structural analysis ii 1

Bachelor of Science in Civil Engineering Structural Analysis II, Construction Materials

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Part I Structural Analysis II

1) Introduction on Structural Design

(Except from “Aims of Structural Design” published by the Institution of Structural

Engineers, August 1969.)

a) Factors in Design

Function : The structure must fulfill its intended function. This function may be

complex, concerned not only with the support of load, but with water tightness,

thermal and sound insulation, fire resistance, and chemical resistance etc. Load itself

may be complex and show not only a single design load, but a wide spectrum of loads

and their combinations occurring with varying frequency. Moreover, as soon as a

structure is built, the influence of weather, ageing, fatigue, use, ill-use, accident and

other causes of deterioration commence to act upon it. The designer seeks to ensure

the continued service ability of his structure during its intended life.

Safety : The structure must be safe. Most structures involve the public; there are few

to whose strength someone does not at some tie entrust his life. The consequences of

collapse are grave and the possibility of collapse must be made remote.

The designer’s estimate of adequate safety is partly analytical and partly a matter of

judgment-the accident-prone structure may be difficult to define, but can be

recognized by the experienced.

Economy : The structure must be of least cost. The client having decided, doubtless

in consultation with his designer, on the standard of strength, durability and other

qualities that he requires, the design will proceed with this intention; the designer is,

in fact, in the market place looking for the best value for his client’s money. Design

decisions, always implicitly and often explicitly, are economic decisions, right down

to the choice of the smallest detail.

The designer is all too seldom able to consider the balance of first cost and

maintenance costs. Clearly, each can be reduced at the expense of each other;

somewhere there must be a minimum total cost. Most owners of structures seem to be

so organized, however, that the two are paid for from different funds and the effect of

one on the other is not revealed. It must be assumed that owners have fiscal and

financial reasons for this.

Other factors to be considered: -

a) statutory requirements to be satisfied

b) physical constraints which may affect the method of construction


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c) time constraints; structures may need to be completed within a very short time

d) environmental impacts to be considered

b) Process of Design

Appreciation of requirements : Design starts with the appreciation of the client’s

requirements; this is the critical stage in the process and calls for the closest

collaboration between the client and the designer.

Formulation of schemes : As many facts and factors as possible are collected and

assimilated; the mind concentrates upon them. By some process of the intellect or

imagination, often in moments of relaxation, concepts of the general structure suggest

themselves, sometimes vague, sometimes surprisingly complete.

Appraisal of schemes : These concepts are then developed approximately and

appraised by the designer. The amount of mathematical analysis at this stage will

depend on his experience; often no more is done than is enough to determine general

dimensions and the designer is likely to concentrate more on functional suitability,

ease of erection and economy.

At the end of this appraisal, the designer returns to the original requirements, perhaps

to the client. The process may have thrown up unexpected possibilities that modify

the requirements; alternatively what was thought feasible may be found to be not so.

The economic aspect may need rethinking.

The most economic solution is usually the one with a straight-forward, clear-cut

method of erection with a minimum of separate process following one another

without mutual interference. In considering cost, it is more useful to concentrate on

operations than on the minutiae of quantities and rates whose apparent precision may

mislead.

Analysis of final scheme : It is now that a full structural analysis is made. This is

frequently referred to as “design”; in fact, the significant design decision have already

been taken and the operation of analysis is much more a process of checking the

scheme for soundness in service and during erection.

Procedure of structural design :

a) Preliminary and feasibility study:

1. Analysis of functional requirements and constraints

2. Planning and scheming; appraisal of different schemes and preliminary design

b) Detailed design:

1. evaluation loads


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2. structural design (size, shape, quality etc)

3. detailing and production of drawings

4. preparation of specification documents

Details and Specifications : Detailing must be seen as an integral and important part

of design; one cannot afford to lose control of the detailed design because:

− bad detailing can leading to disaster-in fact, it is much the more frequent

cause of trouble;

− bad detailing can greatly increase cost;

− it is easy without close supervision for details to creep in which are not

consistent with the rest of the structure and which can render an otherwise

carefully considered erection scheme impossible.

Specifications are part of the design, complementary to the detail drawings, and not

merely a means of regulating contractual relations.

Supervision of Construction : Supervision should be under the designer’s control;

his task is not yet complete when he hands over the bundle of design documents.

Construction is rarely without incident requiring decision, the consequences of which

only the designer can appreciate.

Other follow-up work : Preparation of contract documents, site supervision, contract

administration, sometimes re-designing parts of the works, preparation of as-built

drawing etc.


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c) Loading

Dead Loads

When considering a building as a whole at the preliminary design stage, it is best

and easiest to estimate the total weight of the building in terms of average kPa of

floor area.

Although expressed in terms of floor area, these overall dead-load values include

the weight of the structural floor itself, the roof, walls, shafts, columns, and

possibly the floor surfacing, ceilings, and partitions. However, it should be clear

that they are only rough approximations. They can be varied greatly depending

upon the designer’s skill and his estimate of the materials to be used for the basic

structural system, the floors themselves, the walls, and other permanent portions.

Table : Approximate Weight of Materials (lb/cu ft)

Masonry Water, Snow

Marble, granite 140-165 Water 62.5

Brick masonry 100-150 Snow, fresh 5

Concrete, normal 150 Snow, packed 10 and up

Concrete, lightweight 90-120 Snow, wet 40-50

Metals Miscellaneous

Steel 480 Sand 100-120

Aluminum 165 Glass 160

Brass 530 Asphalt 80-100

Mortar 100

Timber

Redwood 26 Pine 35-40

Douglas fir 32 Oak 54

Live Loads

The actual live load on the floor of a building varies greatly. It is possible to

concentrate a heavy loading such as a safe box or moving equipment over a

rather small area. Then it is necessary to design that small area for a heavy

concentration. However, when there is a large area support by a primary

structural component, the significance of that concentration as compared with the

overall load will be reduced correspondingly.


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Since the actual live load over the floor area of a building can vary greatly with

time of day and occupancy type, it becomes quite difficult to determine a precise

live load in kPa to a specific floor design. Nevertheless, an average design load

value can be assigned when actual or probable type of building occupancy is

known. This would be classified as the basic live load for application when

considering the larger tributary areas.

