Question

A simply supported beam AB of span 8 m carries a uniformly distributed load of 24 kN/m over the left half of span. The ratio of the reactions of left support to right support is

• A. 0.50
• B. 0.33
• C. 3
• D. 2

Hint:

For calculating reactions, always replace a UDL with its equivalent point load.
The magnitude of the point load is the area of the UDL rectangle ($w \times l$).
The location of the point load is at the centroid of the UDL.
Then apply the standard equilibrium equations $\Sigma M = 0$ and $\Sigma F_y = 0$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the ratio of the support reactions ($R_A / R_B$) for a simply supported beam with a UDL on its left half.

Step 2: Key Formula or Approach:
To find the reactions, we use the equations of static equilibrium:
1. Sum of vertical forces is zero ($\Sigma F_y = 0$).
2. Sum of moments about any point is zero ($\Sigma M = 0$).

Step 3: Detailed Explanation:
Let the beam be AB, with support A on the left and support B on the right. The span is $L = 8$ m.
The uniformly distributed load (UDL) is $w = 24$ kN/m and it acts over the left half, from $x=0$ to $x=4$ m.
First, find the total load from the UDL. It can be represented as a single point load for calculation purposes.
Total Load ($W_{UDL}$) = $w \times$ length of load = $24 \text{ kN/m} \times 4 \text{ m} = 96$ kN.
This equivalent point load acts at the centroid of the UDL, which is at half the length of the loaded portion, i.e., at $x = 4/2 = 2$ m from support A.
Now, apply the equilibrium equations. Let $R_A$ and $R_B$ be the reactions at A and B.
Take moments about support A to find $R_B$:
$\Sigma M_A = 0$ (Clockwise moments are positive)
\[ (W_{UDL} \times 2 \text{ m}) - (R_B \times 8 \text{ m}) = 0 \] \[ (96 \text{ kN} \times 2 \text{ m}) = R_B \times 8 \text{ m} \] \[ 192 = 8 R_B \] \[ R_B = \frac{192}{8} = 24 \text{ kN} \] Now, apply the vertical force equilibrium to find $R_A$:
$\Sigma F_y = 0$ (Upward forces are positive)
\[ R_A + R_B - W_{UDL} = 0 \] \[ R_A + 24 \text{ kN} - 96 \text{ kN} = 0 \] \[ R_A = 96 - 24 = 72 \text{ kN} \] Finally, calculate the required ratio:
\[ \text{Ratio} = \frac{R_A}{R_B} = \frac{72 \text{ kN}}{24 \text{ kN}} = 3 \]

Step 4: Final Answer:
The ratio of the reactions of the left support to the right support is 3.