivatives o f (3.6) and (3.7). Various forms are obtained:

velocity:

x = x0 cos out — X0LJ sin ut

= Xu cos(UTF + (/)) — Xu sin (out + </> + TT/2)

= Xcos(ujt + (j>) amplitude o f velocity:

X = Xu acceleration:

x = —XQUJ sin ut — XQU2 sin ujt

= —Xu? sin(cj; + (/>)= Xu? sin (ut + (j) + TT)

= — u?x — — Xsm(ujt + (f)) amplitude o f acceleration:

(3.10)

(3.11)

X = Xuz

The velocity is u times the displacement and leads it by 90°. The acceleration is u? times the displacement and leads it by 180°.

col = 2k

col = 2k

Figure 3.5. Displacement, period, velocity and accelerations

1 6 0 B A N G A S H

Displacement, velocity, acceleration against ut diagrams are shown in Fig. 3.5.

The phase angle indicates the amount by which each curve is shifted ahead, with respect to an ordinary sine curve:

The rotating vector concept: phase-plane solutions The displacement x is given according to Figs 3.6 and 3.7. Three vectors A, B, and C whose relative positions are fixed rotate with angular velocity u. Their angular position at any time t is ut. A and B are at right angles and C leads A by the phase angle </>. A graphical representation is obtained by vertically projecting the vectors onto the x + ut graph.

X = Xu (3.12a)

(3.12b) X = Xu = Xu2

u (3.13a)

C = VA2 + B2 = X)

K_

2

Figure 3.6. The rotating vector

X, X, X X, X,X

X = XG}^'~

X = X(q'

Figure 3.7. Rotating displacements, velocities and accelerations


The velocity and acceleration curves, as well as the displacement can be generated in this manner, using rotating vectors X, X, X. The velocity and acceleration vectors lead the displacement by 90° and 180°, respectively. Their relative position is fixed and they rotate with angular velocity oj:

x — X sin ut = — smut U

x = Xu cos cut = X cos cut

x0 = 0\ (3.13b)

x = u - X sin ut = —X sin ut

3.2.3 Summary and conclusions 3.2.3.1 Example A beam with uniformly distributed mass m, and unit length, supporting a concentrated mass, M, at midspan is illustrated in Fig. 3.8.

Making the assumption that the dynamic deflection curve is o f the same shape as that due to the concentrated load acting statically on the beam, the vertical displacement at a distance x from the left support is

3xLl - 4x3

y = yc (3.14)

max KE (due to distributed mass)

3xL2 - 4x3 \ 2 L/2

0 2g

= l-luLyl

35 2g

dx

max K E (due to concentrated load)

max SE = i Kyi total energy = KE + SE = constant

IW

27 yl

1 . 2 {W + ^ujL) 1

2y'c

g + = constant

dt = y< W + ^luL

^—h+Ky(

g

(3.15)

(3.16)

(3.17)

(3.18)

(3.19)

distributed mass/unit length

W

Figure 3.8. Uniform mass on a beam

1 6 2 BANGASH

This expression is equivalent to Eq. (3.7) except that ±Z o f the total mass o f the beam is lumped with the concentrated mass. The natural frequency is

f 2^\jw + %ujL~2n{^T

( 3 ' 2 0 )

where A S X A X is the static deflection under the concentrated load due to a total load o f W + J$OJL acting at the same point.

3.2.3.2 Estimate of the fundamental frequency of multi-mass systems by Rayleigh's method

The mass o f the beam or structure is assumed negligible, or is 'lumped' as in Eq. (3.20).

Assume again that the dynamic amplitude o f the loads W\, W2^ ... , Wn are approximated by the static deflections o f the structure yx, y2, ... , yn at the load points. Substitute into Eq. (3.16), remembering that the amplitude o f the velocity is

yt = uy

max SE = ^ wiyi by Clapeyren's theorem (3.21)

max K E =

= E

Miy

2

at the equilibrium position

E 2 g 2

Wtyi = u? Wty2

^ 2 g ^ 2

Hence

I Vg< /E Wjyt

E ^ J 2

Vg< \Wiyi + W2y2 + Wly\+W2y1

2 +

(3.22)

(3.23)

(3.24)

The above provides a powerful method for the determination o f the estimate o f the fundamental frequency o f many structural systems o f practical importance:

( i) trusses; the masses and loads are lumped in the joints. The statical deflections may be in this case conveniently found from a Williot diagram (Fig. 3.9), or by calculation. The vertical deflections will usually be o f interest.


Wt of beam W = Mg per unit of length

W , = M,g W 2 = M 2 g W 3 = M 3 g

i yi rya

static deflection line of beam

Figure 3.9. Masses and static deflections

(ii) The shear building. Multi-storey buildings, where the assumption is made that the floor construction is rigid, and the mass of the beams and loads carried can be lumped at the floor levels. The mass of the columns is usually neglected. In this case horizontal deflections at the floor levels will normally be required (Fig. 3.10).

An exact frequency analysis is much more laborious. When the distributed mass of the structure is not quite negligible, the 'static

deflection method' can still be used, with some modifications, compared with Eq. (3.24):

max SE of distributed mass (= PE)

max KE of distributed load

- i

L Wydx o 2

L Mfdx o 2

[L Mu?y2dx

2 r w Jo g

W fdx (3.25)

w,

(a)

Figure 3.10. Lump masses

w,

(b)

164 B A N G A S H

SE and K E for loads Ware as in Eqs (3.21) and (3.22), hence:

1 ^ 1 f 1 u? _ , u? [L , -2Y,Wiyi + - \ o u y d X = -Y,Wiyl + -\QUyl«X

Eg JWDX + £ Weft

(3.26)

(3.27)

/ = 2 ^

The very practical formula now occurs throughout vibration literature in the symbolic forms:

( U 1 ' 19.65 * f

3.127 • M w\ KN) k 157.7 , V

cycles per sec

3.2.4 Multi-degree-of-freedom system 3.2.4.1 Systems of masses, springs and dashpots Consider the following equation grouping:

m\X\ + C\X\ + kx X\ + c2{x\ — x2) + k2(xi — x2) = P\ (t)

m2x2 + c2(x2 -Xi) + k2(x2 - x{) + c3(x2 - x3)

+ k 3 ( x 2 - x 3 ) = P2(t)

mNxN + c N ( x N — xN_\) + k N ( x N — xN_\) = Px(t)

(3.28)

(a)

>(b)

> (c)

(3.29)

Changes due to the multi-degree system to be different from previous other degrees o f freedom (see Fig. 3.11).

These equations may be arranged in the matrix form

Mx + Cx + Kx = Pi (t) • • • = P(t) (3.30)

A

P,(t) P 2 ( t ) P 3 ( t )

cv JT% XT'? JT ki

+

p N ( t )

v77//////?y//////77/

--VWVH

^77777777777777777777777^

| *" x 1 | *" x 2 | *" x 3

Figure 3.11. Multi-degree-of-freedom system

EARTHQUAKES AND BUILDINGS 1 6 5

where

mass matrix M =

M i

Mo

M N

diagonal type (3.31)

all off-diagonal terms = 0

damping matrix C —

c x + c 2 - c 2

- c 2 c 2 + c 3 - c 3

- c N CN

(3.32)

stiffness matrix K

k i + k 2 - k 2

- k 2 k 2 + k 3 - k 3

—kjsf k t f j

displacement vector x =

xN )

(3.33)

load vector P(i) —

p N ( t )

(3.34)

For examples when all masses move in phase with each other

x = X sin(UTF + (j))

For a single degree o f freedom Mx + Kx — 0, it is written as

[K - u?M)X = 0

The evaluation will be the determinant o f [K — u?M] which is

\K - u?M\ - 0

(3.35)

(3.36)

(3.37)

Because o f the phase difference (j)N for all N masses, the existence o f orthogonality for all normal modes o f vibrations exist. The normalized mode shapes would be denoted by (j>N for N massed structures:

1 6 6 B A N G A S H

f <\>\n ' ' ' h n

1 ? ; T 7 — S < h n

1 ? ; T 7 — S

^max

, ^Nn >

(3.38)

The free vibration may be written as:

[K<t>n-u?nM<l>n]

K<f>„ = ojIM4>„ = qn

where

(3.39)

(3.40)

<\>n — the «th mode shape

qn = the inertial force = u?nM<\)n

K(j)n = stiffness equilibrium condition required when these forces are applied as loads on structures.

3.2A.2 Betti's law When the structure is subjected to alternate load systems, the work done by the first set o f loads when moved through the displacements caused by the second set o f loads are identical to the work done by the second set o f loads when moved through the displacements caused by the first set o f loads. Let us say for 'm' mode, the inertial force will be qm and the corresponding deflection § m

according to Betti's law

<t>l<ln = <f>n9m (3.41)

Shape factors must be transposed in order to carry out the appropriate matrix product

.2jJ uA

n(j)l

mM(j)n = u?m(^nM(\)n (3.42)

or

(ujI - oj2

m)4>lM(f>n = 0 (3.43)

3.2.4.3 If any two frequencies are not equal Orthogonality conditions

<f>mM<l)n — 0 um-=£wn (first orthogonality) (3.44)

A second orthogonality condition by premultiplying Eq. (3.44) by cp^

ct>lKct>n = uUlM<t>n (3.45)

EARTHQUAKES AND BUILDINGS 167

Because o f Eq. (3.44) conditions

4>mK(j)n = 0 u m ^ u n (second orthogonality) (3.46)

The two orthogonality conditions, Eq. (3.44) and (3.46), have important considerations when eigenvalue solution methods are adopted and dynamic response is evaluated.

Standard eigenvalue The eigenvector o f a matrix is the vector when premultiplied by the matrix that yields another vector which is proportional to the original vector. This constant of proportionality is called eigenvalue and is expressed as

AX = Xx (3.47)

where A is the square matrix o f order JV and Xis the eigenvector o f order JV. X is the corresponding eigenvalue where

(A-\I)X = 0 (3.48)

lis the identity matrix o f the same order as A. In case o f free vibration, the eigenvalue problem is represented by

KX = XMX (3.49)

in which A = u2. The eigenvalues are both real and positive and eigenvalues can similarly be

obtained using the orthogonality given above. Just take for Xm, Xn (eigenvectors), the corresponding values Xm and Xn and equations are solved for Xm 7^ A„, m ^ n, etc. Hence

\ _ XmKXm

X m ~ XlMXm

The mass orthonormalization for scaling will be

tlMfa = 1 (3.50)

Superposition The basic eigenvectors o f JVth order as in JV-dimensional space can be represented by the superposition o f N independent vectors also o f order JV. The superposition can be expressed as:

m=l

where

£ = a vector o f order JV

1 6 8 B A N G A S H

Xm is as defined above and Wm is a weighting factor

s* = x w m

N

The value of

w m = ( x y l s s

Since Xm vectors are independent, must exist and can be equal to (X)T. This mean the mode shapes obtained are also independent and are used in the representation of vectors 6S. Hence

S8 = WXXX + W2X2 + • • • + WNXN

— </>m\Xl + (/>m2^2 H f" ^mN^N

Wm is written as

XlMXm

Because <p„M4>m = 1, Wm is given by

Wm = $M6s for m= 1, 2, 3, . . . , N

Rayleigh quotient Let K, M and 6S be the stiffness and mass matrices of order N and any arbitrary vector of order N. The scalar factor 5 f , known as the Rayleigh quotient is written as

_ 6TK6 {~6TMS

If 8 = $W then:

N

8TM8 = (W)r$TM$W = iWm)2

m=l

8TK6=(W)T$TK$W

= (W)TAW = JT Xm{Wmf m=\

where $ is a set of normalized modes.


The value o f Sf is given by

s = A i ( ^ ) 2 + X2(W2)2 + • • • + \N{WNf f {WX)2 + {W2)2 + --- + (WN)2

W\ < sf

When a set o f arbitrary vectors 6 are orthogonal to the first s - 1 eigenvectors

(5 T M0 W - 0 /w = 1, 2, . . . , j - 1 (3.51)

and when 5 is in expansion

^ m = 0 ro = 1 , 2 , . . . , 5 - 1 (3.52)

the arbitrary vector 6 used in obtaining or developing the expression for 5 f , the eigenvectors were o f approximate characteristics. The deviation from the true eigenvectors by e6 (e2, taken out) can be written as

6 = <l>m + e6 (3.53)

where

4>m = rath eigenvector

e6 = contribution from all other eigenvectors to 8

6 = J2 (3.54)

Numerator of the value of Sf

STKS = (cj>l + e6T)K(<j>m + eS) (3.55)

Using the orthogonality principle described above

cf>lM8=(6)TKcl)m = 0

<t>T

mK6^(8)TMc/>m^0 ) (3.56)

T T <t>mK<t>m ~ Xm(/)mM(j)m = K,

then substituting into Eq. (3.55)

6TK6 = Xm + e2(a2

lXl + a2

2X2 + ••• + a2

NXN) (3.57)

Denominator of Sf

6TM6 = <f>lM<t>m + e(j)lM6 + e(S)TM(j>m + e2(6)TM6

= 1 + e2{a\ + (£ + ...+ c?N) (3.58)

170 BANGASH

Hence

Sf = Xm + e2{a\{Xx - \n) + « 2 ( A 2 - A w ) + • • • + a2

N(XN - A m ) } (3.59)

3.2.5 Dynamic response of mode superposition Multi-degree-of-freedom system is described by:

Mx + Cx + Kx = F = P (3.60)

As stated, mode superposition is based upon the fact that the deflected shape o f the structure may be expressed as a linear combination o f all the modes:

N

X = faXi + (f>2X2 + <j)3X3 + • • • + (f>NXN = fa*" (3-61) n=\

Xn are the modal amplitudes varying with time. The motion o f the /th mass is

I * I = <l>i\X\ + 4>aX2 H + <t>m^n H 1~ <t>iNXN

where

cj)in = displacement o f the rth mode o f vibration

Equation (3.61) may be written as

x = $ A

where $ is the modal matrix whose columns are the mode shapes so that

$ = { $ 2 $ 2 $ h (- <£ }

X is a vector o f the modal amplitudes

XT = {X\, X2, ... , ... , XN}

They are always in generalized coordinates where x are natural coordinates. Modal analysis is a process o f decomposition. [Eq. (3.60)] using generalized

coordinates so as to obtain a set o f differential equations that are uncoupled, each o f which may be analysed as a single degree o f freedom. Substituting Eq. (3.61) into Eq. (3.60) gives

N N N

M^$nXn + C^2 $nXn+Kj2 ®Xn = P (362) n—l n—l n=l

<&w shape functions are independent o f time. Premultiplying Eq. (3.62) by $ m

for the rath case o f mode

N N N

&mM + 3nC J2 M» + *lK £ * » ^ » = (3-63) «=1 n—l n—l


The first term can be expanded to

(3.64)

Referring to the first or orthogonality conditions every term in this series will be zero except for the case where n = m

N <J>T $ F = <f> M$> X = M X (3.65)

71=1

where Mm is a generalized mass o f the rath mode. The reasoning applies to the stiffness term

N

*T

mK Y, <t>n*n = <t>lMcj>mXm = KmYm (3.66)

where Km is the generalized stiffness. The generalized force Qm will be

Qm = ®lP (3.67)

Damping C is a problem Assume C is proportional to either M or K or a combination o f both. The same effect o f decomposition would be achieved

N

(3.68)

The damping ratios in the N modes are written as f 1 ? £2> • • •» £N a n d the transformed damping matrix is written as

C T R A ~ <t> Ccj) =

'2£iu>i 0

0 2£2u2

0 0

0

(3.69)

where

Q R A = transformed damping matrix.

Then

(3.70)

1 7 2 BANGASH

The values o f </> 1 (obtained from orthogonality condition) = cj)TM:

<j?M(j) = /

Hence

C = M(j)CTRA<j>TM (3.71)

or

C = M fe 2 & J # T > ) M (3.72)

Equation (3.60) can be written in the form

MmXm + CmXm + KmXm = Pm m = 1, 2, 3, . . . , N (3.73)

When a system is assumed to possess proportional damping, the mode superposition method is used. The damping matrix would then be diagonalized by a normal coordinate system. Let the damping matrix C be written as

C = a0M (3.74)

where a 0 is a constant o f proportionality. Premultiply Eq. (3.74) by </3 and post-multiply by cj)m

<t>liC(j)m = a^T

mM^)m = 0 m + n (3.75)

or

Cm = 2^mLOm — <£mC(j)m

a0 = 2£mum (3.76)

= ^ (3.77)

If it is decided to choose a stiffness proportional damping rather than a mass proportional damping, the damping matrix C can be written as

C = axK

4>mC<l)m = 2£mu;m = a0 + axu?m (3.78)

<t>mC<j)n = 2£mu;n = a0 + axu?n (3.79)

A more generalized equation would be

C = a0M + axK + a2M~lKM (3.80)

Damping that leads to satisfying Eq. (3.80) is the Rayleigh damping.

E A R T H Q U A K E S A N D B U I L D I N G S 173

Orthogonality condition is used

4>mC(f)m = 2£mVm = <*0 + a\um + a2^m

<j)JC(/)p = 2£pwp = c*0 + a^u;2 + a 2 o ^

written in a matrix form and solving for them. Alternatively for obtaining an orthogonal damping matrix use:

(3.81)

load vector = < per l Pn{t) ,

(3.82)

3.2.6 Existing codes on earthquake design Some well-known codes are discussed in brief and they are classified under items such as seismic actions, dynamic characteristics, seismic weights, forces, moments, storey drift (P — A effect), seismic factors, site characteristics and building categories.

3.2.6. / Algeria: RPA (1989) Seismic actions/dynamic characteristics

Load combination

G + Q + E or O.&G + E or G + Q+1.2E

G — dead load

Q — live load

E = seismic load

N = number of storeys

h = height of the building

T = 0. IN sec (building frames with shear walls)

T — — ( o t h e r buildings)

Seismic weights, forces and moments

• i i , * (V-Ft)Wkhk Fx, = equivalent lateral force — — E l i wtht

V = ADBQW

174 B A N G A S H

A — seismic coefficient = 0.05-0.25

W = seismic weight

6

Q = 1 + Y P ( l B = 0-20-0.5 l

Ft = 0 T < 0.7sec = 0.07TV < 0.25V for T > 0.7sec

Ft = additional force at the building top

Storey drift/P - A effect

At = Xt - Xt-i with X0 = 0

Xf — lateral displacement at b

A i> 0.0075 x storey height

Seismic factors

A/BC Seismic coefficient

I II III 1 0.12 0.25 0.35 2 0.08 0.15 0.25 3 0.05 0.10 0.15

PF = performance factor = 0.2-0.67

6

Q = 1 + Y, ~ quality factor l

Site characteristics/building categories

Zone Seismicity zone

0 Negligible I Low II Average III High

Building category (BC) Category 1 (500 year return) Category 2 (100 year return) Category 3 (50 year return)


3.2.6.2 Argentina: INPRES-CIRSOC103 (1991) Seismic actions/dynamic characteristics Horizontal seismic spectra

Sa = as + (b-as)— for T<TX

Sa = b

S»=b 2/3

for TX<T <T2

for T > T2

> £ = damping 5%

Sa = <k + ( / A * ~ <h)

=f\b Ti

T2 1 + C / A - D ^ ] [ * { ^ } 2/3-

for T <TX

for Tl<T<T2

for T > T2

£ = damping 5%

T is the fundamental period

fA = amplification factor due to £

= ^ for 0.5% < £ < 5%

£ = relative damping = percentage of critical damping

To = 2TT *

Wi — gravity load at level i

g = acceleration

Uf = displacement at i

Ft — normal horizontal force

for regular buildings level heights equal

T0 = \Wnun

100 V gFn

Alternative empirical formula is: r _ K r°'-ioo

30

L 1 + 300

T0l = fundamental period

1 7 6 B A N G A S H

hn = height of the building

L = length of the building

d = density of the wall

Torsional effects

Mti = torsional moment at level / = (1.5^ + 0A0L)Vt

or Mti = (el-0.01L)Vi

ex = distance between CS at level i and the Une of action of the shear force measured perpendicular to the analysed direction

L = maximum dimension in plan measured perpendicular to the direction of Vt


Wt — seismic weight = Gj + TJLJ

77 -O-I .O

Wt = seismic weight at level i

ht = height of the storey level i above the base level

n — number of levels in the building

GflLi = dead and live loads

Vertical seismic actions

fy seismic zone

0.6 4 0.6 3 0.5 2 0.4 1 0.4 0

Load states

1.32?w ± Ew = actions due to gravitational loads

0.85i? w ± Es = seismic actions


Building separation due to hammering

Y( = separation between adjacent structures = 8t + f$ht

Yi>f0hi+ 1 cm

Yt > 2.5 cm

/ s = factor depending on foundation soil = 0.001-0.0025

f0 — different soils in seismic zones = 0.003-0.010 n

Vi = storey shear force = ^ Fk

n

Mt = overturning moment = a ^ Fk(h* — h*) k=i+l

where

h*= heights at level k and i from the foundation level

/ = 0, 1, 2, 3 , 1 n

V0 = base shear force = CW = C

C =

i = l

i7/ = lateral force — ___

Fy = vertical seismic forces — ± C V 7 d ^

Storey </r/ft/P - A effect Lateral displacements S and storey drift A

A z. = 6t - 6f_i with 60 = 0

Alternatively

_ g r2^

<$ = horizontal displacements at levels

Limiting values for storey drift Non-structural elements attached are damaged

Group A 0 Group A Group B 0.01 0.011 0.014

Non-structural elements attached are not damaged

0.01 0.015 0.019

1 7 8 B A N G A S H

P-A effect

ft=^|>0.08

Pi — total seismic weight at level /

Vt — shear force at storey /

H( = storey height i

— amplification factor for forces and displacements

1

1 ~~~ Anax

/ 3 m a x is the 0 ^ value.

Seismic factors 7 d defines the risk factor:

Group 7 d

A 0 1.4 A 1.3 B 1.0

Reduction factor R A factor for the dissipation o f the energy by inelastic deformation:

U = 1 + ( M - 1 ) ^ (ovT^T,

R = Li for T > Tx

Ii varies from 6 to 1.

