structural design analysis of reinforced concrete buildings

STRUCTURAL DESIGN

ANALYSIS

This work is intended to all students of Mizan – Teppi University, Ethiopia,

Engineering Campus, most especially to my students in Construction

Technology and Management (COTM).

The contents of this stresses professional applications, as the Lecturer is a

Practicing Civil Engineer with innumerable hands on experiences in

Construction Management.

This is a comprehensive reference works that will provide students with

direct approach in Structural Analysis and keep them abreast of New

Practices and Techniques.

Reinforced

Concrete

Buildings

1 (Engr. Roger Salvatierra Tobias)

THE LECTURERTHE LECTURERTHE LECTURERTHE LECTURER Engr. Roger Salvatierra TobiasEngr. Roger Salvatierra TobiasEngr. Roger Salvatierra TobiasEngr. Roger Salvatierra Tobias

E mail : [email protected] Ethio Tel : +251919180034

Registered Civil EngineerRegistered Civil EngineerRegistered Civil EngineerRegistered Civil Engineer License Number 0036565 Professional Regulations Commission www.prc.gov.ph

M. Sc. in ManagementM. Sc. in ManagementM. Sc. in ManagementM. Sc. in Management SLSU, Philippines

Doctor of Philosophy In Technology Management (Ph.D. Doctor of Philosophy In Technology Management (Ph.D. Doctor of Philosophy In Technology Management (Ph.D. Doctor of Philosophy In Technology Management (Ph.D. –––– TM)TM)TM)TM) (Dissertation Agendum On Going) Cebu City, Philippines

Ten(10) years Academe / Teaching ExperiencesTen(10) years Academe / Teaching ExperiencesTen(10) years Academe / Teaching ExperiencesTen(10) years Academe / Teaching Experiences Philippines, Bahrain and EthiopiaPhilippines, Bahrain and EthiopiaPhilippines, Bahrain and EthiopiaPhilippines, Bahrain and Ethiopia

1. Construction Management (Buildings, Roads, Bridges and Water Supply)1. Construction Management (Buildings, Roads, Bridges and Water Supply)1. Construction Management (Buildings, Roads, Bridges and Water Supply)1. Construction Management (Buildings, Roads, Bridges and Water Supply) 2. Contract Administration :2. Contract Administration :2. Contract Administration :2. Contract Administration : Experiences on Project Implementation Based on World Bank Procurement of Works, Goods and Services, Standards. 3. Computer Based Construction Planning, Scheduling, Quantity Surveying and 3. Computer Based Construction Planning, Scheduling, Quantity Surveying and 3. Computer Based Construction Planning, Scheduling, Quantity Surveying and 3. Computer Based Construction Planning, Scheduling, Quantity Surveying and

Estimates :Estimates :Estimates :Estimates : Knowledge on the Application of MS Project (2007 / 2010) Knowledge on the Application of Primavera Project Planner (P3 and P6) Knowledge on Cost Estimates Using MS Excel and Spreadsheet

More than twenty (20) years in Construction Industry :More than twenty (20) years in Construction Industry :More than twenty (20) years in Construction Industry :More than twenty (20) years in Construction Industry : World Bank Philippines, Dammam Kingdom of Saudi Arabia, Tripoli Libya,World Bank Philippines, Dammam Kingdom of Saudi Arabia, Tripoli Libya,World Bank Philippines, Dammam Kingdom of Saudi Arabia, Tripoli Libya,World Bank Philippines, Dammam Kingdom of Saudi Arabia, Tripoli Libya, Riyadh KSA and KFWGerman Govt. Finance Projects.Riyadh KSA and KFWGerman Govt. Finance Projects.Riyadh KSA and KFWGerman Govt. Finance Projects.Riyadh KSA and KFWGerman Govt. Finance Projects.

Experiences in: 1. High Rise Buildings 2. Airports (NewTripoli International Airports in Tripoli Libya–

STRABAG, Germany) 3. Light Railway Transit structures, Manila City Philippines 4. Bridges and Waterworks (1ST LED, Dept. Of Public Works and

Highways, Philippines)


REINFORCED CONCRETE BUILDINGS

Reinforced concrete is concrete in which reinforcing bars or other types of reinforcement have been

integrated to improve one or more properties of the concrete.

