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Notes of Lesson Design of Reinforced Concrete Elements

Department of Civil Engineering

Notes of Lesson

CE6505 - Design of Reinforced Concrete Elements

(R. 2013)

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UNIT I METHODS OF DESIGN OF CONCRETE STRUCTURES

WORKING STRESS METHOD DESIGN

GENERAL PRINCIPLES OF WORKING STRESS DESIGN

(a)

General features

During the early part of 20th

century, elastic theory of reinforced concrete sections outlined in chapter 2

was developed which formed the basis of the working stress or permissible stress method of design of

reinforced concrete members. In this method, the working or permissible stress in concrete and steel areobtained applying appropriate partial safety factors to the characteristics strength of the materials. The

permissible stresses in concrete and steel are well within the linear elastic range of the materials.

The design based on the working stress method although ensures safety of the structures at working orservices loads, it does not provide a realistic estimate of the ultimate or collapse load of the structure in

contrast to the limit state method of design. The working stress method of design results in

comparatively larger and conservative sections of the structural elements with higher quantities of steelreinforcement which results in conservative and costly design. Structural engineers have used this

method extensively during the 20th

century and presently the method is incorporated as an alternative to

the limit state method in AnnexureB of the recently revised Indian Standard Code Is : 4562000 forspecific applications.

The permissible stresses in concrete under service loads for the various stress states of compressive,flexure and bond is compiled in Table 2.1 (Table 21 of IS ; 4562000)The permissible stress in different types of steel reinforcement is shown in table 2.2 (Table 22 of IS 4562000)

The permissible shear stress for various grades of concrete in beams is shown in Table 12.1 (Table 23 o

IS: 4562000)

The maximum shear stress permissible in concrete for different grades is shown in Table 12.2 Table12.2 (Table 24 of IS: 4562000)

In the case of reinforced concrete slabs, the permissible shear stress in concrete is obtained bymultiplying the values given in Table 2.1 by factor k whose values depend upon the thickness of slab

as shown in Table 12.3 (Section 40.2.1.1. of IS; 4562000

)

Table 12.1Permissible Shear Stresses in Concrete (cN/mm2) (Table 23 of IS:4562000)

100 As/ bd Permissible shear stresses in concrete c

N/mm2

M15 M20 M25 M30 M35 M40 &

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ABOVE

0.15 0.18 0.18 0.19 0.20 0.20 0.200.25 0.22 0.22 0.23 0.23 0.23 0.23

0.50 0.29 0.30 0.31 0.31 0.31 0.32

0.75 0.34 0.35 0.36 0.37 0.37 0.38

1.00 0.37 0.39 0.40 0.41 0.42 0.42

1.25 0.40 0.42 0.44 0.45 0.45 0.46

1.50 0.42 0.45 0.46 0.48 0.49 0.49

1.75 0.44 0.47 0.49 0.50 0.52 0.52

2.00 0.44 0.49 0.51 0.53 0.54 0.55

2.25 0.44 0.51 0.53 0.55 0.56 0.57

2.50 0.44 0.51 0.55 0.57 0.58 0.60

2.75 0.44 0.51 0.56 0.58 0.60 0.62

3.00 & above 0.44 0.51 0.57 0.60 0.62 0.63

Note:Asis that area of longitudinal tension reinforcement which continues at least one effective depthbeyond the section being considered except at supports where the full area of tension reinforcement may

be used provided the detailing conforms to 26.2.3.

Table 12.2 Maximum Shear Stress (c, maxN/mm2) (Table 24 of IS: 4562000)

Concrete grade

(cmaxN/mm2)

M151.6

M251.8

M301.9

M352.3

M40 & above2.5

The maximum shear stress permissible in concrete for different grades is shown in Table 12.2 (Table 24

of IS 4562000)

In the case of reinforced concrete slabs, the permissible shear stress in concrete is obtained bymultiplying the3 values in Table 2.1 by a factor k whose values depend upon the thickness of slab asshown in Table 12.3 (Section 40.2.1.1. of IS 4562000)

(b)

General design procedure

In the working stress design, the crosssectional dimensions are assumed based on the basic span /depth ratios outlined in Chapter 5 (Table 5.1 and 5.2) (Section 23.2.1. of IS: 4562000)

The working load moments and shear forces are evaluated at critical sections and the required effective

depth is checked by using the relation:

d = M / Q.b

Where d = effective depth of section

M = working load moment

b = width of section

Q = a constant depending upon the working stresses in concrete and steel, neutral axis depthfactor (k) and lever arm coefficient (f).

For different grades of concrete and steel the value of constant Q is compiled in Table 2.3. The depthprovided should be equal to or greater than the depth computed by the relation and the area of

reinforcement required in the section to resist the moment M is computed using the relation:

Ast= ( M )

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st. j. dThe number of steel bars required is selected with due regard to the spacing of bars and coverrequirements.

After complying with flexure, the section is generally checked for resistance against shear forces bycalculating the nominal shear stress cgiven by v= (V / bd)

Where V = Working shear force at critical section.

The permissible shear stress in concrete (c) depends upon the percentage reinforcements in the crosssection and grade of concrete as shown in Table 12.1

If c< vsuitable shear reinforcements are designed in beams at a spacing svgiven by the relation;

Sv= [ 0.87 fyAsvd / Vus]

Where sv= spacing of stirrups

Asv= crosssectional area of stirrups legs

fy= Characteristics strength of stirrup reinforcement

d = effective depth

Vs= [ Vc.b .d]If v< c, nominal shear reinforcements are provided in beams are provided in beams at a spacing givenby

Sv[ 0.87 fy Ast/ 0.4 b]

In case of slabs, the permissible shear stress if k is a constant depending upon the thickness of the slab.Also in the case of slabs the nominal shear stress (v) should not exceed half the value of cmaxshown inTable 12.2. In such cases the thickness of the slab is increased and the slab is redesigned.

In the case of compression members, the axial load permissible on a short column reinforced with

longitudinal bars and lateral ties is given by

P = (ccAc+ scAsc)

Where scc= permissible stress in concrete in direct compression (Refer Table 2.1)

Ac= crosssectional area of concrete excluding the area of reinforcements.

Ssc= permissible compressive stress in reinforcement

Asc= crosssectional area of longitudinal steel bars.

DESIGN OF SLABS

1. Design example of one way slab

1. Data

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Clear span = 2.5m

Slab supported on load bearing brick walls 230mm thick

Loading: Residential floor, 2 kN/m2

Materials: M-20 grade concrete

Fe415 HYSD bars

2. Allowable Stresses

cbc= 7 N/mm2 Q = 0.91

st= 230 N/mm2 f = 0.90

3. Depth of slab

Assuming 0.4 per cent of reinforcement in the slab, the value of Kt(Figure) Using Fe 415 HYSD

bars, is around 1.25Hence (L/d) = (L/d)basicx Ktx Kc

= (20 x 1.25 x 1)

= 25

d = (2500 / 25) = 100mm

Adopt d = 100mm and overall depth = 130mm

4.

Effective span

Effective span is the least of:

(a) Centre to centre of support = (2.5 + 0.23) = 2.73m

(b)Clear span + effective depth = (2.5 + 0.10) = 2.60m

Effective span = L = 2.60m

5. Loads

Self weight of slab = (0.13 x 25) = 3.25kN/m2

Live load on floor = 2.00kN/m2

Floor finishes = 0.75kN/m2

Total load = w = 6.00kN/m2

Considering 1 m width of the slab, the uniformly distributed load is 6 kN/m2on an effective span

of 2.60m.

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6. Bending moments and shear forces

M = (0.125 w L2) = (0.125 X 6 X 2.6

2) = 5.07KN.m

V = (0.5 w L) = ( 0.5 X 6 X 2.6) = 7.80Kn

7. Effective depth

d = M/ Qb = 5.07 x 106/ 0.91 x 103= 75mm

Effective depth adopted d = 100mm, hence safe.

8. Main reinforcement

Ast= ( M / st. j .d) = (5.07 x 106/ 230 x 0.9 x 100) = 245mm

2

Minimum reinforcement = (0.0012 x 1000) = 156mm2< 245mm

2

Spacing of 10mm diameter bars is given by

S = (1000 ast/ Ast) = (1000 x 79 / 245) = 322mm

Provide 10mm diameter bars 300mm centers (Ast= 262mm2)

9. Distribution reinforcement

Ast= (0.0012 x 1000 x 130) = 156mm2

Provide 8mm diameter bars at 300mm centers (Ast= 167mm2)

10.

Check for shear stress

c= (V / bd) = (7.80 x103/ 20

3x 100) = 0.078 N/mm

2

Assuming 50 percent of reinforcement to be bent up near supports, we have:

(100 Ast/ bd) = (100 x 0.5 x 262 / 1000 x 100) = 0.131

From Table 23 (IS: 4562000), interpolating permissible shear stress for solid slabs is:

(k . c) = (1.30 x 0.18) = 0.234 N/mm2> v.

Hence shear stresses are within safe permissible limits.

11.Check for deflection control

Percentage reinforcement = p1= (100 x 262 / 1000 x 100) = 0.262

For pt= 0.262, Kt= 1.6 (Figure 4 of IS; 4562000)

(L/d)max= (20 x 1.6) = 32

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(L/d)provided= (2600 / 100) = 26 < 32, hence safe.

2. Design example of two way slab for residential floor using the following data:

1. Data

Size of floor 4 m by 5 m, simply supported on all the sides on load bearing walls 230mm thick

without any provision for torsion at corners. Adopt M- 20 grade concrete and Fe415 HYSD.

2. Permissible Stresses

cbc= 7 N/mm2 Q = 0.91

st= 230 N/mm2 j = 0.90

3. Type of slab

Simply supported on all sides without any provision for torsion at corners.

Lx= 4 m

Ratio (LY/ Lx) = 1.25Ly= 5 m

4. Depth of the slab

From span / depth considerations:

Overall depth = D = (short span / 28) = (4000 / 28) = 143mm

Adopt overall depth = D = 150mm

Effective depth = d = (15030) = 120mm

5. Effective Span

Effective span is the least of the following:

(a)Centre to center of supports = (4 + 0.23) = 4.23 m

(b)Clear span + effective depth = ( 4 + 0.12) = 4.12m

Effective span = Lxe= 4.12m

6. Loads

Self weight of slab = (0.15 x 25) = 3.75 kn/m2

Live load on floor = 2.00

Floor finishes = 0.60

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Total services load = w = 6.35 kN/m2

7. Bending Moments

Refer Table 7.1 and read out the moment coefficients for the ratio (Ly/ Lx) = 1.25

x= 0.089, y= 0.057Mx= (xw Lxc

2) = (0.089 x 6.35 x 4.12

2) = 9.60kN.m

My= (yw Lxc2) = (0.057 x 6.35 x 4.12

2) = 6.14kN.m

8. Check for depth

Effective depth

D = M / Q b = 9.60 x 106/ 0.91 x 103= 102.7mm

Effective depth for shorter span = 120mm

Effective depth for long span = (12010) = 110mm

(Using 10mm diameters bars)

9. Reinforcements

Ast= (M / st. j .d) = (9.6 x 106/ 230 x 0.9 x 110) = 387mm

2

Adopt 10mm diameter bars at 200mm centers (Ast= 393 mm2)

Steel for long span = (6.14 x 106

/ 230 x 0.9 x 110) = 270mm2

Provide 10mm diameter bars at 250mm centers (Ast= 315mm2)

10.Shear and bond stresses

Shear and bond stresses in two way slabs are negligibly small and generally within safepermissible limits. The reinforcement details are similar to that of two way slabs designed in

Chapter 7.