Site characteristics/building categories Seismic zones

Zone Seismicity

0 negligible I low II average III high


Building category (BC) Category 1 (500 year return) Category 2 (100 year return) Category 3 (50 year return)

Seismic zone

Zone Risk

0 very low 1 low 2 moderate 3 high 4 very high

Building classifications

Group Classification

A 0 Important centres

A Hotels, stadia, etc. B Private, commercial, industrial buildings C Containers, silos, sheds, stables

Zone b T2

4 0.35 1.05 0.2-0.4 0.35-1.0 3 0.25 0.75 0.2-0.4 0.35-0.1 2 0.16-0.18 0.48-0.54 0.2-0.4 0.5-1.0 1 0.08-0.10 0.24-0.30 0.2-0.4 0.6-1.2 0 0.04 0.12 0.10 1.2-1.6

Vertical seismic coefficient ( C v )

Zone Balcony and cantilevers Roof and large spans

4 1.20 0.65 3 0.86 0.47 2 0.52 0.28 1 0.24 0.13

1 8 0 BANGASH

3.2.6.3 Australia: AS 11704 (1993) Seismic actions/dynamic characteristics Bearing walls and frames where kd = deflection amplification factor

Bearing walls 1.25-4.0 Building frame 1.50-A.O Moment resisting frame 2.0-5.5 Dual system with a special moment resisting frame 4.0-6.5 Dual system with intermediate moment frame (steel or concrete) 4.5-5.0

Torsional effects

edx = Axes+0.05b

^d2 = A2^s ~ 0.056

A$ = dynamic eccentricity factors

e s = eccentricity

A x = 2 . 6 - ^ > 1.4 = 2.6 b

A2 = 0.5

b = maximum dimension at level i


V = total horizontal force (kN) = ZIKCSW

I = occupancy importance factor

= 1.2 essential facilities

= 1.0 other buildings

W = total dead load + 0.25 live load

. . 1 1 ICS „ V = seismic base shear = —— Gg

n

Vi — horizontal shear force = ^ Fx

x—i

n

M0 = overturning moment = a ^ Fthi z=l

i = levels number


a — 0.75 general

a — 1.0 at base

a = 0.5 at top

Base shear distribution

Fx = CyxV

_ Ggshk

x

yX~EUGgrh*

x — i levels

k=\ r<0 .5sec

Gg = gravity load = G + ipQ

G = dead load (kN)

Q = live load (kN)


6X = interstorey drift = k&8xe

8xe — lateral displacement at levels

= a (L\2

=iH (L\2

1 \ 2 i r J Ggi \ 2 * )

P - A effects

To allow for P — A effects, the storey drift is increased by

> 1.0 ( 1 - m )

When m < 0.1 there is no effect

m = stability coefficient Vxhsxkd

Px = total vertical design load

1 8 2 B A N G A S H

Seismic factors

C seismic response factor 1

< 0.12

a = acceleration coefficient = 0.05-0.11

L25a

T = ^ (main direction)

h T — (orthogonal direction)

58

ip = 0.4-0.6

Rf — response modification factor = 1.5-8.0

k = horizontal force factor

= 0.75 (ductile)

= 3.2 (brittle)

Z = zone factor

= 0 for zone A ductile

= 0.09 for zone A non-ductile

= 0.18 for zone 1

= 0.36 for zone 2


S = soil structure resonance factor = 1.5 if not calculated

Building classification Type I Domestic and not more than 2 storeys Type II Buildings with high occupancy (schools, theatres, etc.) Type III Buildings for essential functions (power stations, tall structures,

hospitals, etc.)

Seismic design categories III II I Domestic

as > 0.20 0.10 <as< 0.2 as < 0.10

E D C

D C B

C B A

H 3

H 2

H,


C = static analysis

D = static and dynamic analysis

E = static and dynamic analysis

S = site factor varies from 0.67 to 2.0

3.2.6.4 China: TJI /-78 and GBJ I /-89 Seismic actions/dynamic characteristics

S == total effect of horizontal seismic action

Ti = fundamental period

Sj = modal effect caused by seismic forces of the /th mode

N = number of modes

8n values

Tg Tx > lAlg Tx <\ATg

<0.25 0 .08^ + 0.07 no need to consider

0.3-0.4 0 . 0 8 ^ + 0 . 0 1

>0.55 0.087^ - 0.02

concrete building

Horizontal seismic action Fxji = <*/Y(, W i FyJi = a/Ttf Yfi Wt

Ftfi = a/yt/i:i<l>jiWi

— angular rotation at ith floor jth mode

xj, t — directions in JC, y and angular direction

rt — mass radius of gyration

OLJ — seismic coefficient

i = 1, 2, n

j = 1, 2, m

Torsional effects Modelling with degrees of freedom including two orthogonal horizontal displacements and one angular rotation for each level. The complete quadratic

1 8 4 B A N G A S H

combination (CQC) can be used to obtain the response 'S" (force, moment and displacement) given by

1l pjksJsk

7=1 k=l

Sj, Sfc = effects caused by seismic forces

0.02(1 + A T ) ( A T ) L 5

P j k ~ (1 - A j ) 2 + 0.01(1 + A T ) 2 A T

where

A x = ratio of the periods of the fcth and y'th modes

Seismic weights, forces and moments Base shear force:

i

Fj = horizontal seismic forces at / level

WH

AFn = additional seismic force applied to the top level of the building

H = height from the base

n

Mt = overturning moment = ^ fj{Hj ~ Ht)

^EVK = vertical seismic action force = a V m a x WQq

At ith level (see Fig. 3.12):

v _ WiHt _

Storey drift/P - A effect The elastic relative displacement is:

A [ / e < %}H

where 0 e = elastic drift limitation


n f TF V (

F v i -yf-

4 Figure 3.12. Seismic actions at various levels

Structure frame

Brick infill walls 1/550 Others 1/450 Public buildings 1/800 Frame-shear walls 1/650

Elasto-plastic deformation

AUp = VpAUQ

or

AUy = storey yield displacement

AUQ — elastically calculated storey displacement

7 / p = ampUfication coefficient

Uv < [0P]H

9» Structure

1/30 single storey RC frame 1/50 frame with infill 1/70 frame in the first storey of a brick building

Seismic factors

a = seismic coefficient

a = c w ^ ^ + 0.45) for Tx < 0.1 sec

186 B A N G A S H

« = < W for 0.1 <Tx<Tg

a = \Ti ) < W T g < T l < 3 s e c

a m a x intensity

VI VII VIII IX

0.04 0.08 0.16 0.32

= additional seismic action coefficient

7(, = the mode participation factor of the jth mode

Pjk = coupling coefficient for jth and kth modes

= storey ductility coefficient

£y F

— storey yield strength coefficient = - p

Fy - 2Qyl + (m - 2) x Qy2

m — total number of columns in a storey

Qyl J Qy2 = average yield strength of exterior and interior columns, respectively

Site characteristics/building categories Building classifications

Type A Structures not failing beyond repair. Important structures. Type B Buildings and structures in the main city Type C Structures not included in A, B, D Type D Structures of less importance not likely to cause deaths, injuries or

economic losses.

Tg — characteristic period of vibration

Epicentre Site I II III IV

Near 0.2 0.3 0.4 0.65 Remote 0.25 0.40 0.55 0.85


Structure £ y

0.5 0.4 0.3 0.2 2-4 storey 1.3 1.4 1.6 2.1 5-7 storey 1.5 1.65 1.8 2.4 8-12 storey 1.8 2.0 2.2 2.8 single storey 1.3 1.6 2.0 2.6

3.2.6.5 Europe: /-/ (Oct 94); 1-2 (Oct 94); 1-3 (Feb 95); Part 2 (Dec 94); Part 5 (Oct 94); Eurocode 8

Seismic actions/dynamic characteristics Horizontal seismic action: two orthogonal components with the same response spectrum.

Vertical seismic action:

T < 0.15 sec the vertical ordinates = 0.15 x horizontal

T > 0.15 sec the vertical ordinates = 0.5 x horizontal

T between 0.15 and 0.5 seconds - linear interpolation.

SQ(T) = elastic response spectrum 0 < T < TB

Se(T) = ag-S

Se(T) = agS£B0

SQ(T) = agS£B0

SQ(T) = agSl;B0

2B Tu<T <E

T

Tc

TC<T<TD

TD = T

Figure 3.13. Response spectrum

188 B A N G A S H

B0 = spectral acceleration amplification

KUK2 = exponents which influence the shape of the spectrum for vibration at TD and Tc (see Fig. 3.13)

Design spectrum Here, ag is replaced by a, and Se(T) by Sd(T)

K\ and K2 to K&\ and respectively

TC = TD> 0.2a

Combinations of seismic actions

]C — combination coefficients for variable actions i — 0.5 to 1.0

G and Q are characteristic values of actions

^ E i = ^ 2 i 0 = 0.5-1.0

Design seismic coefficient = 0.2


eu = ±0.05Lt

Mut = eyFt

e u = accidental torsional eccentricity

Lt = floor dimensions perpendicular to seismic action

seismic base shear = Fb = V = Se{Tx)W

W=Wt = total weight

T\ <4TC (fundamental period)

p. — p S i W i

when horizontal displacement is increasing linearly.

, ZTW,

EZJWJ

Sh Sj = displacements of masses Mt and Mj in the fundamental mode shape


Wu Wj = corresponding weights

Z n Zj = heights of masses Mt and Mj9 respectively

d§ = displacement induced by design seismic action

= 9d47i qd = displacement behaviour factor

de — displacement from the linear analysis

jpa = horizontal force on non-structural element

= weight of the element

q2L — B = behaviour factor

= 1.0-2.0 = ?

Storey drift/P - A effect P - A is not considered if the following is satisfied

dx = A in other literature

hf — interstorey height

Ptot = total gravity load at and above the storey

Vtot — total seismic storey shear

For buildings with non-structural elements

< 0.004/*

When structural deformation is restricted

i- < 0.006/*

Where 0.1 < 9 < 0.2 increase the seismic action effects by

'a

0 < 0 . 1 0

(\-0) ^ 0 . 3

1 9 0 BANGASH

Seismic factors

7 = 7! = 0.8-1.5

B0 = 5% viscous damping

£ = damping acceleration factor > 0.7

S = soil parameter = 1 . 0 for 5% damping

q = B = behaviour factor

a x 3 M +^J & — seismic coefficient = - 4-

n - m a% a = — a

ra for non-structural elements

Ti for structural elements

z — height o f the non-structural element

RD = v


Subsoil Class A

Vs = lOOm/sec 5 m

= 400m/sec 10 m

Subsoil Class B

Vs = 200m/sec 10 m

= 350m/sec 50 m

Subsoil Class C

Vs = 200m/sec 20 m

Ground acceleration ag ^ 0.10g

d% = peak ground displacement = 0.5agS • TCTD

Class ^dl Kd2 Kx K2

A 2/3 5/3 1 2 B 2/3 5/3 1 2 C 2/3 5/3 1 2


Si Bo 7B Tc 7b

1.0 2.5 0.10 0.40 3.0 1.0 2.5 0.15 0.60 3.0 0.9 2.5 0.20 0.80 3.0

Building categories versus i?D

I 2.5

II 2.5

III 2.0

IV 2.0

3.2.6.6 India and Pakistan: /S-/893 (1984) and PKS 395-Rev (1986) Seismic actions/dynamic characteristics

T (moment resisting frame, shear walls) = O.ln

T (other buildings) = 0.09H/Vd

n = number of storeys

d = maximum base dimension

H = height of the buildings

Response spectrum

SJg versus T when f = 5%, f = 0, f = 10%, f = 20%

s j g T(sqc) 7'(sec)

0 0.16 0.2 0.1 1.00 1.9 0.2 0.30 1.18 0.3 - 0.80 0.4 - 0.60 0.5 - 0.50 0.6 0.40 0.7 0.15

1 9 2 B A N G A S H

From the above, the average acceleration coefficient SJg is obtained.

Fir — seismic design lateral force at the ith floor level corresponding


V — design base shear = KC(3Ia0 W (India)

V — design base shear = C$ujt (Pakistan)

C s = ZISM7dQ

Z = ACF

F . - y W>ti

W = total load = dead + appropriate live loads

Fj = lateral force at the ith floor = Pt

Wt = gravity load

ht = from the base to the ith floor

n = number of storeys = N

Ft = force in the ith frame to resist torsion

where

Kf = stiffness of the ith frame

rt — distance of the ith frame from the centre of the stiffness

Mt = torsional moment = l.SeV

to the rth mode

KfilFofoCr % W, g

mode shape coefficient (t>ir

India


Pakistan

Wt^Di + nLi

T) = 0.25-0.50

Mt = overturning moment

= ^Fi{hi-hj)+Fi{hn-hj) 1+1

Torsional effects

1.5e + 0.16 or e a 0.16

b = the largest distance or dimension

e = eccentricity

e a = eccentricity for torsional moment

Lt = live load at fth level

Storey driftlP - A effect

A m a x between two floors ^ 0.004 x ht for height > 40 m (India)

g T2Ff

47T2 Wt

A = 6t- 8t_x

P - A effect

S0 = 0 (Pakistan)

Seismic factors

F0 = seismic zone factor

a0 = basic horizontal seismic coefficient

C = 5%

A = 0-0.08

M = material factor = 0.8-1.2

Q = construction factor = 1 . 0

S = 0.67-3.2

a0 = 0.01-0.08

I = importance factor = 1.0-1.5

p = 0.01-0.08

1 9 4 B A N G A S H

For different soil foundations:

7d Height (m)

0.4 up to 20 0.6 40 0.8 60 1.0 90


Zone a 0 ^ 0

V 0.08 0.40 IV 0.05 0.25 III 0.04 0.20 II 0.02 0.10 I 0.01 0.05

3.2.6.7 Iran: ICSRD (1988) Seismic actions/dynamic characteristics Design methodologies: Analyses

(a) Equivalent static (b) Pseudo-dynamic (c) Dynamic analysis using acceleration data

T = 0.09 -^=^>T< 0.06/z 3 / 4

T = 0.08/* 3 / 4 (steel frame)

T = 0.07/* 3 / 4 (RC frame)

2AI FY = vertical seismic action = —— Wv

Ry = reaction coefficient

= 2.4 for steel

= 2.0 for concrete

Wp = Gf + Lt + total


Seismic weights, forces, and moments

V = minimum base shear force = CWX

B Lateral forces

W,h

Ft = additional lateral force at top level

= 0 if T< 0.7 sec

= 0.07 if 7V < 0.25 V N

Mi = Ft(hN - ht) + FAhJ ~ hd / = 0 to iV - 1

n

Mti = J2 eiJFJ + M t a 7=1

M t a = accidental torsional moment

Storey driftIP - A effect

The lateral drift is ^ 0.005/z,. Both lateral forces and torsional moment effects are coupled.

Seismic factors

Wl = Gi + r1Ll

v = 20-40%

A = design base acceleration = 0.35-0.20 5

0.6 < R < 2.0 R = 2.0 | 2.0 | \T

I = 0.8-•1.2

B = 5-8

Site characteristics/building categories 1. High priority buildings 2. Medium priority buildings 3. Low priority buildings

1 9 6 B A N G A S H

Classification (a) Regularity in plan (b) Regularity in elevation

Soil Classification I to IV where

T0 — characteristic period on site

= 0.3-0.7

3.2.6.8 Israel: IC-413 (1994) Seismic actions/dynamic characteristics

T = 0.073/z 3 / 4 (concrete)

= 0.085/* 3 / 4 (steel)

= 0.049/s 3 / 4 (others)

Vertical seismic action

Fr — ±.\ZW cantilevers

= ^min ~ 1.52/5 FT for concrete beams

Modal lateral force = at level /

7m

nn(max)

or

_ g TJFh

rm(max) - ^ ^

where Tm = mth natural period

Seismic weights, forces and moments N


c<>=ir SIZ

or > ' 3 L U

0.3/ low

} i> 0.21 medium

0.1/ for high ductility

F i = x-^N „ R ! ' a t t h e t o P l e v e l FN + Ft

Lateral forces

Ft = 0.07TV < 0.25V

E l i wa

e — torsional accidental eccentricity

= ±0.05L

3.0 >AT = 2.75 ( c

U m a x

c ) > 1.0 ( < W x \

V max + ^min /

£ = multiplying factor applied to lateral load at each stiffness

element to account for torsional effect

= 1.0 + 0 . 6 ^

Modal overturning moment

N

7 = 1 + 1

Modal torsional moment

= Mtim = (di±ei)Vim

di = eccentricity

et — accidental eccentricity

Modal weight

W (E« Wtfaf

4>im = ampUtude at i level of mth mode

1 9 8 B A N G A S H

Storey driftIP - A effect P - A effect

9t > 1.0

N

Vt = ^ Fj (storey shear force)

| V 2

*el,/m

f > e l , , - m ) 2

L m = l

elastic modal drift at that level of i

Vf = modal shear force at i

r N - , 1 / 2

L m = l

Drift limitations

for T < 0.7 sec A i ] i m = min 200

f o r r > 0 . 7 s e c = min ( - ^ ; ^ j

maximum displacement 6^ = ±AT ^ A,-z=0

A,- = computed interstorey displacement

Storey drift

Aim — <5/w fi(i-j)m

Seismic factors

R = steel 4-8, concrete 3.5-7

= concrete load

qt = U D L

4/ = area


Kg — live load factor

= 0.2 (dwellings)

= 0.5 (stores, etc.)

= 1.0 (storage)

/ = 1.0-1.4

Q = seismic coefficient

i? a = spectral amplification factor

Site characteristics/building categories Regular structures Category B < 80 m high Category C < 80 m high

with normalized seismic zones Z < 0.075

Seismic zones Z

S= 1.0-2.0

3.2.6.9 Italy: CNR-GNDT (1986) and Eurocode EC8 Seismic actions/dynamic characteristics Seismic index S — 6, 9 and 12

2.5 > Ra(T) > 0.2K

I II III IV V VI

0.075 0.075 0.10 0.15 0.25 0.30

Structure Maximum height (m)

Frame Frame

5 = 6 5 = 9 No limitation 16.0 11.0 32.0 25.0 10.0 7.0

S= 12

Masonry 7.5 15.0 7.0

Walls Timber

200 BANGASH

Fundamental period T0

T0 > 0.8 sec R = 0.862/T 1^

T0 < 0.8 sec R = 1.0

T0 = 0.1 for a framed structure

Lateral and vertical effects

a = [al + a l ] l / 1

a — single force component

h, v = subscripts for horizontal and vertical

r] = single displacement component

Combination of modal effects

a t o t = combined total force = a ± a p

a p = action due to non-seismic loads such as permanent loads and live load fraction

a t o i = a ± O p ! (non-seismic loads)

= a ± a p 2 (fraction of live load)

R?R = actual displacement for design purpose in elastic situation

= Vp±<fa

7jp = elastic displacement due to non-seismic loads

<t> — 6 if displacements obtained from static analysis

= 4 if displacements obtained from dynamic analysis


Fi = KhiWi

KM = CRRD(3s7iI

Wi = Gi + -yiQi


7 l — ni xr^JV

N

Mti — minimum torsional moment = XD E Fj 7 = 1

D/Ls versus X

D — long dimension

L s = short dimension

2.5 < ^ - < 3.5 A - 0.03 + 0.02 - 2.5

3.5 < ^- X = 0.05 ^s

= resultant vertical force = KylWf

Masonry buildings

Pi = 2 (taking into account ductility)

fii — 2 (taking into account ultimate state)

M v = E FAhJ - hi) j=i+l


U0 = relative displacement between the base and the top of the wall

Kt = total stiffness = f± j J — ^

1 + L 2 ^ U A = panel area

t — panel thickness

b, h = width and height, respectively

G,E = shear and Young's moduli, respectively E ,

2 0 2 B A N G A S H

Seismic factors

I = 1.4 (civil defence)

= 1.2 (structures at risk)

= 1.0 (others)

. . . 5 - 2 C = seismic factor =

F F = foundation factor = 1.0-1.3

(3$ = structural factor = 1.0-1.2

RD = reduction factor = 0.33-1.0

7,- = distribution factor at level /

Site characteristics/building categories Modal plus design spectrum Linear elastic range

-=CRRD0sI g a = horizontal spectral acceleration

g = gravitational acceleration

Shear coefficient Zone I 0.4 Zone II 0.28 Zone III 0.16

3.2.6.10 Japan: BLEO (1981) Seismic actions/dynamic characteristics

Rt = spectral coefficient = 0.4

x = 0.4 (hard soil)

= 0.6 (medium soil)

= 0.8 (soft soil)

J = A(0.02 + 0.18) sec

h — full height of the building

100


ew = (i+o.7A)& Qbi< i.5fi/ g b / = lateral shear strength

_ lateral shear in bracings * total storey shear

Qi = QWi

Qui = DsFesQt [ultimate shear strength (lateral)]

Z>s = structural coefficient

^es —

Seismic intensity Elastic response 0.15-0.25 g Elasto-plastic response 0.30-0.5g

Seismic weights, forces and moments Lateral seismic shear force

Qi = ciwi

Wj = portion of the total seismic weight at the level i

Ct = ZRtAjC()

A 1 _i. ( 1 ^ 2 T

V",- ~'J 1 + 3T

W a, = - ^ = 0-U>

H o = weight above ground floor

C 0 = 0.2 (moderate earthquake motion)

= 1.0 (severe earthquake motion)

q = KW

M i = E F A h J - h i )

or

2 0 4 BANGASH

hj = interstorey height

QB — horizontal seismic shear at the basement = Qp + KWB

WB — weight of the basement

q = horizontal seismic shear force

= KW (for appendages such as penthouse, chimneys, etc.)

