Step 4: Average shear stress

Compute the average shear stress. Match the correct pin to the numbers?
1: Average shear stress of pin A
2: Average shear stress of pin B

Question

Step 4: Average shear stress

Compute the average shear stress. Match the correct pin to the numbers?
1: Average shear stress of pin A
2: Average shear stress of pin B

BlackTom AI · Accepted Answer

First, restating the problem in my own words: In Step 4, you are asked to compute the average shear stress for two pins, A and B, and match them to the given numbers. The data given (or implied) are the shear forces and the cross-sectional areas for each pin.

Option evaluation:
- Option 1: 34 MPa for the average shear stress of pin A.
  To check this, compute (tau_A)avg = V_A / A_A. Here, the force on pin A is V_A = 10.68×10^3 N. The cross-sectional area is A_A = (π/4) d^2 with d = 0.02 m, so A_A = (π/4)(0.02)^2 ≈ 3.1416/4 × 4×10^-4 ≈ 3.1416×10^-4 m^2. Then tau_A,avg ≈ (10.68×10^3 N) / (3.1416×10^-4 m^2) ≈ 34.0×10^6 N/m^2 = 34 MPa. This matches the value given in the solution as 34.0 MPa, so this option is numerically consistent for pin A.
  If you examine the units, N/m^2 is Pa, and 34×10^6 Pa is 34 MPa, which aligns with the typical conversion used in these problems.

- Option 2: 17.7 MPa for the average shear stress of pin B.
  For pin B, use (tau_B)avg = V_B / A_B. The force on pin B is V_B = 12.5×10^3 N. The cross-sectional area is A_B = (π/4) d^2 with d = 0.03 m, so A_B = (π/4)(0.03)^2 ≈ (3.1416/4) × 9×10^-4 ≈ 7.0686×10^-4 m^2. Then tau_B,avg ≈ (12.5×10^3 N) / (7.0686×10^-4 m^2) ≈ 1.77×10^7 N/m^2 = 17.7 MPa. This value is also consistent with the calculation shown, so this option is numerically correct for pin B.

Key conceptual points:
- The average shear stress is computed as tau_avg = V / A, where V is the shear force applied to the pin and A is the cross-sectional area resisting the shear.
- The areas use the given diameters for each pin, with A = π/4 × d^2, so differing diameters lead to different areas and thus different stresses even if the forces are related.
- The units must be kept consistent: forces in newtons and areas in square meters yield stress in pascals (Pa), with MPa as a convenient larger unit (1 MPa = 10^6 Pa).

Potential pitfalls or common mistakes:
- Forgetting to convert the diameter to meters before computing A_A and A_B, which would ruin the stress values.
- Misstating the area formula or using the wrong diameter for each pin, leading to incorrect tau_avg values.
- Confusing the order of magnitude: a miscalculation could yield kPa or GPa if a decimal or exponent is mishandled.

In summary, the calculations shown produce tau_A,avg ≈ 34 MPa and tau_B,avg ≈ 17.7 MPa, consistent with the provided numerical results for the respective pins.

Step 4: Average shear stress Compute the average shear stress. Match the correct pin to the numbers? 1: Average shear stress of pin A 2: Average shear stress of pin B匹配题

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