Question

A hanger is made of two bars of different sizes. Each bar has a square cross-section. The hanger is loaded by three point loads in the mid-vertical plane as shown: the upper bar is $100\,\text{mm\times100\,\text{mm}$ and the lower bar is $50\,\text{mm}\times50\,\text{mm}$. Ignore self-weight and stress concentration. What is the maximum tensile stress (in N/mm$^2$) anywhere in the hanger?} \includegraphics[width=0.25\linewidth]{image27.png}

• A. 15.0
• B. 25.0
• C. 35.0
• D. 45.0

Hint:

Always compute axial force at a section by summing the loads acting below that section. Then divide by cross-sectional area to get direct stress.

Answer 1

The Correct Option is B

Axial force at a section equals the net external load acting below that section.
Upper (100 mm)$^2$ bar:
Load below $=100+100+50=250\ \text{kN}$.
Area $A_u=100\times100=10{,}000\ \text{mm}^2$.
\[ \sigma_u=\frac{250\,000}{10{,}000}=25\ \text{N/mm}^2 \]
Lower (50 mm)$^2$ bar:
Load below $=50\ \text{kN}$.
Area $A_\ell=50\times50=2{,}500\ \text{mm}^2$.
\[ \sigma_\ell=\frac{50\,000}{2{,}500}=20\ \text{N/mm}^2 \]
Therefore, the maximum tensile stress is
\[ \boxed{25.0\ \text{N/mm}^2} \]