For many years, it has been utilized as an economical construction material in one form or another in

buildings, bridges, and many other types of structures throughout the world.

The basic constituent materials—cement, sand, aggregate, water, and reinforcing bars—are widely

available and that it is possible to construct a structure using local sources of labor and materials.

Building code provisions are founded on principles that

1. do not unnecessarily increase construction costs;

2. do not restrict the use of new materials, products, or methods of construction;

3. do not give preferential treatment to particular types or classes of materials, products, or methods

of construction.

American Concrete Institute (ACI) 318 is commonly referred to as the “ACI Code” or the “Code.”

The ACI Code provides minimum requirements for the design and construction of structural concrete

members. The term “structural concrete” refers to all plain and reinforced concrete members used for

structural purposes.

Design philosophies related to reinforced concrete members have changed over the years. Until the early

1960s, the primary design method for reinforced concrete was working stress design. In this method,

members are proportioned so that the maximum elastic stresses due to service loads are

less than or equal to allowable stresses prescribed in the Code.

The strength design method was included for the first time in the 1956 edition of the Code, and it

became the preferred design method in the 1971 Code. The strength design method requires that both

strength and serviceability requirements be satisfied in the design of any reinforced concrete member. In

general, reinforced concrete members are proportioned to resist factored load effects and

to satisfy requirements for deflection and cracking.

I. Foundation Design

Foundation is the base of any structure. Without a solid foundation, the structure would not hold for long.

We have to be very cautious with the design of foundations because our entire structure rests on the

foundation. The job of a foundation is to transfer the loads of the building safely to the ground.

The strength of the foundation determines the life of the structure. As we discussed in the earlier article,

design of foundation depends on the type of soil, type of structure and its load. Higher the load bearing

capacity of the soil, the larger the load it could safely carry.

Foundations are basically divided into Shallow Foundations and Deep Foundations.


We are going discuss the step by step guide to Column Footing Design for a shallow foundation.

I.1 Reinforced Concrete Footings

Footing comprises of the lower end of a column, pillar or wall which i enlarged with projecting courses so

as to distribute load.

Footings shall be designed to sustain the applied loads, moments and forces and the induced reactions and

to ensure that any settlement which may occur shall be as uniform as possible and the safe bearing

capacity of soil is not exceeded.

In sloped or stepped footings, the effective cross-section in compression shall be limited by the area above

the neutral plane, and the angle of slope or depth and location of steps should be such that the design

requirements are satisfied at every section.

Here is a step-by-step guide to Column Footing Design:

Column Footing Plan and Section | Foundation Design

Step 1

Area required for footing


Square = B = (w+w1)/P0

Where, Po = safe bearing capacity of soil

w1 = self weight of footing

w = self weight of footing

For Rectangle = b/d = B/D

A = b x d

Net upward pressure on the footing

q/p = W/A

Step 2

Bending Moment

Critical section for maximum bending moment is taken at the face of the column

For a square footing,

Mxx = q x B/8 (L – a)2

Mxx = q x L/8 (B – b)2

Myy = q x B/8 (L – a)2

Step 3

To fix the depth of the footing shall be greater of the following:

Depth from bending moment consideration

d = √(M/Qb)

where, Q = moment of required factor

Depth from shear consideration

Check for one way shear

Check for two way shear or punching shear

Critical shear for one way shear is considered at a distance ‘d’ from face of the column.

Shear force, V = qB [ ½(B – b) d]

Nominal shear stress, Tv = k . Tc

Tc = 0.16√fck


Step 4

Check for two way shear

Critical section for two way shear is considered at a distance at a distance d/2 from all the faces of the

column.

