In a stay stress calculation, a stay of 2.75 inches diameter is subjected to 450 psi boiler pressure, using an effective area of 26 by 9 inches in the force calculation. What is the resulting stress in psi?

In a stay stress calculation, a stay of 2.75 inches diameter is subjected to 450 psi boiler pressure, using an effective area of 26 by 9 inches in the force calculation. What is the resulting stress in psi?

• A. 17,730 Psi
• B. 15,000 Psi
• C. 12,500 Psi
• D. 20,000 Psi

Answer: (A)

The concept here is that stress in the stay comes from the pressure force divided by the stay’s cross‑sectional area. The boiler pressure acts over the effective area, producing a force the stay must resist, and the stay’s ability to resist that force is limited by its own cross-sectional area. Compute the force first: the effective area is 26 by 9 inches, so A = 26 × 9 = 234 in^2. Force = pressure × area = 450 psi × 234 in^2 = 105,300 pounds. Next, find the stay’s cross-sectional area from its diameter: A_stay = (π/4) d^2 = (π/4) × (2.75 in)^2 ≈ 5.94 in^2. Now stress = force / stay area = 105,300 lb / 5.94 in^2 ≈ 17,730 psi. So the resulting stress is about 17,730 psi.

The concept here is that stress in the stay comes from the pressure force divided by the stay’s cross‑sectional area. The boiler pressure acts over the effective area, producing a force the stay must resist, and the stay’s ability to resist that force is limited by its own cross-sectional area.

Compute the force first: the effective area is 26 by 9 inches, so A = 26 × 9 = 234 in^2. Force = pressure × area = 450 psi × 234 in^2 = 105,300 pounds.

Next, find the stay’s cross-sectional area from its diameter: A_stay = (π/4) d^2 = (π/4) × (2.75 in)^2 ≈ 5.94 in^2.

Now stress = force / stay area = 105,300 lb / 5.94 in^2 ≈ 17,730 psi.

So the resulting stress is about 17,730 psi.