Torsional stiffness

i = l Kxi

1/2

r - I J%y • V Eccentricity ratio

Rex = — < 0.15 Tex

i ^ = - ^ - < 0 . 1 5 ley

For the x and j-directions

Nx Ny

where Nx,Ny = number of resisting elements

Connections

Mu = aMp

a = 1 . 2 - 1 . 3

M p = full plastic moment of the column or beam

Mu — maximum bending strength of the connection

Storey drift/P - A effect Seismic lateral forces = Ft — Qt —

A ^ K or is equal to hf for non-structural elements

for buildings not exceeding 60 m height, i.e. h ^ 60 m


For steel buildings 31 m

i ? e < 0 . 1 5 i ? s > 0 . 6

Qi is increased by Qbi

Seismic factors

Rt = spectral coefficient

At = distribution factor

Z = seismic coefficient = 0.7-1.0

Seismic coefficient K of the basement

H = depth of the basement in metres 20 m)


1 t t j 1 0 - 6 4

Type 1 Hard soil -jr

Type 2 Medium soil —^r

Type 3 Soft soil

r = 0-2.0 sec

Tc for Type 1 soil = 0.4



Z>s = structural coefficients

= 0.25-0.5 (for steel)

= 0.30-0.55 (for RC)

FQ = function of eccentricity

= 1.0 for Re< 0.15

= 1.5 f o r i ? e > 0 . 3

2 0 6 B A N G A S H

Fs = function of stiffness ratio Rx

= 1.0 i ? s > 0 . 6

= 1.5 i ? s < 0 . 3

Buildings Those of 1 to 2 storey: wooden ^ 500 m 2

Special building ^ 100 m 2

RC buildings

2 5 £ ^ w + 7 £ ^ c > ZWtAi for h ? 20m

25 E Aw + 7 E + 10 E > OJSZAiWt for A 31 m

Stee/ buildings (1) Not exceeding 3 storeys

E = summation of total column areas

E ^w — summation of total horizontal cross-sectional areas of walls

h < 13m span < 6m

total floor area < 500 m 2

C 0 = 0.3

(2) Not exceeding 60 m

Qui < Qu

3.2.6. / / Mexico: UNAM (1983) M III (1988) Seismic actions/dynamic characteristics

Design spectra Referring to Fig. 3.14:

for T < T a seconds

a C for ra < T < Th seconds

for T > Tb seconds


Figure 3.14. Design spectra: acceleration versus time

Seismic zone

Zone T Tb r A B C D

c values I 0.2 0.6 l

2 0.08 0.16 0.24 0.48

II 0.3 1.5 2 3 0.12 0.20 0.30 0.56

III 0.6 3.9 1.0 0.16 0.24 0.36 0.64

/ * W-xt V / 2

T (reduced lateral forces) = 0.63 > ' 1

x = displacements

Fi = force at ith level

a = spectral acceleration

Dynamic analysis Modal analysis accepted by the Code:

0 . 8 ^ V O > - q T ~

R = total response = S

S( = modal responses

Seismic weights, forces and moments Seismic loads Dead loads as prescribed (DL) Live loads = 90 kg/m 2 to 180 kg/m 2 (LL)

Load combinations

UIF = ultimate internal forces

= 1.1 (/XL ±LL± ELL ± 0.3ELT)

or = 1.1 (DL ± LL ± 0.3ELL ± ELT)

208 BANGASH

where

ELL = earthquake loads (longitudinal)

ELT = earthquake loads (transverse)

Static method Lateral forces

V0 =

Wiht

c

W0 = Wx = total weight

If T < Tb « - delimiting period, the response spectrum Ft (reduced) will be obtained by changing

If T > Tb

F, (reduced) = W^h, + K2h})

K2 = l.Sr

1 - r

( 1

Torsional effects

e= 1.5es + 0.1L

e = es — 0.1L

es = static eccentricity

e = torsional eccentricity

L = plan dimensions of a storey

The factor of 1.5 is reduced to take into account dynamic modifications of torsional motion.

Overturning moment

N

M( = Fj(hJ - ht) i = 0 to N — I j=i+l


Mi is multiplied by i ? m = 0.8 + 0.2z where

height above ground Z = h


A ^ 0.006/*z (main structural elements)

A ? 0.012 (for partition)

P - A effect When A > 0.08 Vj JV, this effect should be considered

V = calculated shear force

W = total weight of the part of the structure above that storey

Seismic factors

Q' = Q T > ra

E ' = L + ( ^ ) x ( E - L ) T<T,

Q = 4 to 1 for Q > 3

The point of application of the shear force

eT > es - 0.2L if Q = 3

er > - 0 . 1 6 if Q > 3

S/te characteristics/building categories Building classification

Group A Important buildings Group B Ordinary buildings

Seismic zones: soil types

Seismic zones Soil types

I II III

Hill zone, stiff rock, soft clays Transition zone, sandy, silty sands < 20 m Lake bed zone, clays, silty clay

2 1 0 B A N G A S H

sands > 5 0 m

A =

tt — thickness of soil layers i

Gj = shear modulus tons/m2

7,- = unit weight tons/m3

j31 = J^i for Class II V Ii

A < 7 0 0 - 5 5 0 R 0

For outside Mexico

a = oo + (c - ao) for T < Ta

a

3.2.6.12 New Zealand: NZS 4203 (1992) and NZNSEE (1988) Seismic actions/dynamic characteristics Methods

(a) Equivalent static method (b) Modal response spectrum (c) Numerical integration time-history analysis

rc£i»wil/2

T = 2TT

T = 0.042

Ui = 1 . 0 for one storey

= 0.85 for 6 or more storeys

= for fewer than 6 storeys interpolate between 1.0-0.85

ut — lateral displacement at level /

St = lateral displacement in mm at the top of the building

Modal response spectrum

Cl{Tm) = SmCh{Tm,\)RZflj

Sm = response spectrum scaling factor

Tm = modal period



Basic shear force

V = CW

W = total weight

C = Ch(T, fi)RZfL > 0 . 0 2 5 T > 0 . 4 sec

C = C B ( 0 . 4 , fJi)RZfL > 0 . 0 2 5 R < 0.4sec

/i = 6 (structural steel)

= 5 [concrete (RC or prestressed)]

= 4 (masonry)

= 3 - 4 (timber)

Equivalent static force Ft at level i

Wh>

Ft = 0.92V 7

IS* Seismic weight with additional top-force

F t = 0.08F


A = storey drift

j 0 ,0 .2 ,0 .4 ,0 .6 (%)

A m a x = 600 P - A effects At ultimate it should satisfy

(a) T < 0.4 sec (6) A ^ 15 m for T < 0.8 sec (c) /x < 1.5 id)

«t ~ ui-i < Vi

hi - ht-i ^ 7.5 Ej=i Wf

Vt — storey shear N

= * +

2 1 2 B A N G A S H

Seismic factors

Risk factor R given by:

1 = 1.3

11= 1.2

HI = 1.1

IV = 1.0

V = 0.6

C = lateral force coefficient

fL = limit state

= 1.0 (ultimate)

= 1/6 (serviceability)

Z = zone factor

Cb(T,fi) = basic seismic acceleration factor

= 0-1.0 for T = 0-4 for /x = 1.0-0.6

Cb(0.4,M) f o r T < 0 4 s e c

S m = 1 . 0 for the limit state of serviceability

^m 2

KmCW

Km = 0.8 for / x = 1.0

Site characteristics/building categories Building categories

I Important buildings II Holding crowds III Building with highly valued contents IV Buildings not included in any category V Secondary nature

Soil category (1) Rock or stiff soils (2) Normal soil sites (3) Flexible deep soil sites


3.2.6.13 USA: UBC-91 (1991) and SEAOC (1990) Seismic actions/dynamic characteristics Referring to Fig. 3.15:

T = fundamental period = Ct[hN]3/4

hN = h = total height

C t = 0.035 (steel moment-resisting frame)

= 0.030 (RC moment-resisting frame)

= 0.020 (for Rayleigh's formula)

Alternatively

fi = lateral force distribution

Vx = storey shear force at x

N

i—x

Fx is evaluated under seismic weights.

Dynamic method Using modal shapes

[K]-u?[M\{4>} = {0>}

work out u9f and T and finally modal shape [</>].

Spectral accelerations Scaled down by peak one of 0 .3g/ i? w = 12:

5 a l = 0 . 0 3 4 g S a 2 = 0.063 g

S a 3 = 0.061 g 5 a 4 = 0.05g

Figure 3.15. Shear force distribution

2 1 4 B A N G A S H

Seismic weights, forces and moments ZICW

V — basic shear force = R w

i ? w = structural factor = 4-12

„ (V-Ft)Wxhx , , c Fx = \T — — at level x from the base

Uli wtht

Ft = 0.07TV < 0.25 V for T > 0.7 sec

F t = 0 for T< 0.7 sec

r = F t + f > ,

Storey shear force

AT

F, = F t + ^ F , .

Mx = overturning moment, thus: N

Mx = Ft(hN -hx)+J2 Hhi - hx) where x = 0 , 1 , 2 , . . . , - 1 /=JC+1

Torsional moment Torsional (based on UBC-91) irregularities in buildings are considered by increasing the accidental torsion by Ax (amplification factor)

v2

uavg

<5avg = the average displacement at extreme points at level x

The floor and the roof diaphragms shall resist the forces determined by the following formula

where

Wpx = weight of the diaphragm and attached parts of the building

The corresponding Fpx 0J5ZIWpx and ft 035ZIWpx



1 1 * , 0 0 4 * Dnft calculated y> —— x hi

or i> 0 . 0 0 5 / * , for T < 0 . 7 sec

or i> 0 . 0 0 4 / * , for r > 0 . 7 sec

P - A etffec* Secondary moment formula drift Primary moment due to lateral forces

M x s ^> 0 . 1 0 not considered for greater values

Msp VXHX

Subscript x means to the level of storey x.

A f c = -

Elastic storey drift

Aj = inelastic storey drift

= ^f (K<1.0)

or

Aj = RAiQ

N 6t = lateral displacement = ^ Aj

i=i

T<?ta/ displacement

6px = 6ix-0Yp

Spy = Siy + 9Yp

6ix, $iy — lateral displacement in x and y directions

6 = storey rotation

< V o py — total displacement in x and y directions

<5p = total displacement at a selected point P

2 1 6 B A N G A S H

Seismic factors

Z = seismic zone factor

Z = 0.075 Zone I

= 0.15 Zone l lA

= 0.20 Zone IIB

= 0.30 Zone III

= 0.40 Zone IV

/ = occupancy or importance factor = 1.0-1.25


Sesismic zones Soil S*

I Rock like 1.0(50 IIA Dense stiff 1.2(52) IIB III Soft-medium clay 1.5(S3) IV Soft clay 2.0(S4)

S* = site coefficient

Sj type

S* = 1 + 107/

2.57/

0.975 T > 0.39 sec

S2 type

1 + 107/

S3 type

s„ =

1.463

1 + 7 5 7 /

2 .5r

2.288 _

Sa = 2.57/

T > 0.585 sec

0 < T < 0.2 sec

0.2 < T< 0.915 sec

7/ > 0.915 sec

0 < T < 0.15 sec

0.15 < T < 0.39 sec

0 < T < 0.15 sec

0.15 < 7/ < 0.585 sec


3.3 Examples on dynamic analysis of building frames and their elements under various loadings and boundary conditions

This section deals with solved examples of building frames and their elements subjected to dynamic loads. Examples selected are based on their day-to-day occurrence in various building problems. Readers are advised to also study additional materials on dynamic analysis. These examples together with the dynamic analysis in Section 3.2 will hopefully pave the way to finally understanding and tackling earthquake problems.

3.3.1 Example (3.1) A beam of length L is subjected to a uniformly distributed load V as shown in Fig. 3.16. Assuming the vibration deflection curve is of the same form as the static deflection curve, determine the frequency for this beam when

(a) supports are simple (b) supports are fixed.

Take EI as constant.

Solution

Beams with uniform load but with different boundary conditions.

(«) EI—2 = M = - - —

dx 6 4

™ - g - * £ + A + B

(b)

Figure 3.16. A beam under uniform load

2 1 8 B A N G A S H

when the boundary conditions are:

Jo

X = 0 <5 = 0 5 = 0

X = L 6 =

EI 6 =

6 = 24/±7 {

- 2Lx 3 + L 3 x)

6dx = 120EI

I, % 2 d x = — 7 * o 360 V24£7

J_ / f g g L 5 630 / 2 4 £ 7 \ 2 1 7 2tt V 1120^7 31L 9 W / i

= 1 . 5 7 2 ^ (Hz)

( b ) £ 7 ^ = M = M A + ^ - - ^ - x

d6 ay? x2

EI — = MAx + l—--qL— + A dx 6 4

4 r 3

z 24 12

Boundary conditions:

X = 0 £ - 0 A=0 dx

7i = 0

JC = L dx ^

dS _ dx +9* 9 l f + *

6 4

0 = 6 4 A 12

6 = q 2AEI

[7_V + x4 - 2Lx 3]

<5 = 0


I, 62dx C_»_Yil o \ 2 4 £ / / 630

= circular frequency = ^ j° ^ X

j0

L62dx 504EIg

qL<

f = natural frequency = — 2ir

1 /504£/

27r°V qL4

Elg qL*

3 - 5 7 3 ^ (Hz)

3.3.2 Example (3.2) A simply supported beam of weight q/m carries a central load W. Assuming that the vibration deflection curve is of the same form as the static deflection curve, determine deflection tS' in terms of maximum deflection 6max, load W and the inertia effects. Take EI as constant. Find the total central load for the beam in Fig. 3.17.

Solution

A simply supported beam subjected to U D L and a centrally concentrated load:

d2<5 EI —? = Mx = A Wx

dx 2 * 2

Elfx = + A l

S = ^j[^Wx3+Alx + B1]

Boundary conditions:

x = 0 6 = 0 5 , = 0

x = - — = 0 Ax=-k WL2

2 dx 16

2 2 0 B A N G A S H

„ 2 W c <Jlm W c <Jlm

8 "~~ ~ '"" L - -—"J

o max ^

* x * L

-w

Figure 3.17. A beam under concentrated load and self weight

When x = 2

#0 — ^max ~ ~ 12 \2 J + 16 V2 J J_ 7i7

Hence

1_ EI

WU_ 4SEI

WL2 Wx3

16 12 W

4SEI [3L2x - 4x3}

^max

6

j}\IL2-4X2)

6wa^i}L2-4x2)

Kinetic energy due to element dx

qdxv2 qdxcpd1

2g

where v = 8.

2g

Total kinetic energy KE dx

= <tf_ 17

2# , 3 5 ,

inertia effect

17

total central load = W + —qL= W'

W' replaces FFin the above analysis in order to compute '<$'.


3.3.3 Example (3.3) Find the natural frequency of the beam/mass system lying in the plane as shown in Fig. 3.18. Use the following data:

EI = \GJ 6 = Xsinut

Reference coordinates are ® and (2). Use the flexibility method.

Solution

Bending + torsion: beam/mass system

Equations of motion

4Af6 1 0 /n +M620j12 = -*io

4M6l0f21 + M620f22 = -<$20

Displacements due to bending and torsion

FlexibiUty coefficients

f n ~ J ~eT+] ~gT-6Fi{2L }~3EI

/22 = J -EF + J -GT = 6EI ( 2 X 4 L } + 6EJ { 2 L } + 6G7 ( 6 L }

<§io = ^ sinutf 6 1 0 = tcA^ cosutf 6i 0 = — u?X\ sin a;/ = — w26io

S2Q = X2 sin art 6 2 0 = a^cosutf £20 = — w 2 -^ sin art = — w2^2o

All necessary values are substituted, the following determinant is obtained and is made equal to zero:

( l - 4 M a ; 2 / „ ) - M u > 2 / 2 2 = Q

- 4 M w 2 / 2 , 1 - Mu?f22

222 B A N G A S H

Figure 3.18. Beam/mass system in bending and torsion


The solution of the determinant yields

, 2 2 * 4 7 , - , 2 , - L3Mu? 1 ~ 4 l K ~ ~ 9 { K ) ° W H E R E * = —ET~

7C= 1.251 or 0.163

1.2517J7 ML3 or

0T63_E/ ML3

(rad/sec)

/(lowest) = — 0.1637i7

ML3

3.3.4 Example (3.4) A frame lying in a plane with a distributed mass of 1500 kg/m as shown in Fig. 3.19. Using the flexibility method or otherwise, determine the natural frequency ' / ' for this frame. Use the following data:

E = 14 x 10 6 kN/m 2

7 = 2600 x 1 0 _ 6 m 4

G = 6.4 x 10 6 kN/m 2

J = 1800 x 10~ 6 m 4

L = 4m

6 = Xsin u)t

EI = 14 x 10 6 kN/m 2 x 2600 x 10~

GJ = 6.4 x 10 6 kN/m 2 x 1800 x 10"

Use the direct integration process.

Solution

A bent in-plane subject to a distributed mass. Different coordinate system x and y applies:

f mmi , f TtTj A

6 = —u/Xsinut

2 2 4 B A N G A S H

L = 4nrT

1500 kg/m

L = 4m

Figure 3.19. A frame in plane

Flexibility parameters Limits

mx = lxL 0<y<L

m{ = I x L 0 <x <L

m2 = I x L 0 <y <L

m2 = I x L 0 < x < L

TX=\L 0<y<L

TX=\L 0<x<L

T2=\L 0<y<L

T2 = \L 0<x<L

Flexibility coefficients

* [L L2 i t * [L L2 i t 2L 3 L3 0.0021

m Jo EI Jo 4GJ EI AL

t t 2 iL ,2 ! f , 2 i , 00021

Equations of motion:

MiSiofn + M2620f12 = -6io

^ 1 * 1 0 / 2 1 + M2820f22 = -620


Substituting respective values, the following relations are established:

(0.1029u;2 - 1) 0.062a;2

0.1029a;2 (0.041a;2 - 1)

Solving the determinant o f the matrix Xx ^ 0, X2^0 and making K = a/

- -0.1439 ± 0.1718 0.0044

a; = 2.51 (the lowest rad/sec)

/ = ^ = 0.4 Hz

3.3.5 Example (3.5) A single-bay two-storey frame is shown in Fig. 3.20. Derive equations o f motion.

For displacements x \ and x 2 , the tensions in the springs are:

P\ = k\X\ and P2 = k 2 ( x 2 — xx)

The dynamic equilibrium equations are

for mass 1 M\X\ + P\ — P2 = 0

for mass 2 M2x2 + P2 — 0 = 0

Substituting for Pi and P2 gives

for mass 1 Mxxx + (kx + k 2 ) x { - k 2 x 2 — 0

for mass 2 M2x2 — k 2 x x + k 2 x 2 = 0

— W V V - < - i - K W V H - i

\ \ \ \ \

Figure 3.20. (a) A two-storey frame, (b) Idealized mass j spring system

2 2 6 B A N G A S H

In general for a two degree-of-freedom system

Mxxx + knxi + kX2x2 = 0

M2x2 + k2Xxx + k22x2 — 0 (3.83)

where kX2, k2X and k22 are the usual stiffness coefficients for the system. Looking for solutions having the form xx = Xxsinut, x2 = X2sinut; this

implies xx = — Xxuj2 sin ut,x2 — — X2uj2 sin ut. Substituting in Eq. (3.83) for xx, x2, xx and x2, and cancelling sin ut through

out,

Mx(-Xxu2) + kxxXx +kl2X2 = 0

M2(-X2U?) + k2XXx + k22X2 = 0

or

(kn-MlJl)Xl+kl2X2 = 0

k2XXx + (k22 - M2u?)X2 = 0

Eliminating X2/Xx,

(kxx - Mxu2)(k22 - M2u2) - kX2k2X = 0

This is a quadratic in u2, whose two roots give the two natural circular frequencies.

For each root, back substitution into either one of Eqs (3.84) gives X2/Xx, i.e. the mode shape.

In matrix form, Eqs (3.84) may be written

(K - Mu2)X = 0

(3.84)

(3.85)

where

K =

M =

X =

kn ki2

k2i hi.

M{ 0

0 M2

Xi'

X2

(the stiffness matrix)

(the mass matrix)

3.3.6 Example (3.6) A rigidly jointed two-bay two-storey frame with relative masses at two floor levels is shown in Fig. 3.21. Assuming the horizontal members are infinitely


AL

1.5M

1 0 m

1.5M

1 0 m

Figure 3.21. A two-bay two-storey frame

21

5 m

5 m

stiff in comparison with the vertical members, determine two natural frequencies and their corresponding periods for this frame. Use the following data:

EI = 1.50 x 10 6 Nm 2 M= 10 5kg

8 = (Asmut + Bcosut) • D

where

6 = displacement of the frame

u = circular frequency

D = column matrix for the deflected shape of the frame

A, B = constants

Referring to Fig. 3.22, it follows:

6 = -u?6

[[k] - Mu2]]6 = 0.8 = d

2 x 12EI 2 x 12E(2I) Y2JEI

(5) 3

0.672£7

(5) J (5) d

"12 — * 2 1 koA -2 x 12EI

(5Y = -0A92EI

k22 = 2(^-^=+0.l92EI

228 BANGASH

Figure 3.22. Motion of vertical members for Example (A3.6)

+0.672 -0.192

kn kl2

-0.192 +0.192

&21 ^22

EI

[k — Mu2] —» the determinant of this matrix m 0

0 3m

EI kn — mxij? k\2

kn k22 - m2u}2 J

0.672- 1 0 V -0.192

-0.192 0.192- 1.5 x 1 0 V

0.672 - 1 0 V -0.192

-0.192 0.192- 1.5 x 1 0 V

u? = B

w 4-29.6u; 2 + 25.15 = 0

EI

EI = 0

(B)2 — 29.65 + 25.15 = 0


fi=%L = 0 . 3 3 2 H Z 2 7 T

/ 2 = G = 1 . 1 8 5 H z

Ti = ]r = 3 . 0 1 2 Ji

T2 = ]r = 0 . 8 4 4 / 2

3.3.7 Example (3.7) A single-bay single-storey frame with different leg heights is shown in Fig. 3 . 2 3 , determine frequency and acceleration. Take x = Dsinuit where x = displacements - 6; take EI = 2 X 1 0 6 N m 2

^total — + &CD

_ 1 2 £ / _ 1 2 X 2 X 1 0 6 N m 2

k*»~W~ ( 4 ) 3

12FT * A > = — r g - = 3 7 5 0 0 0

( 5 ) J

1 2 X 2 X 1 0 * , 1 9 2 0 0 0 N m 2 =

( 5 ) 3 5 6 7 0 0 0

[*]TOTAI = 5 6 7 0 0 0 N / m

A: = N/m —»• kg as mass

5 6 7 0 0 0 N/m

2 0 0 0 kg

= 1 6 . 8 4 rad/sec

„ 2 7 T 2 7 T „ r = ~ = 7 7 ^ 7 = 0 . 3 7 3 sec w 1 6 . 8 4

/ = ! = - i — = 2 . 6 8 H z J T 0 . 3 7 3

F = 6EI/L2 _ \2EI

L/2 ~ L?