When smaller areas are considered, the effect of a concentrated live board should

be considered as a special case and may be expressed as a single concentrated

load or as a uniform load in kPa, but of a magnitude much higher than the specified

basic live load. Such concentrations are often stated or suggested in building code

specifications.

Although the basic live load does not apply to smaller areas, it also will not be

accurate to apply it without some modification to larger tributary areas. Consider

that, in most building, it would be unusual to load every square foot of a large

tributary area completely, as could be the case for a limited area. Therefore, the

size of a tributary area determines the probability of loading every square meter

to basic live load levels. To simulate this effect, the basic live load can be

reduced when considering large tributary areas.

Wind Loads

Wind pressure on a building surface depends primarily on its velocity, the slope

of the surface, the shape of the surface, the protection from wind offered by other

structures and, to a smaller degree, the density of the air, which decreases with

altitude and temperature, and the surface texture. All other factors remaining

unchanged, the pressure due to wind is proportionate to the square of the velocity

and the density of the air:

P = CD x Q = CD ( ½V2D)

Where P is the pressure on a surface, V is the velocity of the wind, and D is the

density of air, CD is the numerical shape coefficient (called drag coefficient) and

Q is the dynamic pressure moving air, equal to ( ½V2D). The velocity of the

wind assumed in design could be different for structures at seaside and at top of

hill.


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Earthquake Loads

Earthquake loads are specified with the following basic objectives: -

a. To protect the public from loss of life and serious injury

b. To prevent building from collapse and dangerous damage under a

maximum-intensity earthquake.

c. To insure building against any but very minor damages under moderate to

heavy earthquakes.

Equivalent static loads are specified so that the above objectives can be attained

within reason and without excessive cost.

Earthquake resistance calls for energy absorption (or ductility) rather than

strength only. If a building is able to deflect horizontally by several times the

amount envisioned under the basic seismic design load and still maintain its

vertical load-carrying capacity it will be able to absorb earthquakes considerably

heavier than the design earthquake. If such ductility is present, collapse of the

building can be prevented even if the building is seriously damaged. Thus, in

addition to seismic load design, the ductility and plasticity of a building should

be given due consideration.


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Consequences of structural inadequacy

Inadequacy may be of two types: the structure may cease to be serviceable or it

may collapse. The acceptability of the risk will depend on its consequences.

Normally, the loss resulting from unservicability will be much less than that

from collapse.

The response of a structure to the failure of a member may be one of the

following:

(a) the structure will collapse immediately, as with a pin-jointed statically

determinate truss;

(b) the structure will not only collapse but entrain the collapse of adjacent

structures;

(c) the structure may continue to stand, though with reduced strength, due to

redistribution of load. This is the situation of a structure specifically designed to

be “fall safe” and sometimes, but not always, of a statically indeterminate

structure.

The practical consequences of any of these three types of failure will vary

widely and must be assessed in a given case and set against the risk of failure

assessed above.

Clearly, if the failure of a structural member leads to a widespread collapse, the

consequences will be graver than if the collapse were limited to the structure

elements involved. Possibilities of such chain action type progressive failure

must be made very remote.


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2) Relationship of Stress and Strain

a) Stress

Normal Stress

The simplest form of stress is normal stress, which is the stress perpendicular to

the surface on which it act.

where σ is the normal stress,

P is the centric axial load, and

A is the area of the section.

The normal stress is usually expressed in Pascal (Pa), where one Pascal is equal

to one Newton per square meter, that is, 1 Pa = 1 N/ m2

. A Pascal is a very small

unit of stress, so one can usually expect to see stresses expressed in kPa or Mpa.

Shear Stress

Shear stress is stress parallel to the surface on which it acts.

where τ is the shear stress acting on the surface,

V is the force acting parallel to the surface and

A is the area on which the shear stress acts.

The unit for shear stress is same as that for normal stress, ie. Pascal (Pa). However,

the stress direction is parallel to the surface under force. The above formula

assumes that the shear stress is uniformly distributed over the surface.

Force P σ =

Area =

A

Force V τ =

Area =

A

V V


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NORMAL STRESS


10

SHEAR STRESS


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Example 1

A steel bar of rectangular cross-section, 3cm by 2cm, carries an axial load of

40kN. Estimate the average tensile stress over a normal cross-section of the bar.

Solution

The area of a normal cross-section of the bar is

A = 0.03 x 0.02 = 0.6 x 10-3

m2

The average tensile stress over this cross-section is then

Example 2

A steel bolt, 2.5cm in diameter, carries a tensile load of 40kN. Estimate the average

tensile stress at the section a and at the screwed b, where the diameter at the root of

the thread is 2.1cm.

Solution

The cross-sectional area of the bolt at the section a is

π Aa =

4 (0.025)2 = 0.491 x 10

-3 m

2

The average tensile stress at A is then

P 40 x 103

σa = Aa

= 0.491 x 10

-3

= 81.4MN/m2

The cross-sectional area at the root of the thread, section b, is

π Ab =

4 (0.021)2 = 0.346 x 10

-3 m

2

The average tensile stress over this section is

P 40 x 103

σb = Ab

= 0.346 x 10

-3

= 115.6MN/m2

P 40 x 103

σ = A

= 0.6 x 10

-3

= 66.67 MN/m2

2 cm

40kN

Area of normal cross-section = 6cm2

40kN

3 cm


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Example 3

Two pieces of plastic are joined by gluing overlapping areas of 50 by 70 mm, as

shown below. If a tensile force of 780N (V) applied parallel to the glued surfaces

causes the glue to fail, at what shear stress did the glue fail?

Solution

The shear stress acting on the glued surface:

780 τ =

50 x 70 x 10-6

= 222860 Pa

= 222.860 kPa

= 223 kPa

Note that if the stress is to be expressed in Pa, the area must be given in m2.

Force V τ =

Area =

A

V V

V τ


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b) Strain

Normal Strain

A member carrying a tensile load will stretch. The stretch is usually called

deformation, and the symbol for deformation is ∆L. To compare one material to

another, or one sample of the same material to another, it is customary to use strain,

which is the deformation per unit of length.