SF, V = q [ B2 – (b + d)2]

SF, V = q [L x B – (a + d)(b + d)]

Nominal shear stress, Tv = V/2((a+d)(b+d)d) ——- {for a rectangle

Tv = V/4((b+d)d) ——- {for a square

Tv = k . Tc

k = 0.5 + β > 1 ; [Beta β = ratio of sides of the column

Tc = 0.16√fck

Area of steel, Ast = M/((σ)stjd)

II. REINFORCED CONCRETE COLUMN

A column is defined as a compression member, the effective length of which exceeds three times the

least lateral dimension. Compression members whose lengths do not exceed three times the least lateral

dimension, may be made of plain concrete.

A column forms a very important component of a structure. Columns support beams which in turn support

walls and slabs. It should be realized that the failure of a column results in the collapse of the structure.

The design of a column should therefore receive importance.

Supporting the slabs is the main function of the columns… Such slabs are called Simply Supported Slabs.

Simply supported slabs could be either one way slab or a two-way slab. It depends on the dimensions of

the slab.

A column may be classified based on different criteria such as:

1. Based on shape

• Rectangle

• Square

• Circular

• Polygon

2. Based on slenderness ratio


• Short column, ? ? 12

• Long column, ? > 12

3. Based on type of loading

• Axially loaded column

• A column subjected to axial load and unaxial bending

• A column subjected to axial load and biaxial bending

4. Based on pattern of lateral reinforcement

• Tied columns

• Spiral columns

Minimum eccentricity

Emin > l/500 + D/30 >20

Where, l = unsupported length of column in ‘mm’

D = lateral dimensions of column

Types of Reinforcements for columns and their requirements

Longitudinal Reinforcement

• Minimum area of cross-section of longitudinal bars must be at least 0.8% of gross section area of

the column.

• Maximum area of cross-section of longitudinal bars must not exceed 6% of the gross cross-section

area of the column.

• The bars should not be less than 12mm in diameter.

• Minimum number of longitudinal bars must be four in rectangular column and 6 in circular column.

• Spacing of longitudinal bars measures along the periphery of a column should not exceed 300mm.

Transverse reinforcement

• It may be in the form of lateral ties or spirals.

• The diameter of the lateral ties should not be less than 1/4th of the diameter of the largest

longitudinal bar and in no case less than 6mm.

The pitch of lateral ties should not exceed

• Least lateral dimension

• 16 x diameter of longitudinal bars (small)

• 300mm

Helical Reinforcement


The diameter of helical bars should not be less than 1/4th the diameter of largest longitudinal and not less

than 6mm.

The pitch should not exceed (if helical reinforcement is allowed);

• 75mm

• 1/6th of the core diameter of the column

Pitch should not be less than,

• 25mm

• 3 x diameter of helical bar

Pitch should not exceed (if helical reinforcement is not allowed)

Least lateral dimension

• 16 x diameter of longitudinal bar (smaller)

• 300mm

--------------------------------------------------------------------------------------------------------------------------------------------


III. REINFORCED CONCRETE BEAMS

RC beams are cast in cement concrete reinforced with steel bars. Beams take up compressive and add rigidity

to the structure.

Beams generally carry vertical gravitational forces but can also be used to carry horizontal loads (i.e., loads

due to an earthquake or wind). The loads carried by a beam are transferred to columns, walls, or girders,

which then transfer the force to adjacent structural compression members. In Light frame

construction the joists rest on the beam.

A beam bends under bending moment, resulting in a small curvature. At the outer face (tensile face) of the

curvature the concrete experiences tensile stress, while at the inner face (compressive face) it experiences

compressive stress.

A singly reinforced beam is one in which the concrete element is only reinforced near the tensile face and

the reinforcement, called tension steel, is designed to resist the tension.

A doubly reinforced beam is one in which besides the tensile reinforcement the concrete element is also

reinforced near the compressive face to help the concrete resist compression. The latter reinforcement is

called compression steel. When the compression zone of a concrete is inadequate to resist the compressive

moment (positive moment), extra reinforcement has to be provided if the architect limits the dimensions of

the section.

An under-reinforced beam is one in which the tension capacity of the tensile reinforcement is smaller than

the combined compression capacity of the concrete and the compression steel (under-reinforced at tensile

face). When the reinforced concrete element is subject to increasing bending moment, the tension steel

yields while the concrete does not reach its ultimate failure condition. As the tension steel yields and

stretches, an "under-reinforced" concrete also yields in a ductile manner, exhibiting a large deformation and

warning before its ultimate failure. In this case the yield stress of the steel governs the design.