DESIGN OF BEAMS

1. Design of singly reinforced concrete beams; Design a rectangular reinforced concrete beam

simply supported on masonry walls 300mm thick with an effective span of 5 m to support aservice load of 8 kN/m and a dead load of 4 kN/m in addition to its weight. Adopt M 20 gradeconcrete and Fe415 HYSD bars. Width of support of beams = 300mm.

1. Data

Effective Span = L = 5 m

Width of support = 300mm

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Live load = 8 KN/mDead load = 4 KN/m

Material: M20grade concreteFe415 HYSD bars

2. Allowable stresses

cb= 7 N/mm2 Q = 0.91

st= 230 N/mm2 j = 0.90

3. Crosssectional dimensions

Adopt width of beam = b = 300mm

Since the loading is heavy adopt

Effective depth = d = (span / 10) = (5000 / 10) = 500mm

Overall depth = D = (500 + 50) = 550mm

4. Loads

Self weight of beam = (0.3 x 0.55 x 25) =4.125kN/m

Dead load = 4.000kN/m

Live load = 8.000kN/m

Finishes = 0.975kN/m

Total load = w = 17.000kN/m

5. Bending Moment and shear forces

M = 0.125 Wl2= (0.125 x 17 x 5

2) = 53 kNm

V = 0.5 w L = (0.5 x 17 x 5) = 43 Kn

6.

Check for depth

d = M / Q b = 53 x 106/ 0.91 x 300 = 440mm

Effective depth provided = d = 500mm, hence adequate.

7. Main tension reinforcement

Ast= (M / st. j.d) = (53 x 106 / 230 x 0.90 x 500) = 512mm

2

Provide 2 bars of 20mm diameter (Ast= 628mm2)

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8. Shear stress and reinforcement

Nominal Shear stress = v= (Vu/ bd) = (43 x 103/ 300 x 500) = 0.28 N/mm

2

= (100 Ast / bd) = (100 x 628 / 300 x 500) = 0.418

Refer Table (IS; 456) and read out the permissible shear stress in concrete as:

c= 0.25 N/mm2< v

Hence shear reinforcements in the form of stirrups are required since cis nearly equal to v,provide nominal shear reinforcements given by:

Sv= Asv. sv. d / Vs

Using 6mm diameter twolegged stirrups

Sv= (2 x 28 x 0.87 x 415 / 0.4 x 300) = 168mm

Provide 6mm diameter stirrups at 150mm centre up to quarter span length from supports andgradually increased to 300mm centre towards the centre of span.

2. Design a doubly reinforced beam: Design a doubly reinforced concrete beam for a

residential floor of a building to suit the following data:

1. Data

Effective Span = 5 m

Dead load = 8 KN/m

Live load = 12 KN/m

Width of beam = 250mm


Effective depth = 450mm

Cover to compression steel = 50mm.

2. Permissible stresses

cb= 7 N/mm2 Q = 0.91

st= 230 N/mm2 j = 0.90

m = 13 nc= 0.284 d

3. Loads

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Self weight of beam = (0.25 x 0.5 x 25) =3.125kN/m

Dead load = 8.000kN/m

Live load = 12.000kN/m

Finishes etc. = 0.875kN/m

Total service load = w = 24.000kN/m


M = 0.125 Wl2= (0.125 x 24 x 5

2) = 75 kNm

V = 0.5 w L = (0.5 x 24 x 5) = 60 Kn

5. Resisting Moment

Resisting moment capacity of balanced singly reinforced section is computed as;

M1= (Q b d2) = (0.91 x 250 x 450

2) x 10

-6= 46 kNm

Balance moment = M2= (MM1) = (7546) = 29KNm.

6. Tension reinforcement

Ast= (M1/ st. j.d) = (46 x 106 / 230 x 0.90 x 450) = 493mm

2

Additional steel in tension for balanced moment M2is:

Ast2= (M2/ st( ddc) = (29 x 106/ 230 x (45050) = 315mm2

Total tension steel = Ast= ( Ast1+ Ast2) = (493 + 315) = 808 mm2

Provide 3bars of 20mm diameter (Ast= 942mm2)

7.

Compression reinforcement

Asc= [m Ast2( dnc)(1.5 m1) (ncdc)

Where nc= 0.284 d = (0.284 x 450) = 127.8mm

Asc= [ 13 x 315 (450127.8) / 1.5 x 131) ( 127.850) = 916mm2

Provide 3 bars of 20mm diameter (Asc= 942 mm2)

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v= (Vu/ bd) = (60 x 103/ 250 x 450) = 0.53 N/mm

2

= (100 Ast / bd) = (100 x942 / 250 x 450) = 0.83

Refer Table 23 (IS: 456) and read out the permissible shear stress as

c= 0.36 N/mm2< v

Hence shear reinforcements are to be designed to resist the balance shear computed as:

Vs= [ Vcb d] = [ 60( 0.36 x 250 x 450) 10-3

] = 19.5kN

Using 6mm diameter 2 legged stirrips, spacing is:

Sv= Asv. sv. d / Vs

Sv= (2 x 28 x 230 x 450 / 19.5 x 103) = 297mm

Provide 6mm diameter twolegged stirrups at 250mm centre at supports, graduallyincreasing to 300mm centre towards the centre of span.

3. Design of flanged beams: Design a tee beam for an office floor using the following data.

1. Data

Effective Span = 8 m

Spacing of tee beams = 3m

Loading (office floor) = 4 KN/m

Slab thickness = 150mm


2. Permissible stresses

cb= 7 N/mm2 Q = 0.91

st= 230 N/mm2 j = 0.90

m = 13

3. Sectional dimensions

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Effective depth = d (span / 15) = (8000 / 15) = 534mm

Adopt d = 550mm and overall depth = D = 600mm and b = 300mm

4. Loads

Self weight of beam = (0.15 x 25x 3) =11.25kN/m

Live load = (4 x 3) = 12.000kN/m

Floor finish = (0.6 x 3) = 1.80kN/m

Self weight of rib = (0.45 x 0.3 x 25) = 3.37kN/m

Plaster finishes = 1.58 kN/m

Total load = w = 30.00kN/m


M = 0.125 Wl2= (0.125 x 30 x 8

2) = 240 kNm

V = 0.5 w L = (0.5 x 30 x 8) = 120 KN

6. Check for depth

Ast= (M / stj d) = ( (240 x 106/ (230 x 0.9 x 550) = 2108mm

2

Provide 4 bars of 28mm diameter (Ast= 2464 mm2)

7. Effective flange width

Least of the following:

i) bf= [ Lo/ 6 + bw+ 6 Df]

= [ (8000 / 6) + 300 + (6 x 150)] = 2533mm

ii) bf= centre to centre of ribs = 3000mm

Hence bf= 2533 mm8. Check the stresses

Let n = depth of neutral axis

(bfn2/ 2)/2 = (!3 x 2464) ( 550n)

Solving n = 106mm

Level arma[ d(n/3) = [550(106 / 3)] = 514.67 mm

st= (240 x 106/ 2464 x 514.67) = 189 N/mm

2< 230N/mm

2

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cb= [ (189 x 106) / (!3 x 444)] = 3.47 N/mm

2< 7 N/mm

2

Hence the stresses are within safe permissible limits.


Maximum shear force = v = 120kN

v= (Vu/ bd) = (120 x 103/ 300 x 550) = 0.72 N/mm

2

= (100 Ast / b wd) = (100 x2464 / 300 x 550) = 1.49

Refer Table 23 (IS: 456) and read out the permissible shear stress as

c= 0.45 N/mm2< v

Hence shear reinforcements are to be designed to resist the balance shear given by

Vs= [Vcbwd] = [120(0.45 x 300 x 550) 10-3

] = 46kN

Using 6mm diameter 2 legged stirrups, spacing is given by

Sv= Asv. sv. d / Vs

Sv= (2 x 28 x 230 x 550 / 46 x 103) = 154 mm

Provide 6mm diameter 2 legged stirrups at 150mm centre near supports and graduallyincreased to 300mm towards the centre of span.

Limit State Concept:

The structure shall be designed to withstand safely all loads liable to act on throughout its life. It shall also

satisfy the serviceability requirements, such as limitations on deflection and cracking.

The acceptable limit for the safety and serviceability requirements before failure occurs is called a limit state.

The aim of design is to achieve acceptable probabilities that the structural will not become unfit for the use for

which it is intended that is, that will not reach a limit state.

Limit state of Collapse:

The limit state of collapse of the structure or part of the structure could be assessed from rupture of one or more

critical sections and from buckling due to elastic or plastic instability (including the effects of sway where

appropriate) or overturning.

The resistance to Bending

ShearTorsion and

Axial loads at every section shall not be less than the appropriate value at that section

produced by the probable most unfavorable combination of loads on one structure using the appropriate Partial

safety factors.

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Limit State of Serviceability:Deflection

Cracking

The acceptable limits of cracking would vary with the type of structure and environmental.

Characteristic and Design values and Partial Safety factors:

Characteristic value = Minimum yield stress

0.2 percent proof stress

Characteristic Load:

The term characteristic load means that value of load which has a 95 percent probability of not being

exceeded during the life of the structure.

0. L IS 875 (Part 1)

Imposed loads Is 875 (Part 2)

Wind loads IS 875 (Part 3)Snow loads IS 875 IS 1893

Design Value:

Design Strength of Materials fd= fm

Where,

f = Characteristic Strength of the Material

m= Partial Safety factor appropriate to the material and the limitState being considered.

Loads:The design load, Fdis given by

Fd= F rf

Partial Safety factor for material strength:

m 1.5 for concrete

1.15 for Steel

Maximum Strain = 0.0035In concrete at the outermost compression fibre

Area of stress block = 0.36fck.xu

Depth of center of compressive force = 0.42xu from the extreme fibre in compression.

fck= Characteristic compressive strength of concrete and

xu = depth of Neutral axis.

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Maximum Strain in the tension reinforcementfy + 0.002

1.15 Es (0.0037)

Wherefy= Characteristic Strength of Steel and

ES= Modulus of Elasticity of Steel.

fck

Cold worked Deformed bar

Parabolic

Curve0.67 fck

0.67 fck/ m= 0.446fck

0.00350.002Strain

0.42xu

0.36 fckxu b.xu

Stress block Parameters

Strain

0.002

Stress

fYfy

f / 1.15

fY

f

0.446fck

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Steel bar with definite Yield Point

Representative Stressstrain Curve for Reinforcement

fy + 0.0021.15 Es

Strain Diagram

0.0035 = fy + 0.002xu 1.15 Es

dxu

xu = 0.0035

dxu 0.87 fy+ 0.002Es

xu = 0.0035dxu +xu 0.87 fy+ 0.002 + 0.0035

Es

xu = 0.0035d 0.87 fy+ 0.002

Es

Fe 250 xumax = 0.53

Fe 415 d

Fe500

Stress

f / 1.15

xu

d

b

d- xu

0.446 fck

0.42 xu

C = 0.36 fckxu max.b.