Because both ∆ and L have the same units, either mm or m, ε will be unitless although

it is customary to express the units for as m/m.

Shear Strain

If a small square element is cut out of a body where a shear stress acts on one surface,

to maintain equilibrium in the y direction, there must also be an equal shear stress on

the opposite face.

The stresses would form a couple, and in order to maintain rotational equilibrium, an

equal and opposite couple must exist. The combination of forces due to the shear

stress on each face would cause the element to deform as shown:

The shearing strain is

But for small angles, measured in radians, tan γ = γ . Thus,

* Note that ∆s /L has no units and therefore γ has no units.

Normal Deformation, ∆L ∆L Normal Strain, ε =

Original Length, L =

L

∆s

L = tan γ

∆s

L = γ

∆s

L γ

P

∆L L

P


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EXAMPLE 1

A cylindrical block is 40 cm long (L) and has a circular cross-section 10 cm (d) in

diameter. It carries a total compressive load of 90kN (P), and under this load contracts

0.04 cm (∆L). What is the average compressive stress over a normal cross-section and

the compressive strain ?

SOLUTION

The area if a normal cross-section is

π Aa =

4 (0.10)2 = 7.85 x 10

-3 m

2

The average compressive stress over this cross-section is then

P 90 x 103

σ = Aa

= 7.85 x 10

-3

= 11.46MN/m2

The average compressive strain over the length of the cylinder is

∆L 0.04 x 10-2

ε =

L =

40 x 10-2

= 0.01

90kN

90kN

40 cm long

Diameter = 10 cm


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c) Stress – Strain Curves

In a testing machine, incremental force is applied to the specimen. The load and

change of length of the specimen are recorded at the different stages of loading. Stress,

Ơ, and strain, ε, are then calculated from

where P is the applied load and A is the area of the cross-section of the specimen. The

change of the area of the cross-section is usually ignored and the original area is used.

∆ L and L are the change of length and the original length of the specimen.

∆L ε =

L

P σ =

Aa and


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Mechanical Properties of Materials

The stress-strain curves shown are for three different kinds of material;

1. Low-carbon steel, a ductile material with a yield point.

2. A ductile material, such as aluminum alloy, which does not have a yield point.

3. A brittle material, such as cast iron or concrete in compression.

There are several points of interest which can be identified on the curves as follows:

1. Proportional limit: the maximum stress for which stress is proportional to strain.

(Stress at point P.)

2. Yield point: stress for which the strain increases without an increase in stress.

(Horizontal portion of the curve ab. Stress at point Y.)

3. Yield strength: the stress that will cause the material to undergo a certain specified

amount of permanent strain after unloading. (Usual permanent strain ε1 = 0.2

percent. Stress at point YS.)

4. Ultimate strength: maximum stress material can support up to failure. (Stress at

point U.)

5. Breaking strength: stress in the material based on original cross-sectional area at

the time it breaks. Also called fracture or rupture strength. (Stress at point B.)


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Tension test and Stress-Strain Curves

The mechanical properties of materials used in engineering are determined by tests

performed on small specimens of the material. The tests are conducted in

materials-testing laboratories equipped with testing machines capable of loading the

specimens in a variety of ways, including static and dynamic loading in tension and

compression. A tensile-test machine is shown.


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d) Hooke’s Law & Poisson’s Ratio

HOOKE’S LAW

For many materials, the lower end of the stress-strain curve is a straight line. This

behavior was recognized by Robert Hooke (1635-1703) and stated as Hooke’s Law:

“For an elastic body, stress is proportional to strain.”

σ = E ε

Hooke’s Law applied only up to the proportional limit of the material.

The constant E is called the elastic modulus, modulus of elasticity, or Young’s modulus.

E is equal to the slope of the stress-strain curve.

Stress σ E =

Strain =

ε

Since strain is dimensionless, E has the same units as stress, e.g. Pa, MPa. The value

of E for a given material is a constant. Materials with a high modulus of elasticity

have a high resistance to elastic deformation, and are said to be stiff.

HOOKE’S LAW FOR SHEAR

Experiments show that if the stress does not exceed the proportional limit, shearing

stress is also proportional to shearing strain.

Hooke’s law for shear may be expressed as

τ = G γ

Where G is the shear modulus (or modulus of

rigidity) of the material. G has the same units as

stress, e.g. Pa, Mpa.

The shear modulus, G, is another elastic constant for a given material. It measures the

resistance of a material to elastic shearing deformation.

Stress σ

Strain ε

E

1

τ

γ

G

1


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EXAMPLE 1

An aluminum alloy sample is tested in tension. When the stress is 121 MPa the

normal strain is 1.7 x 10-3

m/m. Calculate the modulus of elasticity for this alloy.

SOLUTION

The modulus of elasticity:

σ

E =

ε

121 x 106

=

1.7 x 10-3

= 71.2 GPa

EXAMPLE 2

A 2 m long round bar of polystyrene plastic with a diameter of 25 mm carries a 4kN

tensile load. If the modulus of elasticity of the polystyrene is 3.0GPa, calculate the

longitudinal deformation in the bar.

SOLUTION

The deformation is

∆ = ε L

σ =

E L

P L =

A E

4000 x 2 =

(π / 4 x 252)x 10

–6 x 3 x 10

9

= 5.43 x 10-3

mm


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Poisson’s Ratio

When a load is applied along the axis of a bar, axial strain is produced. At the same

time, a lateral (perpendicular to the axis) strain is also produced. If the axial force is in

tension, the length of the bar increases and the cross-section contracts or decreases.

That is, a positive axial stress produces a positive axial strain and a negative lateral

strain. For a negative axial stress, the axial strain is negative and the lateral strain is

positive.

The ratio of lateral strain to axial strain is called Poisson’s ratio. It is constant for a

given material provided that the material is not stressed above the proportional limit,

is homogeneous, and has the same physical properties in all directions. Poisson’s ratio,

represented by the Greek lowercase letter v (nu), is defined by the equation

The negative sign ensures that Poisson’s ratio is a positive number.

The value of Poisson’s ratio, v, varies from 0.25 to 0.35 for different metals.

