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
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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.

0 = Anchorage length

Calculation:

Mu = 50 kN – m, Vu = 200 kN, ℓ0 = 400 mm

\(\therefore {\ell _d} \le 1.3 \times \frac{{50}}{{200}} + \frac{{400}}{{1000}} = 0.725\;m\)

d ≤ 725 mm

At limiting condition,

d = 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.

Mistake Point:

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

\({\ell _d} \le \frac{{{M_u}}}{{{V_u}}} + {\ell _0}\)