Respuesta :
Answer:
σ1 = 4.38 ksi
σ2 =-1.2 ksi
τ = 2.7903 ksi
Explanation:
diameter (d) = 2 inch
radius (r) = 1 inch
axial tension = 10,000 lb
torque (T) = 300 lb.ft
Determine the principal stresses and the maximum in plane shear stress
normal stress (σₙ) = force / area
σx = \frac{10000}{π x r^{2}}
σx = 3.18 ksi
σy = 0
shear stress (τ) = \frac{Tc}{J}
where c = radius
J = polar moment of inertia = \frac{πc^{4}}{2}
τ = \frac{T}{ \frac{πc^{3}}{2}}
τ = \frac{300 x 12}{ \frac{π x 1^{3}}{2}} = 2.2918 ksi
now that we have σx, σy and τ we have to apply the formula below to find the principal shear stresses
σ1,2 = \frac{σx + σy}{2} ± [tex]\sqrt{(\frac{σx + σy}{2}) ^{2} + τ}[/tex]
σ1,2 = \frac{3.18 + 0}{2} ± [tex]\sqrt{(\frac{3.18 + 0}{2}) ^{2} + 2.29}[/tex]
σ1,2 = 1.5915 ± 2.7903
σ1 = 1.5915 + 2.7903 = 4.38 ksi
σ2 = 1.5915 - 2.7903 = -1.2 ksi
maximum in plane shear stress (τ) = [tex]\sqrt{(\frac{σx + σy}{2}) ^{2} + τ}[/tex]
maximum in plane shear stress (τ) = 2.7903 ksi
The principal stresses and the maximum in-plane shear stress that act at a point are; σ₁ = -3934.53 lb/ft² ; σ₂ = 10,300.73 lb/ft² and τ_max = 7117.63 lb/ft²
What are the principal stresses?
We are given:
Tension; P = 10000 lb
Torque; T = 300 lb.ft
Diameter; d = 2 in
Formula for stress is;
σ = P/A
σ = 10000/(πd²/4)
σ = 3183.1 lb/ft²
Shear stress is gotten from the formula;
τ = Tc/J
τ = (10000 * 1)/(π/2) * 1⁴)
τ = 6366.2 lb/ft²
Principal shear stress is gotten from the formula;
σ₁,₂ = σ_x + σ_y ± √[((σ_x - σ_y)/2)² + (τ_xy)²]
σ₁,₂ = 3183.1 + 0 ± √[((3183.1 - 0)/2)² + 6366.2²]
σ₁,₂ = 3183.1 ± 7117.63
σ₁ = -3934.53 lb/ft² and σ₂ = 10,300.73 lb/ft²
The maximum in-plane shear stress is gotten from;
τ_max = √[((σ_x - σ_y)/2)² + (τ_xy)²]
τ_max = √[((3183.1 - 0) /2)² + 6366.2²]
τ_max = 7117.63 lb/ft²
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