A RCC member of total width 250 mm and effective depth 450 mm is
![A RCC member of total width 250 mm and effective depth 450 mm is](/img/relate-questions.png)
A RCC member of total width 250 mm and effective depth 450 mm is reinforced with Fe 415 bars. If there is a factored shear force of 200 kN and bending moment of 50 kN-m and there is compression confinement the maximum diameter of bars required considering anchorage as 400 mm is
Consider M20, τbd = 1.2 MPa (for Fe 415 bar)
A. 8 mm
B. 9 mm
C. 10 mm
D. 12 mm
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Concept:
Development Length \(\left( {{\ell _d}} \right) = \frac{{0.87\;{f_y}}}{{4\;{\tau _{bd}}}}\)
Also,
\({\ell _d} \le 1.3\frac{{{M_u}}}{{{V_u}}} + {\ell _0}\) (When there is compression confinement)
Where,
Mu = Ultimate MOR at a section considering stress in all bars as 0.87 fy
Vu = Factored shear force at the point of zero moment.
ℓ0 = Anchorage length
Calculation:
Mu = 50 kN – m, Vu = 200 kN, ℓ0 = 400 mm
\(\therefore {\ell _d} \le 1.3 \times \frac{{50}}{{200}} + \frac{{400}}{{1000}} = 0.725\;m\)
⇒ ℓd ≤ 725 mm
At limiting condition,
ℓd = 725 mm
Also,
\({\ell _d} \ge \frac{{0.87\;{f_y}\;\phi }}{{4\;{\tau _{bd}}}} \Rightarrow 725 \ge \frac{{0.87 \times 415 \times \phi }}{{4 \times 1.2}}\)
⇒ ϕ ≤ 9.638 mm
Thus the maximum bar dia that can be provided is 9 mm.
The value of bond stress is given for Fe 415 bar and we need not add 60 % in the given value.
Important Point:
In case of simple support or the places where there is no compression confinement, tension R/F should be limited to a diameter, such that
\({\ell _d} \le \frac{{{M_u}}}{{{V_u}}} + {\ell _0}\)