Cantilever ILD for SF and BM at C

Data: cantilever $AB$ fixed at $A$ (free end $B$). $C$ is 2 m from $A$ with a 50 kN load; $D$ is 2 m beyond $C$; 10 kN/m UDL on $DB$ (4 m). So $B$ is 8 m from $A$. Let $s$ be the distance of the unit load from $A$ and the section at $C$ ($s_C = 2$ m).

ILD for shear at C

A load to the left of $C$ is carried by the fixed support only, so $V_C = 0$. A load to the right of $C$ is transmitted to the section, so $V_C = +1$.

ILD for bending moment at C

For $s<2$: $M_C = 0$. For $s>2$: $M_C = -(s-2)$, a straight line from 0 at $C$ to $-6$ at $B$ ($B$ is 6 m from $C$).

 A (fixed)  C     D           B
 |==========|=====|===========|
 ILD SF:  0  | 1   1   1   1   1
 ILD BM:  0  | 0  -2          -6

Values

The 50 kN load is at the section. For shear take the section just right of $C$ (the load then is on the left of the section); the value just left of $C$ is given as well.

Shear force at C:

• Just right of $C$: $V_C = 10\times 4\times 1 = 40$ kN (UDL on $DB$, ordinate 1).
• Just left of $C$ (50 kN load taken to the right of the section): $V_C = 50\times 1 + 40 = 90$ kN.

Bending moment at C: ordinate at $D$ is $-2$, at $B$ it is $-6$, so the average ordinate under the UDL is $-4$.

$$M_C = 50\times 0 + 10\times 4\times(-4) = -160\ \text{kN m}$$

(Check by statics: UDL resultant $40$ kN acts 4 m from $C$, so $M_C = -160$ kN m, hogging.)

Answer: $V_C = 40$ kN (90 kN just left of the 50 kN load); $M_C = -160$ kN m (hogging).