Question

A simply supported beam of 4m span shown in the figure. The cross section of the beam is 120 mm wide 240 mm deep. At a section, 1m from left support, the minor principal stress at a point 50mm above the neutral axis is ?

• A. $0.01N/m{m}^2$
• B. $0.50N/m{m}^2$
• C. $2.5N/m{m}^2$
• D. Zero

Answer 1

The correct option is A $0.01N/m{m}^2$

By symmetry $R_A=R_B=\frac{4}{2}=2kN$

At a section 1 m from left support

SF = 2 kN

BM = $2\times 1=2kNm$


Moment of inertia,
$I=\frac{120\times {240}^3}{12}=1.38\times {10}^{8}m{m}^4$

At 50 mm above NA

$\sigma =\frac{M}{I}y=\frac{2\times {10}^6}{1.38\times {10}^8}\times 50=0.724N/m{m}^2$

$\tau =\frac{VA\overline{y}}{Ib}=\frac{(2\times 1000)(70\times 120)(120-35)}{1.38\times {10}^8\times 120}=0.086N/m{m}^2$

Stress element

${\sigma }_{minor}=-\frac{0.724}{2}+\sqrt{{(-\frac{0.724}{2})}^2+(0.086{)}^2}=0.01N/m{m}^2$