T = 0.87 f Ast

d-0.42 xu

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0.87 fy + 0.002

Es

0.0035 = 0.87fy + 0.002xumax Es

dxumax

xu max = 0.0035

dxu max 0.87 fy+ 0.002Es

xumax = 0.0035dxumax +xumax 0.87 fy+ 0.002 + 0.0035

Es

Singly Reinforced Sections

Maximum depth of Neutral Axis

0.0035 = 0.87fy + 0.002xumax Es

dxumax

xumax = 0.0035

dxumax +xumax 0.87 fy+ 0.002 + 0.0035Es

xumax = 0.0035d 0.0055 + 0.87 fy

fy xumaxd

Xumax =

250 0.53 0.53d

415 0.48 0.48d

500 0.46 0.46d

xu

b

d

0.0035

C = 0.36 fckxu max.b.

0.42 xu

0.87 f Ast = T

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Es

Mild Steel

xumax = 0.53

d

Fe415

xumax = 0.48

d

Fe500

xumax = 0.46

d

C = T

0.36 fckxu b = 0.37 fyAst

Xu = 0.87 fyAst

0.36 fckbd

Xu = 0.87 fyAst

d 0.36 fckbd

Moment of resistance = {Total Compression or Total Tension}

Level armMR= 0.36 fck. Xu.b ( d0.42 xu)

= 0.36 fckxu bd ( d0.42 xu)d

MR= 0.36 fckxu bd2(10.42 xu )

d d

MR= 0.87 fyAst( d0.42 xu)= 0.87 fyAstd (10.42 xu )

d

= 0.87 fyAstd (1- 0.42 0.87 fyAst

0.36 fck. bd

= 0.87 fyAstd (1fyAst )bd fck

C= T

0.36 fckxu b = 0.87 fyAst

Xu = 0.87 fyAst

0.36 fckb

xu = 0.87 fyAst

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d 0.36 fck. bd

MR= 0.87 fyAst(d0.42 xu)= 0.87 fyAstd (1- 0.42 xu )

d

= 0.87 fyAstd (1- 0.42 0.87 fyAst )

0.36 fckbd

= 0.87 fyAst d (1- Ast fy )

Bd fck

MR= 0.36 fckxu max. b ( d0.42 xu)= 0.36 fckxumax bd

2(10.42 xu )

d

= 0.36 fckxu bd2(1- 0.42 xumax

d d= 0.36 fckxumax (1- 0.42 xumax) bd

2

d d

= 0.36 xumax (1- xumax ) fckbd

2

d d

Plimt= 41.4 fck xumax

fy d

Limiting Percentage of Steel

Plimit= 41.4 fck xumaxfy d

Moment of Resistance at limiting Condition

Grade of steel Xumax / d Limiting Moment of Resistance

Fe250 0.53 0.149 fckbd

Fe415 0.48 0.138 fckbd

Fe500 0.46 0.133 fckbd

Minimum and Maximum Percentage of Steel

Pmin= Ast x 100 = 85 %bd fy

For Fe 250 Pmin= 0.34%

Fck fy250 415 500

20 1.75 0.96 0.76

25 2.19 1.20 0.95

30 2.63 1.44 1.14

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Fe 415 Pmin= 0.20%

Fe 500 Pmin= 0.17%

UNIT II LIMIT STATE DESIGN FOR FLEXURE

A simply supported beam 250mm wide is 450mm deep to the centre of the tension reinforcement.

Determine the limiting moment of resistance of the beam section and also the limiting area of

reinforcement. Use M20 concrete and Fe 415 steel.

b = 250mm M20 & Fe 415

d = 450mm

Mulim= 0.138 fck bd2

Ast limit = 0.0957 x 250 x 450 = 1077 mm2

100Mu= 0.36 xumax (1- 0.42 xumax ) bd

2

d d

= 0.36 x 0.48 (1-0.42 x 0.48) bd2(20)

= 0.138 fck bd2

= 139.725 x 106

Nmm.

139.725 x 106= 0.87 fyAstd (1- Ast fy )

bd fck

139.725 x 106= 0.87x 415 fy x450 (1- Ast 415 )

250x450x20

= 162472.5 Ast (1-1.844 x 10-4

Ast)

29.967 Ast2= 16472.5 Ast+ 139.725 x 10

6= 0

5421.71 3277.86

2

Pt lim= 41.4 fck xumaxfy d

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A singly reinforced beam 250mm wide is 400mm deep to the centre of the tensile reinforcement.

Determine the limiting moment of resistance of the beam section and also the limiting area of

reinforcement. Use M20 concrete and the 250 steel.

Solution:

Given Data:B = 200mm

D = 400mm

Mulim = 41.4 fck/ fy xumax / d

M20 & Fe 250

Mu = 0.149 fckbd2

= 0.149 x20x200x4002

Mulimt= 95.36 x 106Nmm

Plimt= 41.4 x 20 x 0.53 = 1.755%

250

Ast lim = 1.755 x 200 x 400

100

= 1404 mm2

Use 20mm # No. of Bars = 1404 = 4.47

314.16

Use 16mm # 7 Nos.

Design a R.C.C. beam to rsist an applied of 50 kNm. Assuming width is 230mm. Use M20 & Fe 415

grade.

Solution:i) Data:

Applied moment = 50kNmFactored moment = 1.5 x 50 = 75 kNm

Breadth is restricted to = 230mm

M20 fck= 20N/mm2

Fe 415 fy = 415 N/mm2

ii) Maximum depth of Neutral axis

xumax = 0.48d

Ast = 1071.93 mm2

d

b

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xumax = 0.48d

iii) Moment of resistance

Mulim = 0.36 xumax (10.42 xumax) bd2fck

d d= 0.36 x 0.48d (1-0.42 x 0.48) 230x20 x d

2

= 0.138 fckbd2

75 x 106= 0.138 x 20 x 230 d2d = 343.73mm

6 31

Assume d = 25 + 12 = 37 mmD = d +d =373.73 mm

= 400mm

iv) Area of steel required

0.87 fy Ast = 0.36 fck xmax .b

Ast = 0.36 fckxumax .b

0.87 fy

= 0.36 x 20 x 177.12 x 230

0.87 x 415= 812.38 mm

2

Use 20mm # Nos = 3 Nos. Ast p = 942.48 mm2

v) Check for reinforcement:

Main reinforcement =

As = 0.85

bd fyAst = 0.85 bd

fy

= 0.85 x 230 x 369

415= 173.83 mm

2

Max. Reinforcement = 0.0460

= 0.04 x 230x400

= 3680 mm2

Astmim< Astp< Astmax

DOUBLY REINFORCED SECTION

1.Calculate the ultimate moment of resistance (or) factored moment of resistance of RCC beam of

rectangular section 300mm wide and 400mm deep for the following.

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Ast = 6 / 16mm #

Asc = 2/ 16mm #

M20 & Fe 250 grade

Effective cover = d = 33mm

Solution:

Given Data:

Width b = 300mm

Effective cover = d = 33mm

Effective depth = 367mm

Ast = 6x / 4 (16)2= 1206.37mm2

Asc = 2 x / 4 (16)2= 402.12

M20 fCK = 20 N /mm2

Fe 250 fy = 250 N/mm2

ii) Maximum depth of Neutral axis:

xumax = 0.53

d

xumax = 0.53d = 0.53 x 3.67 = 194.51mm

iii) To find the fsc C.G. 1.2 P96)Strain = Esc = 0.0035 (xumaxd)

Xumax

= 0.0035 (194.5133) / 194.51.

E = Stress

StrainStress = fsc = Esc Strain

= Esc

= 2X105X 0.002906

= 581.20 N/mm2

Alternatively

0.87fy + 0.0020

Es

0.0035 = Esc

xumax xumaxd

Asc fsc = 0.87 fy Ast2

2 - 16#

6 - 16#

300

367

33

xumax

0.0035Esc

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Asc 0.87 fy = 0.87fy Ast2

Asc = Ast2

Ast = Ast1+ Ast2

Ast1= AstAst2

= 1206.36402.12Ast1= 804.25

G 1.1 (P-96)

0.36fck xu b = 0.87 fy Ast

Xu = 0.87 fy Ast1 = 0.87 x 415x804.250.36 fck .b 0.36 fck.b

Xumax = 194.51 mm

Xu xumax under reinforced section

Moment of resistance tension side steel

Mu1= 0.87 fy Ast1d (1Ast1fybd fck)

= 0.87 x 415x804.25x367 (1-804.25 x415

300 x 367 x20)

Mu1= 90.41 x 106

Nmm.

Moment of resistance compression side steel

Mu2= fsc Asc (dd)

Mu2= 0.87fy Asc (d-d)

= 0.87 x 250x402.12 (40033)

= 32.098 x 106Nmm.

Total Limiting Moment of Resistance

Mulim= M1+ M2

0.36 fck xu

Asc fsc

0.87 f Ast

0.446 fck

dd

d

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= 122.51 kNm.

Doubly Reinforced Section

Find the moment of resistance of a beam 250mm x 500mm. If reinforcement with 2/12 FF in compressive

zone and 4 / 20mm # in tension each at an effective cover of 40mm. Use M20 and Fe 415 grade.

Given Data:

Ast = Ast1+ Ast2

Ast = 4 x / 4 (20)2

= 1256.64 mm2

Asc = 2 x / 4 (!2)2= 226.195 mm2

M20 & Fe 415

0.87 fy Ast2= Asc. Fsc

Esc = 0.0035 (xumaxd)Xumax

Esc = 0.8188 x 0.0035

Esc = 0.002866

fsc = Esc

= 0.002866 x 2x105

= 573.20 N/mm2(or) 0.87fy = 361.05 N/mm

Whichever is less

0.87fy Ast2= Asc 0.87 fyAst2= Asc

Ast1= AstAst2= 1030.445 mm2

0.36fck xu.b = 0.87 fy Ast

Xu = 0.87 fy Ast1 = 206.69mm

0.36 fck b

Mulim1= 0.87 fy Ast1d (1- Ast1fy

bd fck)

= 138.61 kNm

Mu2= fsc Asc (d-d)

= 42.46 kNm

Mulim= Mu1+ Mu2

= 181.07 kNm

2 - 12#

4 - 20#

250

500

40mm

40mm

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Design a rectangular beam of effective span 5m superimposed load is 75 kN/m. Size of beam is

restricted to 300 x 600mm. Use M20 and Fe 415 grades.