An over-reinforced beam is one in which the tension capacity of the tension steel is greater than the

combined compression capacity of the concrete and the compression steel (over-reinforced at tensile face).

So the "over-reinforced concrete" beam fails by crushing of the compressive-zone concrete and before the

tension zone steel yields, which does not provide any warning before failure as the failure is instantaneous.

A balanced-reinforced beam is one in which both the compressive and tensile zones reach yielding at the

same imposed load on the beam, and the concrete will crush and the tensile steel will yield at the same time.

This design criterion is however as risky as over-reinforced concrete, because failure is sudden as the

concrete crushes at the same time of the tensile steel yields, which gives a very little warning of distress in

tension failure.


Types of beam construction and RCC design of Doubly reinforced beam…

RC beam construction is of two types:

• Singly reinforced beam

• Doubly reinforced beam

Singly reinforced beam

A singly reinforced beam is a beam provided with longitudinal reinforcement in the tension zone only.

Doubly reinforced beam

• Beams reinforced with steel in compression and tension zones are called doubly reinforced beams.

This type of beam will be found necessary when due to head room consideration or architectural

consideration the depth of the beam is restricted.

• The beam with its limited depth, if reinforced on the tension side only, may not have enough

moment of resistance, to resist the bending moment.

• By increasing the quantity of steel in the tension zone, the moment of resistance cannot be

increased indefinitely. Usually, the moment of resistance can be increased by not more than 25%

over the balanced moment of resistance, by making the beam over-reinforced on the tension side.

• Hence, in order to further increase the moment of resistance of a beam section of unlimited

dimensions, a doubly reinforced beam is provided.

Besides, this doubly reinforced beam is also used in the following circumstances:

• The external live loads may alternate i.e. may occur on either face of the member.

For example:

• A pile may be lifted in such a manner that the tension and compression zones may alternate.


• The loading may be eccentric and the eccentricity of the load may change from one side of the axis

to another side.

• The member may be subjected to a shock or impact or accidental lateral thrust.

Design procedure for doubly reinforced beam

Step 1

Determine the limiting moment of resistance for the given c/s (Mu lim) using the equation for singly

reinforced beam

Mulim = 0.87.fy.Ast1.d [1 – 0.42Xumax]

Or

Balanced section

Ast1 = (0.36.fck.b.Xumax)/(0.87fy)

Step 2

If factored moment Mu > Mulim, then doubly reinforced beam is required to be designed for additional

moment.

Mu – Mulim = fsc.Asc (d – d’) [fsc value from page no. 70]

Step 3

Additional area of tension steel Ast2

Ast2 =Asc.fsc/0.87fy

Step 4

Total tension steel Ast, Ast = Ast1 + Ast2

-------------------------------------------------------------------------------------------------------------------------------------------

IV. RC Staircase Design

RCC Structures are nothing but reinforced concrete structures. RCC structure is composed of building

components such as Footings, Columns, Beams, Slabs, Staircase etc.

These components are reinforced with steel that give stability to the structure. Staircase is one such

important component in a RCC structure.


In this article, we will discuss different types of staircases and study the dog-legged reinforced cement

concrete staircase design.

Stairs

Stairs consist of steps arranged in a series for purpose of giving access to different floors of a building.

Since a stair is often the only means of communication between the various floors of a building, the

location of the stair requires good and careful consideration.

In a residential house, the staircase may be provided near the main entrance.

In a public building, the stairs must be from the main entrance itself and located centrally, to provide quick

accessibility to the principal apartments.

All staircases should be adequately lighted and properly ventilated.

Various types of Staircases

• Straight stairs

• Dog-legged stairs

• Open newel stair

• Geometrical stair

RCC Dog-legged Staircase design


In this type of staircase, the succeeding flights rise in opposite directions. The two flights in plan are not

separated by a well. A landing is provided corresponding to the level at which the direction of the flight

changes.