230 BANGASH

B 1 unit C 1 unit

/ /

/ 5 m

W W V

T 1

L / / C

/ /

Figure 3.23. Single frame with different height of legs

When

L? , \2EI

k = ^T

D = VA2 + B2 = 20 mm = 0.02m

x = Dsinujt = 0.02 sin ut = 0.2 sin 16.84?

dx — = x = 0.2 x 16.84 cos 16.84/ dt

= 3.368 cos 16.84?

dx2

—Tr = x= -3.368 x 16.84sin 16.84? dt2

= -56.7 sin 16.84?

3.3.8 Example (3.8) A typical rigidly jointed frame is shown in Fig. 3.24. The beam BCEG is assumed to be infinitely stiff. Assuming the frame is vibrating in a horizontal direction,


Total 5 x 10" kg

6 m

W " V \

I 5 m

G

3 m ( I

H V \ \ W

Figure 3.24. A four-legged frame

determine the natural frequency for this frame. Take

8 = X sin ut

£ = 200 x 10 6 kN/m 2

7 = 0.04 m 4

Solution A four-legged rigidly jointed frame:

E K = E ^AB + E *CD + E ^EF + E ^GH

_ 12£(37) 12£(27) 12E7 12E/

(6 ) 3 + ( 5 ) 3 + " ( 4 7 + 737

= 0.990577

= 0.9905 x 200 x 10 6 x 0.04

= 784.8722 x 10 6

8 = Xsinut

8 = —uX sin ut = UJ 8

MS + k6 = 0

Substituting the values o f 8, 8 and M, the above equation becomes

-Mo? 8 + k8 = 0

(-Mu2 + k)8 = 0

-Mu2 + k = 0

2 3 2 B A N G A S H

u — 784.8722 x 10 6

50000

= 0.036 x 10 3rad/sec

ui / = — = 5.73 Hz

3.3.9 Example (3.9) Figure 3.25 shows a multi-storey frame with masses and second movement o f areas (circled). With the indicated modes o f vibration and the displacement 6 = (A sin ut - B cos cut), determine natural frequencies and periods. Take E = 200GN/m 2 ; / = 0.01 m 4 . Reference coordinates are © and (2) as shown in the figure.

Solution

Multi-storey frame

Stiffness coefficients:

(a) (b)

Figure 3.25. A two-bay two-storey frame


V (4.5)3 ) (4.5)3

k n = &21

8 = Acosut — Businut

8 = — Au? smut — Bu? cos ut = —u28

•: 8 = -u28

Substituting various values in the equation of motion

M8 + k8 = 0

M{-u28) + k8 = 0

or

k8 - u2M8 - 0

k - Mu2 = 0

The determinant of which is given below:

k u — u Mi

<-21

* 1 2

k22 - u2M2

0

or

1.6 x 1 0 6 - 7 x 1 0 V -1.6 x 106

-1.6 x 106 3.2 x 106 - 14 x I O V

98u4 - 44.8 x 103a;2 + 2.56 x 106 = 0

u2 = k

k = 396 or 67

ui = 19.75 u2 = 8.186rad/sec

A = ^ = 3.142 Hz f2 = 1.3 Hz

= 0

Ti = - = 0.318 sec T2 = - = 0.77 sec / 2

234 BANGASH

3.3.10 Example (3.10) A single-bay three-storey building with data given below is subjected to a horizontal dynamic load shown in Fig. 3.26. Calculate displacement at three levels as a multi-degree-of-freedom system.

Data

cj> =

Ul •

M =

Equations of motion in matrix form

Mx + KxP(t)

This equation is coupled due to the stiffness matrix. To uncouple equations, use an orthogonality relationship

(f>jM(f>j = 0. If i = j

2 . 5 x 1 0 4 k g

2 . 5 x 1 0 4 k g

2 . 5 x 1 0 4 k g

Figure 3.26. A frame of a multi-degree-of-freedom system


then

<f)J M 4>i = Lz (say a constant)

l\ = <£[ m <t>,

/ 2 5 0 0 0 0 0 \ / 1 \

= (1 0.533 0.155) 0 25000 0 0.533

V 0 0 2 5 0 0 0 / \ 0 .155/

= (25000 13 325 3875) 0.533

\0 .155 /

= 32702.35 =• Li = 180.84 k g 1 / 2 m

2 X L2 = 4>2 M (f>2

= (1 - 1.51 - 1.24)

/ 25 000 0 0 \ / 1 \

-1 .51 0 25000 0

\ 0 0 2 5 0 0 0 / \ - 1 . 2 4 /

/ 1 \ = (25000 - 3 7 2 5 0 - 3 1 0 0 0 )

= 120442.5

L\ = $ M 03

-1.51

V-1.24/

L 2 = 347.048 k g 1 / 2 m

/ 25 000 0

= (1.0 - 3 . 2 4.5)

0 \

0 25000 0

\ 0 0 2 5 0 0 0 /

( L°\

(25000 - 8 0 0 0 0 112500) - 3 . 2

787250 =• L 3 = 897.27 k g 1 / 2 m

( L°\ - 3 . 2

V 4 . 5 /

2 3 6 B A N G A S H

The following can now be written

1 u

which gives

M 4>i M 4>i

and

because

&

M zj = 0

M 4>i = o

z =

/ 0 U 021 031 \

L\ L2

012 022 032

L\ L2 L3

013 023 033

V Lx L2 L3 /

1.127\

-3.60

5.071 /

x 10 - 3 m k g l / 2 m

kg -1/2

/5 .53 2.88

2.95 -4 .35

\0 .857 -3.573

From the equation of motion

K = Mu2

Hence zTKz = zTMzu2. As zTMz = 1, then

zTKz = 7a;2

At this stage, it is necessary to uncouple the equation of motion by introducing a check of coordinates by writing

x — z • q

x = z • q

Mx — Kx = P( tj

M z # — X z # = p^ tj


By premultiplying by z T , the result is

q-u2q = zTP{t)

/ 2 0 0 O \ 0 13.10 0

\ 0 0 35.26/

/ 2 0 \

rad/sec

(t) 0

V O/

/5 .53

sin 0.37/(kN)

2.95 0.857 1 / 2 0 \

2.88 -4 .35 -3.573 | I 0 I sin0.37/

\ 1.127 -3 .60 5.071 / \ 0 /

/110.6 \

57.6 s in0 .3r ( z [kg- 1 / 2 ]10 - 3 P [ ^ ] 1 0 - 3 ) [kg 1 / 2 /ms 2 ]

\ 22 .54 /

Now we have the set of uncoupled equations:

( h \ / 2 . 0 0 0 \ ( q , \ (110.6 \

92 0 13.10 0

\0 0 35 .26/

# 2

w 57.6

V 22 .54 /

sin 0.3 T

This is a set of three SDOF equations subjected to p0 sin OJT motion in undamped and particular integral solution is given by

<7IW

*i (0 =

=

sin a;?

110.6

4.0 x 2.5 x 10 4

57.6

171.69 x 2.5 x 10 4

22.54 1243.36 x 2.5 x 10 4

sin 0 .3?= 1106 sin 0.4? x 1 0 ~ 3 k g 1 / 2 m

sin 0.3? = 1.34 sin0.4? x 1 0 ~ 5 k g 1 / 2 m

sin 0.3? = 7.25 sin 0.3? x 10~ 7 kg 1 / 2 m

2 3 8 B A N G A S H

Now xt = z • qt

fxi(t)\ /5 .53 2.88 1.127

2.95 -4 .35 -3 .60

\ 0.857 -3.573 5.079

/1 .106 \

*2(0 = U(')/

X\ max

%2 max

x 3 max

x 10 - 3 0.0134 sin0.3/x 10 _ 3 (m)

V 0.000725/

xx{t) = 6.115sin0.3/ x 10~ 6m

(maximum displacement when sin 0.3/ = 1)

x2(t) = 3.20 sin0.3/ x 1 0 _ 6 m

tnn ^maximum displacement when 0.3r = ^

x3(t) = 0.923 sin0.3/ x 10~ 6 m

(maximum displacement when t — mx 5.23 sec)

6.115 x 1 0 _ 6 m

3.20 x 1 0 _ 6 m

= 0.923 x 1 0 _ 6 m

3.3.11 Example (3.11) A single-bay four-storey building frame with data is shown in Fig. 3.27. Calculate frequencies and modes of this frame with a multi-degree-of-freedom system.

Data

stiffness matrix =

/ 1 -1

0

-l

2

-1

0

0

-1

2

-1

0 \

0

-1

2 /

x 10 9 N/m

flexibility matrix =

/ 4 3 2 1 \

3 3 2 1

2 2 2 1

VI 1 1 \ )

x 10 _ 6 (m/kN) = x lO" 9 m/N


1 x 1 ( f N/m

1 x 1 C f N/m

1 x 1 0 y N/m

1 x 1 0 9 N/m

M 2 = 2 x 1 0 4 kg

M 4 = 4 x 1 0 4 k g

Figure 3.27. A single-bay four-storey building frame of a multi~degree-of-freedom system

M =

H=FxM=

n 0 0 o \

0 2 0 0

0 0 3 0

\ o 0 0 4 /

(4 6 6 4 \

3 6 6 4

2 4 6 4

u 2 3 4 /

10 4kg

1 0 _ 5 m 2

(a) Stodola method Based on equation of motion

X = FMX

240 BANGASH

the initial mode shape is

i

*? = i

Hence

1 HXf

But is unknown, hence

X = X ur

Thus, it follows:

xt n x\ n IT? *i 3 xl n xt

1 20 1 96.5 1 15.98 1 15.90 1 1 19 0.95 15.5 0.939 14.98 0.937 14.90 0.937 1 16 0.8 12.6 0.764 12.11 0.758 12.03 0.757 1 10 0.5 7.3 0.442 6.938 0.434 6.884 0.433

Hence, the first mode shape is

/ i \

0.937

0.757

\ 0.433 J

01 =

and

4 xh 1

X 4 i 10- 5 x 15.9

.". u>\ = 79.3rad/sec

for the second mode shape, (i) Calculate sweeping matrix

= 6239.3

0 i = / - 0 i <£[M

0 T M 0 !


or for higher modes

fii

Qm

= 1 J _

M i

Qm-n ~ <A 4>lM

M AJ <t>mM<t>„

Mi = (10 0.937 0.757 0.433)

( \ 0 0 0 \

0 2 0 0

0 0 3 0

\ 0 0 o 4 ;

/ 1 . 0 \

0.937

0.757

V 0.433 /

(10 0.937 0.757 0.433)

fii =

5.226 x 10 4kg

/ 1 0 \

0.937

0.757

\ 0 .433/

/ 1 . 0 1.874 2.271 1.732 \

0.937 1.756 2.123 1.624

0.757 1.418 1.719 1.312

\0 .433 0.812 0.984 0 .748/

/ 0.808 -0.358 -0 .434 -0.331 \

-0.279 0.664 -0.407 -0.311

-0.145 -0.279 0.671 -0.259

V-0.083 -0.155 -0.188 0.357/

/ l 0 0 0 \

0 2 0 0

0 0 3 0

\ 0 0 0 4 /

x 10 4kg/m 2

(ii) Calculate matrix H2 (which eliminates contribution of first mode from the assumed initial shape of the second mode which converge to a true shape

H2 = H1Qn

242 B A N G A S H

For higher shapes

(A 6 6 4 \ 0.808 -0.358 --0.454 -0.339 \

3 6 6 4 -0.179 0.664 --0.407 -0.311

2 4 6 4 -0 .145 -0.279 0.679 -0.251

^1 2 3 -0.083 -0.155 --0.188 0.857 j

/ 0.965 0.306 -0.904 -1.268 \

0.143 0.664 -0 .47 -0.932 x 10" -5

-0 .302 -0.306 0.778 0.016

^-0.317 -0.463 0.013 1.722 j

(iii) N o w de

/ i \

i

- i

x\ — H2X2

x\ n xl x\ x\ n 4 n x\

1 3.443 1 2.455 1 2.42 1.0 2.427 1 2.437 1

1 2.219 0.644 1.451 0.591 1.415 0.585 1.423 0.586 1.432 0.587 - 1 -1.402 -0.407 -0.327 -0.336 -0.756 -0.312 -0.736 -0.303 -0.729 -0.299 - 1 -2.515 -0.730 -1.877 -0.765 -1.912 -0.790 -1.952 -0.804 -1.926 -0.811

02 =

/ 1 \

0.587

-0.299

V-0.811)

2 _ ^2 _ 1 W 2 ~ X\ ~ 10- 5 x 2437

u>2 = 202.56 rad/sec


(b) Hozler method Evaluate second mode shape:

(i) ti>2 assumed = 200 rad/sec, u% = 40 000. (ii) Displacement of mx is taken as unity (i.e. Xx = 1). Thus, new displace

ment of the next pass in the chain, m2, is:

k2

now with mass

OJ2 M~" Xm — Xm-n ' ~j~ 3 MrXr

K " > r = l

Take u>2 = 200 (uj = 40000). Then it follows:

Level X M,Xr £ MTXt

M t X "

1 1 10 4 10 4 0.4 2 0.6 12 x 10 4 22 x 10 4 0.88 3 -0 .28 - 8 4 0 0 13600 0.544 4 -0.824 -32960 - 1 9 3 6 0 -0.774 5 0.05 mm OK

Displacement at level 5 should be 0 at next iteration. Thus: 2

Level X MTXT £ MTXT y-J2M*X*

1 1 10 4 10 4 0.41 2 0.589 1.178 x 10 4 21780 0.893 3 -0 .30 -9000 12780 0.524 4 -0.824 -32974.8 - 2 0 194.8 -0.828 5 -0.0046 mm

Both Stodola and Hozler's iterations converge to the same second mode shape:

( 1 x

0.589

- 0 . 3

^ -0.824 /

UJ2 = 202.56 rad/sec

2 4 4 B A N G A S H

202.56

0.05 0 0.00046

Figure 3.28. Diagram for interpolation

We can interpolate to get better approximation for u (see Fig. 3.28):

u = 202.34 rad/sec

3.4 Response spectra 3.4.1 Introduction The main cause of the structural damage during earthquake is its response to ground motions which are, in fact, input to the base of the structure. To evaluate the behaviour of the building under this type of loading condition a knowledge of structural dynamics is required. The static analysis and design can now be changed to a separate time-dependent analysis and design. The loading and all aspects of responses vary with time which results in an infinite number of possible solutions at each instant during the time interval. For a structural engineer the maximum values of the building response are needed for the structural design.

The response may be deflection, shear, equivalent acceleration, etc. The response curves are generally similar with a major variation occurring in the vertical ordinates. The variations occur with the magnitude of the earthquake and location of the recording instruments. Accelerations derived from actual earthquakes are surprisingly high as compared with the force used in designs and the main reason is the effect of different degrees of damping.

The recorded earthquake ground accelerations have no doubt similar properties to those of non-stationary random functions but owing to a lack of statistical properties related to such motions artificially generated accelerograms are used which are flexible for any duration. A typical artificially generated accelerogram prepared by the California Institute of Technology, USA is shown in Fig. 3.29.

3.4.2 Fourier spectrum The frequency content of a function (accelerogram) can be exhibited by a standard method known as the Fourier spectrum. A typical single-degree-of-freedom oscillator is shown in Fig. 3.30 and is subjected to a base acceleration x, applied force Mx and the response is 6.


1 5 - i (ft/sec 2)

o 9

10

2 - i (ft/sec)

CD

2 J

r o

"i r 2 0 3 0

Time (sec)

r 40 50

" T -10 2 0 30

Time (sec) 4 0 50

l o H 0 £ CD

8 0.5 H Q. ( 0

§ 1.0-1 P 1.5

(ft)

\ 7 ^

r i 10 2 0 30

Time (sec) 40 50

Figure 3.29. nology)

Artificially generated accelerogram (California Institute of Tech-

The equation of motion is

M6 + K6 = -Mx (3.86)

where 6 is the relative displacement and x is the base acceleration. The vibratory response is given for the relative displacement at time by

1 V 8(t, of) = - x(r) sinu;(r - r ) dr (3.87)

w Jo

m ( r ) where T is the period.

2 4 6 B A N G A S H

M x -

J V777777777777777/7,

/777777777777777777T7

Figure 3.30. A single-degree-of-freedom oscillator

The total energy is:

E = \M82 +\K82

= \M Q XSINCTRRDR +Q XCOSARRDR

The duration of x from t = 0 to t — t\, the square root of twice the energy per unit mass at that time t\ is

(3.88)

(3.89)

\2E(t, u>) lYf* . . . A \ 2 /ft .. A V Y — — — = I xsmourarj +1 XCOSCRDR 1

1/2

(3.90)

where

u • 2tt

T 2nf

A plot as a function of u or 71 or/is the Fourier amplitude spectrum. In a normal situation the earthquake spectra is plotted as a function of T. On this basis Hudson and Housner (435) produced a plot for the Fourier spectrum of the ground acceleration component along with the recorded one at Taft, California (Fig. 3.31).

At the end t — t\^ E(t\,w) was the excitation value. The maximum value of the energy E(tm,u) will likely occur at tm < tx. If t\ is changed to tm, the value of E{tm,u) when plotted as a function of period or frequency, becomes the energy response spectrum. If 2E(tm^uj)/M as a velocity is plotted as a function of period or frequency, it is called a maximum velocity response spectrum. This was then plotted for the Taft earthquake in California in Fig. 3.31. One can visualize that the amplitude of the Fourier spectrum is somewhat larger.

If the oscillator in Fig. 3.31 is subjected to viscous damping £ and un — w\J 1 - £ 2 , then its response is

8(t, 0 = — f X(T) e-^n{t-T\smu;n(t - r) DR <*>n JO

for f < 0.2.

(3.91)


3 •o O E

"D C to

o

2

Response spectrum Fourier spectrum

Period (sec)

Figure 3.31. Response spectrum based on Fourier spectrum of ground acceleration component S69E. Recorded 21 July 1952 at Taftt California

When UJ = u m the maximum value occurs at tm, and then it is known as the displacement response spectrum S<i and is generally plotted as a function of period T for several values of £. The maximum velocity |5(f m ,a; ,f) | is called the velocity response spectrum Sy. The absolute acceleration spectrum is 5 a and is given by:

S a = ( ^ ) S d = " 2 S d = ( y ) 2 S d (3-92)

The pseudo-velocity spectrum Spv is derived when using the maximum displacement for zero kinetic energy and maximum strain energy \KS\

\M{8f=\KSl (3.93)

The maximum relative velocity 6 would then be

* = ^ | s d = ( y ) s d = S p v (3.94)

Using the above-mentioned analytical expressions and recording certain well-known earthquakes, various researchers (384-524) have produced response spectra. Figures 3.29 to 3.45 give these spectra plotted for a number of earthquake zones. In some cases various methods have been compared with those recorded from actual earthquakes.

248 B A N G A S H

T, sec

Figure 3.32. Response spectra for elastic systems (May 1940 El Centro Earthquakes, NS component) (after Blume, Newmark and Corning, 1960)

0 0.5 1.0 1.5 2.0 2.5 3.0 Undamped natural period (sec)

Figure 3.33. Velocity response spectrum of the S80E component of ground acceleration. Recorded at Golden Gate Park, San Francisco


Ap rmpia ril 13

, Wa? 194$

shing ) N 8 0

ton E

J J 1

n = 0.0

I ? n = O.C

V 4 y

-~T—

j T , J —H

0 0.5 1.0 1.5 2.0 2.5 3.0 Natural period (sec)

Figure 3.34. Acceleration response spectrum of the N80E component of ground acceleration. Recorded at Olympia, Washington

2 5 0 B A N G A S H

Period (sec)

Figure 3.36. Combined plot of design spectrum for acceleration, velocity and damping as a function ofperiod and damping (20% g acceleration at zero period)

, i i i i 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 1 2.2 2.4 2.6 2.8

1.0 2.0 3.0 Period of vibration, T (sec)

Figure 3.37. Average acceleration response spectrums (El Centro, 1940). (US Atomic Energy Report TID-7024, August 1963)

1.4

£ = 2 % '

£ = 5 % / \

§ = 1 0 %

\ * = 2 0 %

1.0 2.0 3.0 Period of vibration, 7 (sec )

Figure 3.38. Average displacement response spectrum (El Centro, 1940) (US Atomic Energy Report TID-7024, August 1963)

400 4 0 0

2 0 0

.04 .06 .08.1 .2 4 6 8 10 2 0 .4 .6 .8 1 2

Period (sec)

Figure 3.39. Response spectra for Imperial Valley earthquake, 8 May 1940

g 1.2 a?

CD

§ 0.8-o OS Q. CO

T3

£ 0.4-o CD

a o?

