Overhang beam: max +/- BM at C under 4 m UDL

Data: overhanging beam $ABD$ ($AB = 20$ m, overhang $BD = 5$ m), section $C$ at 6 m from $A$. A UDL of 15 kN/m and 4 m length rolls from left to right. Let $x$ be the position of a unit load from $A$.

ILD for $M_C$

$R_A = \dfrac{20-x}{20}$, $R_B = \dfrac{x}{20}$.

• $0\le x\le 6$: $M_C = R_B\times 14 = 0.7x$
• $6\le x\le 25$: $M_C = R_A\times 6 = 0.3\,(20-x)$

 A o------C------------------o B--------D
        /\ 4.2                    \ -1.5
 ILD: 0 /  \_______________  0 ____ -1.5

Maximum positive moment

For a UDL shorter than the span, the maximum positive moment occurs when the load is placed so that the section divides the load in the same ratio as it divides the span (loaded length on the left of $C$ : total $= 6 : 20$):

$$c = 4\times\frac{6}{20} = 1.2\ \text{m (left of } C),\qquad 4 - 1.2 = 2.8\ \text{m (right of } C)$$

So the UDL covers $x = 4.8$ m to $8.8$ m. End ordinates: $0.7(4.8) = 3.36$ and $0.3(20-8.8) = 3.36$ (equal, as required).

$$\text{Area} = \frac{(3.36+4.2)}{2}\times 1.2 + \frac{(4.2+3.36)}{2}\times 2.8 = 4.536 + 10.584 = 15.120\ \text{m}^2$$ $$M_C^{+} = 15\times 15.120 = 226.8\ \text{kN m}$$

Maximum negative moment

The negative ILD exists only on the overhang ($B$ to $D$). Place the 4 m load at the free end, from $x = 21$ m to $25$ m, where the ordinates are $-0.3$ and $-1.5$:

$$\text{Area} = \frac{(-0.3)+(-1.5)}{2}\times 4 = -3.60\ \text{m}^2, \qquad M_C^{-} = 15\times(-3.60) = -54.0\ \text{kN m}$$

Answer: maximum positive moment at $C$ = 226.8 kN m (UDL from 4.8 m to 8.8 m); maximum negative moment at $C$ = -54.0 kN m (UDL on the overhang, 21 m to 25 m).