Question

A beam of L meters span simply supported at end carries a central load W. The overall depth of beam section is 300 mm with horizontal flanges each of 200 mm x 20 mm and a web of 260 mm x 10 mm. If the maximum shear stress is limited to 45 N/mm${}^2$ and maximum bending stress is limited to 150 N/mm${}^2$ , then the values of centrally applied load W and span L respectively are:

• A. 250.23 kN, 4.242 m
• B. 239.77 kN, 2.864 m
• C. 220.57 kN, 2.142 m
• D. 280.52 kN, 5.121 m

Answer 1

The correct option is B 239.77 kN, 2.864 m

$I_{NA}=\frac{200\times {300}^3}{12}-\frac{190\times {260}^3}{12}=1.717\times {10}^{8}m{m}^4$

Let point load at midspan = W N

$\therefore \text{Maximum shear force}=\frac{W}{2}N$

$\therefore$ ${\tau }_{max}=\frac{Va\overline{y}}{Ib}=45N/m{m}^2$

$a\overline{y}=$ Moment of area above NA portion about NA

$=(200\times 20)\times (130+10)+10\times 130\times 65$

$=644500m{m}^3$

$\therefore$ ${\tau }_{max}=\frac{W}{2}\times \frac{644500}{1.717\times {10}^8\times 10}=45$

$\Rightarrow W=239.77kN$

$M_{max}=\frac{WL}{4}$

$\because {\sigma }_{max}=\frac{M_{max}}{Z}$

$\Rightarrow 150=\frac{WL}{4}\times \frac{y}{I}$

$\Rightarrow 150=\frac{239.77\times {10}^3\times L\times {10}^3\times 150}{4\times 1.717\times {10}^8}$

$\Rightarrow L=2.864$ m