Procedure for Dog-legged Staircase design

Based on the direction along which a stair slab span, the stairs maybe classified into the following two

types.

1. Stairs spanning horizontally

2. Stairs spanning vertically

Stairs spanning horizontally

These stairs are supported at each side by walls. Stringer beams or at one side by wall or at the other side

by a beam.

Loads

• Dead load of a step = ½ x T x R x 25

• Dead load of waist slab = b x t x 25

• Live load = LL (KN/m2)

• Floor finish = assume 0.5 KN/m

Stairs spanning Longitudinally

In this, stairs spanning longitudinally, the beam is supported ay top and at the bottom of flights.

Loads

• Self weight of a step = 1 x R/2 x 25

• Self weight of waist slab = 1 x t x 25

• Self weight of plan = 1 x t x 25[(R2 + T2)/T]

• Live load = LL (KN/m2)

• Floor finish = assume 0.5 KN/m

For the efficient design of an RCC stair, we have to first analyze the various loads that are going to be

imposed on the stair.

The load calculations will help us determine, how much strength is required to carry the load. The strength

bearing capacity of a staircase is determined on the amount of steel and concrete used.

The ratio of steel to concrete has to be as per standards. Steel in the staircase will take the tension

imposed on it and the concrete takes up the compression.


--------------------------------------------------------------------------------------------------------------------------------------------

V. SIMPLY SUPPORTED SLABS

Simply Supported Slabs Are Supported On Columns

Simply supported slabs don’t give adequate provision to resist torsion at corner to prevent corner

from lifting.

The maximum bending moment will be given if the slabs are restrained. But atleast 50% of the

tension reinforcement provided at the mid span should extend to the support. The remaining 50%

should extend to within 0.1Lx or Ly at the support as appropriate.

RC Slab Design depends on the on the dimensions of the slab after which the slab is termed as a

one-way slab or a two-way slab…

In the design of RC Slab structures, Column Design and Beam Design are to be done before we

start with RCC Slab Design…

Basic Rules followed in the design of simply supported Slab :


PRACTICAL ANALYSIS / CALCULATIONS


SEISMIC DESIGN

Total Dimension of building: Given Data: From NSCP

Width = 5.10 mts S = 1.00

Length = 7.00 mts I = 1.00

Unit wt. of concrete = 24.00 Kn/m^3 Rw = 10.00

hn = 3.30 mts Z = 0.40

h1 = 0.00 mts Ct = 0.08

h2 = 3.30 mts

Weight Level2 = 0.00 Kn

Weight Level1 = 171.36 Kn

Total weight = 171.36 Kn

Compute Base Shear:

V = (ZIC/Rw)Wt

= 26.54 Kn

C = (1.25S)/(T^2/3)

= 3.87

T = Ct(hn^0.75)

= 0.18 < 0.70 sec

Eccentricty:

e = 0.35

Mt = 0.35 Kn.m

Compute Level Forces

LEVEL hx Wxhx Fx

2.00 3.30 0.00 0.00

1.00 3.30 565.49 26.54

ΣWh 565.49 26.54

Distribution of Lateral forces

FRAME R d d^2 Rd^2 (MtRd^2)/dΣRd^2

Wx

0.00

171.36

R/ΣR Direct + Torsion


DESIGN OF SQUARE FOOTING

f'c = 21.0000 Mpa Width "L1" = ? mm

fy = 228.0000 Mpa length "L2" = ? mm

Dead load = 10.0000 KN Depth "d" = ? mts

Live load = 25.0000 KN Es = 200000.0000

Allow. soil

Pressure

= 250.0000 Kpa Wt. Soil = 15.6000 Kn/m^3

Unit wt. of

concrete

= 23.5000 Kpa Column Width

"b"

= 300.0000 mm

Column depth

"d"

= 300.0000 mm

Solution:

Initial Estimate of Footing Depth:

Area of

Footing

= (Ll+Dl)/Allow. Soil Pressure

= 0.1400 m^2

Depth = 20%(Length) + 75mm concrete covering

= 75.0280 mm Say

adopt

= 350.0000 mm

Effective Soil Bearing Capacity

qe = qe-Σφh

= 250.0000 Kpa

Dimension of footing:


Area of footing = Unfactored load/qe

= 0.1400 m^2

Footing Dimesion = 0.3742 x 0.3742 SAY

= 0.9000 x 0.9000 mts

Depth of Footing:

qu = Factored Load / Area of Footing

The linked image cannot be display ed. The file may hav e been mov ed, renamed, or deleted. V erify that the link points to the correct file and location.


