This document provides information on the structural design of a simply supported reinforced concrete beam. It includes: - A list of students enrolled in an elementary structural design course. - Equations and diagrams showing the forces and stresses in a reinforced concrete beam with a singly reinforced bottom section. - Limits on the maximum depth of the neutral axis according to the grade of steel. - Examples of analyzing the stresses and determining steel reinforcement for a given beam cross-section. - A design example calculating the dimensions and steel reinforcement for a rectangular beam with a factored uniform load.

1. GUIDED BY: Prof. Sunil Jaganiya
Prof. Pritesh Rathod
Sub : Elementary Structural Design
NAME ENROLL NO.
Patel Jimi 131100106029
Patel Milind 131100106035
Patel Nirmal 131100106036
Patel Viraj 131100106040
Patel Yash 131100106042
Shah Ashit 131100106051
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2. r.c.c as per is 456-2000 2

3.  In singly reinforced simply supported beams reinforcing steel
bars are placed near the bottom of the beam where they are
most effective in resisting the tensile stresses.
3

4.  Reinforcement in simply supported beam
COMPRESSION b
STEEL REINFORCEMENT D d
TENSION
SUPPORT SECTION A - A
CLEAR SPAN
4

5. ( IS : 456 – 2000 , P.69 )
0.0035 0.446 fck
0.42Xu
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6.  Xu max = Maxi. Depth of Neutral axis
 b = width of beam
 d = effective depth
 Z = d-0.42xu = lever arm
 D = overall depth
 Xu = depth of N.A
 Ast = area of steel
 Resultant force of compression C = average stress X area
= 0.36 fck b xu
 Resultant force of tension T = 0.87 fy Ast
 Force of compression should be equal to force of tension,
0.36 fck b xu = 0.87 fy Ast
xu =
Where Ast = area of tension steel
0.87 fy Ast
0.36 fck b
6

7.  Moment of resistance with respect to concrete= compressive force x lever arm
= 0.36 fck b x z
 Moment of resistance with respect to steel = tensile force x lever arm
= 0.87 fy Ast x z
Maximum Depth Of Neutral Axis (IS 456 – 2000 , P.70)
fy ( N/mm2 ) Xu max
250 0.53 d
415 0.48 d
500 0.46 d
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8.  If Xu = Xu max then it is balanced section
 If Xu < Xu max then it is under reinforced section
 If Xu > Xu max then it is over reinforced section
Xu max Xu < Xu max
Xu > Xu max
Balanced section Under reinforced Over reinforced
b
D
8

9.  Mu with respect to steel ( for under reinforced section)
Mu = T X Z ( IS 456-2000, P.96, Cl. G1.1 (b) )
where , Z = d - 0.42 xu
= 0.87 fy Ast x (d - 0.42 xu ) T = 0.87 fy Ast
= 0.87 fy Ast d ( 1- fy Ast / fck b d )
 Mu with respect to concrete ( for balanced section)
Mu = C x Z ( IS 456-2000, P.96, Cl. G1.1 (c) )
= 0.36 fck b xu ( d – 0.42 xu ) where , C = 0.36 fck b xu
= 0.36 fck b xu d (1 – 0.42 xu / d)
= 0.36 fck b xu/d d2 (1 – 0.42 xu / d)
= 0.36 xu/d (1 – 0.42 xu / d) fck b d2
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10.  Since the maximum depth of neutral axis is limited, the
maximum value of moment of resistance is also limited.
 Mu lim with respect to concrete = 0.36 fck b x Z
= 0.36 fck b Xu (d – 0.42 Xu)
 Mu lim with respect to steel = 0.87 fck Ast (d – 0.42 Xu)
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11. Grade of
concrete
Grade of steel
Fe 250 steel Fe 450 steel Fe 500 steel
General 0.148 fck bd 0.138 fck bd 0.133 fck bd
11

12. a) Analysis of a section
b) Design of a section
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13. Analysis of section13

