Question

If the maximum shear stress in a rectangular beam cross section is 120 N/mm$^2$, then the average shear stress is

• A. 40 N/mm$^2$
• B. 80 N/mm$^2$
• C. 90 N/mm$^2$
• D. 180 N/mm$^2$

Hint:

Memorize the relationship between max and average shear stress for common shapes:
- Rectangle: $\tau_{max} = 1.5 \times \tau_{avg}$
- Circle: $\tau_{max} = \frac{4}{3} \times \tau_{avg} \approx 1.33 \times \tau_{avg}$
- I-Beam: $\tau_{max} \approx \tau_{avg}$ (in the web, where most shear is carried)

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the average shear stress in a rectangular beam, given the maximum shear stress.

Step 2: Key Formula or Approach:
For a beam with a rectangular cross-section, the shear stress distribution is parabolic, with the maximum value at the neutral axis and zero at the top and bottom fibers.
The relationship between the maximum shear stress ($\tau_{max}$) and the average shear stress ($\tau_{avg}$) for a rectangular section is:
\[ \tau_{max} = 1.5 \times \tau_{avg} \] The average shear stress is simply the total shear force ($V$) divided by the cross-sectional area ($A$): $\tau_{avg} = V/A$.

Step 3: Detailed Explanation:
We are given the maximum shear stress:
\[ \tau_{max} = 120 \text{ N/mm}^2 \] We need to find the average shear stress, $\tau_{avg}$. We can rearrange the formula:
\[ \tau_{avg} = \frac{\tau_{max}}{1.5} \] \[ \tau_{avg} = \frac{120 \text{ N/mm}^2}{1.5} \] \[ \tau_{avg} = 80 \text{ N/mm}^2 \]

Step 4: Final Answer:
The average shear stress is 80 N/mm$^2$.