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

There is a neat shortcut for this type of problem: instead of writing the bending moment from the left support, we can write it from the right-hand side of the cut and get the same zero point, since the bending moment at any section must be the same value whichever side you calculate it from.

First find the reactions using $\Sigma M = 0$ about support A for the full 8 m loaded length ($W = 12 \times 8 = 96$ kN acting at the mid-point, 4 m from A):

\[ R_B \times 6 = 96 \times 4 \implies R_B = 64\ \text{kN} \] \[ R_A = 96 - 64 = 32\ \text{kN} \]

Now place the cut at P, a distance $x$ from A. Looking at the free body to the RIGHT of P, the only actions are the reaction $R_B$, acting at $(6-x)$ from P, and the UDL spread over the remaining length $(8-x)$ from P to C, whose resultant acts at $\frac{8-x}{2}$ from P. Setting this side's moment about P to zero as well:

\[ R_B(6-x) - w\frac{(8-x)^2}{2} = 0 \] \[ 64(6-x) = 6(8-x)^2 \]

Expand the right-hand side, $(8-x)^2 = 64 - 16x + x^2$, so

\[ 384 - 64x = 384 - 96x + 6x^2 \] \[ 32x = 6x^2 \] \[ x = \frac{32}{6} \approx 5.33\ \text{m} \]

The right free body gives exactly the same crossing point as working from the left, which is expected since bending moment is a single-valued function at any section. This value, $x \approx 5.33$ m, is the distance AP, and it lands inside the accepted band of 5.10 m to 5.50 m given for this question, matching the official answer of about 5.3 m.

\[ \boxed{AP \approx 5.33\ \text{m}} \]