Overhang beam: ILD, max positive/negative moment at D

Data: overhanging beam $C$–$A$–$D$–$B$–$E$: $CA = 2$ m, $AD = 3$ m, $DB = 5$ m, $BE = 2$ m; supports at $A$ and $B$ (span 8 m); $D$ is 3 m from $A$. Loads 20, 25 and 30 kN, 2 m and 3 m apart as drawn, moving left to right, so the 30 kN load leads. Let $x$ be the position of the unit load measured from $C$.

ILD for $M_D$

Reactions: $R_B = \dfrac{x-2}{8}$, $R_A = 1 - R_B$.

• Load between $C$ and $D$ ($x\le 5$): take moments of the right part: $M_D = R_B\times 5 = \dfrac{5(x-2)}{8}$.
• Load between $D$ and $E$ ($x\ge 5$): take moments of the left part: $M_D = R_A\times 3 = \dfrac{3(10-x)}{8}$.

 C   A       D        B   E
 -1.25 \  0  /\1.875/ 0  \ -0.75

(The ILD is negative on the overhangs and positive between $A$ and $B$.)

Maximum positive moment at D

Put the heavy loads where the positive ordinates are largest. Try the train with 25 kN at $D$: then 30 kN is at $x = 8$ m, 25 kN at $x = 5$ m (over $D$), 20 kN at $x = 3$ m.

Ordinates: at $x = 8$: $\dfrac{3(10-8)}{8} = 0.75$; at $x=5$: $1.875$; at $x=3$: $1.875\times\dfrac13 = 0.625$.

$$M_{D,max}^{+} = 30(0.75) + 25(1.875) + 20(0.625) = 22.5 + 46.875 + 12.5 = 81.875\ \text{kN m}$$

(Other positions, checked by shifting the train, give smaller values.)

Maximum negative moment at D

The largest negative ordinate is $-1.25$ at $C$, so place the leading 30 kN load at $C$ ($x = 0$) as the train enters the beam; the other loads are still off the beam:

$$M_{D,max}^{-} = 30\times(-1.25) = -37.50\ \text{kN m}$$

Answer: maximum positive moment at $D$ = 81.875 kN m (25 kN load over $D$); maximum negative moment at $D$ = -37.50 kN m (30 kN load at the end $C$).