For concrete it may be as low as v = 0.1 and for rubber as high as v = 0.5.

E, G, and v Relationships

It can be shown that the three elastic constants E, G, and v are not independent of each

other for an isotropic material. An isotropic material has the same properties in all

directions. The relationship between elastic constants is given by the equation

E G =

2(1 + v)

For most materials, v is in the neighborhood of ¼.

lateral strain lateral strain v =

axial strain = -

axial strain


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MECHANICAL PROPERTIES OF SELECTED MATERIALS

Material Yield Strength

MPa

Ultimate

Strength

MPa

Young’s

Modulus

MPa

Shear

Modulus

MPa

Poisson’s

Ratio

METALLIC MATERIALS

Aluminum

2014-T6

6061-T6

Copper – Hard

Cast Iron

Gray

Malleable

Magnesium – Extruded

Steels

0.2% Carbon – Hot-rolled

0.8% Carbon – Hot-rolled

Stainless 18-8

Structural – 230G

Structural – 400G

NONMETALLIC MATERIALS

ABS Plastic

Concrete

Low Strength

High Strength

Vinyl – Rigid

Wood – Parallel to Grain

Douglas Fir – Dry

Red Oak - Dry

365

240

300

50

220

240

240

480

1140

230

400

-8

56

-44

58

-32

415

260

330

170

-690

340

340

410

830

1300

400

600

41

-21

-45

48

-51

-48

73

69

110

100

170

45

207

207

190

200

200

2.1

21

34

3.0

13

13

12

12

28

26

40

40

17

80

80

86

77

77

0.34

0.34

0.35

0.25

0.32

0.29

0.29

0.30

0.30

0.20

0.20


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EXAMPLE 1

Determine the value for G for tin if E is known to be 20GPa and v is 0.36.

SOLUTION

E

G =

2(1 + v)

20 x 109

=

2(1 + 0.36)

= 7.35 x 109 Pa

EXAMPLE 2

A table of mechanical properties gives E = 96 GPa and G = 36 GPa for a titanium

alloy, but does not give the value for v.

Calculate v.

SOLUTION

E

v =

- 1

2G

96 x 109

=

- 1

2 x 36 x 109

= 1.3333 - 1

= 0.3333


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Axially loaded members

From the definition for strain,

δ

ε =

or δ = εL

L

From Hooke’s law,

σ = E ε

When the stress and strain are caused by axial loads, we have

P δ

= E

A L

PL

δ =

AE

δ

L

P

P


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Shear Stress in Bending

Consider a simply beam with a concentrated load at mid-span. If we cut the beam at

any transverse cross-section, a shear force V exists at the section to maintain

equilibrium. The shear force V is distributed in the form of vertical shear stresses

acting over the face of the section.

The vertical shear stresses are not uniform in magnitude over the face of the section.

It can be shown that the shear stresses:

(a) are highest at the neutral axis of the beam,

(b) are zero at the free surfaces (i.e. the top and bottom surfaces of the beam),

and

(c) varies with the distance from the neutral axis.

An important feature of the vertical shear stresses is that they give rise to

complementary horizontal shear stresses. At any point in a beam, the horizontal and

vertical shear stresses are always numerically equal in magnitude.

Vertical shear stresses developed at the section and acting over the face of the section

provide the internal resisting shear, V


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Horizontal shear stresses in a beam

The numerical magnitudes of the horizontal and vertical shear stresses at any point in

a beam are always equal.

(a) Loaded beam (b) Stresses acting on a portion of the beam.

(c) Horizontal shear stresses acting

on the upper and lower faces of an

elemental area in the beam. The

stresses produce a rotational moment.

(d) The rotational moment produced by the

horizontal shear stresses is exactly balanced by

a rotational moment associated with vertical

shear stresses. The rotational moment, hence

stresses, must be equal at the point.

(e) Horizontal shear-stress distribution

in a rectangular beam.

(f) Vertical shear-stress distribution in

rectangular beams.


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GENERAL SHEAR STRESS FORMULA FOR BEAMS

The shear stress (at the “imaginary cut”) is given by

VQ

τ =

It

where V = Shear Force at the section

I = Moment of inertia of the entire cross-section

t = Width of the section (at the “imaginary cut”)

and Q = Apy

Ap= The cross-sectional area above the “imaginary cut”

(the shaded area fghj in the diagram)

y = Distance from Neutral Axis to Centroid of the shaded area Ap

Note:

(a) The vertical shear stress at a point is equal to the horizontal shear stress.

(b) Using the area below the “imaginary cut” for Ap should give the same result.

For Rectangular Section (t = b), the maximum shear stress (at the Neutral Axis) is


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EXAMPLE 1

If the maximum transverse shear force in a

beam with a section 40 by 90 mm deep

is 18kN, determine the value for the

maximum longitudinal and transverse shear

stresses in the beam.

SOLUTION

Both the longitudinal and transverse shear stresses have the same value at any point in

the beam. For a beam with a rectangular section, the maximum value occurs at the

neutral axis.

b h3

I = 12

40 x 903

= 12

= 2.430 x 106 mm

4

Q is the first moment about the neutral axis of the area from the plane where the stress

is evaluated to the outside fibers, and is shown shaded.

Q = y A

45 =

2 x 40 x 45

= 40 500 m3

V Q τ =

I t

18 000 x 40 500 x 10-9

=

2.430 x 106 x 10

-12 x 40 x 10

-3

= 7.5000 x 106 Pa

= 7.5000 MPa

40

45

45

y

N. A.


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3) Analysis of Structure

a) Forces system (Force & Moment)

Force has been defined as the action of one body on another. It is a vector

quantity because its effect depends on the direction and on the magnitude of the

action and because it may be combined according to the parallelogram law of

vector combination.

Moment of a force is the rotational tendency of the force to move a body in the

direction of the force application. The magnitude of moment of a force to rotate a

body above an axis normal to the plane of the body is proportional to both the

magnitude of the force and to the moment arm, which is the perpendicular

distance from the axis to the line of action of the force.

Forces and moments are vector quantities and may be resolved into components;

that is to say, a force or a moment of a certain magnitude and direction may be

replaced and exactly represented by two or more components of different

magnitudes and in different directions.