The linked image cannot be display ed.

The linked image cannot be display ed. The


The linked

image

cannot be

display ed.

The file

may hav e

been

mov ed,

renamed,

or deleted.

V erify that

the link

points to

the correct

file and

location.

The linked image cannot be display ed. The file may hav e been mov ed, renamed, or

deleted. V erify that the link points to the correct file and location.

T

h

e

l

i

n

k

e

d

i

m

a

g

e

c

a

n

n

o

t

b

e

d

i

s

p

l

a

y

e

d

T

h

e

f

i

l

e

m

a

y

h

a

v

e

b

e

e

n

m

o

v

e

d

,

r

e

n

a

m

e

d

,

o

r

d

e

l

e

t

e

d

V

e

r

i

f

y

t

h

a

t

t

h

e

l

i

n

k

p

o

i

n

t

s

t

o

t

h

e

c

o

r

r

e

c

t

f

i

l

e

a

n

d

l

o

c

a

t

i

o

n

The linked image cannot be display ed. The file

The linked image cannot be

display ed. The file may hav e

been mov ed, renamed, or

deleted. V erify that the link

points to the correct file and

location.


= 0.0698 Mpa

Based on wide-beam shear:

Vu = qu*Ashaded

Vu = 62.7778 * ( 300.00 - d)

b/2 = 150.0000 mm

(L1)/2-b/2)-d = 300.0000 - d

Vu = ФVc

ФVc = 1/6sqrt(f'c)bwd

Vc = 584.2784 d

Therefore :

Depth = 7.5571 say

=

8.3071

Required Steel Area:

Mu = 30.3750 Kn.m

Mu = Rubd^2

Ru = 4.9622 Mpa

ρ = 0.85f'c/fy(1-SQRT(1-2Ru/0.85f'c)

= 0.0783

The linked image

cannot be display ed.

The file may hav e been

mov ed, renamed, or

deleted. V erify that the

link points to the

correct file and location.

The linked image cannot be display ed. The file may hav e been mov ed, renamed, or deleted. V erify that

the link points to the correct file and location.

The linked image cannot be display ed. The file may hav e been mov ed, renamed, or deleted. V erify that the link points to the correct file and location. The linked image cannot be display ed. The file may hav e been mov ed, renamed, or

deleted. V erify that the link points to the correct file and location.


file may hav e been mov ed, renamed, or

deleted. V erify that the link points to the


The

linked

image

cannot

be

display e

d. The

file may

hav e

been

mov ed,

renamed

, or

deleted.

V erify

that the

link

points to

the

correct

file an…

The linked image

cannot be

display ed. The file

may hav e been

mov ed, renamed,

or deleted. V erify

that the link points

to the correct file

and location.

The linked image


The file may hav e

been mov ed,

renamed, or deleted.

V erify that the link

points to the correct

file and location.

The linked image


The file may hav e

been mov ed,

renamed, or deleted.

V erify that the link

points to the correct

file and location.


The linked image cannot be display ed. The file may hav e been

mov ed, renamed, or deleted. V erify that the link points to the correct

file and location.

The linked image cannot be display ed. The file may hav e been mov ed, renamed, or deleted. V erify that the link points to the correct file and

location.

The linked image

cannot be

display ed. The

file may hav e

been mov ed,

renamed, or

deleted. V erify

that the link

points to the

correct file and

location.


0.0261

ρmin = 1.4/fy

= 0.0061

ρmax = 0.75ρb

= 0.0362

ρb = 0.85βf'c/fy(600/600+fy) Use ρ = 0.0061

= 0.0482

As = ρbd

= 41.4887 mm^2

Using 16 mm dia.