2 5 2 B A N G A S H

Elastic response Displacement for \i = 3 Maximum ground motions

Frequency, Hz

Figure 3.40. Inelastic design spectra (Newmark and Hall, 1982)

, — . — • — i • • • -i 0.1 0.5 1

Natural period (sec)

300

200

100

50 H

20

10

= 0.05 = 5

Newmark-Hal l inelastic displacement

/ Newmark-•Hall elastic

S = 10 sees 20 30 40

0.1 0.5 1 Natural period (sec)

Figure 3.41. Mean inelastic acceleration and displacement response for different strong ground-motion durations (Lai and Biggs, 1980)


o § o

T3

300

. 100

50

10

La-Biggs Newmark-Hal l inelastic inelastic x ^ _ d i s p l a c e m e n t displacement / Newmark-Hal l

elastic

0.1 "—[—*•

0.5 Natural period (sec)

Figure 3.42. Comparison of Lai-Biggs and Newmark-Hall inelastic spectra (from Lai and Biggs, 1980)

50.0

0.01 0.02 0.05 0.1 0.2 5.0 10.0 20.0 50.0 100.0

Figure 3.43. Inelastic yield spectra for the S90W component of El Centro, the Imperial Valley earthquakes (El Centro, 18 May 1940). Elastic-plastic systems with 5% damping (Riddle and Newmark)

254 B A N G A S H

0.1 0.2 0.5 1 2 5 1 0 2 0 5 0 1 0 0

Frequency (cycles/sec)

Figure 3.44. NRC horizontal design spectra (1.0g horizontal ground acceleration )

2 5 6 B A N G A S H

3.4.3 Design examples Conversion factors Imperial SI lft 0.3048 m lkip 1000 lbf 1 lbf 4.448 N = 0.4536 kg m/s2

lft kip 1.356 kNm

3.4.3. / EXAMPLE (3.12) AMERICAN PRACTICE

A portal frame built in steel is shown in Fig. 3.46. The horizontal girder 25 ft (7.62 m) is infinitely stiff to prevent significant rotation at the top of the columns that are 15 ft high. The total weight inclusive of self weight is 1.18 kips/ft. Using the following data and the design spectrum developed by Housner (Fig. 3.47), calculate the response of the structure for the earthquake design spectrum with and without 8 in concrete block masonry infill wall:

Steel:

I (column) = 56.4 in 2

Es = 30 x 10 6 lb/in2 (200GN/m 2 )

Concrete block masonry:

EM = 1500 kip/in2

Ey = shear modulus of elasticity = 0.4isM

Masonry weight = 100 lbf/ft2

Using the US code:

K — stiffness = 2 2(12)(30 x 106)(56.4) x 10

(15 x 12) 3

i -3

= 6.963 kip/ft

Floor weight

Steel girder

15ft

Steel columns

25 ft

Figure 3.46. A steel portal frame


100

Period (sec)

Figure 3.47. Housner design spectra

2TT = 0.658 « 0.66 sec

Using 5% damping, from Housner's spectrum, Sv — 10 in/sec. Thus:

5 a = 0 SY = 95.24 in/sec2

0.66 V I x 95 .24= 1.05 in 2TT )

Sd is the relative displacement between the top and bottom of the column. Now:

floor mass x acceleration seismic coefficient

floor weight

Sa 95.24 n „ a - ~ 0.25

g 386

2 5 8 B A N G A S H

The total shear force in the two columns at the ground level is equal to floor mass x acceleration of the mass.

Base shear

or

V = 0.250 x 1.18 x 25 - 7.375kips

V = KSd = 6.963 x 1.05 = 7.31 kips

The difference between the two methods for the base shear is due to the difference between 5 a and the true floor acceleration.

Force-deflection equation

* + A = 1 5 f t = 1 8 0 i n Ac P 3E^I AE\

(my 1.2 x 180

7.62(300)3 (7.62 x 300) x 0.4 x 1500 3 x 1500 x

12

2.337 x 10~ 4 in/kip =

P 1 K - stiffness = 4279 kip/in

Ac ~ 2 . 3 3 7 x 1 0 - 4

One-half weight of the wall = \ x 1 5 x 2 5 x 1 0 0 lbf/ft2 x 1 0 ~ 3 = 1 8 . 7 5 kips

KT = total spring constant = 6 . 9 6 3 + 4 2 7 9 = 4 2 8 6 kip/in

FLTOTAI = 1 - 1 8 x 2 5 + 1 8 . 7 5 = 48 .25k i p s

25 = 185.17rad/sec

2TT

3 8 6

0 . 0 3 3 9 » 0 . 0 3 4 " 185.17

V — SaW = 0.24 x 48.25 = 11.58kips

Using Housner's diagram (Fig. 3.47):

Tn = 0.03

^ = 0 . 4 5 ^ / 8 6 0

5 a = 0.24g and ^ = 0 .0021^


3.4.3.2 Example (3.13) American and other practices Using the following codes with respective conditions and criteria, calculate the base shear force for the framed buildings shown in Fig. 3.48. Use various codes given below:

(a) US code UBC-91 (b) Iran code ICSRDB 1988 (c) Australia code 1995 (d) India/Pakistan codes 1994 (e) Israel code 1994 (f) Japan code 1994 (g) Mexico code 1995 (h) European code 1995

US code UBC-91

V = base shear force

Z = zone 3 = 0.3

7 = 1 . 0

£ W = 3 x 800 + 700 = 3100 kips

i ? w = special moment resisting frame = 1 2 (RC frame)

S = rock = 1.0

ZICW

1 ft = 0 .3048 m 1 kip = 1000lbf 1 Ibf = 4 .448 N = 0.4536 kg m/s 2

1 ft kip = 1.356 KNm

2 2 ft

2 2 ft

W 4 = 7 0 0 kips

W 3 = 8 00 kips

W 2 = 8 00 kips

W 1 = 8 00 kips

4 8 ft

77777" "77777" "77777

2 2 ft » « 2 2 f t — * j

4 @ 12ft = 4 8 f t = 14.63 m

Figure 3.48. A multistorey building frame using various codes

260 B A N G A S H

r=l-25S 1.25 x 1 1.25 x 1.0 T 2 / 3 [0.03 x ( 4 8 ) 3 / 4 ] 2 / 3 (0.55) 2/ 3

_ _ 0.3 x 1.0 x 1.862 x 3100 1 A A ^ n r t . N V = — = 144.305 kips (642 kN)

Iran code ICSRDB 1988

T = 0.07'HV 4 = 0.07 x (14.63) 3 / 4 = 0.524

^(metres) = 0.3048 X 48 = 14.63 m

T0 = soil type 1 = 0.3

4 4 4 8

W = 3100kips = 3100 x 10 3 x - ^ r = 13789kN

/ = importance factor = 1.0

R = behaviour coefficient = 7.0

A = design base acceleration = 0.35

(T \ 2 / 3 / 0 3 \ 2 / 3

B = response coefficient = 2 U = ( ) = 0 6 8 9

where 0.6 < B < 2.0

. ABI 0.35 x 0.689 x 1.0 A A O C

C — seismic coefficient = —— = - 0.035 R 7.0

V = base shear = CW = 0.035 x 13 789 = 475 kN

Australia code 1995

V = total seismic base shear = ^j^- Gg

Rf

Gg= 13 789kN

Vmia = 0.01Gg= 137.89 kN 2.5/ai „ 2 . 5 x 1 x 0 . 1 0

^max = = x 13789 = 575kN

y = l x 0.25 x 0.67 x , 3 7 8 9 = 3 8 6 s k N

6

where a = 0.10

/ = 1.0


: 0 . 2 5 1

or

S(soil) = 0.67

Rf = 6

Kd = 5 (for RC structures)

C = seismic design coefficient

_ 1.25a _ 1.25 x 0.10 _ 1.25 x 0.10

" T 2 / 3 ~ (0.252) 2 / 3 ~ 0.498

^ hn 48 x 0.3048 A „ 1 0 T = = — = 0.318 sec 46 46

^ 4 8 x 0 . 3 0 4 8 , A , . T , R = — = 0 . 2 5 2 sec (adopt this value)

5 8

India/Pakistan codes 1994

Total weight of the building frame 22 W = 13 789 kN

P (soil) = 1.0

I (importance factor) = 1.0

K (performance factor) = 1.0

T (in x-direction) = O.IAZ = 0.1 x 4 = 0.4 sec

. v 0 .09# 0.09 x 14.63 A ^ R (in ^-direction) = —-=— = — = 0.36 sec

Vd V13.41 d = 44 ft = 13.41m

Adopt R = 0.4 for maximum dimension, then C = 0.9, a 0 = 0.08:

V = base shear = KCpia0 W

= 1 x 0.9 x 1 x 1 x 0.08 x 13789 = 992.8kN

Israel code 1994

T (for concrete frame) = 0 . 0 7 3 # 3 / 4 = 0.073(14.63) 3 / 4 = 0.546 sec

22 W= 13 789kN

K — reduction factor = 5.5

I = importance factor = 1 . 0

Z = acceleration factor = 0.1

262 B A N G A S H

S (soil) = 1.2

Ra(T) = spectral amplification factor

But

Ra{T) > 0.2K = 0.2 x 5.5 = 1.1 < 2.246

Adopt Ra(T) = 2.246.

Q = seismic coefficient = RJZ _ 2.246 x 1 xO.l

K J7s = 0.041

SIZ 1.2 x 1 x 0.1 = 0.0295 (adopted)

V3K V3 x 5.5

V = base shear force

= Q £ JP", = 0.0295 x 13 789 » 407 kN

On the basis of Q = 0.041

— ZR^AjCq

Z = 1 . 0 (for Tokyo)

C 0 = (moderate earthquake) = 0.2

T = /j(0.02 + 0.l7) = 14.63(0.02 + 0.1 x 0) = 0.29 sec

(7 = 0 since no steel components are present).

Soil profile I: rock

Tc = 0.4

Rt = 1.0

W <*i = 177 = 0-305

0.041 0.0295

x 4 0 7 « 565kN

japan code 1994 n


1 + 1 + 3 7 -

27/ = 1 +0 .467 = 1.467

1.0 x 1.0 x 1.467 x 0.2 = 0.2934

Y, W = 3100kips = 3100 x 10 3

2240 =1384tonne

Qt=V= 1384 x 0.2934 = 406 tonne

Mexico code / 9 9 5

Total weight £ W = 1384 tonne

C — seismic coefficient

Q' = reduction factor for the moment-resisting frame

= 0.8(2 = 0.8(0.4) = 3.2 (along x-direction)

= 0.8(3) = 2.4 (along ^-direction)

Base shear

V0 = 0.4(1384)/3.2 = 173 tons (along x-direction)

V0 = 0.4(1384)/2.4 = 230.67 tons (along ^-direction)

European code 1995

Fb — seismic base shear = V = S&(T{)W

£ = 0.7

5 = 1 . 0 (for soil)

« g = 0.1g

V = aBSZB0W = 0.1 x 1.0 x 0.7 x 2.5 x 13789 = 2413kN

or

T = 0.29 sec

TB<T<TC

0.15 < 0.29 < 0.60

S d (7 i ) = agSZBQ

where

2 6 4 B A N G A S H

3A.3.3 Example (3.14) Algeria and Argentina practices A five-storey RC building is shown in Fig. 3.49 and is adopted as an office building. Using the seismic coefficient method, determine seismic forces, storey shear forces and overturning moments. Use Algeria regulations RPA-88 and Argentina seismic code INPRES-CIRSOC 103.

Algeria code

T = 0.lN = 0.l x 5 = 0.5sec

A — seismic coefficient = 0.15

D — dynamic amplification factor for firm soil (T — 0.5 sec) = 1.42

B (RC frame) = behaviour factor = 0.25

6

Q = quality factor = 1 + ]P Pq = 1 + 0.05 = 1.05

q=\

V = base factor = ADBQW

= 0.15 x 1.42 x 0.25 x 1.05 x (5 storeys x 1000 kN)

= 279.56 kN

A P CP m

B 3 E3 E3 E3 B Eh<-*.o

c b M M M T±L

3 . 5 - V - 3 . 5 H V - 3 . 5 - 4 — 4 . 5 ^ 3 . 5 -

Levels:

1 and 2 F A , Ac = 0 .3048 m x 0 .508 m l B = 0 .3048 m x 0 .6096 m

All other levels:

l A , l c = 0 .3048 m x 0 .4064 m l B = 0 .3048 m x 0 .5080 m

T 4.5

4.5

F 4"

Roof level 18.0

4th floor 3.5

14.5 y

3rd floor 3.5 1 1 . 5 V

2nd floor 3.5

7 ' 5 V

1st loor 3.5

4 ' ° V

Groun d level

4.0

^ r — 4 . 5 4 . 5 — ^

Figure 3.49. A multi-storey frame - Algeria and Argentina codes [Example (3.14)]


Table 3.6. Lateral forces for the five storeys

Level WK (kN) hK (m) WKhK FK(kN) VK(kN) MK(kNm)

ROOF LEVEL 1000 18.0 1 8 0 0 0 9 1 . 4 9 9 1 . 4 9 0 4TH FLOOR 1000 14.5 1 4 5 0 0 7 3 . 7 0 165.2 3 2 0 . 2 1 5 3RD FLOOR 1000 11.0 1 1 0 0 0 55 .91 2 2 1 . 1 1 8 9 8 . 4 1 5 2ND FLOOR 1000 7.5 7 5 0 0 3 8 . 1 2 2 5 9 . 2 3 1672.3 1ST FLOOR 1000 4 .0 4 0 0 0 20 .33 2 7 9 . 5 6 2579 .6 BASE 0.0 - - 4 0 6 6 . 3 6

Lateral forces (see Table 3.6):

F AV-Fx)WKhK

K T l x wtht

Ft = 0 for T = 0.5 sec < 0.7 sec

N

VK = storey shear = Ft + FK

i=K

N

MK = Ft(hN-hK)+ J2 W i - h K ) i=K+l

N

WKhK = 55000 i=l

Sample FK calculations for the

Roof: Hjjjjjj x (V = 279.56) = 91.49kN

4th floor: x 279.56 = 73.70 kN

Sample VK calculations (Ft = 0) for the

2nd floor: 279.56 - 20.33 = 259.23 kN

1st floor: 279.56 kN

Calculations for similar other floors.

Sample MK calculations for the

Roof: MK = 0

4th floor: 91.49(18.0 - 14.5) = 320.215 kNm

3rd floor: (91.49 + 165.2)(14.5 - 11.0) = 898.415kNm

2nd floor: (91.49 + 165.2 + 259.23)(7.5 - 4.0) = 1672.3 kNm

266 BANGASH

Argentina code (a) Static load (Analysis done in kg units as suggested by the code):

seismic weights = Gt + = 4 kN/m 2 + 0.25 x 8 kN/m 2 = 6 kN/m 2

Total mass load = 6 x 166.5 « 1000 kN

as before. Total seismic weight (kg)

5 x 1000 x 1 kg ^ W = 0.009805 = 5 ° " 5 X 1 0 k g

r)i (for the building) = 0.25

r)j (for roof) = 0

seismic zone 3, soil type 1 (rock)

plan area = 9(4.5 + 4 x 3.5)

= 166.5 m 2

G = 4 kN/m 2 L, = 8 kN/m 2

total number of floors = 5

L = length = 18.5 m

There is no information about the density of the wall and is therefore taken to be zero, i.e. d = 0:

T 0 e = fundamental period of vibration = L 1 + 30J

18.0 / 3 0 „

T o o t e + 2 = 0 " 3 4 3 s e c

(b) Seismic coefficient A 5% damping for seismic zone 3, soil type 1 reinforced concrete-framed building with:

as = 0.25 b = 0.75 Tx = 0.20 sec T2 = 0.35 sec

In this case Tx < T < T2, hence

5 a = horizontal seismic spectra pseudo acceleration


The seismic coefficient C is calculated as

°~ R

= Q ' 7 5 * 1 , 3 = 0.195 (for RC structures with ductility)

7 d = risk factor Group A = 1.3

R = reduction factor = fi = 5 (for T > T{)

V0 = base shear force = CW

= 0.195 x 50.995 x 10 5 = 9.944 x 10 5kg

Comparison with the Algeria code

V = 279.56 kN = 28 512.239 kg = 0.2 x 10 5 kg

the equivalence of

C = ADBQ = 0.056 < 0.195

and the factor for

0.2 x 10 5

The tabulated values, assuming a = 1.0 and a = 0.9 for all levels and the base, respectively, are to be multiplied by 49.72. The frame is subjected to heavy lateral loads and overturning moments and this building in Argentina will have to have a more robust design.

(c) Torsional moment Table 3.6 is extended to include torsional moments. A column is created in Table 3.6 on the basis of:

Mtf = ( 1 . 5 ^ + 0 . 1 0 1 , ) ^

= 0.10LK,

= 0.1 x 18.5K,

= 1.85K/

ex = distance from the CS at level i and the line of action of the

shear force measured perpendicularly, assumed to be zero

all units in MKS, i.e. kg units. The modified V column x 1.85 will give the Mti values.

2 6 8 B A N G A S H

(d) Storey drifts, lateral displacements and storey distortions

Kt = stiffness at ith level = l2EIt/H?

Ht = 3.5 m storey height

Ec = 2.lx 10 9kg/m 2

It — cross-sectional area moment of inertia for the storey column

(e) Storey drift A, Given by:

Level Vt x 10 4kg

Roof 0.9331

4th floor 1.8649

3rd floor 2.2551

2nd floor 2.6439

1st floor 2.8512

Base -

(f) I for columns For floors 1 and 2:

IA = Ic = ± x 0.3048(0.508)3 = 3.3298 x 10~ 3 m 4

/ „ = ^ x 0.3048(0.6096)3 = 5.7539 x 1 0 _ 3 m 4

All other floors

/ A = / c = i (0.3048)(0.4064)3 = 1.7048 x 10~ 3 m 4

IB = i (0.3048)(0.5080)3 = 3.3298 x 10" 3 m 4

Total

I = 6 x IA + 6lB + 6IC

(g) Kj values for floor I Grid A and C:

UEIj _ 12 x 2.1 x 10 9 x 3.3298 x 10~ 3

H] ~ (4) 3

= 1.3111 x 10 6 kg/m


Grid B:

h • J * \ 5.7539 x l O - 3 * -f- x ^ ( G n d A) = — — — — 5 - x 1.3111 x 10 6

IA 3.3298 x I O - 3

= 2.266 x 10 6kg/m

i£,(total)floor = 12 x 1.3111 x 10 6 + 6 x 2.266 x 10 6

= 29.3292 kg/m x 10 6

5V-A,- = storey drift = ~ (total)

A,(m) Mtt(kg/m)

Roof 0.00196 1.726 x 10 4

4th 0.00392 3.450 x 10 4

3rd 0.00474 4.172 x 10 4

2nd 0.0056 4.891 x 10 4

1st 0.0049 5.275 x 10 4

Base - -

(h) Sample calculations for A ( / H f

Roof level = = 0.00056 (m)

Floor A,/Ht

4 0.0011 3 0.0014 2 0.0016 1 0.001123

(i) Kj values for all other floors

\2EIi 1

Grid A and C:

12 x 2.1 x 10 9 x 1.7048 x 10~ 3 , 1 a 6 i . = = 1.002 x 10 6kg/m

(3.5) J

2 7 0 B A N G A S H

Grid B:

7 B / / A x (Grid A) = ^ x ^ x 1.002 x 10 6

= 1.9571 x 10 6kg/m

^(total) = 12 x 1.002 x 10 6 + 6 x 1.9571 x 10 6

= 23.767 x 10 6kg/m

(;) A, (sample calculations)

t „ 5 x 2.8512 x 10 4

1st floor = — — — — — 7 - = 0.0049 29.3292 x 10 6

(k) Sample calculation for the lateral displacement <5,

2nd floor = (0.0049 + 0.0056) = 0.0016 (m)

(/) Sample calculations for /?, for P — A

Floor 1:

ft - 0.00123 f f f ? " f f - 0.2200 H l 2.8512 x 10 4

All results are summarized in Table 3.7. The maximum relative storey drift is 0.0056 < 0.014 as allowed by the code. The

conditions established for a P — A affect /?,• > 0.08. Most values exceed this value, and it is necessary to carry out a P - A analysis, using a spectral modal analysis. The modal shear is used.

Table 3.7. Summary of results for Example (3.14)

Level Ai(m) 6t (m) At/Hi Wt X 105kg ^WtXlO^kg P - A

relative coeff.

drift

ROOF 0 .00196 0 . 0 2 1 1 2 0 .00056 10.199 10 .199 0 . 0 6 1 2

4 0 . 0 0 3 9 2 0 . 0 1 9 1 6 0 . 0 0 1 1 10 .199 2 0 . 3 9 8 0 .1203

3 0 . 0 0 4 7 4 0 . 0 1 1 5 2 4 0 . 0 0 1 4 10 .199 30 .597 0 .1679

2 0 .0056 0 .0105 0 . 0 0 1 6 10 .199 4 0 . 7 9 6 0 .2469

1 0 .0049 0 .0049 0 .00123 10 .199 50 .995 0 .2200

BASE


3.4.3A Example (3.15) American practice ( a ) A three-storey shear wall building is to be designed against earthquake

effects using the US Uniform Building Code (UBC). Calculate the base shear, total lateral shear force and overturning moment at each level. Assume the storey height is 12ft. Use the following data:

7 = 1 . 0 Z a = 1.0 5 = 1 . 5

# = 1 . 3 3 T = 0.15 sec

Each floor weight = 1000 kips. (b) If the shear wall is of reinforced masonry with peak ground acceleration

— 0.4 (for seismic coefficient C s = 2AJR) and Ay = 0.4 (for lower value of base shear C s = \2AyS/RT2^) recalculate base shear, lateral shear force and overturning moment at each level when S — S\ = 1.0 for rock-like formation.

