The maximum bending stress produced in the pole is 73.34 psi and the maximum shear stress produced in the pole is 3.81 psi.
How to find bending stress and shear stress?
Given data:
Diameter (d) = 10 inches
Length (L) = 20 feet
(converting to inches: 20 feet × 12 inches/foot = 240 inches)
Force (F) = 300 pounds
r = d/2
= 10 inches / 2
= 5 inches
(convering to inches: 5 inches × 2.54 cm/inch = 12.7 cm)
For a solid circular cross-section, the moment of inertia is:
I = πr⁴/4
I = π × (5 inches)² × (5 inches)² / 4
I = 785.3982 inches⁴
The maximum bending stress occurs at the base of the pole, where the bending moment is maximum. The formula for bending stress is:
σ = (M/I) × (y/2)
where:
M = bending moment
y = distance from the neutral axis to the outer edge of the pole (in this case, y = r)
The bending moment is:
M = FL
M = 300 pounds × 240 inches
M = 72,000 pound-inches
Substituting the values:
σ = (72,000 pound-inches / 785.3982 inches⁴) × (5 inches / 2)
σ = 73.34 psi (pounds per square inch)
The maximum shear stress occurs at the neutral axis, where the shear force is maximum. The formula for shear stress is:
τ = (V/A)
where:
V is the shear force
A is the cross-sectional area
The shear force is:
V = F
V = 300 pounds
The cross-sectional area of the pole is:
A = πr²
A = π * (5 inches)²
A = 78.54 square inches
Substituting the values:
τ = (300 pounds / 78.54 square inches)
τ = 3.81 psi
Therefore, the maximum bending stress produced in the pole is 73.34 psi and the maximum shear stress produced in the pole is 3.81 psi.