Solution:

Given Data: l = 5m

Breadth b = 300mm

Over all depth D = 60mm

Assume d = 40mm

D = 60040 = 560mm

Fck = 20 N/mm2 fy = 415 N/mm

2

Load Calculation:

Dead Load(0.3 x 0.6 x 1 x 25) = 4.5 kN/m

Live Load = 75 KN/m

---------------------Total = 79.5 KN/m

Factored Load = 1.5 x 79.5 = 119.25 KN/m

Factored Moment = wl2/ 8 = 372.66 kNm.

i) Considering singly reinforced balanced section

xumax = 0.48

d

xumax = 0.48 x 560 = 268.80mm

Mulim= 0.36 xumax (1-0.42 xumax) bd2fck

d d

= 0.36 x 0.48 (1-0.42 x 0.48) bd2fck

= 0.138 bd2

fck

= 256.66 x 106Nmm.

Note: Mu = 372.66 kNm

Mulim= 256.66 kNm

Mulim < Mu

Design the given section as doubly reinforced section.

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ii) Design of Doubly reinforced Sectionxu = 0.87fy Ast1

d 0.36 fck.bd

Ast1= 0.36 fck b xu = 1608.11 mm2

0.87 fy

Esc = 0.0035 (268.8040)

Xumax

= 0.0035 (268.80d)268.80

= 0.002979

E = f /e

fsc = Esc. Esc

= 0.002979 x 2.1 x 105

fsc = 625.59 N/mm2

(or)0.87fy = 0.87 x 415 = 361.05

Fsc = 361.05 mm2

M2= MuMulim= fsc. Asc (dd)

116 x 106= 361.05 Asc (56040)

Asc = 617.86 mm2

Total Ast = Ast1+ Ast2

= 1608.11 + 617.86

= 2225.97mm2

M2= MuMulim= 116 x 106Nmm

Mulim= 0.87 fy Ast d (1-Ast fyBd fck)

256.66 x 106= 0.87 x 415 x Ast x560 (1- Ast 415

300 x 560x20)

256.66 x 106= 202188 (1-1.235 x 10

-4)

256.66 x 106= 20218824.97 Ast2

24.97 Ast2202188 + 256.66 x 106= 0

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Ast28097.24 + 10.2787 x 106= 0

-b b24ac2a

8097.24 4944.742

Ast1= 1576.25 mm2

A reinforced concrete beam 300mm x 600mm is to be designed for a factored moment 3.25 x 108.

Calculate the reinforcement needed. Use M20 and Fe 415. Effective cover is d = 37.5mm.

Solution

Given Data:

Factored Moment = 3.25 x 108Nmm

M20 & Fe 415

xumax = 0.48

d

xumax = 0.48 x (60037.5)

= 270mm

Mulim= 0.36 xumax (1-0.42 xumax ) fck bd2

d d

= 0.138 fck bd2

= 261.98 kNm.

Doubly reinforced section is required

Mulim= 261.98 kNm

Mu = 325 kNm

Mu2= MlimMu = 63.02 kNm.

Ast1=

0.36fck xu.b = 0.87 fy Ast

300

600mm

562.50

xumax

0.0035 Esc

Xumaxd

d

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Ast1= 0.36 fck xumax. B = 1615.29 mm2

0.87 fyTo calculate Asc

Esc = 0.0035 (xumaxd)Xumax

Es = f / e

fsc = Es esc

= 2.1 x 105x 0.0030139

= 632.92 (or) 0.87fy 361.05 N/mm2

fsc = 361.05 N/mm2 whichever is lesser.

Ast2= fsc Asc / 0.87 fy

= 0.87fy Asc

0.87 fy

Mu2= fsc Asc (dd)

Asc = 332.47 mm2

Ast = Ast1+ Ast

2

Ast = 1947.76 mm2

FLANGED BEAMS

A Tbeam floor consists of 150mm thick R.C slab cast monolithic with 300mm wide beams. The

beams are spaced at 3.5m c/c and their effective span is 6m. If the superimposed load on the slab 5

kN/m2. Design an intermediate beam. Use M20 & Fe 250 grades.

Given data:

Thickness of R.C Slab = Df= 150mm

Width of web (beam) bw= 300mm

Spacing = 3.5 m c/c

Span = l = 6m

Superimposed load = 5KN/m2

On slab

M20 & Fe 415

Ast2= Asc

3.5m c/c

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Note:

Overall depth of T beam = span15

Breadth of web = bw= span + 80mm

30D = span = 6000 = 400mm

15 15

Width of flange = bf= lo + bw+ 6 Df

6= 6000 + 300 + 6 x 150

6

= 2200mm

(Or)C/c of beam 3500mm

bf= 2200mmi) Load Calculation:

Dead weight of slab = 3.75 KN/m2

Superimposed load on the slab = 5.00 KN/m2

Total Load = 8.75 kN/m2

Load /m run on the slab 8.75 x 3.5 = 30.625 KN/m

Dead weight of beam = 1.875 KN/m.

Factored Load = 32.5 x 1.5 = 48.75 kN/m

Factored moment = Mu = wl2/ 8

= 48.75 x 62

8

= 219.38 kNm.

AssumeMu = 0.87fy Ast d (1-Ast fy

bd fck)

6m

300mm

bf

Df

bw

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219.38 x 106= 0.87 x 415 Ast x 360 (1- Ast x 415

2200 x 360 x 2)

= 129978 (1-0.000026 Ast)

= 129978 Ast3.38 Ast23.38 Ast

2129978 Ast + 219.38 x 106= 0

Ast238455.03 Ast + 64905325.44 = 0

-b b24ac2a

= 38455.03 38455.0324 x 64905325.442

= 38455.03 34916.59

2

ii) Check the depth of Neutral axis:

Xu = 0.87 fy Astd 0.36 fck bf

xu = 0.87 x 415 x 1769.22

0.36 x 20 x 2200= 40.33mm

iii) Determination of number of bar:Ast = 1769.22mm

2

Using 25mm # bar

No. of bar = 1769.22/ 4 (25)2

= 3.60 say 4 Nos.

iv)Check for reinforcement:

Minimum reinft.

As = 0.85

bwd fy

As = 0.85 bwd

Fy

= 0.85 x 300 x 360415

= 221.21 mm2

Astmin< Astprovided< Astmax.

Calculate the amount of steel required in a T beam to develop a moment of resistance of 300 kNm at

working loads. The dimensions of beams are given in figure. Use M20 & Fe 415 grade.

750mm

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Mu = 300 x 1.5 = 450kNm

. Neutral axis lies within the flange

Mu = 0.87 fy Ast d (1-Ast fy

Bfd fck)

450 x 106= 0.87 x 415Ast x 500 (1- Ast x 415

750 x 500 x 20)

Ast = 2985. 47mm2

Check the depth of Neutral axis:

xu = 0.87 fy Ast

d 0.36 x 20 x 750

= 199.61mm.

Xu > Df

Hence our assumption is wrong.

Neutral axis lies outside the flange:

Df = 100 = 0.20d 500

Mulim= 0.36 xumax (1-0.42 xumax ) fck bd2

d d

+ 0.45fck (bfbw) Df(d - Df )2

xumax = 0.48

d

xumax = 0.48 x 500 = 240mm

Mulim= 0.36 x 0.48 (1-0.42 x 0.48) 20 x 200 x 5002

+ 0.45 x 20 (750200) 100 (500100 / 2)

Mulim = 3.58 x 108Nmm.

100mm

200mm

470mm

500mm

70mm

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Mu > Mulim

But Mu = 4.5 x 108Nmm

Design as doubly reinforced Section

Find Ast

Mulim= 0.87fy Ast1d (1- Ast1fy )

bd fck

Ast1= 2482mm2

Find Ast 2

MuMulim= fsc Asc (d-d)

Ast2= Asc = 592.59mm2

Total Ast = Ast1+ Ast2

Asc = Ast2

Find Number of bars

Check reinforcement

Min As = 0.85

bwd fy

Max 0.04 bwD.

Depth of NA

0.36 fck xu . bw+ 0.446 fck (bfbw)Yt

= 0.87 fy Ast

Yt= (0.15xu + 0.65 Df)

Corners of the slab are not Held down

(2Way slab)

Design a twoway slab for a room 5.5m x 4m clear in size, if the superimposed load is 5kN/m2

. Use M20 & Fe415 grade.

5.50m

4m

300mm

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L = 5.5 = 1.375 < 2 Two way Slab

B 4

Design Data:

M20 fck20 N/mm2

Fe 415 fy415 N/mm2

xumax = 0.48

d

xumax = 0.48d

Estimation of thickness of slab:

Span = 4000 = 3.5 X 0.8

D D

D = 142.86mm

40mm/m Span = 40 x 4 = 160mm

Overall depth = 1/30 x short span

= 1/30 x 4000= 133.33mm

Provide overall thick of Slab = 140mm

Assuming an effective cover = 20mm

Effective depth = 120mmEffective Span

Shorter Span lx

a) 4 + 0.3 = 4.3m

b)

4 + 0.12 = 4.12m

lx = 4.12m

Longer span (ly)

a) 5.5 + 0.3 = 5.8m

b) 5.5 + 0.12 = 5.62m

ly = 5.62m

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Load Calculation:

Dead weight of slab = 3.5 kN/m2

(0.14 x 25)

Superimposed load = 5.0 kN/m2

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

8.5 kN/m2

Load / m run = 8.5 kN/m

Factored load = Wu= 1.5 x 8.5 = 12.75 kN/m.

Maximum B.M along shorter span

Mx = dx w/ x2

Maximum B.M along longer span

My = y w/x2

1.3 1.36 1.4ly = 5.62 = 1.36 x 0.093 0.099

lx 4.12 y 0.055 0.051

x = 0.0966

y = 0.0526

0.006 = ?

0.1 0.06

0.004 = ?

0.1 0.04

Mx = x w/x2

= 0.0966 x 12.75 x 4.122

= 20.91 kNm

My = y w/x2

= 0.0526 x 12.75 x 4.122

= 11.38 kNm.

Check for depth

Mux = 0.36 xumax (1-0.42 xumax ) bd2fck

0.0990.0966

0.093

0.006

?

1.3 1.3

1.4

0.050.0526

0.055

1.3 1.36

1.4

0.004

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d d

20.91 x 106= 0.36 x 0.48 (1-0.42 x 0.48)bd

2fck

d = 100.52mm < 120mm Hence Safe

Reinforcement details: Assuming 10mm # bar,

Ast (shorter span)

Mu = 0.87fy Ast (d) (1- Ast fy

bd fck)

Mux = 20.91 x 106= 0.87 x 415 Ast (175) (1- Ast 415

1000 x 175 x 20)

Ast

Muy = 0.87fy Ast d (1- Ast fy

Bd fck)

11.38 x 106= 0.87 x 415 x 165 (1 - Ast fy

1000 x 165 x 20)

Asty=

Spacing of reinforcement:

Shorter span:

SV = 1000 A

Ast

Longer Span

Check for Spacing 3d (or) 300mm

Hence Provide

Check for shear

VU= W lx2

Nominal Shear Stress = = Vubd

100 Ast = 0.47%Bd

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Check for development length

Mx1= M1= 0.87 fy Ast d (1- Ast fy

Bd fck)

Ld 1.3 M1 + LO

VDesign a simply supported roof slab for a room 8m x 3.5m clear in size. If the superimposed load is

5kN/m2. Use M20 & Fe 415.

i) Design Data :

M20 & Fe 415

Fck20 N/mm2Fe 415415 N/mm2

Xumax = 0.48

D

Xumax = 0.48d

ii) Estimation of Slab Thickness:

d = span

BV x MF

Simply Supported 20

xu = 0.87 fy Ast dd 0.36 fck.bd

Ast = 0.36 fck b xu0.87 fy

= 0.36 x 20 b x 0.48 d

0.87 x 415

Ast = 0.00957 bd

8m

3.5m

300mm

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100 Ast = 100 x 0.00957 bd

bd bd

100 Ast = 0.9572

bd

M.F = 1

d = Span = 3500 = 175mm

BV x MF 20 x 1

(or) 40mm / m run = 140mm

Assuming effective cover = 25mm

D = 175 + 25 = 200mm

iii) Effective Span:

i)

c/c bearing = 3.5 + 0.3 = 3.8mii) Clear span + d = 3.5 + 3.675m

Leff = 3.675m.