Bar:

N16 = As/Ab

= 0.2065 say 8 pcs

Development Length:

lb = .02Abfy/sqrt(f'c)

= 199.9700 mm or

lb = .06dbfy

= 218.8800 mm

therefore use ld = 218.8800 mm

Furnished ld = 225.0000 > 199.9700 OK


Verify if dowels of column bars extension are necessary:

Actual bearing Strenght:

Pu = 1.4Dl + 1.7Ll

= 56.5000 Kn

Permissible Bearing Strenght:

Ф0.85f'cA1 = 1124550.00

= 1124.5500 Kn

But this may be multiplied by SQRT(Aftg/Ac) <=2

= 0.9000

Permissioble Bearing Strenght:

= 1012.0950 > 56.5000 (No Need)

THEREFORE ADOPT:

0.9 X 0.9 FOOTINGWITH EFFECTIVE DEPTH TO TOP BARS OF 365MM,

WITH 8 PCS.16MM DIA REBARS ON BOTHWAYS AND AT LEAST 2

COLUMN BARS MUST BE EXTENDED INTO THE FOOTING



file may hav e been mov ed, renamed, or

deleted. V erify that the link points to the


The

linked

image

cannot

be

display

ed.

The

file

may

hav e

been

mov ed

,

renam

ed, or

delete

d.

V erify

that

the

link

points

to the

correct

file

and

locatio

n.


DESIGN / INVESTIGATION OF COLUMN

Given:

f'c = 21.0000 Mpa

fy = 228.0000 Mpa

Dead load = 10.0000 KN

Live load = 25.0000 KN

Bar diameter = 16.0000 mm

Solve External axial load "P"

Pu = 1.4Dl+1.7Ll

= 56.5 Kn

FROM NSCP CODE:

Pu = Ф0.80[0.85f'c(Ag-Ast)+FyAst]

Where Ф=0.70

Ast = 0.02 Ag

56.5 = 0.70*0.80[0.85f'c(Ag-Ast)+FyAst]

Ag = 4575.0173 mm^2 17.85

0.357

TRY: 4.56

Width"b"= 300.0000 mm

Depth "d"= 15.2501 say 22.053

= 300.0000 mm

Ast = 91.500347 mm^2

Number of Bars = 0.4553162 say

= 4.0000 pcs

Check for maximum steel ratio:

Actual Ast = 803.84

Actual Ag = 90000

ρg = Ast/Ag

= 0.0089316 < 0.02 OK

Spacing of Ties Adopt:

3 @ 75 mm

3 @ 100 mm

rest @ 200 mm

width

depth


DESIGN OF REINF. CONC. BEAM

Given:

f'c = 21.0000 Mpa Width "b" = 200.0000 mm

fy = 228.0000 Mpa Depth "d" = 300.0000 mm

Dead load = 3.5000 Kpa L = 4.4000 mts

Live load = 2.4000 Kpa Es = 200000.0000

Meq = 12.0000 Kn-m

Solution:

Wu = 1.4Dl+1.7Ll

Wu = 8.9800 Kn/m

Compute "-" Moment @ support

Compute "+" Moment @ support

-Mdl = 1/12(Wdl*L^2) +Mdl = 1/24(Wdl*L^2)

= 5.6467 Kn-m = 2.8233 Kn-m

-Mll = 1/12(Wll*L^2) +Mll = 1/24(Wll*L^2)

= 3.8720 Kn-m = 1.9360 Kn-m

Mu = 1.4Mdl+1.5Mll Mu = 1.4Mdl+1.5Mll

= 13.7133 Kn-m = 6.8567 Kn-m

Compute Design Moment (Md)

a.) Md = 1.4Mdl+1.7Mll

= 31.2180 Kn-m

b.) Md = 0.75(1.4Mdl+1.7Mll+1.87Meq)

= 40.2435 Kn-m

c.) Md = 0.90Mdl+1.43Meq

= 20.6448 Kn-m

Therefore Adopt:

Md = 40.2435 Kn-m

Check if Singly Reinforced or not:

