Question

A timber power­line pole is 10 in. in diameter at its base where it is solidly embedded in concrete. The pole extends 20 ft vertically upward from its base and is subjected to a horizontal pull of 300 lb at its top. Calculate the maximum bending stress and the maximum shear stress produced in the pole.

Answer 1

To calculate the maximum bending and shear stresses on a timber powerline pole, one uses mechanics of materials formulas involving the moment due to the applied force, the moment of inertia, and the material thickness.

Step-by-step explanation:

The calculation of maximum bending stress and shear stress in a timber powerline pole subjected to a horizontal force involves using principles of statics and mechanics of materials. The bending stress (also known as flexural stress) can be found using the formula σ = My/I, where M is the moment caused by the applied force, y is the distance from the neutral axis to the outer fiber (radius of the pole), I is the moment of inertia of the pole's cross-sectional area, and σ is the bending stress. The shear stress (τ) can be calculated using the formula τ = VQ/It, where V is the shear force, Q is the first moment of area about the neutral axis, I is the moment of inertia and t is the thickness of the material.

For a pole with a 10-inch diameter at the base and a 20 ft vertical height subjected to a 300 lb horizontal pull at the top, we would assume the pole's cross-section is circular to compute the moment of inertia. Since the pole is circular, I = πd^4/64, and Q can be found for a semi-circle since the maximum shear occurs at the neutral axis. Calculations would require conversion of units from inches to feet or to consistent SI units for calculations.

Answer 2

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.