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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
Material: M20grade concreteFe415 HYSD bars
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
4. Bending Moment and shear forces
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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8. Shear stress and reinforcement
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
Material: M20grade concreteFe415 HYSD bars
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
5. Bending Moment and shear forces
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.
9. Shear stress and reinforcement
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
i) Least lateral dimension of the column = 400mm
ii) 16 x diameter of longitudinal bar = 16 x16 = 256mm.
iii) 48 x diameter of ties = 48 x 6 = 288mm
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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i) Least lateral dimension of the column = 400mm
ii) 16 x diameter of longitudinal bar = 16 x20 = 320mm.
iii) 48 x diameter of ties = 48 x 6 = 288mm
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:
Diameter of the column D = 425mm
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.
Diameter of the column D = 450mm
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.