For part ( a )

C = 0.12

S= 1.5

C s = 0.12x 1.5 = 0.18 > 0.14

and

C s = 0.14

£ W = 3 x 1000 = 3000 kips

base shear —V — ZJCSKW = 1 x 1 x 0.14 x 1.33 x 3000

= 558.6 kips (2485 kN)

since T < 0.75 sec

Ft = 0 at top

For lateral forces (Table 3.8):

C = isVt

= 0.171 > 0 . 1 2 15%/0l5

use

272 BANGASH

Table 3.8. Results for part (a) of Example (3.15)

Floor Lateral force Storey shear Storey moment

(kips) (F) (kips) (V) (kipft) (M0)

3 28 0 0 2 19 280 3 360 1 9 470 9000 Base 0 560 15720

Table 3.9. Results for part (b) of Example 3.15

Floor Lateral force Storey shear Storey moment

(F) (kips) (V) (kips) (kipft)

3 340 0 0 2 230 340 4080 1 120 570 10920 Base 0 560 19200

For part (b)

AY = 0.4

R — 3.5 for masonry wall

Aa>3

S = S\ = seismic coefficient

= 1.0

= rock-like formation

Referring to Table 3.9:

C s = HM = 0 .23 m M 3.5

C s = = 0.49 > 0.23 1.2AWS

V = 0.23 x 3000 = 690 kips (3069 kN)

3.4.3.5 Example (3.16) American practice Determine the base shear, storey shear, overturning moment and allowable inter-storey displacement for nine-storey building with moment-resisting steel


frame for an office in California as shown in Fig. 3.50. Use the US UBC-91 code and compare with UBC-85:

Data

storey height = 13 ft = hx

total load = 0.1 kip/ft2 (all levels)

soil profile type 2, i.e. S2 — 1.2

seismic coefficient C = — o r C — *

15VT T2/3

Ductile moment frame and zone location: zone 4:

ZICW base shear V = ZJKCSW or

Rw = 12

For zone 4:

* a = l

7 = importance factor = 1.0 (for the office building)

K - ductile value = 0.67

1 C = 15VT

T = Cth]j4 = 0.035 x (117) 3 / 4 = 1.25 sec (UBC-91)

T = 0.1 JV = 0.1 x 9 = 0.9 sec (UBC-85)

S2 = 1.2

c = ( 1 . 2 5 x 1 2) = 2 3 8 ( U B C _ 9 1 )

(1 .25) 2 / 3 V 7

C = — = 0.07 < 0.12 (UBC-85) 15^0^9

C = \ = = 0.061 < 0.12 15VT55

C s = 0.07(1.2) = 0.084 < 0.14

2 7 4 B A N G A S H

///// T777T /////

- 100f t -

/////

117ft

///// ELEVATION

170 ft

PLAN

Figure 3.50. Moment-resisting steel-framed building


Comparison with UBC-85 leads to:

uj = 0.1(170)(100) = 1700 kips/floor

W = 9 x 1700= 15 300 kips

Z = 1.0

7 = 1.0

y ZICW Ix 1 x 2.38 x 15300 ( U B C _ 9 1 )

12

= 3034.5 kips

C 2.38 n M C

= 0.2 > 0.075 Rw 12

V = 1.0 x 1.0 x 0.67 x 0.07 x 1.2 (15 300) (UBC-85)

= 861.1 kips

(A) Based on UBC-85 sample calculations Vertical distribution T > 0.7 sec, i.e. 0.9 or 1.25 sec:

Ft = 0.07 x 0.9 x 861.1 kips = 54.25 kips

0.25 x V = 215.3 > 54.25

Ft = 54.2 kips

V — Ft = 861.1 — 54.2 = 806.9 kips

, , r- {V-Ft)Wxhx Fx = lateral forces = v „ v x x

„ „ 806.9(1700)(117) e A f % ^ B „ . F > + F < = 994.5 x 103 ; + * . 2 = 2 1 5 .6 k i p S

„ 806.9(1700)(104) f « = 994.5 x 103

The results are summarized in Table 3.10.

Storey shear

Vx = Ft + f^Fi i—x

V9 = 215.6 kips

V% = 215.6 + 143.4 = 359 kips

2 7 6 B A N G A S H

Table 3.10. Individual and cumulative loading (UBC-85)

Level hx (ft) Wx (kips) Wxhx (kip ft)

9 1 1 7 1700 1 9 8 9 0 0 8 104 1700 1 7 6 8 0 0 7 9 1 1700 1 5 4 7 0 0 6 78 1700 1 3 2 6 0 0 5 6 5 1700 1 1 0 5 0 0 4 5 2 1700 8 8 4 0 0 3 3 9 1700 6 6 3 0 0 2 2 6 1700 4 4 2 0 0 1 1 3 1700 2 2 1 0 0

1 5 3 0 0 £ 9 9 4 . 5 X 1 0 3

Overturning moment n

Mx = Ft(hn -hx) + J^ F i ( h i ~ hx) i=x

M9 = 215.6 x 13 = 2803ftkips

M 8 = 215.6(213) + 143.4(13) = 7469ftkips

Inter-storey displacement

A < 0.005Kh = 0.005 x 0.67 x 13 = 0.04355 ft

= 0.0133m

(B) Based on UBC-91 sample calculations

Ft = 0.07 x 1.25 x 3034.5 kips = 265.52 kips < 0.25 x 3034.5

= 758.625 kips

V — Ft — 3034.5 - 265.52 = 2768.98 w 2769

2769(1700)(117) F 9 + F t ~ 994.5 x 103 + 2 6 5 - 5 2

= 553.8 kips+ 265.52 = 819.32 kips

2769(1700)(104) F « ~ 994.5 x 10^ - 4 9 2 - l k l P s

Storey shear

V9 = 819.32 kips

F 8 = 819.32 + 492.1 = 1311.42kips


Table 3.11. Comparative results for Example (3.16)

Level Ft + Ft(kips) Vx(kips) Mx (ft kips)

9 215.6 (819.32) 215.6 (819.32) 2803 (10651) 8 143.6 (492.1) 359.0 (1311.42) 7469 (27700) 7 125.5 (430.6) 484.5 (1772.62 13768 (50343) 6 107.6 (369.1) 592.1 (2141.72) 21466 (62383) 5 89.7 (307.6) 681.8 (2449.32) 30329 (195604) 4 71.7 (246.06) 754.5 (2695.38) 40125 (143 876) 3 53.8 (184.54) 807.3 (2880) 50619 (180917 2 35.9 (123.03) 843.2 (2893) 61 581 (219558) 1 17.9 (61.515) 861.1 (2955) 72775 (258998)

861.1 (3033.865)

Overturning moment

M9 = 8.19.32 x 13 = 10651.16ftkips

M 8 = 819.32(2 x 1 3 ) + 492.1

= 26 223.32 ft kips

A = allowable interstorey displacement (for 0.7 sec or greater)

^ x s t o r e y height n or 0.004 x storey height

^ x 13 = (0.001 m) ? 0.004 x 13 = 0.052ft (0.016m)

Table 3.11 gives the final comparative results:

3.4.3.6 Example (3.17) American practice A two-bay two-storey shear wall building is in a seismic environment. Using the following data and assuming that the natural frequencies fall into the constant acceleration range of the response spectra, calculate storey forces, overturning moments and displacement.

Data

storey height = h

floor weight = Ws

Ww = weight of shear wall/unit length

2 7 8 B A N G A S H

Floor is asumed to be infinitely rigid:

/ fx = natural frequency =

\.2EI/h3

1.54/?MW + 1.125MS

Floor is assumed to be infinitely flexible:

/]», = •

/ B 2 — '

where

OAEI/h3

0.514/?MW + 0.281M S

OAEI/h3

0.514/iMw + 0.562M,

(shear walls 1 and 3)

(shear wall 2)

g

w g

Use the Rayleigh principle and assume that the shear wall mass is lumped at floor levels.

The deformation shape (see Fig. 3.51) is given by

d>(z) z 4 4z 3 z 2

^max i - I f.

3 U4 L3 + I? )

Specific deformation shape = (f){z) = \[(z*/\6hA) - (z3/2h3) + (3z2/2h2)} for this building.

The combined mass at the floor level is treated as M s w and M w = 0 in the above equations, i.e. Ww = 0 and is now Wm to replace Ws:

swA

JPswBi =

\A25W^/g

= i/1.19 „ 1 , shear wall 1 and 3 0.281 Wm/g Wsw/g

8 h -

Figure 3.51. Diagram for deformation shape calculation


0-4/J//A3 / Q o , EI/h3

0.562^sw/g V ' W™/8 W = K — = i/0.84 shear wall 2

or

Deformation equation ^> (Z ) applies when:

z = h <j>{z) = 0.354 fl.OOl Z = 2/? </>(Z) = 1.00 > { < t > l i ~~ \0 .35 J

for the first case W — Wm

for the first case Bi W = \ WSVf

for the first case B 2 W = \ WSVI

The lateral force at each level

= M(Z)<T3(Z) X 1.204 x 5 .

F(z,t)mm = 1.204 Y ( / » ( Z ) 5 a

^(2/2, O i n a x = 1-204 — X 1 .0^ = 1.204 — Sa (level 2)

F(h, r ) m a x = 1.204 — X 0.354S a = 0.426 — 5 a (level 1) g g

F(t) = base = 0

Storey shears

level 2: 0

W level 1: 1.204 — 5 a

base: (1.204 + 0.426) — Sa = 1.630 — 5 a

280 B A N G A S H

Overturning moment

level 2: 0

W

g level 1: 1.204 — W a

base: 2.834 — hSa

W g

lateral displacements = 1.204<^(z)5d

1

(fi)2

Note that f\ varies for each case.

Storey displacement

z = 2h

z = h

0(z)1.2O45a

A Bi B 2

1.2045a 1.2045a 1.2035,

fi f2

J B, / B 2

0.4265 a 0.4265a 0.4265 a

fi / | 2

3.4.3 J Example (3.18) American practice For the collapse level, the earthquake spectral design requirements are:

S a = 0.60g

Sy = 21.5 in/sec

Sd = 11.9in

Take 1 in = 25.4 mm. The elastic response spectrum for a 5% damped system ductility = 2%.

Correct the inelastic yield spectrum:

velocity region = - = ^ = 0.5

, • • l l

acceleration region - .. 4 = . --- 0.58

1 1 n , displacement region = — = - = 0.5


In summary (1) S a (inelastic) = .Sa (elastic) x 0.58 = 0.348 g

acceleration region (2) Sy (inelastic) = Sy (elastic) x 0.5 = 10.75 in/sec

velocity region (0.273 m/sec) (3) Sd (inelastic) = Sd (elastic) x 0.5 = 5.95 in/sec

displacement region (0.151m/sec) (4) The constant displacement line on three-way log paper. Draw the ground

acceleration Une on three-way log paper. Identify point on ground acceleration line corresponding to period of 0.03 sec, i.e. point 'a'. Identify then point *b' on the constant acceleration line. Draw elastic spectrum similarly for 10% damping.

The plot and data are shown in Fig. 3.52.

Period (sec)

Figure 3.52. Spectral velocity versus period

2 8 2 B A N G A S H

3.4.3.8 Example (3.19) American practice Using the following data and the Uniform Building Code (UBC) of the USA, design the beam-column connection and analyse the concrete ductile frame (Fig. 3.53).

Data All beams 12 in x 28 in All columns 16 in x 27 in Working loads:

Wd = dead load = 700001bf

/ ^ = 40001bf/in2

fy = 600001bf/in2

Z = zone 3 = 0.3

/(occupation factor) = 1.25

C = seismic response = 1 . 0 for rock-like soil

i ? w = system quality factor = 12 full ductile moment-resisting frames

T = 0.15 sec

C 1.25

= 4.43 > C = 2.75 (0.15) 2 / 3

W 5 x ^ s

5

4

3

2 storey height = 14' 1 4 ' - 0

24'

Figure 3.53. Concrete ductile frame


If T < 0 . 7 seconds

Ft = horizontal force = 0

Fi = lateral earthquake forces in which

Ft = 0 at the ith level

Ft = QV = OMQWs

Initial value to start with is: n

] P ( 1 + 2 + 3 + 4 + 5 ) Ws (i = 5 at the topmost floor)

All floors need to be analysed for lateral forces, storey shears and storey moments.

Brief design notes and design formulae UBC and ACI Relevant terms are

Ws — weight of one storey

i ? w = system quality factor

W = total vertical weight of building = nW$

S = site factor

C = seismic response = ( 1 . 2 5 / 7 7 ) 3

n = number of storeys

Step I Determine earthquake zone, select UBC seismic coefficients Z , / , C, S, i ? w . Determine period T by UBC Method A or B.

Step II Compute

ZTC n

V — —— W and V = Ft + V Ft

i t = 0 when T = 0 .7s

F t = 0 . 9 7 R C < 0 . 2 5 F

Tn - Ct(hwf4

2 8 4 B A N G A S H

Step III

Tabulate base lateral force and each storey force Ft to Fn:

F . =

w& y

Find each storey shear and moment:

_{V-Ft)Wxhx

x ~ £ win

Step IV Carry out a structural frame analysis to determine all shears and moments in the frame beams, columns, and shear walls.

Step V Revise where necessary the size and main reinforcement of the moment-resistant frame members: beams, and beam-columns (beam-column when

P u > vVio). Step VI Use strong column-weak beam concept, plastic hinges in beams and not columns:

£ M c o l > 6 / 5 M b m at joint.

Seismic beam shear forces:

n r / M~h + M p

+

r R 1 . 4 0 + 1.7X Beams: VL = — - — - — - — H 0.75

v M+L + M~R L4D + VJL = £ ° ' 7 5 2

Seismic column shear:

_ M p r l + M p r 2

^ = beam span

M p r = probable moment of resistance

L = left ] > subscripts

R = right J

A = column height

Hinges at beam + ^ M " ) ^ > 6/5(<T5AC + 0M")bm

M n = nominal moment strength


Step VII Design longitudinal reinforcement.

(a) Beam-columns or columns:

0.01 < Pg < ^ < 0.06

For practical considerations pg < 0.035:

/Wn < ~ > ^ # (for +M) > (for —M) Jy Jy Jy

(fy is in lb/in2 and p ? 0.025). (b) Beams: at joint face > \ M~ at that face

Mn or M~ at any section > \ M a m a x at face

The above are material properties as adopted in Example (3.19).

Transverse confining reinforcement (a) Spirals for columns:

A „R >0AS(4l.M' fyh V ch /fyh

whichever is the greater. (b) Hoop reinforcements for columns:

f Ash > 0.09shc ^

Jyh

> 03sh V ^ c h / fyh

A% — gross area

^ c h — c o r e a r e a to outside of spirals

hc — column dimension

s — \ of smallest cross-sectional dimension or 4 in, whichever is smaller. Use standard tie spacing for the balance of the length where Ash = area of transverse steel.

(c) Beams: Place hoop reinforcement over a length = 2h from face of columns. Maximum spacing: smaller of s = \ d, Sdh main bar, 24db hoop, or 12 in. If joint confined on all four sides, 50% reduction in confining steel and increase in minimum spacing of ties to 6 inches in columns is allowed. Use the standard size and spacing of stirrups for the balance of the span as needed for shear.

2 8 6 B A N G A S H

Beam-column connection (joint)

Available nominal shear strength > applied Vu.

Confined on all faces:

Vn < 2 0 v 7 U j

Confined on three faces or two opposite faces:

Vn < 1 5 ^ j

All other cases:

Vnl2y/fUi

A} — effective area at joint reduced by 25% for lightweight concrete. Horizontal shear is calculated if 1.25/y applies for tensile reinforcement.

Development length, normal-weight concrete bar sizes #3 to 1 hooks the largest of:

^ d h > / y 4 / 6 5 v / / T > 8 4 > 6 i n

£d = 2.5£dh for 12 in or less concrete below straight bar

£d = 3.5£dh f ° r > 12 in in one pour

If bars have 90° hooks, £ d = £ d h . For lightweight concrete, adjust as in the A C I Code. Bar designations and dimensions are given in Tables 3.12 and 3.13.

Table 3.12. Reinforcement designation and bar areas used in the USA

Bar size Nominal cross- Weight Nominal British or designation sectional area LB PER FT diameter European codes

FSQ. in) (in) equivalent

# 3 0 . 1 1 0 .376 0 .375 A

# 4 0 .20 0 .668 0 .500 # 5 0 .31 1.043 0 .625 # 6 0 .44 1.502 0 .750 SEE EUROCODE 2 # 7 0 .60 2 .044 0 .875 B S 8 1 1 0 # 8 0 .79 2 .670 1.000 FOR BAR # 9 1.00 3 .400 1.128 DIAMETER

# 1 0 1.27 4 .303 1.270 DESIGNATION

# 1 1 1.56 5 . 3 1 3 1 .410 # 1 4 2 .25 7 .650 1.693 # 1 8 4 .0 13 .600 2 .257 r


Table 3.13. ASTM metric reinforcing bars*

Bar size

designation Nominal dimensions Bar size

designation

Mass (kg/m) Diameter (mm)

1 0 M 0 .785 11.3 1 5 M 1.570 16.0 2 0 M 2 .355 19.5 2 5 M 3.925 25 .2 3 0 M 5.495 29 .9 35 M 7 .850 35.7 4 5 M 11.775 43 .7 55 M 19.625 56.4

* A S T M A 6 1 5 M GRADE 3 0 0 IS LIMITED TO SIZE NOS. 1 0 M THROUGH 2 0 M ; OTHERWISE 500 M P A FOR ALL

THE SIZES. CHECK AVAILABILITY WITH LOCAL SUPPLIERS FOR NOS. 4 5 M AND 55 M .

Design shear wall Fuh > 2Acy^/fl; use two reinforcement curtains in wall. If wall fc > 0.2ff

c, provide boundary elements.

Available:

Vn = Acy(as^c + pnfy)

for / * w / 4 > 2 . 0 , a s = 2.0

for hw/£w = 1.5, as — 3.0

Interpolate intermediate values of /z w /^ w . Maximum allowance:

vn = %Acy\ff[ for total wall

vn = 10Acp\/f[: for individual pier

py = 0.0025

Acy = net area of cross-section

= thickness x length of section

(j) = 0.6

Vu = ct>Vn

Acp — cross-sectional area of the individual pier

288 BANGASH

Compressive strength:

/ ' c = 20MPa

/ c = ^ 5 0 . 0 4 3 ^ M P a

(j> = 0.9 for beam

cj) = 0.07 or 0.75 for columns

Design calculations

Rw

Rw = 12

1= 1.25

1.25

( Z for zone 3 = 0.3)

C = J 2 / 3

5 = 1.0

T = 0.7 sec

/ t = 0

Floor load (refer to Table 3.14):

Ws=rVD+WL = 70 000 lbf + 80 000 lbf = 150 000 lbf

or

(3.13.83 kN) + (359.467kN) = (673.297 kN)

Hence, base shear V = 129 585 lbf.

Table 3.14. Data for loadings

Floor Q F=CtWs

(M) Storey shear (M)

Storey moment (lbf ft;

5 c5 = 5WS/15WS = 0.333 45960.0 0 0 4 C4 = 0.267 36850.8 45960.0 643440.0 3 c3

= 0.200 27603.6 82800.0 1802640.0 2 c2

= 0.133 18 356.4 110400.0 3 348240.0 1 C\ = 0.0066 911.0 128760.0 5150880.0 Base 6 = 0 0 129585.0 6965070.0


D = 70000 lb f L = 80000 lb f

M A

/n = 2 4 '

V A = 35 000 Ibf (D) L = 8 0 0 0 0 Ibf (L)

Figure 3.54. Moment diagram for the design calculation

The moment at each level is given in Table 3.14. Seismic beam shear forces (Fig. 3.54):

y ^ M p r L + < r R | ^ IAD + ML

^MA + MB | Q 7 5 IAD + ML

where £ is the beam span, and M p r = probable moment of resistance. Now:

v M p

+

r L + M p r R ^nc\AD+ML = i 0 7 5 2

where L, R = left and right

d = 27 - 2 . 5 = 24.5 in

^ s = 4 in 2

4 0.0136 < 0.025 (allowed by AC318)

r 12 x 24.5

Mn = Asfy(d-^j

Asfy 4 x 60000 a = -— = = 5 9m

0 . 8 5 / ^ 0.85 x 4000x 12

Mn = 4 x 60000 ^24.5 - ~^j= 5172000Ibf

290 B A N G A S H

This value can be obtained from structural analysis of these frames. Hence:

5172000 + 5172000 A neflA x 60000 + 1.7 x 75000 \ V h = 24712 + 0J5{ 2 )

= 35916.67 + 0 . 7 5 ( 8 4 0 Q 0 +

2

1 2 7 5 0 0 ) 2

= 35916.67 + 79312.5

= 115229.17M

V* = 2v7 M = 2 V 4 0 0 0 x 12 x 24.5 = 37188.385 lbf

According to the ACI318-95 code Vc can be zero under certain conditions. Now:

Vn at dn from the face of the support (nominal shear strength)

= 115 2 2 9 . 1 7 ^ 1 2 ~ 1

2

2

8 / 1 2 ) = 92 826.7 lbf

K = Vn - Vc = 92826.7 - 37 188.385 = 55638.3151bf

Vc = seismic column shear force = ^ p r l ~|" ^ p r 2

h In the US, generally the No. 4 (#4) hoop reinforcement is used:

#4 bar size, As = 0.2, nominal diameter = 0.375 in

AY = 2 x 0 . 2 = 0.40 in2

_ Avfyd 0.40 x 60000 x 24.5 S = -TT = 55638J15 = 1 0 - 5 7 m

Four closed hoops are placed at 10^ in c/c on the critical section. When S is increased to

- = — - = 12.25 m 2 2

the code recommends curtailing stirrups at

^ = 18 594.193 lbf 2

Confinement of reinforcement in the beam at the beam-column joint (Fig. 3.55) gives:

^column — (7 + 7)12

5172000

14 x 12

30 785.714 lbf


V column

- = T

column size 16" x 24"

joint *r

V column

Figure 3.55. Schematic of the beam-column joint

Vn at a joint = Asfy - V^i^

= 4.0 x 60000-30785 .714

= 209 214.299 Ibf

Vn is allowable when < 15 ^/f^Aj

where

Aj = area at the joint = 16 x 24 = 384 in2

Vu = allowable when 0.85(15v /4000 x 384) = 309 650.23 Ibf

actual Vu = 129 585 Ibf (permitted)

column d = 24 - 2.5 = 21.5 in

Vc at the Aj plane = 2y/fc x 16 x 21.5 = 2V^000 x 16 x 21.5

= 43 512.941 Ibf

V- 309650.23 Vc =—— 43 512.941 0.85 c 0.85

320 781.45 Ibf

Ayfyd _ 0.4 x 60 000 x 21.5 V. ~ 320781.45

Maximum allowable

21 5 S = d/4 = —- = 5.375

0.643

(Thus, 9 #4 bars)

2 9 2 B A N G A S H

Adopt 3 in {i.e. 4 in diameter}

9 84 = 8 x - = 9in 8

4 x 24 = 24 x ^ = 12 in (minimum)

Thus, use four hoops plus 2 #4 cross-ties at 3 in c/c. The length of such hoops:

4 = 2A = 2 x 28 = 56 in

Reinforcement in the column at the beam-column joint Whichever Ash is greater should be adopted:

Ash > 0.009Shcf'Jfyh

or

Ash>03Shc(^-l)^ \ A c h J ^yh

hc — column core

= 24 - (2 x 1.5 + 2 x 0.5) = 20in

Try S = 3 in.

