Question

A 3m long concrete bar is subjected to a concentrated load as well as udl are shown in figure. The cross-section of beam is given as 200mm x 200mm. Considering linear stress distribution across the cross-section, the modulus of rupture (expressed in Mpa)

• A. 8.25

Answer 1

The correct option is A 8.25
By symmetry

$R_A=R_B=\frac{1}{2}(2+10+10+2)$

$R_A=R_B=12kN$

$\text{B.M.at A}=0$

$\text{B.M.at C}=12\times 1-2\times 1\times 0.5$

= 11 kN -m

B.M. at mid of CD = $12\times 1.5-2\times 1\times 1-10\times 0.5$

= 11 kN-m

B.M. at D = 11 kN-m

B.M. at B=0

Maximum BM = 11 kN-m
= $11\times {10}^{6}N-mm$


Now maximum bending stress

$\sigma =\frac{M_{max}}{I}\times y$

$=\frac{11\times {10}^6}{\left(\frac{200\times {200}^2}{6}\right)}=8.25N/m{m}^2$

= 8.25 MPa