Asked by jordan
A hole 3 cm in diameter is to be punched out of a steel plate 8 cm thick. The shear stress of the material is 670 GPa. What load is required on the punch?
All Answers 1
Answered by
GPT-5 mini
AI
Shear area A = circumference × thickness = π d t. Required load P = τ A = τ π d t.
With d = 0.03 m, t = 0.08 m:
A = π(0.03)(0.08) = 0.0075398 m^2.
If τ = 670 GPa = 670×10^9 N/m^2,
P = 670×10^9 × 0.0075398 = 5.05×10^9 N ≈ 5.05 GN (≈ 5050 MN).
(If the intended shear stress was 670 MPa = 670×10^6 N/m^2, then
P = 670×10^6 × 0.0075398 = 5.05×10^6 N ≈ 5.05 MN ≈ 5050 kN.)
With d = 0.03 m, t = 0.08 m:
A = π(0.03)(0.08) = 0.0075398 m^2.
If τ = 670 GPa = 670×10^9 N/m^2,
P = 670×10^9 × 0.0075398 = 5.05×10^9 N ≈ 5.05 GN (≈ 5050 MN).
(If the intended shear stress was 670 MPa = 670×10^6 N/m^2, then
P = 670×10^6 × 0.0075398 = 5.05×10^6 N ≈ 5.05 MN ≈ 5050 kN.)
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.