^ = 0.09 x 3 x 2 0 ( ^ ) = 0 . 3 6 i „ ;

or

^ = 0 . 3 x 3 x 2 0 ( ^ - 1 ) — = 0 . 8 9 , ^

Now, *S (maximum allowable) = \b — \ small dimension or 4in

— I x 16 = 4 in

Therefore, use #4 hoops at 3 in c/c, and place the confining hoops in the column on both sides of the joint over a distance £ 0, using the largest of

( a ) depth of the member = 24 in (b) \ x clear span = \ x 24 x 12 = 48 in (c) 18 in

Thus: £q = 48 in; #4 hoops and cross-ties at 3 in c/c spacing (Fig. 3.56).


-vA-

#4 @ 3 in c-c

16 sp @ 3 in

4 No. 9

No. 4 @ 3 in c-c

« -18sp @ 3in =54in » | « 9 in-*--«-9 in—»-J

• 2 in

v #4 @ 3 in c-c

-*-hc=20 i n - *

8 No. 9

Figure 3.56. Reinforcement in the column at the beam-column joint

3.4.3.9 Example (3.20) European practice

The four-storey three-span concrete frame o f a regular building is shown in Fig. 3.57. Adopting Eurocode 8 Parts 1-5 and the following data, design this frame for an earthquake environment.

Data Concrete density (specific weight) = 24 kN/m ; concrete class C30

E m = Young's modulus = 30.5 GPa (30.5 x 10 6 kN/m 2 )

294 BANGASH

lio

11

12

113

14

15

16_

17

18_

19_

20

hi

hi

hi

hi

/ / / / / "77777" "77777" "77/77"

3 @ 5 . 0 m

Frame Dimensions

Figure 3.57. Frame dimensions for Example (3.20)

v = Poisson's ratio = 0.2

fck/lc — design compressive strength = 30/1.5 = 20 MPa (20MN/m 2 )

r R d = design shear strength = 0.30 MPa (0.3MN/m 2 )

hc = 3.0 m

hx = 4 . 0 m

All columns 1-2 levels = 500 x 500; all beams at 1-2 levels = 400 x 400. All columns 2-3 levels = 500 x 500; all beams at 2-3 levels = 400 x 400. All columns 3 - 4 and 4 - 5 levels = 400 x 400. All beams 3 - 4 and 4 - 5 levels =500 x 500.

Eurocode class Steel class S400

fyd = design tensile strength = fy^/js

= 400/1.15 = 348 MPa (348MN/m 2 )

Design loads

g = dead load (floor, slabs, beams, finishes etc.) = 6 kN/m 2

G = 6kN/m 2 x 5.0m = 30kN/m

Imposed or live loads

qk = 2.0 kN/m 2

g k = 2 x 5 = 10kN/m


Design seismic action Design response spectrum derived from:

I = 1.0; d% — effective peak ground acceleration = 0.35 g more than El Centro.

S = site coefficient = 1.0

B0 = amplification factor = 2.5

TB = 0.4 sec

q = behavioural factor (Eurocode 8) = 3.5 (assumed)

Se(T) = 1 . 0 ( 0 . 3 5 ^ ) ( 1 . 0 ) ( 2 . 5 ) ( ^ ) 2 / 3 ( ^ )

/ 0 . 4 \ 2 / 3

= 0 .25 , ( T J (g<0.25g)

Loads and inertial forces

tpi = i>2 = 0-2 - the loads become quasi-static permanent live loads Q or (30 +

0.2 x 10) = 32 kN/m

floor = 32(3 x 5.0) = 480 kN

Equal at all floors:

Floor2 = 4 8 0 = W2

Floor 3 = 480 = W3

Floor 4 = 480 = W4

Floor 5 = 480 = W5

(Note that h( = interstorey height)

Seismic coefficient = 0.25

A - . U f . U T,Wt . (1 + 1 + 1 + 1 ) ^ 7,- = distnbution factor = ht ' = ht ZWthi ' (4 + 7 + 1 0 + 1 3 ) FT,

4 = _ / j . = 0.1176471/1,

2 9 6 B A N G A S H

183.7 KN Floor 4

141.4 Floor 3

98.85 Floor 2

56.55 Floor 1

/////

Figure 3.58. Horizontal forces at each level offrame

Horizontal forces at each level The method has been demonstrated in other examples and the horizontal forces are computed (Fig. 3.58).

Load combination for structural analysis Eurocode 8 specifies at least two load combinations: (a) with the live load as the main variable action and (b) with the seismic load as the main action:

(a) Sd = S(1.35G+l.5Qk) i > load combinations

(b) Sd = S(G + J2^Qk + E)

The frame has been analysed for these two loading cases using an IBM PC and the above data with the Program STIFF. Design moments are based on the linear analysis of the building frame. Redistribution is permitted (see Fig. 3.59).

(A) Beams From seismically considered situations, all beams shall be equally reinforced to cover 40 mm or 4 cm beams of 500 x 500.

MRd = the ultimate bending moment at support section:

positive = 240 kNm > 145

negative = 355 kNm > 265

See Table 3.15 for data.

Table 3.15. Beam loading

Main reinforcement Eurocode British code

A T SUPPORT 4 DIA. 3 2 TOP 4 T 3 2

3 DIA. 2 5 BOTTOM 3 T 2 5

A T M I D SPAN 2 DIA. 3 2 TOP 2 T 3 2

3 DIA. 2 5 BOTTOM 3 T 2 5


265 257 257 265

All values (kNm)

Figure 3.59. Loading cases for structural analysis

Design shear forces Conditions of static equilibrium at the beams subjected to vertical loads and end moments give design shear forces. The end moments must be from the actual reinforcement placed in beams. Eurocode 2 suggests the following beam analysis to evaluate moments and reactions. Each section at the end shall have two values of shear force - the maximum and the minimum value corresponding to positive and negative moments yielding at the hinges. The algebraic ratio between the maximum and the minimum values of the shear force at a section shall be

298 B A N G A S H

M A

Dead + semi-permanent load Dead + semi-permanent load

(i)

1 10 (ID

t v A , d

V - V , M A + M / B V A , d ~ V A , d + "

F . V - V . M A + M B

S * V A , d - V A , d + :

Figure 3.60. Isolated moments and shears or reactions (Eurocode 2 requirements)

denoted by For the purposes of design the value of £ shall not be taken as smaller than - 1 (Fig. 3.60):

u or p — 32 kN/m

*W = , 2 x 1 • 355 + 240 = +199 kN ViMn 2 5 -3 .9 kN

€ = ^ = - 0 . 0 1 9 6 « 0

Resistance to shear (contribution of concrete) = 0 Vertical stirrups area

A™-Q.9d\fyJ = 13.83 cm 2 /m = 1383mm 2/m

Now:

2 dia. 10-125 Asw = 1570mm 2 /m (limited to a metre from the support) (2T10-125) 2 legs of 10 mm dia.

Vmzx < ^ R d i = 6 r R d • K • d = 6 x 0.3 x 500 x 460 x 10~ 3

= 414kN

f'max = 199 kN < 414 kN (permitted)

(B) Columns

Critical columns are on the 2nd and 3rd floors, specifically the two at the centre. A reference is made to the loads and bending moment diagrams in Figs 3.59 and 3.60 based on unfavourable combinations of loads. Only partially critical results


are indicated from the linear structural analysis. For regular structures o f three storeys or higher, the column moments due to the lateral forces alone shall be multiplied by a dynamic magnification factor.

For a planar frame such as this

^ = 0 .67^+0 .85 (1.3 < v < 1.8)

where Tx is the fundamental period o f the structure. Tx (Eurocode 8) is in the range o f 0.4 to 0.6. Hence

u = 0.6(0.6) + 0 . 8 5 = 1.21 > 1.3

Therefore u = 1.3 is retained.

Relative strength between columns and beams at joint This requirement o f Eurocode 8 is to be adhered to. Thus:

Mb = sum o f beam moments = ( M R d + M^d)

= (240 + 355) = 595 kNm

Mc = sum o f column design moments

= 1.3(155 + 231) = 501.8 kNm

Reinforcement

Adverse combinations o f axial force N and bending moment M can be adopted:

upper column

7V = 320kN

M = 1.3 x 177 = 230.1 kNm

lower column

Af = 565kN

M = 1.3 x 227 = 295.1 kNm

The section is therefore checked. Actual reinforcement provided in the section (avoiding traditional column calculations) is

8 dia. 25 and 12 dia. 20 Eurocode

(8T25) (12T20) British standard

As = 3932 m m 2 As = 3760 m m 2

Total A^ = 7692 m m 2

3 0 0 B A N G A S H

The corresponding ultimate bending moments:

upper column

M R d u = 265 kNm > 230.1 kNm

lower column

M R d L = 392 kNm > 295.1 kNm

Design shear forces

These are due to the contribution of the bending moments:

upper column

V = 1.3(167 + 177)/3 = 149.07kN

lower column

Md — design moment

Coefficient Be(l < Be < 2) takes into account the axial load. Neglecting the reinforcement contribution the two cases are again con

sidered as:

upper column

V= 1.3(210 + 227)/3 = 189.37kN

Shear resistance

N = 0.\Agfci

320 > 0.1 x 0.4 2 x 1670 = 26.72

535 > 0.1 x 0 . 5 2 x 1670 = 41.75

Vcd = 2rmbwdBe

M 0 = decompression moment — —rW

1 0.4 1 + — x 320 x — = 1.085 < 2

265 6

2 x 300 x 0.4 2 x l . 0 8 5 = 104kN


lower column

* - I + ( i 5 r 5 M T ) = 1 + 5 E x 5 3 ' x T = M 1 4

Vcd = 2 x 300 x 0.5 2 x 1.114 = 167kN

The shear force carried by the concrete is no different from the total shear force computed. Therefore, no change in reinforcement is needed.

Stability for a planar structure The stability index is based on the secondary effects o f storey shears and storey moments. Every floor condition has to be satisfied:

c9 = ^ < 0 . 1 0 = §4 Vh Vtothi

where

9 = deformability index ^ 0.20

V = seismic design shear force across the storey

A e l = elastic inter-storey drift due to actions

q = behaviour factor

h = floor height

W = vertical load above the storey in consideration o f 9 ^ 0.20

The above means no inelastic analysis is needed for the 0.10 < 9 < 0.20 elastic static approach, although conceptually inappropriate, it can provide extra strength:

W = 4x480 = 1920 kN

q = 2.0

V = Se(Tl)W = 0.27 x 1920 = 518.4kN

h = 3 m or 4 m

Take h = 3 m from the method given for the storey drift A e l = dj = 0.006:

0 =

1 9 2 Q

5

X

1 8

Q

4 ° ^ 6

3

X 2 0 = 0.0148148 « 0.014815 < 0.10 (permitted)

With 4 m height h:

9 = \ x 0.014815 = 0.011 < 0.10 (permitted)

3 0 2 BANGASH

Take the average:

« . f t M g l i + 0 1 0 = 0.0.3 < 0.10 (permitted)

Thus:

A 0 0 1 0 , / X A / \ A = h (m) A e l (m)

(1) = 0.010/2 x 4.0 = 0.02 0.006

(2) = 0.010/2 x 3.0 = 0.015 0.0055

(3) 0.015 0.0050

(4) 0.015 0.0040

The deformability condition is satisfied.

Note that for serviceability verification, the elastic drift A e l resulting from the application of horizontal forces or from the dynamic action shall at any storey satisfy the condition:

A ^0.010 , e q

For certain buildings, the indicated limits may be increased by 50% where significant damage is expected. Where the limits are exceeded separation of nonstructural elements occurs. To take the drift without restraint:

A = 0.35A e l • q

A = 0.35 x 0.006 x 2.0 = 0.0042m

Maximum limit established by Eurocode 8 is:

0.025 , ^max '

0.025 x 3 = 0.0375 m

0.025

2.0 or

2.0 x 4 = 0.05 m (this value adopted)

The values in the previous table are correct and do not exceed A m a x . Tensile reinforcement shall not be less than:

Xir > 7- = 7?£ = 0 0 0 3 5 > ° - 0 0 6 (Permitted) Jyk Jyk. 4UU

The final structural details for Example (3.20) are given in Fig. 3.61.


4 0 0 x 4 0 0

4 0 0 x 4 0 0 -

5 0 0 x 5 0 0 -

T 100 T 1 0 - 1 0 0 - 1 m length ^ W T T 1 0 0 ^ T 1 0 - 100 - 1m length

T 1 0 - 1 2 5 T 1 0 - 2 5 0

4T32 4T32 I 30 x 5 0 I

3T25 3T32 3T25

T 4 T 3 2 1500 I

- L 3T25

T 1 0 - 1 0 0

r - 5 0 0 x 5 0 0

4 0 0

4 0 0

5 0 0

500

8T25 12T20

Figure 3.61. Final structural details

3.4.3.10 Example (3.21) British practice For the response-time history using direct integration, the equation for the dynamic equilibrium, ignoring initial damping, is

where M is the mass, K is the structure stiffness and F(t) is the time-dependent force. Develop the direct integration procedure that is generally met in earthquake analysis. A two-dimensional rigidly jointed steel frame with rigid support provides a single portal frame of 4m height (Fig. 3.62). The horizontal girder to the two vertical members is infinitely stiff and carries a distributed mass of

F(t) = 1 0 k N -B 5000 kg total

4 m

AI ID ///// -77777

- rigidly jointed girder

Figure 3.62. Response analysis of a portal frame

3 0 4 B A N G A S H

500 kg. At the girder level an earthquake force of lOkN for 3 seconds exists. Using the following data, develop the response-time history for this frame:

/ = second moment of area = 2 x 10 3 cm 4

E — Young's modulus for steel = 200 kN/mm 2

Initial time-step = 0.015 sec

c = damping of 10%

Construct the response of this portal over a period of 3 seconds indicating factors relating to frequency and amplitude of the dynamic response:

S v (average acceleration at time t = 0) =

Sy

fy/Stt=o.5 - 8y/Stt=_0,5

At

St Sy St

s 2 y

fit1

(average velocity at t = +0.5) =

(average velocity at t = —0.5) =

y,=i -2yt=0 + yt=_l

yt=i - y<=o At

yt=o - yt=-\ At

(At)2

At general time t — r. Hence

M ^ + i - 2 ^ + A - ! = p { t )

(At)2 V '

(3.95)

(3.96)

(3.97)

(3.98)

(3.99)

where yT = amplitude of displacement at t = r and similarly others in the above forward difference equations. For example at time t = r + 1

j r + i = yT 2-

K(Atf M

Now:

T = natural period = 27ry — K.

-yT-i+F(t)7

M

(At)2

M (3.100)

(3.101)

In order to calculate displacement at t = + 1 , one must obtain yt=-\. One has to know the initial velocity 8y/8t — 0, which is:

yt=+i - yt=-i

or

2At

yt=-\ = yt=+i - 2 A ' _6y_

(3.102)

(3.103)


The starting equation by making r = 0 becomes:

y\ = Jo 2 - K(At)2

M yx - 2At

fy (At)2

M (3.104)

By including damping c = 10%, i.e. c(Sy/6t) in the generalized equation, then:

Equation (3.104) becomes

cAt yT+\ l

2M

+ F(t\

K(At)2

M

(At)2

- J r - l cAt 2M

M

yT+i is computed from Eq. (3.105):

K (2 columns) = 2 X ] 2 E I = 1.5 x 10 6 Nm

(3.105)

. jM I 5000 = 0.3635 sec

When c = 10%, the maximum time-step interval = 0.036 « 0.04 sec. The value of At = 0.015sec is acceptable. y T + i from Eq. (3.105) is given by:

>V(1.9325) -yT_i (0.974) + 4 . 5 x lO~4F(t) 1.026

For the initial case:

yi = 2.25 x 10~4(1.9325) - O x 0.974 + 4.5 x 10

1.026

- 4 0.0008624 m

yi

J4

0.0008624(1.9325) - 2.25 x 10~ 4 x 0.974 + 4.5 x 10"

1.026

= 0.0018493 m

0.0018493 x 1.9325 - 0.0008624 + 4.5 x 10~ 4

1.026 0.0031198 m

These calculations are continued for other values of y. Figure 3.63 shows the cyclic nature of the response. The period of vibration is the same as the natural period of this frame. With damping of 10%, the displacement will have a static value reaching up to 4 seconds.

3 0 6 B A N G A S H

2.5 T = time seconds

Figure 3.63. Displacement-time relationship

3.5 Seismic design of tall buildings in Japan 3.5./ General introduction Kiyushi Muto in his report ( M U T O Report 73-1-1, May 1973) laid out a comprehensive method for the planning and design o f tall buildings (H < 45 m and H > 45 m) in Japan. In order to obtain horizontal seismic coefficients, Tables 3.16 to 3.18 give preliminary calculations based on an old Japanese code. In the same section, the author has modified the work by including the effects o f vertical acceleration as seen in Figs 3.64 to 3.66 from the Kobe earthquake.

The Muto Institute has provided a flow chart (Fig. 3.67) which indicates the AIJ guidelines for the seismic analysis and design o f tall buildings in Japan. In the planning o f tall buildings it is suggested that preliminary design planning must be checked and analysed against several earthquakes. Dynamic analysis is performed using a rigorous vibration model in conjunction with structural stiffness. A series o f computer programs were developed (FAPP I to FAPP IV) for steel structures. Figures 3.68 and 3.69 give a brief illustration o f what these computer programs can do. Some examples in Table 3.18 indicate the response control in actual tall buildings.

After developing a sound technique for planning, testing various buildings and obtaining a criterion for the seismic analysis o f tall buildings, the Muto Institute took up the case o f the Shinjuku Mitsui Building (SMB) in Tokyo, a 55-storey, 225 m high office building, as shown in Fig. 3.70. Typical floor areas


Table 3.16. Seismic load in Japan (horizontal coefficient)

BUILDING CODE FOR H < 4 5 M

SEISMIC FORCE

fi=kWt

WHERE k = SEISMIC COEFFICIENT

Wj = ith FLOOR LUMPED LOAD

SEISMIC COEFFICIENT

k = K 0 Z S WHERE A: 0 = 0.2 FOR H < 1 6 M

K0 = 0.2 + i FOR H > 1 6 M

I = INCREMENT = 0 . 1 EVERY 4 M OVER 1 6 M

Z = ZONING FACTOR = 1.0, 0 .9 , 0.8

S = SOIL AND CONSTRUCTION FACTOR

= 0.6 , 0 .8 , 1.0, 1.5 (ROCK -»SOFT SOIL)

Fundamental seismic coefficient K,, Zoning factor Z

are 58.4 m x 44.5 m. The building has sufficient stiffness against strong winds and typhoons, which frequently occur in Japan. The stiffness o f this building was assessed initially by a value given to the fundamental natural period Ti = 0.1 JV where N is the number o f storeys. Hence T\ = 0.1 x 55 = 5.5 seconds. Various layouts and schemes were considered but T{ was not sufficient in the transverse direction. The final suitable design layout is shown in Fig. 3.71.

3 0 8 B A N G A S H

Table 3.17. BRI standard for H>45m

SEISMIC FORCE

F = C I W

C i ^ C o - Z ' I

WHERE C 0 = BASIC SHEAR COEFFICIENT

C 0 = 0 . 2 5 FOR T < G + 1.75 SEC

Q 35c = — — F O R T > C + 1.75SEC

T 7 = FUNDAMENTAL NATURAL PERIOD

G — SOIL FACTOR

= - 0 . 7 5 , 0 , 0 .5 , 0 .75

S — CONSTRUCTION FACTOR

= 0 .9 , 1.0

Z = ZONING FACTOR

= 1.0, 0 .9 , 0.8

/ = IMPORTANCE FACTOR

LATERAL SEISMIC FORCE

ft

WHERE

1 / / =

ifi +-lfi+3fi aCxWx

2fi=pCxWx W x

zfi=lCxW

Wx — ith FLOOR LUMPED LOAD

W= TOTAL LOAD

x = HEIGHT OF ITH FLOOR

a = ( 2 - T)/2(T < 2 ) , 0 ( R > 2 )

p = R / 2 ( R < 2 ) , ( 1 2 - R ) / I O ( R > 2 )

7 = O ( R < 2 ) , ( R - 2 ) / I O ( R > 2 )


Table 3.18. Response control of actual tall buildings

Building Earthquake Class* Maximum storey Maximum

drift** (cm) acceleration (gal)

Total Slitted

wall***

Top Middle

KASUMIGASEKI BUILDING EL CENTRO I 0.7 0.2 100 6 0

1940 ( N S ) I I 1.3 0.5 130 9 0

I I I 2 .4 1.0 2 4 0 180 HEIGHT - 127 M TAFT I 0.4 0.2 8 0 6 0 STOREYS - 3 6 1952 ( E W ) I I 1.0 0.5 130 120

I I I 1.9 1.1 2 6 0 2 2 0

TOKYO WORLD TRADE CENTRE EL CENTRO I 0.7 0.3 130 7 0 BUILDING I I 1.3 0.7 160 9 0

I I I 2.3 1.5 2 4 0 190 HEIGHT - 152 M TAFT I 0.4 0.2 8 0 7 0 STOREYS - 4 0 I I 0.8 0.7 140 120

I I I 1.6 1.5 2 8 0 2 4 0

T H E K E I O PLAZA HOTEL EL CENTRO I 0.6 0.2 100 6 0

I I 1.2 0.5 170 100

I I I 2.5 1.0 3 4 0 2 1 0 HEIGHT - 170 M TAFT I 0.5 0.2 9 0 7 0 STOREYS - 4 7 I I 1.0 0.4 170 140

I I I 1.9 0.9 3 5 0 2 9 0

SHINJUKU MITSUI BUILDING EL CENTRO I 0.4 0.1 6 0 4 0

I I 0.7 0.2 1 1 0 8 0

I I I 1.4 0.5 2 2 0 160 HEIGHT - 2 1 2 M TAFT I 0.4 0.2 6 0 50 STOREYS - 55 I I 0.7 0.3 120 9 0

I I I 1.4 0.6 2 4 0 180

* I - 100 GAL, I I - 2 0 0 GAL, I I I - 4 0 0 GAL

* O N TYPICAL STOREY HEIGHT

* STOREY DRIFT OF SLITTED WALL DUE TO SHEAR DEFORMATION

The El Centro earthquake was considered as the basis for the animated earthquake response o f the SMB building. A comparative study with other earthquakes o f a linear and nonlinear nature is also given (see Figs 3.72 to 3.74).

