Question

An ISLB 350@ 486 N/m beam is laterally supported throughout. The following properties are given for the I - section
$I_x=13158.3\times {10}^{4}m{m}^4,$ shape factor $=1.12,f_y=250MPa,{\gamma }_{mo}=1.1$
Assume the section to be a compact section and a case of low shear. The design bending strength will be kNm.

• A. 191.39

Answer 1

The correct option is A 191.39

Elastic section modulus, $Z_e=\frac{13158.3\times {10}^4}{\frac{350}{2}}=751.9\times {10}^{3}m{m}^3$
Plastic section modulus, $Z_P=S.F.\times Z_e=1.12\times 751.9\times {10}^3$
$=842.13\times {10}^{3}m{m}^3$
$M=\begin{array}{c}\beta Z_P.\frac{f_y}{{\gamma }_{m_0}}\\ 1.2Z_e.\frac{f_y}{\gamma m_0}\end{array}|$ Which ever is less
$\because \beta =1$ for compact section
$\therefore M={\left\{\begin{array}{c}1\times 842.13\times {10}^3\times \frac{250}{1.1}\times {10}^{-6}=191.39kN-m\\ 1.2\times 751.9\times {10}^3\times \frac{250}{1.1}\times {10}^{-6}=205.06kNm\end{array}\right\}}_{minimum}$
So, M = 191.39 kNm