Load Calculation:

Dead weight of Slab = 5 kN/m2

(0.2 x 25)

Superimposed load = 5 kN/m2

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

Total = 10 KN/m

2

Load / m run = 10 kN/m.

Factored load = Wu = 1.5 x 10 = 15kN/m

Factored moment = Mu = Wul2/ 8 = 15 x 3.675

2/ 8

= 25.32 kNm

iv) Check for depth:

For balanced Section

Mulim = 0.36 xumax (1-0.42 xumax ) bd2fck

d d

25.32 x 106= 0.36 x 0.48 (1-0.48 x 0.42) 1000 d

2x 20

D = 95.78mm < 175mm

Hence Safe

v)Area of Steel reinforcement

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Mu = 0.87 fy Ast d ( 1- Ast fy

Bd fck)

25.32 x 106= 0.87 x 415 x Ast x 175 (1- Ast 415

1000 x 170 x 20)

v) Spacing of Main reinft. / m run

Using 12mm #

Sv = 1000 A

Ast

vii) Check for Spacing

i) 3d (or) 300mm

viii) Distribution bar:

Ast = 0.12 x bD

100= 0.12 X 1000 X 200

100

ix) Spacing of distribution:

i) 5d (or) 450mm

x) Check for development length at supports;

Ld = s = 12 x 0.87 x 415

4 Tbd 4 x 1.2

= 902.625

Ld = 300.875mm

3Provide 310mm

Check for Shear

Shear Stresses in slab are within the permissible limit, shear reinforcement are not necessary.

Near support main bar is bent up at l / 7 from the face of the wall

Near intermediate beam the reinft. Is bent up at l / 7 and projected over the beam at l / 4 from the center.Check

for Shear

Vu = 15 x 3.675 = 27.5625 kN2

Nominal Shear Stress = = Vu = 27.56 x 1000 = 0.1575 N/mmbd 1000 x 175

100 Ast = 100 x / 4 (12)2 x 1000 / 260 = 0.25%

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Bd 1000 x 175

Tc= 0.36 N/mm2

Tcmax= 2.8 N/mm2

(0.1575 N/mm2) Nominal Shear Stress

(0.36 N/mm2) Permissible Shear Stress

(2.8 N/mm2) Maximum Shear Stress.

3500300 300

12#@ 260mm

Bottom Plan

Section

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UNIT III LIMIT STATE DESIGN FOR BOND, ANCHORAGE SHEAR & TORSION

DESIGN FOR TORSION

INTRODUCTION

Torsion when encountered in reinforced concrete members usually occurs in combination with flexure shear.

Torsion in its pure form (generally associated with metal shafts) is rarely encountered in reinforced concrete.

The interactive behavior of torsion with bending moment and flexural shear in reinforced concrete beams is

fairly complex, owing to the no homogeneous, nonlinear and composite nature of the material and the presenceof cracks. For convenience in design, codes prescribe highly simplified design procedures, which reflect a

judicious blend of theoretical considerations and experimental results.

These design procedures and their bases are described in this chapter, following a brief review of the general

behavior of reinforced concrete beams under torsion.

EQUILIBRIUM TORSION AND COMPATIBILITY TORSION

Torsion may be induced in a reinforced concrete member in various ways during the process of load transfer in

a structural system. In reinforced concrete design, the terms equilibrium torsion and compatibility torsion arecommonly used to refer to two different torsioninducing situations.

In equilibrium torsion, the torsion is induced by an eccentric loading, and equilibrium conditions alone sufficein determining the twisting moments. In compatibility torsion, the torsion is induced by the application of anangle of twist and the resulting twisting moment depends on the torsional stiffness of the member.

In some (relatively rare) situations, axial force (tension or compression) may also be involved.

It must be clearly understood that this is merely a matter of terminology, and that it does not imply for instance,

equilibrium conditions need not be satisfied in cases of compatibility torsion.

There are some situations (such as circular beams supported on multiple columns) where both equilibrium

torsion and compatibility torsion coexist.

EQUILIBRIUM TORSION

This is associated with twisting moments that are developed in a structural member is maintain static

equilibrium with the external loads, and are independent of the torsional stiffness of the member. Such torsion

must be necessarily considered design. The magnitude of the twisting moment does not depend on the torsional

stiffness of the member, and is entirely determinable from statics alone. The member has to be designed for the

full torsion, which is transmitted by the member to the supports. More ever, the end(s) of the member should be

12#@ 260mmc/c

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suitably restrained to enable the member to resist effectively the torsion induced. Typically, equilibrium torsion

is induced in beams supporting lateral over hanging projections, and is caused by the eccentricity in the loading

(Figure). Such torsion is also induced in beams curved plan and subjected to gravity loads, and in beams where

the transverse loads are eccentric with respect to the shear centre of the crosssection.

(a)Beam supporting a lateral

overhanging

(c) twisting momentDiagram

(b)free body of beam

Compatibility Torsion

This is the name given to the type of torsion induced in a structural member rotations (twists) applied at oneor more points along the length of the member. It twisting moments induced are directly dependent on the

torsional stiffness of the member. These moments are generally statically in determine and their analysis

necessarily involves (rotational) compatibility conditions; hence the name compatibility torsion. For example,in the floor beam system has shown in figure, the flexure of the secondary beam BD results in a rotation Batthe end B. As the primary (Spandrel) beam ABC is monolithically connected with the secondary beam BD at

the joint B., compatibility at B implies an angle of twist, equal to Bin the spandrel beam ABC, and a bendingmoment will develop at the end b of beam BD. The bending moment will be equal to, and will act in a direction

Column

Cantilevered Shell Roof

Beam subjectedto equilibrium

torsion

Total torque = T

T / 2

T / 2

T / 2

T / 2

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opposite to the twisting moment, in order to satisfy static equilibrium. The magnitude of Band the twisting /bending moment at b depends on the torsional stiffness of beam ABC and the flexural stiffness of beam BD.

The torsional stiffness of a reinforced concrete member is drastically reduced by torsional cracking. This results

in a very large increase in the angle of twist, and, in the case of compatibility torsion, a major reduction in theinduced twisting moment. For this reasons, the code (CL.40.1) permits the designer to neglect the torsional

stiffness of reinforced concrete members at the structural analysis stage itself, so that the need for detailed

design for torsion in such cases does not arise at the design stage. With reference to figure, this implies

assuming a fictitious hinge (i.e., no rotational restraint) at the end B of the beam BD, and assuming a continuous

support (spring, support, actually)at the joint D. Incidentally, this assumption helps in reducing the degree ofstatic indeterminacy of the structure (typically, a grid floor), thereby simplifying the problem of structural

analysis. Thus, the code states:

In general, where the torsional resistance or stiffness of members has not been taken into account in the analysis

of a structure no specific calculations for torsion will be necessary [CL40.1 of the code].

Of course, this simplification implies the acceptance of cracking and increased deformations in the torsional

member. It also means that during the first time loading, a twisting moment up to the cracking torque of the

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plain concrete section develops in the member, prior to torsional cracking. In order to control the subsequent

cracking and to impart ductility to the member, it is desirable to provide a minimum torsional reinforcement,

equal to that required to resist the cracking torque. In fact one of the intentions of the minimum stirrupreinforcement specified by the code (CL. 25.5.1.6) is to ensure some degree of control of torsional cracking of

beams due to compatibility torsion.

If, however, the designer chooses to consider compatibility torsion in analysis and design, then it is important

that a realistic estimate of torsional stiffness is made for the purpose of structural analysis, and the requiredtorsional reinforcement should be provided for the calculated twisting moment.

Estimation of Torsional stiffness

Observed behavior of reinforced concrete members under torsion (see also section 7.3) shows that the

torsional stiffness is little influenced by the amount of torsional reinforcement in the linear elastic phase, and

may be taken as that of the plain concrete section. However, once torsional cracking occurs, there is a drastic

reduction in the torsional stiffness. The postcracking torsional stiffness is only a small fraction (less than 10percent) of the precracking stiffness, and depends on the amount of torsional reinforcement, provided in theform of closed stirrups and longitudinal bars. Heavy torsional reinforcement can, doubt, increase the torsional

resistance (strength) to a large extent, but this can be realized only at very large angles of twist (accompanied by

very large cracks).

Hence, even with torsional reinforcement provided, in most practical situations, the maximum twisting moment

in a reinforced concrete member under compatibility torsion is the value corresponding to the torsional cracking

of the member. The cracking torque is very nearly the same as the failure strength obtained for an identicalplain concrete section.

In the usual linear elastic analysis of framed structures, the torsional stiffness kt(torque per unit twist T/ ) of abeam of length l is expressed as

KT= GC / l

Where GC is the torsional rigidity, obtained as a product of the shear modulus G and the geometrical parameter

C of the section (Ref. 7.1). It is recommended in the Explanatory Handbook to the code (Ref.7.2) that G may betaken as 0.4 times the c is a property of the section having the same relationship to the torsional stiffness of a

rectangular section as the polar moment of inertia has for a circular section

UNIT IV LIMIT STATE DESIGN OF COLUMNS

Axially Loaded Columns

A column forms a very important component of structure. Columns support beams which a turn support wallsand 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 great importance.

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

dimension. Compression members whose lengths do not exceed three times their least dimension are classifiedas pedestals.

RCC columns concrete has a high compressive strength and a low tensile strength. Hence theoretically concrete

should need no reinforcement when it is subjected to compression. Reinforcements are provided in order toreduce the size of columns. Through a column is mainly a compression member, it is liable to some moment

due to eccentricity of loads or transverse loads or due to its slenderness. Such moments may occur in any

direction and so it is necessary to provide reinforcement near all faces of column. This reinforcement forms thelongitudinal steel. In order to maintain the position of the longitudinal reinforcement and also to prevent their

buckling which may cause splitting of concrete, it is necessary to provide transverse reinforcements in the form

of lateral ties or spirals at close pitch. The transverse reinforcement also assists in confining the concrete.

Classification of columns: A column may be classified on the basis of its shape, its slenderness ratio, the

manner of loading and the type of lateral reinforcement provided. A column may have a section which may be

square, rectangle, circular or a desired polygon.

