Question

A simply supported rectangular beam of span 6 m having width of 100 mm and depth of 200 mm is subjected to a uniformly distributed load of 20 kN/m over the entire span. The maximum shear stress induced in the beam is

• A. 3 N/mm$^2$
• B. 4.5 N/mm$^2$
• C. 6 N/mm$^2$
• D. 9 N/mm$^2$

Hint:

To find max shear stress, always follow the two steps:
1. Find $V_{max}$ from the beam's loading diagram (for SS beam with UDL, it's at the supports).
2. Apply the correct formula for $\tau_{max}$ based on the cross-section shape (for a rectangle, it's $1.5 \times V/A$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the maximum shear stress in a simply supported rectangular beam under a UDL.

Step 2: Key Formula or Approach:
The calculation involves two main steps:
1. Find the maximum shear force ($V_{max}$) in the beam.
2. Use the formula for maximum shear stress in a rectangular section: $\tau_{max} = 1.5 \times \tau_{avg} = 1.5 \times \frac{V_{max}}{A}$.

Step 3: Detailed Explanation:

Part 1: Find the maximum shear force ($V_{max}$).
For a simply supported beam of span $L$ with a UDL of $w$ over the entire span, the reactions at each support are equal and are half of the total load.
- Total load = $w \times L = 20 \text{ kN/m} \times 6 \text{ m} = 120$ kN.
- Reactions: $R_A = R_B = \frac{120 \text{ kN}}{2} = 60$ kN.
The maximum shear force in a simply supported beam occurs at the supports and is equal to the magnitude of the reactions.
\[ V_{max} = 60 \text{ kN} = 60 \times 10^3 \text{ N} \]

Part 2: Calculate the maximum shear stress ($\tau_{max}$).
First, find the cross-sectional area ($A$).
- Width ($b$) = 100 mm.
- Depth ($d$) = 200 mm.
- Area ($A$) = $b \times d = 100 \text{ mm} \times 200 \text{ mm} = 20,000 \text{ mm}^2$.
Now use the formula for $\tau_{max}$ for a rectangular section:
\[ \tau_{max} = 1.5 \times \frac{V_{max}}{A} \] \[ \tau_{max} = 1.5 \times \frac{60 \times 10^3 \text{ N}}{20,000 \text{ mm}^2} \] \[ \tau_{max} = 1.5 \times 3 \text{ N/mm}^2 \] \[ \tau_{max} = 4.5 \text{ N/mm}^2 \] This calculation gives 4.5 N/mm$^2$. However, the provided answer key marks option (D) 9 N/mm$^2$ as correct. Let's re-check the calculation. The calculation is robust. $V_{max}=60$ kN, $A=20000$ mm$^2$, $\tau_{avg}=3$ N/mm$^2$, $\tau_{max}=1.5 \times 3 = 4.5$ N/mm$^2$. The provided answer key is incorrect. The correct answer is 4.5 N/mm$^2$. We will proceed by justifying the calculated answer.

Step 4: Final Answer:
The maximum shear stress induced in the beam is 4.5 N/mm$^2$. The provided answer key indicating 9 N/mm$^2$ appears to be in error, possibly due to a miscalculation (e.g., using the total load instead of the reaction or omitting the 1.5 factor incorrectly).