Considering firstly the two-dimensional case shown below, the force F may be

replaced by the two components Fx and Fy provided that :

Fx = F cos θ and Fy = F cos θ

y

x

Fy

Fx

F

θ


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b) Types of structures and Support conditions

Type of structures :


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Support conditions :

Structures are either partially or completely restrained so that they cannot move freely

in space. The restraints are provided by supports that connect the structure to some

stationery body, such as the ground or another structure. There are generally three

types of supports and they are : -

b) Pin or hinged support

c) Roller support

d) Fixed support

Pin or Hinged support:-

A hinge support is represented by the symbol below :

It can resist force in any direction but cannot resist a moment.

Roller support :-

A roller is represented by the symbol below :

It can resist only force normal to the supporting surface through the center of the

connecting pin.

Fixed end support :-

A hinge support is represented by the symbol below :

It can resist force in any direction and moment about the fixed end. Therefore, it can

prevent the end from both translation and rotation.


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c) Equilibriums

A structure is considered to be in equilibrium if it remains at rest when subjected to a

system of forces and moments. If a structure is in equilibrium, then all its members

and parts are also in equilibrium.

For a structure to be in equilibrium, all the forces and moments (including support

reactions) acting on it must balance each other. For a plane structure subjected to

forces in its own plane, the conditions for equilibrium can be expressed by the

following equations of equilibrium :

Σ Fx = 0 Σ Fy = 0 Σ Mz = 0

The third equation above states that the sum of moments of all forces about any point

in the plane of the structure is zero.

Equations of Condition

Sometimes internal hinges are present within a structure. An internal hinge cannot

transmit moment. Therefore, the bending moment at a hinge is zero.

The condition that the moment is zero at a hinge provides an additional equation for

analyzing the structure. Such equations are commonly called equations of condition.

Determinate and Indeterminate Structures

A structure is externally determinate if all the support reactions can be obtained by

static, i.e.

No. of Support Reactions = No. of Equations (includes Equilibrium & Conditions)

A structure which is not determinate is called indeterminate.

Degree of Statically Indeterminacy = No. of Support Reactions – No. of Equations


33

d) Free-Body Diagrams

1. Free-body diagrams make use of the concept that if a whole structure is in

equilibrium, any part of the structure is also in equilibrium.

2. Free-body diagrams can be plotted for various parts of a structure and also

for the entire structure.

3. When plotting a free-body diagram, it is important to indicate all the

possible forces and moments acting on the “cut” structure concerned.

4. Internal forces common to two free-bodies (on opposite sides of the “cut”)

should be denoted as equal in magnitudes but in opposite direction.

Free-body diagrams are very useful in finding the support reactions and determining

the internal forces in structures. The use of free-body diagrams is an important tool in

structural analysis and stress analysis.

D B C

E A

D B

C

E A


34

Internal Forces at Cut Section of a Structure


35

e) Example

EXAMPLE 1

A beam ABCDE has a pinned support at A and a roller support at D. It carries three

concentrated loads as shown. Determine the reactions.

SOLUTION

Support A can provide two components of reaction (VA and HA). Support D, being a

roller, provides only one reaction (VD), which will act vertically (that is, at right

angles to the direction of motion of the rollers). These reactions are shown on the free

body diagram.

(1)To determine HA

(∑H = 0) There are no horizontal loads; thus,

HA= 0

(2)To determine VD

Take moments about A:

(∑MA= 0)

+ (10 × 1) + (20 × 3.5) + (5 × 5.3) – (VD × 4.5) = 0

∴(VD = +23.67 kN (i.e. 23.67 kN upwards)

A B

10kN 5 kN 20kN

1m 2.5m 1m 0.8m

VA VD

10kN 5 kN 20kN

HA


36

A B

C

D

(3)To determine VA

(∑V=0)

+VA + VD = 10 + 20 + 5

+ VA+ (+23.67) – 35 = 0

∴VA = + 11.33kN

(4)Check by taking moments about D:

∑MD = + (VA × 4.5) – (10 × 3.5) − (20 × 1) + (5 × 0.8)

= + (+11.33 × 4.5) – 51 = 51 – 51 = 0

∴correct

This last calculation provides a useful check on the mathematical accuracy of the

previous two calculations.

EXAMPLE 2

Beam ABCD has a pinned support at A and a roller support at C. It carries two

concentrated loads of 30kN each and a uniformly distributed load of 5 kN/m over the

right-hand half, as shown. Determine the reactions.

30kN

5 kN/m

30kN

1m

2m 2m 2m


37

30kN

5 kN/m

30kN

A B

C

D

SOLUTION

Free body diagram,

(1)To determine HA

(∑H=0) There are no horizontal loads.

∴HA = 0

(2)To determine VC

Take moments about A:

Note that the moment of the U.D. load is the resultant total U.D. load

(5 ×××× 3 = 15 kN) multiplied by the distance from A to the line of action of that

resultant (i.e. 6 – 1.5 = 4.5 m)

(∑MA = 0)

+ (30 × 2) − (VC × 4) + (5 × 3 × 4.5 ) + (30 × 6)= 0

∴VC = + 76.875 kN

(3)To determine VA

(∑V =0)

+VA − 30 + VC − (5 × 3) − 15= 0

+VA − 30 + (+76.875) − 15 − 30 = 0

∴VA= − 1.875 kN (i.e. 1.875kN downwards)

(4)Check by taking moments about C:

∑MC= + (VA × 4) − (30 × 2) + (5 ×3 ×0.5) + (30 ×2)

= + (−1.875 × 4) −60 + 7.5 + 60 = 0

∴correct


38

2 kN/m

2 kN/m

5 kN/m

15m

9m

2 kN/m

2 kN/m

5 kN/m

EXAMPLE 3

Determine the reactions at the support for the frame shown.

SOLUTION

Free-Body Diagram The free-body diagram of the frame is shown in the right hand

side above. Note that the trapezoidal loading distribution has been divided into two

simpler, uniform, and triangular, distributions whose areas and centroids are easier to

compute.

