Design RCC Beam Using Flexure PDF
Design RCC Beam Using Flexure PDF
Design RCC Beam Using Flexure PDF
and 6m apart masonry walls (centre to centre). The beam has to carry, in addition to its own weight, a distributed live load of 10
kN/m and a dead load of 5 kN/m. Design the beam section for maximum moment at midspan. Assume Fe415 grade of steel.
Solution:
1. Consideration of grade and cover
Exposure condition = Severe (Table 3, IS 456:2000)
Grade of Concrete= M35 (Table 5, IS 456:2000)
fck ≔ 35 MPa
Clear Cover= 40 mm (Table 16, IS 456:2000)
2.Trial section:
b ≔ 300 ⋅ mm
6000
D ≔ ――mm = 600 mm
10
Let us assume 16 mm main bar and 8mm stirrups
16
d ≔ D - 40 mm - 8 mm - ― mm = 544 mm
2
2. Effective Length (Cl 22.2, IS456:2000)
a) Lcc ≔ 6 m
b) Lcc - Sw + d = 6.314 m
⎛ 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾Mu ⎞
⎜1 - 1 - 4.598 ⋅ ――― ⎟
⎜⎝ fck ⋅ b ⋅ d 2 ⎟⎠
Ast_required ≔ fck ⋅ b ⋅ d ⋅ ―――――――――= ⎛⎝1.245 ⋅ 10 3 ⎞⎠ mm 2
2 ⋅ fy
Numbers of Bars
Ast_required
n ≔ ―――――― 2
= 3.966
((20 ⋅ mm))
3.14 ⋅ ――――
4
Let us consider 4 numbers of 20 mm bars
2
((20 mm))
Ast_provided ≔ 3.14 ⋅ ―――― ⋅ 4 = ⎛⎝1.256 ⋅ 10 3 ⎞⎠ mm 2
4
Detailing:
Minimum Amount of reinforcement(Cl 26.5.1.1,
IS 456:2000):
d
As_min = 0.85 ⋅ b ⋅ ―
fy
342
As_min ≔ 0.8 ⋅ 300 ⋅ ―― mm 2 = 197.783 mm 2
415
Max Amount of reinforcement(Cl 26.5.1.1, IS
456:2000):
Spacing:
((300 - 2 ⋅ ((40 + 8)) - 4 ⋅ 20))
s ≔ ―――――――― mm = 41.333 mm
4-1
Minimum Spacing
a)16mm
b)20+5=25 mm OK
Check for Strength:
⎛ kN ⎞ kN kN kN kN
DL ≔ ⎜25 ⋅ ―― ⎟ ⋅ b ⋅ D + 5 ―― = 8 ―― LL = 10 ―― w ≔ 1.5 ⋅ DL + 1.5 ⋅ LL = 27 ――
⎝ m3 ⎠ m m m m
Leff 2
Mu ≔ w ⋅ ――= 121.5 kN ⋅ m
8
⎛ ⎛⎝Ast_provided ⋅ fy⎞⎠ ⎞
MuR ≔ 0.87 ⋅ fy ⋅ Ast_provided ⋅ ⎜1 - ―――――⎟ ⋅ d = 132.578 kN ⋅ m Which is grater than Mu = 121.5 kN ⋅ m
⎝ ⎛⎝fck ⋅ b ⋅ d⎞⎠ ⎠
hence OK
Check for Deflection:
Ast_required
fs ≔ 0.58 ⋅ fy ⋅ ―――― = 238.638 MPa
Ast_provided
Ast_provided
pt ≔ ―――― ⋅ 100 = 1.224 kc ≔ 0.9
b⋅d
pc ≔ 0 kc ≔ 1
Lcc
―― = 17.544 Less than 18, Hence OK
d
Check for Under reinforced section:
xu ⎛⎝0.87 ⋅ fy ⋅ Ast⎞⎠
―= ―――――
d ⎛⎝0.36 ⋅ fck ⋅ b ⋅ d⎞⎠
⎛⎝0.87 ⋅ fy ⋅ Ast_provided⎞⎠
――――――― = 0.351
⎛⎝0.36 ⋅ fck ⋅ b ⋅ d⎞⎠
xulimit
――= 0.48 oK
d