Depending on the slenderness ratio, column may be short or a long column. The slenderness ratio of a column is

the ratio of the effective length of the column to its least lateral dimension. A column whose slenderness ratio

exceeds 12 is a long column. A column whose slenderness ratio does not exceed the above limit is a shortcolumn.

Based on the manner of loading, column may be classified into

i) Axially loaded columns

ii) Columns subjected to axial load and unaxial bendingiii) Column subjected to axial and biaxial bending

Columns may also be classified based on the type of lateral reinforcement provided. On this basis, columns are

classified into

(i)

Tied columns in which separate or individual ties are provided surrounding the longitudinalreinforcement. The load on it. The object of stipulating a

(ii) Spirally reinforced columns in which helical bars are provided surrounding the longitudinalreinforcement.

Longitudinal reinforcement (or main steel) is provided to resist compressive loads along with concrete. As per

I.S. 456 a reinforced concrete column shall have longitudinal steel reinforcement and the cross sectional area ofsuch reinforcement shall not be less than 0.8% nor more than 6% of the crosssectional area of the columnrequired to transmit all the loading. The object of stipulating a minimum percentage of steel is to make

provision to prevent buckling of the column due to any accidental eccentricity of a maximum percentage of

steel is to provide reinforcement within such a limit to avoid congestion of reinforcement which would make it

very difficult to place the concrete and consolidate it. This may be best realized from the following twoexamples. Consider two columns 450mm x 450mm. Reinforcement required at 0.8% of gross area = 0.8 / 100 x

4502= 1620mm

2.

This may be provided by four bars of 25mm diameter with an area of 1963mm2(Figure a)

Reinforcement required at 6% of the gross area

= 6 x 4502= 12150mm

2

100

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Even if the bigger diameter bars selected. Say 32mm. diameter bars;

We will require 16 bars of 32mm. diameter providing a total area of804 x 16 = 12864mm

2. (Figure b). The difficulty of placing concrete

between the 16 bars of 32mm. diameter with the overall size of

450mm x 450mm. may be quite apparent. Practically the

Maximum percentage of steel may be limited to 4 percent of thegross area so as to ensure a good and sound concrete. (a) (b)

I.S. RECOMMENDATIONS REGARDING LONGITUDINAL REINFORCEMENTS:

The I.S. 456 code has stipulated the following:

(a)The crosssectional area of longitudinal reinforcement shall be not less than 0.8 percent, of the grosscross sectional area of the column.

(b)In any column that has a larger crosssectional area than that required to support the load, the minimumpercentage of steel shall be based upon the area of concrete required to resist the direct stress and not upon

the actual area.

(c)

The minimum number of longitudinal bars provided in a column shall be four in rectangular columns andsix in circular columns.

(d)The bars shall not less than 12mm. in diameter.

(e)A reinforcement concrete column having helical reinforcement shall have at least six bars of longitudinalreinforcement within the helical reinforcement.

(f) In a helically reinforced column, the longitudinal bars shall be in contact with the helical reinforcement and

equidistant around its inner circumference.

(g) Spacing of longitudinal bars measured along the periphery of the column shall not exceed 300mm.(h)In case of pedestals in which the longitudinal reinforcement is not taken into account in strength

calculations, nominal longitudinal reinforcement not less than 0.15 percent of the crosssectional areashall be provided.

Note: Pedestal is a compression member the effective length of which does not exceed three times the least

lateral dimension.

Sprial or

HelicalReinforcement

Transverse

Reinforcement

(Links)

4-25mm

(main bars

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Figure R.C.Columns

TRANSVERSE REINFORCEMENT

The longitudinal reinforcement should be laterally tied by transverse links to provide a restraint against outwardbuckling of each of the longitudinal bars. I.S. 456 code stipulates that the diameter of longitudinal bars shall not

be less than 12mm. and that the diameter of the transverse reinforcement shall not be less than onefourth ofthe diameter of the main rods and in no case less than 5mm. in diameter. The ends of transverse links should be

properly anchored. Figure (a) & (b) show how transverse reinforcement are provided in R.C. Columns.

Arrangements of transverse reinforcementI.S. recommendations

1. If the longitudinal bars are not spaced more than 75mm. on other side, transverse reinforcement only to

go round the corner and alternate bars for the purpose of providing effective lateral supports.

2. If the longitudinal bars spaced at a distance of not exceeding is times the diameter of the tie are

effectively tied in two directions, additional longitudinal bars in between these bars need to be tied in

one direction by open ties (see figure)

3. Where the longitudinal reinforcing bars in a compression member are placed in more than one row,

effective lateral support to the longitudinal bars in the inner rows may be assumed have been provided

if:

i) Transverse reinforcement is provided for the outer most rows.

ii) No bar of the inner row is closer to the nearest compression face than three times the diameter of the

largest bar in the inner row (figure).

4. Where the longitudinal bars in a compression member are grouped (not in contact) and each group

adequately tied with transverse reinforcement then the transverse reinforcement the compressionmember as a whole may be provided on the assumption that each group is a single longitudinal bars for

the purpose of determining the pitch and the diameter of the transverse reinforcement. The diameter of

such transverse reinforcement need not, however, exceed

20mm (See the figure)

Ag = D2 Ag = D

2

D

D

D

D

D

D

D

D

B B

D

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Ag = D2

Ag = BD Ag = BD

Ag = 0.785D2 Ag = 0.865D

2

Ag = 0.823D2

(a) (b)

Diameter

(a)

SPACING OF TRANSVERSE LINKS:

This shall not exceed the least of the following

(a)The least lateral dimension of the column

(b)Sixteen times the diameter of the smallest longitudinal reinforcement rod in the column.

(c)Fortyeight times the diameter of the transverse reinforcement.

D

D

D

< 75mm < 48w

>3

Individual

Groups

Transverse

Reinforcement

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DIAMETER OF TRANSVERSE LINKS;

The diameter of the transverse links shall not be less than

(i) Onefourth the diameter of the largest longitudinal bar.

(ii) 5mm

COVER:

The minimum cover to column reinforcement equals 40mm or diameter of bar whichever is greater.

EFFECTIVE LENGTH OF A COLUMN:

The effective length of a column is not necessarily its actual length. It depends on the degree of fixity of the

ends of the columns. The table on page gives the effective length corresponding to the unsupported length l of

the column from floor to floor or between properly restrained supports.

SHORT AND LONG COLUMNS:

A column will be considered as short when the ratio of the effective length to its least lateral dimension is lessthan or equal to 12. When this ratio is exceeds the column will be considered as a long column.

SLENDERNESS LIMITS FOR COLUMNS:

The unsupported length between end restraints shall not exceed 60 times the latest lateral dimension of the

column.

If in any given plane, one end of a column is unstrained, its unsupported length I, shall not exceed (100b2

D)

Where, b = width of that cross section, andD = depth of the crosssection measured in the plane under consideration.

MINIMUM ECCENTRICTY:

All columns shall be designed for minimum eccentricity equal to,

Unsupported length of column + Lateral dimension , subject to a minimum of 20mm where500 30

bi-axial bending is considered, it is sufficient to ensure that eccentricity exceeds the minimum about an axis.

Note:In case the minimum eccentricity requirements govern, bending about one axis alone at a time should beconsidered. Bending simultaneously about both axes should not be considered, i.e. this should not be regarded

as a case of bi-axial bending.

Effective Length of Compression Members [I.S. 456]

Degree of end restraint of compressive member Theoretical Recommended

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Value of

effective

length

Value of

effective

length

Effectively held in position and restrained against rotation at both ends. (i.e.

both ends are fixed)

0.50 l 0.65 l

Effectively held in position at both ends, restrained against rotation at one

end (i.e., fixed at one end and hinged at the other end.)

0.70 l 0.80 l

Effectively held in position at both ends but not restrained against rotation

(i.e., both ends are hinged)

1.00 l 1.00 l

Effectively held in position and restrained against rotation at one end, and

other restrained against rotation but not held in position.

1.00 l 1.20 l

Effectively held in position and restrained against rotation at one end, and

the other partially restrained against rotation but not held in position.

- 1.50 l

Effectively held in position at one end, but not restrained against rotation,

and at the other end restrained against rotation but not held in position.

2.00 l 2.00 l

Effectively held in position and restrained against rotation at one end but not

held in position nor restrained against rotation at the other end (i.e., fixed atone end and free at the other end.)

2.00 l 2.00 l

AXIALLY LOADED SHORT COLUMNS;

The ultimate compressive load for an axially loaded short column is determined on the following assumptions.

i) The maximum compressive strain in concrete is 0.002.

ii) Strain in concrete is equal to strain in steel

iii) Stressstrain relation for steel is the same in compression or tension.

For an absolutely axially loaded short column, at ultimate stage, the ultimate compressive load is resisted partly

by concrete and partly by steel. Thus, at ultimate stage,

Ultimate load = PU= PUC+ PUS

Where, PUC= Ultimate load concrete = 0.45fckAC

PUS= Ultimate load on steel = 0.75 fyAsc

AC= Area of concrete

ASC= Area of longitudinal Steel

This relation is applicable for the ideal condition of axial loading. In the practical conditions the loading is

never absolutely axial and there will always be some eccentricity which cannot be avoided. Hence we may

consider the possibility of a minimum eccentricity of 0.05 times the lateral dimension and assume a 11%

reduction in the ultimate strength of the column.

On this basis, the ultimate load for an axially loaded short column is taken as,

PU= 0.40 fckAc+ 0.67 fyAsc

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Let Ag= Gross sectional area of the column.

Ag= Ac+ Asc

P = 0.40 fCK(AgAsc) + 0.67 fyAsc

P = 0.40 fckAg+ (0.67 fY0.40 fck) Asc

If P = percentage of steel provided = Asc

x 100

Ag

Then, Pu= 0.40 fck(AgP / 100 Ag) + 0.67 fyp / 100 Ag

Pu = 0.40 fck+ P/ 100 (0.67 fy0.40 fck)Ag

Problem 1: A short column R.C.C column 400mmx 400mm is provided with 8 bars of 16mm diameter. If

the effective length of the column is 2.25m, find the ultimate load for the column. Use M20 concrete and

Fe 415 steel.

Solution:

Size of the column: 400mm x 400mm

L = 2.25m

Minimum eccentricity is greater than the following:

i) L + b = 2250 + 400 = 4.50 + 13.33 = 17.83mm

500 30 50 30

ii) 20mm

emin= 20mm

0.05b = 0.05 x 400 = 20mm

eminhas not exceed 0.05b

Gross area of the section = Ag= 400 x 400 = 160000mm2

Area of Steel = Asc= 8 x 201 = 1608mm2

Area of concrete = Ac= 1600001608 = 158392mm2

Since eminhas not exceed 0.05b, the ultimate load is given by,

Pu= 0.40 fckAc+ 0.67 fyAsc

Pu= 0.40 x 20 x 158392 + 0.67 x 415 x 1608

= 1267136 + 447104 = 1714240N = 1714.24kN

Problem 2: A short column 450mm x 450mm is reinforced with 8 bars of 20mm diameter. The effective

length of the column is 2.75m. Find the ultimate load for the column. Use M20 concrete and Fe

250 steel.