Mumax = ФAsmaxFy(d-a/2) Asmax = ρmaxbd

= 2169.9800 mm^2

a = (AsmaxFy)/(0.85f'cb) ρmax = 0.75ρb

= 138.5870 = 0.0362


ρb = 0.85βf'c/fy(600/600+fy)

= 0.0482

Mumax = 102.7290 > 40.2435

Therefore the beam is Singly Reinforced

(If Mumax > Md)

Compute Area of Reinforcement:

Mu = ФRubd^2

Solve for Ru:

Ф = 0.9000

b = 200.0000

d = 300.0000

Ru = 2.4842 Mpa

Solve for ρ:

ρ = 0.85f'c/fy(1-SQRT(1-2Ru/0.85f'c)

= 0.0118

ρmin = 1.4/fy

= 0.0061

ρmax = 0.0362

As = 0.0066bd

As = 396.0000 mm^2

Using 16 mm dia. Bar:

N16 = As/Ab

= 1.9705 say 4 pcs

Check Actual Strain:

Єy = fy/Es c = a/β

= 0.0011 = 60.3973

Єs = (0.003(d-c))/c a = Asactfy/(0.85*f'cb)

= 0.0119 = 51.3377

Єs > Єy

There steel yield first

Check Web Reinforcement

ФVc/2 = (Ф(1/6)(f'c^0.5)*b*d)/2

= 19.4759 Kn


Vueff = 0.5WuL-Wud

= 17.0620

ФVc/2 > Vueff

Still provide Minimum Shear Reinforcement

Therefore Adopt the following cross-section

TOP BARS 16 MM DIA REBARS

BOTTOM BARS 16 MM DIA REBARS

4 2 2

2 2 4

400 m

m

200

AT LEFT AT RIGHT ATSUPPORT SUPPORT MIDSPAN


DESIGN OF CONCRETE SLAB

(-)Mc (+)Md (-)Mc (+)Md

Dl Ll Dl Ll

0.0330 0.0200 0.0280 0.0610 0.0230 0.0300

1.4Dl 3.4400 0.9890

1.7Ll 4.8160 1.2900

Mu 5.6760 2.7520 2.6230 0.7597

-0.05656 -0.02694 -0.02566 -0.00735

0.0052 0.0025 0.0024 0.0007

0.0062 0.0062 0.0061 0.0061

620.0000 620.0000 610.0000 610.0000

Middle Strip 182.3226 182.3226 182.3226 182.3226 185.3115 185.3115 185.3115 185.3115

Column Strip 273.4839 273.4839 273.4839 273.4839 277.9672 277.9672 277.9672 277.9672

Code=3t

Middle Strip 150.0000 150.0000 150.0000 150.0000 150.0000 150.0000 150.0000 150.0000

Column Strip 150.0000 150.0000 150.0000 150.0000 150.0000 150.0000 150.0000 150.0000

300.0000 300.0000

use "ρ"

0.0362 0.0362

0.0062 0.0061

620.0000 610.0000

Sp

acin

g

21.0000

228.0000 228.0000

0.0077

-0.02225

0.0020

2.2790

100.0000

12.0000 12.0000

74.0000 74.0000

0.0061 0.0061

-0.08365

21.0000

2.4000

2.0000 2.0000

2.4000

6.8800

25.0000

6.8800

6.2500

Use (mm)

Eff. Depth "d" mm

ω

ρact

ρmin=1.4fy

ρmax

As=ρbd

Bar diameter "db" mm

f'c

fy

Thickness "t" mm

SLAB (S-1)

Two-way

2.0000

SHORT SPAN DIRECTION

(+)Ms

L 5.0000 2.5000

LONG SPAN DIRECTION

m=la/lb

COEFFICIENT

(+)Ms

M=

cw

l^2

DL kn/m

LL kn/m

Wu=1.4Dl+1.7Ll

L^2 (m)

8.2560

Therefore Adopt:100 mm thk slab with 10 mm diameter Rebars spaced at 100 mm on centers bothways


Recommended Resources for further study

1. Soil Mechanics and Foundation Engineering, by author S. K. Garg, published by Standard

Publishers (931 pages)

structural design analysis of reinforced concrete buildings

Documents