Shear Stress

Example: (continued)

ΣM_C = 0: (R cosβ)c(cosθ) – (R sinβ)c(sinθ) – (P cosα)d(sinθ) – (P sinα)d( cosθ) = 0

Solve for R. R = P (d/c) [ (cosα sinθ + sinα cosθ) / (cosβ cosθ – sinβ sinθ ) ]

Use trig identities. R = P (d/c) [ sin(α+θ) / cos(β+θ) ]

→ ΣF_x = 0: – R sinβ – C_x + P cosα = 0

↑ ΣF_y = 0: – R cosβ + C_y – P sinα = 0

Solve for C_x and C_y.

C_x = P cosα – R sinβ , P cosα – P (d/c) [ sin(α+θ) / cos(β+θ) ] sinβ

C_y = P sinα + R cosβ , P sinα + P (d/c) [ sin(α+θ) / cos(β+θ) ] cosβ

Note: Both C_x and C_y check dimensionally.

The resultant force on the pin is √ ( C_x² + C_y² )

The cross sectional area of the pin is: π e² / 4 = 0.196 in²

So the average shear stress in the pin, S, is S = 4 √ ( C_x² + C_y² ) / π e²

For the given data: C_x = -125.6 lb C_y = 1502.8 lb S = 1508 lb

S = 7681 psi = 7.7 ksi (result)

Note: With the symbolic expressions “what if” testing can show how the result for

average shear stress depends on the variables such as c and d or others.

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