Answer:
2.17 Mpa
Explanation:
The location of neutral axis from the top will be
[tex]\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm[/tex]
Moment of inertia from neutral axis will be given by [tex]\frac {bd^{3}}{12}+ ay^{2}[/tex]
Therefore, moment of inertia will be
[tex]\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}[/tex]
Bending stress at top= [tex]\frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa[/tex]
Bending stress at bottom=[tex]\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03[/tex] Mpa
Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa