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Solution:

Size of the column; 450mm x 450mm, l = 2.75m

Minimum eccentricity is the greater of the following:

i) L + b = 2750 + 450 = 5.50 + 15 = 20.50mm

500 30 50 30

ii) 20mm

emin= 20.50mm

0.05b = 0.05 x 450 = 22.50mm

emin< 0.05b

Gross area of the section = Ag= 450 x 450 = 202500mm2

Area of steel = Asc= 8 x 314 = 2512mm

2

Area of concrete = Ac = 2025002512 = 199988mm2

Since e< 0.05b, the ultimate load is given by,

Pu= 0.40fckAc+ 0.6 fyAsc

= 0.40 x 20 x 199988 + 0.67 x 250 x 2512

= 1599904 + 420760 = 2020664 N = 2020.664 kN

Problem 3: A reinforced concrete short column 400mm x400mm has to carry an axial load of 1200kN.Find the area of steel required. Use M20 concrete and Fe 415 steel.

Solution:

Gross Area of the column section = Ag= 400 x 400 = 160000mm2

Area of Steel = Asc

Area of concrete = Ac= (160000Asc)mm2

Pu= 0.40 fckAc+ 0.67 fy= 1800 x 103

= 0.40 x 20 (160000Asc) + 0.67 x 415 Asc= 1800 x 103

= 1280008 Asc + 278.05 Asc= 1800 x 103

270.05 Asc= 52000

Asc= 1926mm2.

Provide 4 bars of 20mm and 4 bars of 16mm

Actual area of steel provided = (4 x 201) = 2060mm2

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Lateral Tiles

Diameter of ties shall be not less than,

i) 5mm

ii) diameter of the larger size bar = (20) = 5mm

Provide 5mm ties.

Spacing of lateral ties shall not exceed 416mm

i) Least lateral dimension of the column = 400mm

ii) 16 x diameter of smallest size of bar = 16 x 16 = 256mm

iii) 48 x diameter of ties = 48 x 6 = 288mm

iv) 300mm 4.20mm 6mm ties @

250mm c/c

Provide 6mm ties @ 250mm c/c. 6mm ties @

250mm

c/c

Problem 4: Find the area of steel required for a short reinforced concrete column 400mm x 425mm to

carry an axial load of 1195KkN.Use M20 concrete and Fe 415 steel.

Solution:

Gross area of the column section = Ag= 400 x 425 = 170000mm2

Area of Steel = Asc

Area of concrete = Ac= (17000Asc) mm2

Ultimate load = Pu= 1.5 x 1195 = 1792.5 x 103

Pu= 0.40 fckAc+ 0.67 fyAsc= 1792.5 x 103

270.05 Asc= 432 x 103

Asc= 1599.7mm2

Provide 8 bars of 16mm diameter

Spacing of ties shall not exceed 400mm


ii) 16 x diameter of longitudinal bar = 16 x16 = 256mm.


iv) 300mm

v) Provide 6mm 2 250mm c/c

6mm tiles@

250mm c/c

8 -16mm

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Problem 5: A reinforced concrete column is 450mm x 400mm and has to carry a factored load of 1800kN.

The unsupported length of the column is 2m. Find the area of reinforcement required. Use M20 concrete

and Fe 250 steel.

Solution:

Size of the column: 450mm x400mm

Factored Load Pu= 1800kN, L = 2m = 2000mm

Fck= 20N/mm2 fy= 250N/mm

2

Let the area of Steel be Asc

Area of concrete = Ac= 450 x 400Asc= (180000Asc) mm2

In the direction of the longer lateral dimension,

emin= l + D = 2000 + 450 = 4 + 15 = 19mm

500 30 500 30

In the direction of the shorter lateral dimension

emin= l + D = 2000 + 400 = 4 + 13.3 = 17.3mm

500 30 500 30

But eminshall be at least 20mm

emin = 20 = 0.05 emin has not exceed 0.05

b 400 b

Pu= 0.40 fckAc+ 0.67 fyAsc= 1800 x 103

Hence,

0.40 x 20 (18000Asc) + 0.67 x 250 Asc= 1800 x103

1440 x 1038 Asc+ 167.5 Asc= 1800 x 10

3

159.5 Asc= 360 x 103

Asc= 2257mm2

Provide 8 bars of 20mm diameter (2512mm2).

Lateral ties

Diameter of ties shall not less than,

i) x diameter of longitudinal bars = 1.4 x 20 = 5mm

ii) 5mm

Provide 6mm ties425mm

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ii) 16 x diameter of longitudinal bar = 16 x20 = 320mm.


iv) 300mm

v) Provide 6mm @ 280mm c/c

Problem 6: A reinforced concrete column of 2.75m effective length carries an axial load of 1600kN.

Design the column using M20 concrete and Fe 415 steel

Solution:

Assuming that the minimum eccentricity is less than 0.05 times the lateral dimension of the column,

Ultimate load = Pu= 0.40 fckAc+ 0.67 fyAsc

Ultimate load = Pu= 1.5 x 1600 = 2400kN

Assuming 2 % steel, Asc = 0.02 Ag

Ac = 0.98 Ag

Pu= 0.40 x 20 x 0.98 Ag+ 0.67 x 415 x 0.02 Ag= 2400 x 103

7.84 Ag+ 5.561 Ag= 2400 x 103

13.401 Ag= 2400 x 103

Ag= 179091.11mm2

Providing a square section

Side of the square = b = 179091.11 = 423.2mm

Provide 425mm x 425mm

Minimum eccentricity is the greater of

i) 20mm

L + b = 2750 + 450 = 5.50 + 14.17 = 19.67mm500 30 50 30

emin= 20mm

0.05b = 0.05 x 425 = 21.25mm

Emin< 0.05b

Gross area of the column section Ag= 425 x 425 = 180625mm2

400mm

8-20mm 6mm tiles @

@280mm c/c

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Area of steel =Asc

Area of concrete = Asc= (180625Asc) mm2

Ultimate load Pu= 0.40 fckAc+ 0.67 fyAsc= 2400 x 103

1445 x 1038 Asck + 278.05 Asc = 2400 x 10

3

270.05 Asc= 955 x 103

Asc= 3536.4mm2

Provide 8 bars of 25mm diameter (3928mm2)

Lateral ties

Diameter of lateral ties shall be not less than

i) 5mm

ii)

x diameter of longitudinal bar = 16 x 25 = 400mm

iii) 48 x diameter of the = 48 x 8 = 384mm

iv) 300mm

v) Provide 8mm ties @300mm fe

CONTINUOUS COLUMNS

Often in multistoried structures, a column continues up through a floor from one storey to another. In such a

cases the main bars of the column must be first continued up either within or outside the reinforcement of thefloor beam which frames into the column. When the main bars continue up outside the reinforcement of the

beam, it is necessary that the width of the column should be at least 80mm more than the width of the beam.

Sometimes the column sizes in plan may be smaller above the floor than below it. In such cases the main bars of

the column will have to bent inwards at the floor level, or alternatively these main bars may be stopped just

below the floor level and separate lap bars may be provided for connecting the part of the column above and

below the floor.

425

825mm

8 mm tiles

@ 300mm c/c

425

300

300

300

425

8 mm tiles

@ 300mm c/c

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SPIRALLY REINFORCED CIRCULAR COLUMNS:

These are circular columns, which are reinforced with closely and uniformly spaced spiral reinforcement

in additional to longitudinal steel. Columns of circular section are usually spirally reinforced. Sometimes

separate loops may also be provided in place of the spiral. The continuous spiral is adopted in preference to

separate loops. A column with helical reinforcement shall have at least six bars as longitudinal reinforcement.

The strength of a column with helical reinforcement satisfying the requirement given below shall be taken as

1.05 times the strength of similar member with lateral ties.

The ratio of the volume of helical reinforcement to the volume of the core shall not be less than 0.36

[Ag/ Ak1] fck/ fy

Where,Ag= Gross area of the section

Ak = area of the core of the helically reinforced column measured to the outside diameter of the helix.

fck= Characteristic compressive strength of concrete,

(28 days strength of concrete)fy= Characteristic strength of the helical reinforcement not exceeding 415 N/mm2

Pitch of helical reinforcement (I.S. 456): Helical reinforcement shall be of regular formation with the turns of

the helix spaced evenly and its ends shall be anchored properly by providing one and half extra turns of the

spiral bar. The pitch of the helical turns shall be not more than 75mm. nor more than one sixth of the corediameter of the column, or less than 25 mm. nor less than three times the diameter of the steel bar forming the

helix.

Diameter of helical reinforcement: The diameter of the helical reinforcement shall not less than onefourth thediameter of the largest longitudinal bar and in no case than 5mm.

Figure (a) Figure (b)

Since the strength of a helically bound circular column has a strength equal to 1.05 times the strength of similar

column with lateral ties.

Ultimate strength of the column with helical reinforcement = Pu= 1.05(0.4 fckAc+ 0.67 fyAsc)

D = Columndiameter

D k= Corediameter

Core diameter

Pitch

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Note: The above equation is valid provided the following condition is satisfied,

Volume of helical reinforcement 0.36 (Ag/ Ak1) fck/ fyVolume of core

Problem 7: Determine the safe axial load for a short column 400mm in diameter, reinforced with 6 bars

of 25mm diameter. It is provided with 8mm diameter helical reinforcement at a pitch of 45mm Use M20

concrete and Fe 415 steel.

Solution:

Diameter of the column D = 400mm

Clear cover to longitudinal bars = 40mm

Area of longitudinal Steel = 6 x / 4 x 252= 2945mm2

Diameter of the core = 4002 x 40 + 2 x 8 = 336mm

Area of the core Ak= / 4 x 3362= 88668mm

2

Diameter of the column corresponding to the centre of helical bars = dh= 3368 = 328mm

Gross area of the column Ag= / 4 x 4002= 125664mm

2

Area of concrete Ac= 1256642945 = 122719mm2

Ultimate load for the column

Pu= 1.05 (0.4 fckAc+ 0.67 fyAsc)= 1.05 (0.4 x 20 x 122719 + 0.67 x 415 x 2945) = 1890640 N

Safe load for the column

= 1890640 / 1.50 = 1260427N = 1260.427Kn

Check for validity of the formula used

Consider one pitch length of the column

Length of helix per pitch length= ( dh)

2+ P

2

= ( x328)2+ 452= 1031.42mm

Volume of the helix per pitch length = 50 x 1031.42 = 54571mm2

Volume of the core per pitch length = 88668 x 45 = 3990060mm2

Ratio of volume of helical steel to volume of core

= 51571 = 0.013.3990060

This should be 0.36 (Ag/ Ah1) fck/ fy

0.36 (125664 / 8868 1) 20 / 415

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0.007Hence, the provision of the helical reinforcement is satisfactory.

Problem8: Determine the safe axial load for a short column 425mm in diameter, reinforced with 6 bars of

22mm diameter. It is provided with 8mm diameter helical reinforcement at a pitch of 40mm diameter.