+ → ∑ Fx = 0

Ax + 2(15) = 0

Ax = − 30 kN

+ ↑ ∑ Fy = 0

Ay −2(9)− (3)(9) =0

AY =31.5 kN

+ ∑MA=0

MA – 2 x 15 x 15/2 – 2 x 9 x 9/2 – 3/2 x 9 x 9/3 x 9 = 0

MA = 387 kNm

Checking Computations

∑MB = -30 x 15 – 31.5 x 9 + 387 + 2 x 15 x 15/2 + 2 x 9 x 9/2 + 3/2 x 9 x 9/3 = 0

=> checked OK

1

2


39

4 x 10m = 40m

30kN

30kN

15 m

HL

VL VR

L R

EXAMPLE 4

Determine the truss reaction forces.

SOLUTION

+ ∑ML =0 = 30(20) +30(30) – 40V R VR = 37.5N↑

→ + ∑H = 0 = HL HL =0

↑ + ∑V = 0 = VL + VR –30- 30 VL = 22.5 N↑

Check

+ ∑ Mc = 0 = V L (20) + 30(10) –VR (20)

= 750 - 750 = 0 = > checked OK


40

4) Beams

a) Calculations of Beams Reactions

Since the forces applied to a beam are in one plane, there are three equilibrium

equations for determining the support reactions. The equations are

where A is any point in the plane of loading.

Once the support reactions are known, internal forces and stresses in the beam may be

calculated.

Equations of Condition

Sometimes hinges or pinned joints are introduced into beams and frames. The

condition of hinges or pinned joints can provide equations for determining the support

reactions. This additional equation is called an equation of condition.

Feature of hinges or pinned joints :

1.A hinge is capable of transmitting only horizontal and vertical forces.

2.No moment can be transmitted at a hinge. (ie moment above joint = zero)

3.The bending moment at a hinge is zero.

The hinge is a particularly convenient location for “separation” of the structure into

free-bodies for computing support reactions.

∑Fx = 0 ∑Fy = 0 ∑MA = 0


41

F

L

L / 2

Hinge

C A B

a

F

F / 2 F / 2

A B

F / 2

F / 2

F a / 2

C B


42

INTERNAL FORCES IN A BEAM

(Shear Force, Bending Moment and Axial Force)

Before a structural member can be designed, it is necessary to determine the internal

forces that act within the member. In general, the internal forces for a beam section

will consist of a shear force V, a bending moment M, and a normal force (axial force)

N. For beams with no axial loading, the axial force N is zero.

Sign Convention (considering a small segment of the member):

Shear Force V: Positive shear ends to rotate the segment clockwise

Moment M : Positive moment bends the segment concave upward

(so as to “hold water”)

Axial Force N : Tension is positive.

An important feature of the above sign convention (often called the beam convention)

is that it gives the same (positive or negative) results regardless of which side of the

section is used in computing the internal forces.

Positive sign convention


43

Procedure for finding V, M and N at a beam section

1. Compute the Support Reactions

2. Draw a Free-Body Diagram of the beam segment.

Keep all internal loading on the member in their exact location.

Draw a free-body diagram of the beam segment to the left or right of the section.

(Although the left or right segment could equally be used, we should select the

segment that requires the least computation.)

Indicate at the section the unknowns V, M and N acting in their positive

directions.

3. Use the Equations of Equilibrium to determine V, M and N,.

If the solution gives a negative value for V, M or N, the actual force or moment

acts in the negative direction.

4. Check the calculations using the opposite beam segment.


44

EXAMPLE 1

Determine the shear V and the moment M at section P of the overhanging beam as

shown below.

SOLUTION

(a) First to determine reactions RB and RC. Taking moments about point C

10RB = 8(12) + 4(10)(5) − 10(3)

RB = 26.6 kN

Taking moments about point B,

10RC = 4(10)(5) + 10(13) − 8(2)

RC = 31.4 kN

Check by taking the summation of vertical forces,

8 + 4(10) + 10 = 26.6 + 31.4

58 = 58 (Ok)

(b) V and M at P using left free body,

V= sum of upward forces on ABP

= + 26.6 − 8 − 4(4) = +2.6 kN

M=sum of clockwise moments of forces on ABP about P

= 26.6(4) − 8(6) −4(4)(2) = +26.4kN m

(c) V and M at P using right free body,

V=sum of downward forces on DCP

= +10 + 4(6) − 31.4 = +2.6 kN (check)

M=sum of counterclockwise moments of forces on DCP about P

= 31.4(6) − 10(9) − 4(6)(3) = +26.4 kN m (Check)

2 m 10 m 3 m 4 m

A B C D

P

8 kN 10 kN 4 kN/m

RB Rc


45

EXAMPLE 2

Determine the shear V and the moment M at section P of the cantilever.

SOLUTION

(a) Determine reactions MA and RA.

Taking moments about point A,

MA = 36(2.5) + 4(5.5)(5.5/2) = 150.5 kNm

Taking the summation of vertical forces,

RA = 36 + 4(5.5) = 58 kN

A check may be made by taking moments about point C, although it does not seem to

add much to an independent check. If done, it is

10.5 +36(3) + 4(5.5)(5.5/2) = 58(5.5)

319 = 319 (OK)

(b) V and M using left free body,

V = sum of upward forces on ABP

= + 58 −36 − 4(3.5) = +8 kN

M = sum of clockwise moments of forces on ABP about P

= -105.5 + 58(3.5) −36(1) − 4(3.5)(3.5/2) = −8 kNm

(c) V and M using right free body.

V = sum of downward forces on CP

= + 4(2) = +8kN (Check)

M = sum of counterclockwise moments of forces on CP about P

= − 4(2)(1) = −8kN.m (Check)

A P

RA

B C

3.5 m

36 kN

4 kN/m 2.5 m

5.5 m

MA


46

b) Shear and Bending Diagrams

By the methods discussed before, the magnitude and sign of the shear forces and

bending moments may be obtained at many sections of a beam. When these values are

plotted on a base line representing the length of a beam, the resulting diagrams are

called, respectively, the shear diagram and the bending moment diagram.

