Question

For the beam given below, determine the resultant internal loading acting on the cross section at C. Use the following assumption: The beam is weightless. Support A is a roller, and support B is a pin. The cross section C is 1 m away from A. (Tips: Internal loading includes normal force, shear force, and moment). . 100 kN 50 kN/m ol B 2m 1m 2m

Answer 1

25 kN/m 12.5 kN/m Mc Nc Vc 0.5 m 1 m

According to an above figure, we have

w / (1 m) = (50 kN/m) / (2 m)

w = (25 kN/m²) (1 m)

w = 25 kN/m

Magnitude of the resultant of distributed load is equal to the area under the loading curve (triangle) and acts through the centroid of this area.

Then, we have

F = (1/2) (25 kN/m) (1 m)

F = 12.5 kN

Which acts [(1 m) / 2] = 0.5 m from C.

For an equilibrium conditions, we have

_$\sum$ F_x = 0 ;   - N_C = 0

N_C = 0

_$\sum$ F_y = 0 ; V_C - (12.5 kN) = 0

V_C = 12.5 kN

Therefore, the resultant internal loading acting on the cross section at C which will be given as -

_$\sum$ M_C = 0 ;   - M_C - (12.5 kN) (0.5 m) = 0

- M_C - (6.25 kN.m) = 0

M_C = - 6.25 kN.m

{ The negative sign indicates that M_C acts in the opposite direction