The Muto Institute has provided graphic displays for the displacements, storey shears and overturning moments for the SMB. These are shown in Figs 3.75 to 3.79.

Figure 3.64. A devastated building site at Kobe

312 B A N G A S H

Figure 3.66. Hanshin (Kobe) motorway flyover twisted and buckled and adjacent structures damaged


TALL BUILDING E x c e e d i n g 45 m

L

GUIDE LINES OF A U For s e i s m i c d e s i g n

of tai l bu i ld ing

L _ —

EXAMINATION BOARD in J . B . C .

5)

MINISTER APPROVEMENT of const ruc t ion

6 )

- F r e e a p p r o a c h to s t ructura l d e s i g n

- S ta t i c load - B a s e s h e a r fo rce - S h e a r fo rce d is t r ibut ion

- S ta t ic ca lcu la t ion - O p e n f r a m e - W a l l e d f r a m e - B r a c e d f r a m e

- D y n a m i c d e s i g n cr i te r ia - S t i f fness e v a l u a t i o n - E x p e c t e d e a r t h q u a k e w a v e s - D a m p i n g - D a m a g e contro l - S a f e t y j u d g e m e n t

- E x a m i n a t i o n b o a r d

- Min is t ry of C o n s t r u c t i o n

- S p e c i a l p e r m i s s i o n

Figure 3.67 (pages 313-315). Seismic design procedure of a tall building in Japan (Muto Institute)

314 B A N G A S H

SEISMIC LOAD

1)

STATIC CAL

2 )

STRUCT DESIGN CRITERIA

5)

EQKE

6)

RESPONSE OR DAMAGE

CONTROL S t r e s s D e f o r m a t i o n

8 )

END

10 )

EQKE RESIST CONFIRM I

STRUCT TEST

VIBRATION TEST N a t u r a l v ibra t ion Exc i t ing v ibra t ion

EQKE OBSERVATION

Figure 3.67. (Continued)

3 1 6 B A N G A S H

FAPP I (open frame)

2D \

2B

FAPP II (walled frame)

Slitted wal l -

Panel zone

Beam

-Column

Column (M, Q, N ) -

Panel (Q)

1/ i •> Beam (M,Q)

T T

Rigid zone of beam Panel (Q)

h ^ \

Wall (M, Q, N)

Beam (M,Q)

C o l u m n ^ (M, Q, N)

FAPP III (braced frame)

Brace -

Open frame Walled frame Braced frame

Figure 3.68. The FAPP computer program for a two-dimensional coupled system

Figure 3.69. The FAPP IV computer program for a three-dimensional system

3 1 8 B A N G A S H

Figure 3.70. SMB: Shinjuku Mitsui Building (55-storey building in Tokyo)

Figure 3.71. Plan and sectional elevations of the SMB building

3 2 0 B A N G A S H

Mark Earthquake Damping ocMax Mark Earthquake Damping a M a x

El Centro (NS) 0.030 100. El Centro (NS) 0.030 100. El Centro (NS) 0.030 100. El Centro (NS) 0.030 100.

Taft (EW) 0.030 100. « , Taft (EW) 0.030 100.

Tokyo 101 (NS) 0.030 100. Tokyo 101 (NS) 0.030 100.

Senda i501 (NS) 0.030 100. Senda i501 (NS) 0.030 100. Senda i501 (NS) 0.030 100. Senda i501 (NS) 0.030 100.

A

ii v

\ \ X

^ \ 5

\

55

50

40

30

20

10

1

\ \ X

\ i \ \

\ i

Y

|,| 1 \

1 1 1 \ ^ 1

t \ I k \ \

\ \ \ ^

50

Gal

A C C E L E R A T I O N

100 0 500 1000

Ton

S H E A R

150C

Figure 3.72. Linear earthquake response (earthquake intensity: 0.1 g)


Mark Earthquake Damping a M a x Mark Earthquake Damping a M a x

El Centro (NS) 0.030 100. El Centra (NS) 0.030 100. El Centro (NS) 0.030 100. El Centra (NS) 0.030 100. Taft (EW) 0.030 100. Taft (EW) 0.030 100. Tokyo 101 (NS) 0.030 100. Tokyo 101 (NS) 0.030 100. Sendai501 (NS) 0.030 100. Sendai501 (NS) 0.030 100. Sendai501 (NS) 0.030 100. Sendai501 (NS) 0.030 100.

55

50

40

o 30 oo

20

10

\ \ \

\ \ I t

\\ \ 1

))

j] \ \ I \ i ]

/ / / J \ 1 I1

V r A * '1 !

.y,—Y~-

/.' J

/: (/

0.20

cm

55

50

40

30

20

10

0.40

S T O R E Y DRIFT

Figure 3.73. Comparative study of earthquakes

50 000 100000

Ton/m

O V E R T U R N I N G M O M E N T

w \ \ \ v \ w

V \

\ \ V

V \ l \ \ X

\ \ \ I — U N ,

1500C

3 2 2 B A N G A S H

Mark Earthquake Damping ocMax

El Centro 1940 (NS) 0.030 100.

El Centro 1940 (NS) 0.030 300.

El Centro 1940 (NS) 0.030 500.

Mark Earthquake Damping a Max

El Centro 1940 (NS) 0.030 100.

El Centro 1940 (NS) 0.030 300.

El Centro 1940 (NS) 0.030 500.

\ v \ \ \

\ \ X

1 1

\ \ i \

\

1 \ \ \ \

1 i i ! \

I /

1 I t

\ < /

\ \

\

\ \

\ \

_ \ \

._ .1 \ S s

.V.

4000

Ton

S H E A R

55

50

45

40

35

30

25

2 0

15

10

5

1

t ** % \

t ~~\

i i i 1

i

/ " " ~ T

1 \

—f. „

K % i \ 4~ — ~ I I

i

/ /

/ /

/ /

/ /

* /

/ " " / "

/ / /

/ ' / — — 4: 8000 0 1.0

C m

S T O R E Y DRIFT

2.0

Figure 3.74. Non-linear earthquake response (El Centro earthquake, intensity: 0.1, 0.3, 0.5 g)


DISPLACEMENT

0.40 sec

STOREY SHEAR OVERTURNING MOMENT

2 0 0 0 t 0 .40sec • • 1 . 0 x 1 0 5 t . m 0 .40sec

- 4 0 0 0 0 4000 S H E A R

- 3 . 0 0 3.0 O V T M

E N L A R G E D BY

50 T I M E S

0 40 cm

0.80 sec 2 0 0 0 t 0.80 sec 1 . 0 x 1 0 5 t . m

x 1 0 5

0.80 sec

- 4 0 0 0 0 4000 S H E A R

- 3 . 0 0 3.0 O V T M

x 1 0 5

E N L A R G E D ( 1.20sec BY

50 T I M E S

0 4 0 c m

2000 t 1.20 sec 1 . 0 x 1 0 5 t . m 1.20 sec

- 4 0 0 0 0 4000 S H E A R

- 3 . 0 0 3.0 O V T M

x 1 0 5

E N L A R G E D BY

50 T I M E S

0 ^ 0 cm

1.60 sec 2000 t 1.60 sec 1 . 0 x 1 0 5 t . m 1.60 sec

- 4 0 0 0 0 4000 S H E A R

- 3 . 0 0 3.0 O V T M

x 1 0 5

E N L A R G E D BY

50 T I M E S

0^40 cm

2.00 sec 2000 t 2 .00sec 1 . 0 x 1 0 5 t . m 2 .00sec

- 4 0 0 0 0 4000 S H E A R

- 3 . 0 0 3.0 O V T M

x 1 0 5

Figure 3.75. Animated earthquake response 1

3 2 4 B A N G A S H

ENLARGED BY

50 TIMES

0 40 cm

DISPLACEMENT

2.40 sec


2 0 0 0 t 2 .40sec • 1 . 0 x 1 0 5 t . m 2 .40sec

- 4 0 0 0 0 4000 SHEAR - 3 . 0 0 3.0

OVTM x 1 0 5

2.80 sec 2 0 0 0 t 2 .80 sec 1 . 0 x 1 0 5 t . m 2.80 sec

- 4 0 0 0 0 4000 SHEAR - 3 . 0 0 3.0

OVTM x 1 0 5

ENLARGED , 3 .20sec BY

50 TIMES

0 40 cm

2000 t 3.20 sec 1 . 0 x 1 0 5 t . m 3.20 sec

-4000 0 4000 SHEAR 3.0 0 3.0 x 1 0 5

O V T M

ENLARGED BY

50 TIMES

o"~40 cm

3.60 sec 2 0 0 0 t 3.60 sec 1 . 0 x 1 0 5 t . m 3.60 sec

- 4 0 0 0 0 4000 SHEAR - 3 . 0 0 3.0

OVTM x 1 0 5

4.00sec 20001 4 .00sec 1 . 0 x 1 0 5 t . m 4.00 sec

- 4 0 0 0 0 4000 SHEAR - 3 . 0 0 3.0 x 1 0 5

O V T M



E N L A R G E D BY

50 T I M E S

DISPLACEMENT

4.40 sec


2 0 0 0 t 4 .40sec 1 1 1 . 0 x 1 0 5 t . m 4 .40sec

- 4 0 0 0 0 4000 S H E A R

3.0 0 3.0 O V T M

4.80 sec 4.80 sec 1 . 0 x 1 0 5 t . m

- 4 0 0 0 0 4000 S H E A R

5.20 sec

5.60 sec

2 0 0 0 t 5.20 sec 1 . 0 x 1 0 5 t . m

- 4 0 0 0 0 4000 S H E A R

2 0 0 0 t 5.60 sec 1 . 0 x 1 0 5 t . m

- 4 0 0 0 0 4000 S H E A R

6.00 sec 2 0 0 0 t 6.00 sec 1 . 0 x 1 0 5 t . m

- 4 0 0 0 0 4000 S H E A R

x 1 0 5

4.80 sec

3.0 0 3.0 O V T M

x 1 0 5

5.20 sec

- 3 . 0 0 3.0 O V T M

x 1 0 5

5.60 sec

- 3 . 0 0 3.0 O V T M

x 1 0 5

6.00 sec

3.0 0 3.0 O V T M

x 1 0 5


3 2 6 B A N G A S H

E N L A R G E D BY

50 T I M E S

0 4 0 cm

DISPLACEMENT

6.40 sec

E N L A R G E D BY

50 T I M E S

oTiocm

E N L A R G E D BY

50 T I M E S

0*40 cm

E N L A R G E D BY

50 T I M E S

0^40 cm


2 0 0 0 t 6 .40sec • » 1 . 0 x 1 0 5 t . m 6 .40sec

- 4 0 0 0 0 4000 S H E A R

7.20 sec 2 0 0 0 t 7.20 sec 1 . 0 x 1 0 5 t . m

- 4 0 0 0 0 4000 S H E A R

7.60sec 2 0 0 0 t 7.60 sec 1 . 0 x 1 0 5 t . m

- 4 0 0 0 0 4000 S H E A R

. 8.00 sec 2 0 0 0 t 8.00 sec 1 . 0 x 1 0 5 t . m

x 1 0 5

6.80 sec

3.0 0 3.0 O V T M

x 1 0 5

7.20 sec

- 3 . 0 0 3.0 O V T M

x 1 0 5

7.60 sec

- 3 . 0 0 3.0 O V T M

x 1 0 5

8.00 sec

- 4 0 0 0 0 4000 S H E A R

- 3 . 0 0 3.0 x 1 0 5

O V T M



DISPLACEMENT

E N L A R G E D BY

50 T I M E S

0 40 cm

8.40sec


2 0 0 0 t 8 .40sec 1 1 1 . 0 x 1 0 5 t . m 8 .40sec

E N L A R G E D BY

50 T I M E S i—i

0 40 cm

E N L A R G E D BY

50 T I M E S I—1

0 4 0 c m

E N L A R G E D BY

50 T I M E S

0 40 cm

E N L A R G E D BY

50 T I M E S I—i

0 40 cm

8.80sec

- 4 0 0 0 0 4000 S H E A R

- 3 . 0 0 3.0 O V T M

x 1 0 5

9.20 sec 2 0 0 0 t 9.20 sec 1 . 0 x 1 0 5 t . m 9.20 sec

- 4 0 0 0 0 4000 S H E A R

- 3 . 0 0 3.0 O V T M

x 1 0 5

9.60 sec 2 0 0 0 t 9.60 sec 1 . 0 x 1 0 5 t . m 9.60 sec

- 4 0 0 0 0 4000 S H E A R

- 3 . 0 0 3.0 O V T M

x 1 0 5

10.00 sec 2 0 0 0 t 10.00 sec 1 . 0 x 1 0 5 t . m 10.00 sec

- 4 0 0 0 0 4000 S H E A R

- 3 . 0 0 3.0 O V T M

x 1 0 5


3 2 8 B A N G A S H

3.5.2 Resimulation analysis of SMB based on the Kobe earthquake using three-dimensional finite element analysis

The earthquake o f 17 January 1995 that occurred in the area o f Kobe in southwestern Japan, also known as the Hanshin or H y o g o - K e n Nanbu earthquake, caused extensive damage and numerous casualties. The magnitude o f this earthquake was M = 7.2. Various universities and institutes in Japan reported crust deformations, the foreshocks, with around 6000 events immediately after these shocks, irregular recordings o f electromagnetic signals o f both high and low frequency and disturbances in seawater temperature.

For the first time it became clear how important the peak vertical acceleration is in conjunction with peak horizontal acceleration. Figures 3.80 and 3.81 produced by the Seismological Society o f Japan give acceleration and velocity response spectra compared with others including those o f the El Centro earthquake. The shear wave velocity was 110 m/sec on average. The Kobe amplification factor was taken as 2.5 for the current analysis. For the input parameters o f the finite element analysis, the following values were used for the Kobe data.

3.5.2./ Dotal

Maximum horizontal acceleration 818 gal

Maximum vertical acceleration 447 gal

CO

I

1000.0

500.0

200.0

100.0

50 .0

20 .0

10.0

5.0

2.0

: Pc rt Island • f /

_ AJ

Kltahama

Kob

OSi nagasaki Et Kakogama ^joogam.

ika 3 B >\ _

FY

M X-

Osakc Akou

City (33) • mjf u m

» /7 Am ele

Z.....M

agasaki ated

: / •

• •

_ / * •

4 . •

Ka

0

rtogata ILT • / / m

/ \ m

• m J V —

: / y w

/ • • •

• • •

•

•

i i WW i I„ M i l

- V = H

- V = H/2

5.0 10.0 20 .0 50 .0 100.0 200 .0 500.0 1000.0

Peak horizontal acceleration (cm/s/s)

Figure 3.80. Plots of peak horizontal acceleration versus peak vertical acceleration (courtesy of Japan Seismological Society, Tokyo, 1995, Fig. 4)


h = 0.05 5 0 0 0 , , , , , , ,

CO

0.05 0.1 0.2 0.5 1.0 2.0 5.0

Period T (sec)

h = 0.05

1 5 0 0 I ^ r—r—I 1 1 *

0.05 0.1 0.2 0.5 1.0 2.0 5.0

Period T (sec)

Figure 3.81. Acceleration and velocity response spectra (courtesy of Japan Seis-mological Society, Tokyo, 1995, Fig. 6)

Maximum vertical velocity 40 kine

Maximum horizontal velocity 90 kine

Maximum horizontal displacement 11 cm

Maximum vertical displacement 21 cm

3.5.2.2 Data II In the finite element dynamic equations given in the Appendix, the damping ratio is 10%. Isoparametric four-noded solid elements are used for the finite element discretization. The following gives additional data:

1. SBM 55-storey frame and floors floor: four-noded solid elements (55 500) frame: four-noded line elements (220 000)

2. Soil-structure interaction 2400 nodes on ground surface 2200 nodes on ground-structure interface

Reference is made to the soil-structure analysis in the Appendix. The finite element mesh scheme with boundary conditions defined is shown

in Fig. 3.82. The step-by-step time integration analysis is adopted as part o f this dynamic finite element analysis (see the Appendix). The loadings given by the Muto Institute in Fig. 3.83 are adopted as the input data to the three-dimensional finite element analysis.

For reasons o f economics in the overall performance o f time-consuming solution procedures, the entire structure is divided into seven N o . 8 substructures. The loading on one substructure from the Muto seismic analysis is given in

| Computer C R A Y - 3

3 3 0 B A N G A S H

Figure 3.82. (a) Finite element mesh scheme for the SMB Building, (b) Building substructure (total No. of 20 substructures)

Fig. 3.83. All columns are 500 x 500mm box-type fully welded and rigidly connected to a girder. The transverse framing consists o f wall frames, open frames and braced frames. Dead weight is assumed to be 115 kg/m 2 (kN/m 2 ) . A non-linear earthquake of 0.5 g was assumed for this study.

Due to the inclusion o f vertical acceleration simultaneously with the horizontal acceleration, the approach is completely changed. If this building were located in the Kobe region during the earthquake, the damage to the building

3 3 2 B A N G A S H

floors and frames would be in the ratio o f three times the damage if vertical accelerations were not ignored. The deformation demonstrated by the Muto Institute for the animated earthquake responses in their graphic form given in Figs 3.75 to 3.79 would be on average four times more. A comparative study o f the displacement-time function graph (Fig. 3.84) indicated a marked difference for the Kobe earthquake.

The SM Building is taken as an example for the three-dimensional dynamic finite element analysis in which material and geometric non-linearity are considered. Figure 3.85 shows ductility models for the various materials adopted. These models are included in the ETABS program (Extended Three-dimensional Analysis o f Building Systems) developed by Wilson et al. at the University o f California. The building is idealized as independent frames and sub-frames interconnected by a flooring system adopted by the Muto Institute. Axial, bending and shear deformations are included within each column made o f 500 x 500 welded box-type members. Beams, girders and vertical panel elements allow discontinuity. The loadings evaluated accurately by the Muto Institute for the SMB have been modified to include the effects o f both horizontal and vertical accelerations. Non-linear earthquakes o f 0.5 g are considered for Kobe and other earthquakes locations, assuming the SMB is sited in these areas.

Figure 3.86 shows a comparative assessment o f modes based on the El Centro and Kobe earthquakes. In all cases vertical accelerations are included simultaneously with horizontal accelerations. Table 3.19 shows a comparative safety assessment o f the SMB subjected to various earthquakes. The effects on this building due to a Kobe-type earthquake are enormous. The damage is predicted. Owing to massive outputs, only a Table (3.19) summarizes the strength reduction and excessive storey drift under various earthquakes.

Table 3.19. SMB - comparative safety assessment

Earthquake* Strength reduction Storey drift factor Storey shear strength factor of design on top of design reduction factor from stresses** driffi the design factor^

Longitudinal Transverse Longitudinal Transverse Longitudinal Transverse

EL CENTRO 0 .76 0 .87 1.80 1.82 0 .71 0 .73

TAFT 0 .72 0 .83 1.73 1.74 0 .84 0 .85

SAN-FERNANDO 0 .89 0 .88 1.65 1.70 0 .81 0 .82

KOYNA 0 .84 0 .83 1.36 1.56 0 .79 0 .80

MEXICO 0 .79 0 .81 1.71 1.75 0 .77 0 .80

TOKYO 0 .73 0 .75 1.69 1.73 0 .78 0 .81

K O B E 0 .92 0 .89 1.93 1.94 0 .83 0 .69

* NON-LINEAR EARTHQUAKE (0 .5 G), ELASTO-PLASTIC SPECTRA, 5 % DAMPING AND VERTICAL ACCELERATION

INCLUDED.

* DESIGNER REPORTED.

3 3 4 B A N G A S H

S T E E L F R A M E W I T H SL ITTED WALL

Curvature ductility

Figure 3.85 (above and facing). Ductility models for the SMB Building

R E I N F O R C E D C O N C R E T E F R A M E


3 3 6 B A N G A S H

55

50

45

40

35

Elcentro

Kobe with inclusion of vertical acceleration

- 1 . 5 - 1 . 0 - 0 . 5 0.5 1.0

- 0 . 1 - 0 . 0 8 - 0 . 0 6 0 0.06

Maximum relative displacement

0.08

1.5

_ A / 0.1 / R

max

Figure 3.86. SMB - comparative assessment using the El Centro and Kobe earthquakes

It is concluded that vertical acceleration is important and cannot now be ignored. Since the SMB is not actually in the K o b e region, the building is safe and the existing design featues are adequate even if the vertical acceleration parameters are included. In the K o b e region this type o f building would receive superficial damage but would not collapse due to specific layout techniques adopted by the Muto Institute, Japan.

ch 3 -earthquake and buildings

Documents

terms o f

collapse o f buildings

history o f ground shaking

list o f countries

frequency content o

entire base o f

lethal earthquakes

s e c t i o n