Question

An overhang beam ABC is loaded with uniformly distributed load (UDL) throughout its entire span as per the figure below. The intensity of UDL is 12 kN/m. The bending moment at an intermediate point P (between A and B) is found to be zero. The length (in m) of the portion AP is (rounded off to two decimal places).

Hint:

Find R_A and R_B by taking moments about the supports, then set the bending moment expression M(x) = R_A x - w x^2/2 to zero for x between A and B.

Answer 1

Correct Answer: 5.3

Step 1: Set up the beam and find the reactions.
Beam ABC has a pin support at A, a roller support at B located 6 m from A, and a free overhanging end C located 2 m beyond B, so the total length AC is 8 m. A uniformly distributed load of intensity \(w = 12\ \text{kN/m}\) runs over the entire 8 m length. The total load is
\[ W = w \times L = 12 \times 8 = 96\ \text{kN} \]
acting through the centre of the 8 m span, which is 4 m from A.

Step 2: Take moments about A to find the reaction at B.
\[ R_B \times 6 = W \times 4 = 96 \times 4 = 384 \]
\[ R_B = \frac{384}{6} = 64\ \text{kN} \]

Step 3: Find the reaction at A.
\[ R_A = W - R_B = 96 - 64 = 32\ \text{kN} \]

Step 4: Write the bending moment equation between A and B.
Measuring \(x\) from A, for \(0 \le x \le 6\), the bending moment at a section is the reaction at A times \(x\) minus the moment of the UDL acting over that length:
\[ M(x) = R_A x - w\frac{x^2}{2} = 32x - 6x^2 \]

Step 5: Solve for the point P where the bending moment is zero.
Setting \(M(x) = 0\) other than the trivial root at the support itself:
\[ 32x - 6x^2 = 0 \implies x(32 - 6x) = 0 \implies x = \frac{32}{6} = 5.333\ \text{m} \]
This point is the point of contraflexure between A and B, where sagging changes to hogging.

Final Answer:
The length AP is about 5.33 m, which lies inside the accepted range for this question, 5.10 m to 5.50 m, matching the official key value of about 5.3 m.
\[ \boxed{AP = \frac{16}{3}\ \text{m} \approx 5.33\ \text{m}} \]