Use M20 concrete and Fe 250 steel.

Solution:


Clear cover to longitudinal bars = 40mm

Diameter of the core = 4252 x 40 + 2 x 8 = 361mm

Diameter of the column corresponding to the centre of helical bars = dh= 3618 = 353mm

Area of longitudinal Steel Asc= 6 x 380 = 2280mm2

Gross area of the column section Ag= / 4 x 4252

= 141862.5mm2

Area of concrete Ac= 141862.52280 = 139582.5mm2

Area of the core Ak= / 4 x 3612= 102353.9mm

2

Ultimate load for the column

Pu= 1.05 (0.4 fckAc+ 0.67 fyAsc)

= 1.05 (0.4 x 20 x 139582.5 + 0.67 x 250 x 2280) = 1573488 N

Safe load for the column= 1573488 / 1.50 = 1048992N = 1048.9927kN

Check for validity of the formula used

Consider one pitch length of the column

Length of helix per pitch length= ( dh)

2+ P

2

= ( x353)2+ 402= 1110mm

Volume of the helix per pitch length = 50 x 1110 = 55500mm2

Volume of the core per pitch length = 102353.9 x 40 = 4094156mm2

Ratio of volume of helical steel to volume of core= 55500 = 0.136.

4094156

This should be 0.36 (Ag/ Ah1) fck/ fy

0.36 (141862.5 / 1023353.9 1) 20 / 250

0.01Hence, the provision of the helical reinforcement is satisfactory.

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Problem9: Design a circular column to carry an axial load of 1500kN. The column has an effective length

of 2.50m. Use M20 concrete and Fe 415 steel.

Solution:

Let the diameter of he column be D.

Gross sectional area of the column = Ag= D2/ 4

Providing 2 % steel

Asc= 0.02 Ag

Area of concrete Ac= Ag0.02 Ag= 0.98 Ag

Ultimate load Pu= 1.5 x 1500 = 2500kN

Assuming the column to be short, and the minimum eccentricity does not exceed 0.05 D.

Ultimate load = Pu= 1.05 (0.4 fckAsc+ 0.67 fyAsc) = 2250 x 103

N

0.4 fckAc+ 0.67 fyAsc= 2250 x 103/ 1.05

13.401 Ag= 2250 x 103/ 1.05

Ag= 159902.78mm2

D2/ 4 = 159902.78

Provide a diameter of 450mm for the column

Ratio of effective length to the lateral dimension of the column

= l / D = 2500 / 450 = 5.6 (less than 12)

This is a short column.

Minimum eccentricity. This is the greater of

i) 20mm

L + D = 2500 + 450 = 5 + 15= 20mm500 30 500 30

emin= 20mm

But 0.05 D = 0.05 x 450 = 22.5mm

emin< 0.05D

Hence, the ultimate load for the column is given by

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Pu= 1.05 (o.4 fckAc+ 0.67 fyAsc)

Gross area of the column section Ag= / 4 x 4502= 159043.13mm

2

Area of steel = Asc

Area of concrete = Ac= 159043.13 - Asc

Ultimate load Pu= 1.05 [0.4 x 20 (159043.13Asc) + 0.67 x 415 Asc]

= 2250 x 103N

Asc= 3223.5mm2

Provide 8 bars of 25mm diameter (3927.2mm2)

Check for validity of the formula used.


Providing 8mm helical at a pitch of 45mm

Diameter of the core = 450(2 x 40) + (2 x 8) = 386mm

Diameter of the column corresponding to the centre of helical reinforcement

= 3868 = 378mm

Length of helix per pitch length = ( x 378)2+ 452= 1188.37mm.

Volume of helix per pitch length = 50 x 1188.37 = 59418.8mm3

Volume of the core per pitch length = / 4 x 3862x 45 = 5265953.3 mm3

Ratio of volume of helical steel volume of the core

= 59418.8 / 5265953.3 = 0.011

This should be 0.36 (Ag/ Ak1) fck/ fy

We know, Ag= 159043.13mm2

And Ak= / 4 x 3862= 117021.18mm

2

0.36(Ag/ Ak1) fck/ fy

= 0.36 (159043.13 / 117021.18 -1) 20 / 415 = 0.0062

Hence the design is satisfactory.

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Problem 10: Figure shows the plan and part section of a four storeyed building of flat slab construction.

Design an interior column to the following particulars.

Height of each floor = 3.50m

Plinth height above the ground level = 0.50m

Thickness of wall = 250mm

Columns are 400mm x 400mm

Thickness of floor slabs = 150mm

Depth of foundation = 1.25m

Ignore moment transmitted to column from slab. Use M20 concrete and Fe 415 steel.

Solution:

For each floor

DL of slab = 25 x 0.15 = 3.75kN/m2

Floor finish = 1.00 kN//m2

Total D.L for floor = 4.75 KN / m2

DL of walls for 1 floor = 0.25 x 3.5 x 20 = 17.5 kN/m.

Figure (a) Figure (b)

Load transmitted to the column at its base

5m 5m

5m

5m

5m

5m

Ground Level

Plinth Level

0.5m

0.15m

0.15m

0.15m

0.15m400 x 400

Column

3.50m

3.50m

3.50m

3.50m

0.5m

1.25m

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DL of floor : (4.45 x 5 x 5) 4 = 475kn

DL of walls : [17.5 (5 + 5)] 3 =525kn

DL of column : 0.4 x 0.4 x 15.25x25 = 61kn

Live load from 4 floors

(30% reduction) : 0.7 [4 x 5 x 5 x 4] = 280kn

Total load of the column=Pu= 1.5 x 1341 = 2011.5 Kn

Ultimate load of the column leff= 0.65 l = 0.65 x 3.50 = 2.275m

Minimum eccentricity: This taken as the greater of the following

i) 20mm

ii) leff + b = 2275 + 400 = 17.88mm.

500 30 500 30

emin= 20mm

But 0.05 b = 0.05 x 40 = 20mm

emin has not exceed 0.05b

Hence, the ultimate load is given by

Pu= 0.4 fckAc+ 0.67 fyAsc

Gross area of the column section Ag= 400 x 400 = 160000mm2

Area of steel = Asc

Area of concrete = Ac= 160000 - Asc

Ultimate load Pu= 0.4 x 20 (160000Asc) + 0.67 x 415 Asc= 2011.5 x 103N

1280 X 1038 Asc+ 278.05 Asc= 2011.5 x 10

3

270.05 Asc= 731.5 x 103

Asc= 2709mm2

Provide 8 bars of 22mm diameter.

Lateral ties

Diameter of lateral ties shall not be less than,

i) 5mm

ii) diameter of longitudinal bars = x 22 = 4.4mm

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Providing 6mm diameter ties

Spacing of lateral ties

The spacing of lateral ties shall not exceed,

i) Least lateral dimension of the column : 400mm

ii) 16 x diameter of longitudinal bars :16 x 22 = 352mm

iii)

48 x diameter of ties : 48 x 6 = 288mm

Providing 6mm ties @250mm c/c.

ANALYSIS AND DESIGN OF AXIALLY LOADED COLUMNS BY THE USE OF CHARTS:

Charts have been made by the Bureau of Indian standards (SP 16 : 1980) for designing columns in

accordance with the equation Pu= 0.4fckAc+ 0.67 fyAsc.

See charts 1, 2 and 3. In the chart, in the lower portion, Pu/ Aghas been plotted against the percentage of

steel for various grades of concrete. Suppose the sectional area of the column is known, we can determine Pu

/ Agand we can get from the chart the percentage of steel required. In the upper portion, Pu/ Agis plotted

against Pufor various values of Ag.By using the upper and the lower portions of the chart, calculations are

considerably minimized. These charts will be of great advantage in selecting sizes of columns in the

preliminary design stage of multistoried buildings.

Problem 11: Design an axially loaded short column to carry an axial load of 1650kN. Use M20 concrete

and Fe 415 steel.

Solution:

Let us provide a column size of 400mm x450mm

Gross area of the column section = Ag= 400 x 450mm2

Ultimate load Pu= 1.5 x 1650 = 2475Kn

Pu = 2475 x 103 = 13.75 N/mm

2

Ag 400 x 450

Referring to chart 2

Corresponding to PU = 13.75 N/mm2

Ag

Percentage of steel required = Pt= 2.15%

Asc = 2.15 x 400 x 450 = 3870mm2

100

Provide 8 bars of 25mm diameter (3927mm2) and provide 8mm ties @ 300mm c/c.

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Problem 12: A short R.C.C.column 425mm x 500mm in section carries an axial load of 1600kN. Find the

area of steel reinforcement required. Use M20 concrete and Fe 250 steel.

Solution:

Gross area of the column = Ag= 425 x 500mm2

Ultimate load Pu= 1.5 x 1600 = 2400kN

Pu = 2400 x 103 = 11.3N/mm

2

Ag 425 x 500

Referring to chart 1

Corresponding to PU = 11.3N/mm2

Ag

Percentage of steel required = Pt= 2 %

Asc = 2 x 425 x 500 = 4250mm2

100

Provide 12 bars of 22mm diameter (4560mm2) and provide 8mm ties @ 300mm c/c.

EXERCISE1. A short R.C.C. column 450mm X 450mm is provided with 8 bars of 18mm diameter. If the effective

length of the column is 2.50m, find the ultimate load for the column. Use M20 concrete and Fe 415

steel. (2168.7kN)

2. A short R.C.C. column 475mm x 475mm is reinforced with 8 bars of 25mm diameter. The effectivelength of the column is 3m. Find the ultimate load for the column. Use M20 concrete and Fe 250 steel.

(2431.5kN)

3. A short reinforced concrete column 450mm x 450mm has to carry an axial load of 1400kn. Find the area

of steel required. Use M20 concrete and Fe 415 steel (1778mm2)4. A reinforced concrete column has an effective length of 2.80m. It carries an axial load of 1800kN.

Design of column using M20 concrete and Fe 415 steel. (provide approximately 2% steel)(450mm x 450mm Asc= 4000mm

2)

5. Determine the safe axial load for a short circular column 450mm in diameter reinforced with 6 bars of

25mm diameter. It is provided with 8mm diameter helical reinforcement at a pitch of 45mm. Use M20

concrete and Fe415 steel. (1447.3kN)6.

Determine the safe axial load for a short circular column 450mm in diameter, reinforced with 6 bars of

20mm diameter. It is provided with 8mm diameter helical reinforcement at a pitch of 40mm. Use M20

concrete and Fe 250 steel. (1651.5Kn)

COMBINED AXIAL LOAD AND UNAXIAL BENDING

As mentioned earlier, a compression member shall be designed for a certain minimum eccentricity of the load.

It is always necessary to ensure that a column section is designed for a moment which is not less than the

moment due to the minimum specified eccentricity.

Assumptions: The following assumptions are made

a) Plane section normal to the axis of the member remains plane after bending. This means that the strain at

any point of the cross section is directly proportional to the distance from the neutral axis.

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b) The design stressstrain relationship for concrete is taken as indicated earlier.c) The tensile strength of concrete is ignored.

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