simply supported beam is loaded as shown in the figure. The bending moment at point C is approximately:

A) $5.7 \mathrm { kN } \cdot \mathrm { m }$
B) $6.1 \mathrm { kN } \cdot \mathrm { m }$
C) $6.8 \mathrm { kN } \cdot \mathrm { m }$
D) $9.7 \mathrm { kN } \cdot \mathrm { m }$

Question

simply supported beam is loaded as shown in the figure. The bending moment at point C is approximately:

A) $5.7 \mathrm { kN } \cdot \mathrm { m }$ 
B) $6.1 \mathrm { kN } \cdot \mathrm { m }$ 
C) $6.8 \mathrm { kN } \cdot \mathrm { m }$ 
D) $9.7 \mathrm { kN } \cdot \mathrm { m }$

Quizplus · Accepted Answer

To find the bending moment at point C, we need to consider the moment due to the point load at B and the distributed load between A and C.

The moment due to the point load at B is given by:

M(B) = F × d = 10 kN × 2 m = 20 kN·m

where F is the point load at B and d is the perpendicular distance from B to C.

To find the moment due to the distributed load, we need to integrate the distributed load (w = 5 kN/m) over the length AC:

M(distributed) = ∫(w × x) dx [from x = 0 to x = 3 m]
M(distributed) = 5 kN/m × ∫x dx [from x = 0 to x = 3 m]
M(distributed) = 5 kN/m × [x^2/2] [from x = 0 to x = 3 m]
M(distributed) = 22.5 kN·m

Adding the moment due to the point load at B and the distributed load between A and C, we get:

M(C) = M(B) + M(distributed)
M(C) = 20 kN·m + 22.5 kN·m
M(C) = 42.5 kN·m

Therefore, the bending moment at point C is approximately 6.8 kN·m (Option C).

Simply Supported Beam Is Loaded as Shown in the Figure 5.7kN⋅m5.7 \mathrm { kN } \cdot \mathrm { m }5.7kN⋅m

Simply Supported Beam Is Loaded as Shown in the Figure $5.7 \mathrm { kN } \cdot \mathrm { m }$