14.  For a limiting section 200 mm x 300 mm effective ,
determine the following , if it is reinforced with an effective
cover of 50 mm
take M-20 conc. And Fe 250
Sol :
b = 200 mm
d = 300 mm
fck = 20 N / mm2
the given section is
the given section is balanced section
therefore , xu = xu max (IS 456 – 2000 , P 70)
b = 200
d= 300
Ast
14

15.  For Fe – 250 ,
xu max = 0.53 d = 0.53 x 300 = 159 mm
xu = xu max = 159 mm
1. Maxi . Comp. stress in concrete = .446 x fck
= .446 x 20
= 8.92 N / mm2
2 . Lever arm Z = d – 0.42 x
= 233.22 mm
3 . Total tension T = 0.87 fy Ast
Ast ;
xu = (IS 456 – 2000 , P.96)
Ast = 1052.6 mm2
0.87 fy Ast
0.36 fck b
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16. T = 0.87 fy Ast
= 0.87 x 250 x 1052.6
= 228.96 kN
Total compression C = 0.36 fck b xu
= 0.36 x 20 x 200 x 159
= 228.96 kN
for limiting section C = T .
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17. Design of section17

18.  Design a rectangular RC beam having width 250mm, simply
supported with effective span of 4.5 m, it is loaded wit a udl
of 20 kN/m including self weight . Use M20 concrete and Fe
415 steel. Check the beam for maximum and minimum steel
and deflection Solution :
factored udl w = 1.5 x 20 b = 250
= 30 kN/m
Mu = wl2 / 2
= 75.94 kN/m
Design as a balanced section ,
fy = 415 N/mm
Mu lim. = 0.138 fck bd2
75.94 x 106 = 0.138 x 20 x 250 x d2
Ast
D
18

19.  d= 331.75 mm
 provide d = 340 mm
 Therefore , effective size of beam is 250 mm x 340 mm
to find Ast :
pt = 50 fck / fy ( 1- (1- 4.6 Mu / fck bd2)1/2 )
pt = 0.896%
Ast = pt bd /100
= 761.6 mm2
provide 4 – 16 dia. (Ast = 804 mm2)
effective cover = 40 mm
D = d + 40
= 380 mm
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20.  Minimum steel in beam
As / bd = 0.85 / fy
As = 174.09 mm2 ( mini. Required)
Ast > As therefore, ok
Maximum steel in beam
Ast = 0.4 bD
= 0.4 x 250 x 380
= 3800 mm2
804 mm2 < 3800 mm2 therefore, ok
{ IS 456-2000 P.42 }
{ IS 456-2000 P.42 }
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21.  Check for deflection
actual l/d = 4500/340 = 13.23
Pt = 100 Ast / bd = 0.945 %
Maxi l/d = 20 x M.F = 19.5
actual l/d < maxi. l/d therefore, ok
o check for development length
o Ld <= 1.3 M1 / V + L0
o d= 340 mm
o 12 = 12 * 16 = 192 mm
o taking larger of two values L0 = 340 mm
o S F at support = wl/2 = (30 * 4.5) / 2 = 67.5 kN
o Ast = 804 mm2
o M1 = 0.87 fy Ast d ( 1 – (fy Ast / fck bd))
o M1 = 79.35 * 106 N.mm
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22. 1.3 M1 / V + L0 = 1.3 *(79.35 * 106 / 67.5 * 103 ) + 340
= 1868 mm
Ld = (0.87 fy) / 4 * bd
= 1203.5 mm
1203.5 <= 1868 mm it is ok
Check for shear
v = Vu / bd
= 67.5 * 103 / (250 * 340)
= 0.794 N/mm2
For fck = 20 N/mm2 Ast = 804 mm2 pt =0.94% c =
0.62 N/mm2
for D = 380 K= 1
’c = K * c
= 0.62
v > ’c it is not ok
Therefor, increasing in Depth (d).
{ IS 456 - 2000 P.73
T.19 }
{ IS 456-2000 P.42 }
{ IS 456 - 2000 P.72 Cl 40.2.1.1 }
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