The 150-lb man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb>ft. Determine the maximum internal bending moment. Assume that the water exerts a uniform distributed load upward on the bottom of the boat

Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):

∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0

⇒ q' = (W + q*L) / L

⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft

⇒ q' = 13 lb/ft (+↑)

The free body diagram of the boat is shown in the pic.

Then, we apply the following equation

q(x) = (13 - 3) = 10 (+↑)

V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)

M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)

The maximum internal bending moment occurs when x = 7.5 ft

then

M(7.5) = 5(7.5)² = 281.25 lb*ft

AFOKE88 AFOKE88 · Answer 2 · 2022-03-03T12:19:14+01:00

The maximum internal bending moment of the given balance system is; 281 lb.ft

How to find bending moment?

We are given;

Weight of man; W_m= 150 lb

Weight per linear foot of the boat: q = 3 lb/ft

Length of boat; L = 15 m

Summing forces about the y-axis from the given diagram of the boat gives;

∑Fy = 0

⇒ q'*L - W - q*L = 0

⇒ q' = (W + q*L)/L

⇒ q' = (150 lb + (3 lb/ft * 15 ft))/15 ft

q' = 13 lb/ft

Thus;

Net weight per linear foot;

q(x) = (13 - 3) = 10

Shear force is;

V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)

Bending moment is;

M(x) = ∫10x dx

M(x) = 5x² (0 ≤ x ≤ 7.5)

The maximum internal bending moment occurs when x = 7.5 ft.

Thus;

M(7.5) = 5(7.5)² = 281 lb.ft

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