Shear and moment diagrams are very useful to a designer, as they allow him to see at

a glance the critical sections of the beam and the forces to design for Draftsman like

precision in drawing the shear and bending moment diagrams is usually not necessary,

as long as the significant numerical values are clearly marked on the diagrams.

The most fundamental approach in constructing the shear and bending moment

diagrams for a beam is to use the procedure of sectioning. With some experience, it is

not difficult to identify the sections at which the shear force and moments should be

calculated directly. The shape of the shear and bending moment diagrams between

these sections are also readily identified after some experience and can be sketched in.


47

EXAMPLE 1

Draw the shear and moment diagrams for the beam shown.

∑Mc =0

−(AY × 8) + 2 × 8 × 4 + 6 × 3 = 0

∑Fy = 0

10.25 − 2 × 8 −6 + CY = 0

CY= −10.25 + 16.00 + 6.00 = 11.75MN

For 0 ≤ x ≤ 5 m:

∑Fy = 0

10.25−2 × x – V = 0

V =10.25− 2×(MN)

∑MO = 0

x −(10.25×x)+2×x×

2 +M =0

M =10.25x − x2(MNm)

For 5 ≤ x ≤ 8 m:

∑FY = 0

10.25 −2 × x − 6−V=0

V=10.25 − 6 −2x

=4.25− 2x(MN)

∑MO=0

x −(10.25×x)+2×x×

2 +6(x-5)+M =0

M =10.25x − 6(x−5) − x2(MNm)

64.00+18.00 Ay =

8

82.00 =

8 =10.25MN


48


49

EXAMPLE 2

Draw the shear and moment diagrams for

the beam shown.

∑MC = 0

50×6−BY×4=0

50×6 By=

4

=75.000kN

∑FY=0

−50+75.000+Cy = 0

CY=50.000-75.000

= - 25.000kN

For 0 ≤ x ≤ 2 m

∑Fy=0

-50-V=0

V=-50.00(kN)

∑ MO = 0

50 × x + M = 0 M = -50.00x(kNm)

For 2 ≤ x ≤ 6 m

∑Fy= 0

-50+75.00-V=0

V=-50.00 + 75.00

=25.00(kN)

∑MO=0

50 × x - 75.00(x-2) + M=0

M = -50.00x + 75.00(x-2)(kNm)


50


51

EXAMPLE 3

Sketch the Shear Force diagram and Bending Moment diagram for the cantilever

shown below.

SOLUTION

Even though there is no loading on AB and therefore no shear force or bending

moment, the distance x will still be measured from A rather than B.

Shear force

0<x<2 Q=0

2<x<5 Q=-5000-2000(x-2)

5<x<8 Q=-5000-6000-3000=-14000N

Bending moment

0<x<2 M = 0

(x-2) 2<x<5 M = -5000(x-2)-2000(x-2)

2

= -1000(x-2)(x+3)Nm

(parabolic distribution)

1 5<x<8 M = -5000(x-2)-6000(x-3

2 )-3000(x-5)

= -14000x+46000Nm (linear distribution)


52

SHAPES OF SHEAR & BENDING MOMENT DIAGRAMS

A. Beams under Point Loads only

1. Shears are constant along sections between point loads.

2. The shear diagram consists of a series of horizontal lines.

3. Moment vary linearly between point loads

4. The moment diagram is composed of sloped lines.

B. Beams under Uniformly Distributed Loads (UDL) only

1. A Uniformly Distributed Lad produces linearly varying shear forces.

2. The shear diagram consists of sloped line or a series of sloped lines.

3. A UDL produces parabolically varying moments

4. The moment diagram is composed of 2nd –order parabolic curves.

Characteristic of shear force and bending moment diagrams of beams under the

following different loading cases:

1. Section with No Load:

Shear Diagram is a Horizontal Straight Line

Moment Diagram is a Sloping Straight Line

2. At a Point Load:

There is a Jump in the Shear Diagram

3. At a Point Moment :

There is a Jump in the Moment Diagram

4. Section under UDL:

Shear Diagram is Sloping Straight Line(1 st order)

Moment Diagram is a curve (2 nd order parabolic)

5. Section under Linearly Varying Load:

Shear Diagram is a Curve (2 nd order)

Moment Diagram is a curve (3 rd order)

The Curve of the Bending Moment Diagram is 1 order above the curve of the

Shear Diagram.

Maximum and Minimum Bending Moments occur where the Shear Diagram

Passes through the X-axis(i.e. at points of zero shear).


53

Sharp of Shear and Bending Moment diagrams


54


55

c) Deflections

The qualitative deflected shape (also called “elastic curve”) of a beam is simply an

approximate exaggerated sketch of the deformed beam due to the given loading. The

deflected shape is useful in understanding the structural behavior.

A sketch of the deflected shape can be made by making use of the bending moment

diagram:

Where the moment is positive, the deflected shape should be concave upward.

Where the moment is negative, the deflected shape should be concave downward.

Where the moment is zero, there is a point of zero curvature.

At a roller or pin support, the vertical deflection must be zero and the member may

rotate at that support.

At a fixed support, the rotation must also be zero


56

Principle of Superposition :


57

EXAMPLE 1

Use the methods of superposition to find the deflection at the free end of the

cantilever beam as shown.

SOLUTION

Two basic loadings are used as shown in diagrams (b) and (c) above.

The deflection at A is equal to the sum of the deflections under the Point Load and the

Uniform Distributed Load.

P L3 w L

4

δA = 3 EI

+ 8 EI

2 (9)3 0.5 (9)

4

δA = 3 EI

+ 8 EI

896 δA =

E I


58

EXAMPLE 2

Use the method of superposition to find the deflection at the middle of the simply

supported beam shown below:

SOLUTION

Two basic loadings are used. The deflection at point M is equal to the sum of the

deflections due to the point load and UDL.

P a 5 w L4

∆M = 48 EI

(3L2 – 4a

2) +

384 EI

10 (3) 5 (1.5)(12)4

∆M = 48 EI

(3(12)2 – 4(3)

2) +

384 EI

652 ∆M =

E I


59

5) TRUSSES

A truss is a structure composed of straight slender members joined together at their

end points. The members are usually formed into triangular patterns to produce an

efficient light-weight, load bearing structure.

