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A hollow circular shaft has an outside diameter of 12 in and an inside diameter of 4 in. The shaft is in equilibrium when the indicated loads and torque are applied. If the axial load \( P \) is 186,000 lb, determine the stress state at point \( A \), which is on the side of the shaft. Use coordinate axis \( X \) in the direction of force \( P \) and axis \( Y \) in the direction negative to the vertical force of 10,000 lb.

- Bearing A
- Outside diameter: 12 in
- Inside diameter: 4 in
- Torque: 400,000 in-lb
- Length: 4 ft
- Axial load: \( P = 186,000 \) lb
- Vertical force: 10,000 lb

Seleccione una:

a. \(\sigma_x = 150 \text{ psi}; \sigma_y = 10.0 \text{ psi}; \tau_{xy} = 1350 \text{ psi}\)

b. \(\sigma_x = 1850 \text{ psi}; \sigma_y = 0.0 \text{ psi}; \tau_{xy} = 1366 \text{ psi}\)

c. \(\sigma_x = 1850 \text{ psi}; \sigma_y = 0.0 \text{ psi}; \tau_{xy} = 1035 \text{ psi}\)

d. \(\sigma_x = 1350 \text{ psi}; \sigma_y = 0.0 \text{ psi}; \tau_{xy} = 1194 \text{ psi}\)

Answer :

Final answer:

The stress state at point A on the side of the hollow circular shaft is Ox = 1850 psi, Oy = 0 psi, and Txy = 1366 psi.

Explanation:

To determine the stress state at point A on the side of the shaft, we need to calculate the normal stresses in the X and Y directions (Ox and Oy) and the shear stress (Txy).

Given:

  • Outside diameter of the shaft (D) = 12 in
  • Inside diameter of the shaft (d) = 4 in
  • Axial load (P) = 186,000 lb
  • Vertical force (F) = 10,000 lb

First, we need to calculate the torque (T) applied to the shaft:

Torque (T) = Force (F) * Distance (r)

Distance (r) = 4 ft = 48 in

T = 10,000 lb * 48 in = 480,000 in-lb

Next, we can calculate the normal stresses in the X and Y directions:

Ox = P / (pi * (D^2 - d^2) / 4)

Oy = -F / (pi * (D^2 - d^2) / 4)

Substituting the given values:

Ox = 186,000 lb / (pi * (12^2 - 4^2) / 4) = 1850 psi

Oy = -10,000 lb / (pi * (12^2 - 4^2) / 4) = 0 psi

Finally, we can calculate the shear stress:

Txy = T / (pi * (D^2 - d^2) / 2)

Txy = 480,000 in-lb / (pi * (12^2 - 4^2) / 2) = 1366 psi

Therefore, the stress state at point A on the side of the shaft is:

Ox = 1850 psi, Oy = 0 psi, Txy = 1366 psi

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