Question

A continuous beam 250 mm x 450 mm carries 6 numbers of 12 mm diameter longitudinal bars as shown. The factored shear force at the point of inflection is 200 kN. Check if the beam is safe in bond. Assume M15 mix with ${\sigma }_{ck}=15N/m{m}^2$ and mild steel with ${\sigma }_y=250N/m{m}^2$. A clear cover of 25 mm canbe assumed. The design bond stress for mild steel bars in M15 concrete is specified to be 1.0 N/mm${}^2$

• A. 652.5

Answer 1

The correct option is A 652.5

Assume stirrups are provided of 6 mm$\varphi$
Effective depth
$d=450-25-6-\frac{12}{2}=413$ mm
For safety in bond
$L_d\le \frac{M}{V}+L_0$
Calculation $L_d$
$L_d=\frac{0.87f_y\varphi }{4{\tau }_{bd}}$
$=\frac{0.87\times 250\times 12}{4\times 1}=652.5mm$
Calculation of depth of NA
C = T
$0.36f_{ck}b{x}_u=0.87f_y{A}_{st}$
$0.36\times 15\times 250x_u=0.87\times 250\times 6\times \frac{\pi }{4}(12{)}^2$
$x_u=109.327$ mm

Limiting depth of NA for Fe250 grade of steel.
$x_{ulim}=0.53d$
$=0.53\times 413$
= 218.89 > 109.327 mm
Hence beam is under reinforeced.

Calculation of MOR
$M=0.36f_{ck}b{x}_u(d-0.42x_u)$
$0.36\times 15\times 250\times 109.327(413-0.42\times 109.327)$
$=54.18\times {10}^3$kN/mm
$L_0=$ anchorage length
=greater of d or 12 $\varphi$ from the point of inflection
$max\{\begin{array}{cc}413& \\ 12\times 12& =144mm\end{array}$
$L_0=413$ mm
Hence, $\frac{M}{V}+L_0=\frac{54.18\times {10}^3}{200}+413$
$683.9mm\nless 652.5mm$

Hence, beam is safe in bond