The joint connections are usually formed by bolting or welding the ends of the

members to a common plate, called a gusset plate, or by passing a large bolt or pin

through each of the members. In the analysis of trusses, the members are commonly

assumed to be connected at the joints by frictionless pins. Since no moment can be

transferred through a frictionless pin joint, truss members are assumed to carry only

axial force.

Planar trusses lie in single plane and are often used to support roofs and bridges.

Space trusses, which are three-dimensional structures, are also in common use.


60

CONVENTONAL TYPES OF BRIDGE AND ROOF TRUSSES


61

Space Trusses

A space truss consists of members joined together at their ends to form a stable

three-dimensional structure.


62

ANALYSIS OF TRUSSES

A truss is completely analyzed when the support reactions and the forces in all the

members are determined. The members forces may be analyzed by:

(1) the method of joints (considering the equilibrium of a joint),

(2) the method of sections (considering the equilibrium of a section of the truss),

or a combination of the two methods.

The analysis of a truss is based on the following assumptions:

1. Members are connected to joints by frictionless hinges.

2. Loads and reactions are applied only at the joints.

3. Members are straight and carry only axial load.

Although members in real trusses are commonly connected to gusset plates by welded

or bolted connections, the forces obtained under the above assumptions are reasonably

accurate if the members are relatively slender and the connections are such that the

centroidal axes of the members meeting at a joint are concurrent at a point.

Sign Convention

The convention is to label a tension force positive and a compression force negative.

Alternatively, we add a T after the numerical value of a force to indicate tension and a

C to indicate compression.


63

(a) Methods for Analysis of Plane Trusses

The member forces in a statically determinate truss can be found by making use of the

Equations of equilibrium. The process is to consider different free-body diagrams of

Parts of the structure.

The two common techniques are:

(1) Method of Joints

In this method, a free-body of each joint is considered, one joint at a time.

Two independent equilibrium equations are available for each joint.

You should work each time with at a joint with only two unknown member

forces Once the unknown forces at one joint are determined, they become known forces

for other joints.

(2) Method of Sections

In the method, a larger part of the structure is taken as a free-body.

All three equations of equilibrium are available and three unknown bar forces

can be determined.

This method is particularly useful when only certain bar forces are required.

In practice, it is often convenient to use a combination of the two methods. The

key is to choose the most convenient free-body diagrams.

(a) Section 1.1 (b) Section 2.2

Method of Joints Method of Sections


64

(1) The Method of Joints

If a truss is in equilibrium, then each of its joints must also be in equilibrium. Hence, the method of joints consists of satisfying the Equilibrium conditions :

ΣFx = 0 andΣFy = 0

for the forces exerted on the pin at each joint of the truss.

Procedure for Analysis:

First determine the reactions at the supports by considering the equilibrium of the

entire truss. Then draw the free-body diagram of a joint having at least one known

force and at most two unknown forces. (If this joint is at one of the supports, it may be

necessary to know the external reactions at the support.) By inspection, attempt to

show the unknown forces acting on the joint with the correct sense of direction. The x

and y axes should be oriented such that the forces on the free-body diagram can be

easily resolved into their x and y components, and then apply the two force

equilibrium equations ΣFx = 0 and ΣFy = 0. It should also be noted that if one of

the axes is oriented along one of the unknown forces, the other unknown can be

determined by summing forces along the other axis. Solve for the two unknown

member forces and verify their correct directional sense.

Continue to analyze each of the other joints, where again it is necessary to choose a

joint having at most two unknowns and at least one known force, realize that once the

force in a member is found from the analysis of a joint at one of its ends, the result

can be used to analyze the forces acting on the joint at its other end. Strict adherence

to the principle of action, equal but opposite force reaction, must, of course, be

observed. Remember, a member in compression” pushes” on the joint and a member

in tension “pulls “ on the joint.


65

(2) The Method of Sections

If the forces in only a few members of a truss are to be found, the method of sections

generally provides the most direct means of obtaining these forces. The method of

sections consists of passing an imaginary section through the truss, thus cutting it into

two parts. Provided the entire truss is in equilibrium, each of two parts must also be in

equilibrium; and as a result, the three equations of equilibrium may be applied to

either one of these two parts to determine the member forces at the “cut section”

Procedure for Analysis

Free-Body Diagram Make a decision as to how to “cut “ or section the truss

through the members where forces are to be determined. Before isolating the

appropriate section, it may first be necessary to determine the truss’s external

reactions, so that the three equilibrium equations are used only to solve for member

forces at the cut section. Draw the free-body diagram of that part of the sectioned

truss which has the least number of forces on it. By inspection, attempt to show the

unknown member forces acting in the correct sense of direction.

Equations of Equilibrium Try to apply the three equations of equilibrium such that

simultaneous solution of equations is avoided. In this regard, moments should be

summed about a point that lies at the inter-section of the lines of action of two

unknown forces; in this way, the third unknown force is determined directly from the

moment equation. If two of the unknown forces are parallel, forces may be summed

perpendicular to the direction of these unknowns to determine directly the third

unknown force.


66

Applying the Method of Sections

In applying the method of sections, two decisions are made:

(1) Choosing the free-body,

(2) Choosing the points for taking moments about.

In choosing the free-body, remember that the “cut” does not need to be a straight

line.

In choosing a point for taking moment, remember that it can be any point in the

plane. It does not have to be at a joint or a support.


67

Zero-Force Members

Truss analysis using method of joints is greatly simplified if one is able to first

determine those members that support no loading. These Zero-force members may be

necessary for the stability of the truss during construction and to provide support if the

applied loading is changed. The zero-force members of a truss can generally be

determined by inspection of the joints.

CASE 1 If no external load is applied to a joint that consists of two bars, the

force in both bars must be zero.

CASE 2 If no external load acts at a joint composed of three bars --- Two of which are

collinear, the force in the bar that is not collinear is zero.

ΣFy = 0 requires F3 = 0

ΣFx = 0 requires F1 = F2


68

Zero-Force Members

EXAMPLE

EXAMPLE


69

EXAMPLE

EXAMPLE

structural analysis ii 1

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