Overhang beam ILD for shear force at C

Data: beam $ABD$: $A$ hinge, $B$ roller, $AB = 20$ m, overhang $BD = 5$ m. Section $C$ is 6 m from $A$. Let the unit load be at $x$ from $A$.

Reactions

$$R_B = \frac{x}{20}, \qquad R_A = 1 - \frac{x}{20} = \frac{20-x}{20}$$

These are valid for $0\le x\le 25$ (for $x>20$, $R_A$ is negative).

Shear at C

• Load between $A$ and $C$ ($0\le x<6$): take the right part: $V_C = -R_B = -\dfrac{x}{20}$. At $x = 6^-$: $-0.30$.
• Load between $C$ and $B$ ($6<x\le 20$): take the left part: $V_C = R_A = \dfrac{20-x}{20}$. At $x = 6^+$: $+0.70$; at $B$: 0.
• Load on the overhang ($20\le x\le 25$): $V_C = R_A = \dfrac{20-x}{20}$, which is negative: $-0.25$ at $D$.

 A o----C--------------------o B----D
 ILD V_C:  0 \-0.30 | +0.70 \       0 .. -0.25

The ILD is a line from 0 at $A$ down to $-0.30$ just left of $C$, a jump of 1 to $+0.70$ just right of $C$, then a straight line falling to 0 at $B$, continuing to $-0.25$ at the free end $D$.

Use

• Maximum positive shear at $C$: load only the positive part, $6<x<20$; for a UDL $w$, $V = w\times\tfrac12\times 14\times 0.70 = 4.9\,w$.
• Maximum negative shear: load $A$ to $C$ and the overhang; $V = -w\,(\tfrac12\times 6\times 0.30 + \tfrac12\times 5\times 0.25) = -1.525\,w$.