A singly reinforced beam has a width of 400 mm and an effective d
A. The depth of the neutral axis is 151.8 mm.
B. The limiting Moment of Resistance is 620.84 kN-m
C. The section is under-reinforced.
D. The area of tensile reinforcement is 1211 mm<sup style="">2</sup>.
Please scroll down to see the correct answer and solution guide.
Right Answer is:
SOLUTION
Concept:
Limiting MOR, Mu, lim = 0.36 fck b xu, lim (d – 0.42 xu lim)
Where,
xu, lim = 0.48 d (for 4e 415 steel).
Calculation:
Given:
b = 400, d = 750, fck = 20, fy = 415, Working BM = 200 kN-m.
∴ Design BM, Mu = (1.5 × 200) = 300 kN-m.
Also, xu, lim = (0.48 × 750) = 360 mm
∴ Mu, lim = (0.36 × 20 × 400 × 360) × (750 – 0.42 × 360) = 620.84 kN-m
So, Mu (= 300) < Mu, lim (= 620.84) i.e. the section is under-reinforced.
Now; (0.36 × 20 × 40 xu) (750 – 0.42 xu) = Mu = 300 × 106
∴ xu = 151.79 ≈ 151.8 mm
i.e. depth of Neutral axis = 151.8 mm
again, C = T ⇒ 0.36 fck b xu = 0.87 fy Ast
⇒ \({{A}_{st}}=\frac{\left( 0.36\times 20\times 400\times 151.8 \right)}{\left( 0.87\times 415 \right)}=1210.788~m{{m}^{2}}\)
⇒ Ast ≈ 1211 mm2
Area of Tensile Reinforcement = 1211 mm2
Important Point:
|
|
xu, lim |
Mu, lim |
|
Fe 250 steel |
0.53 d |
0.148 fck bd2 |
|
Fe 415 steel |
0.48 d |
0.138 fck bd2 |
|
Fe 500 steel |
0.46 d |
0.133 fck bd2 |