Problem 8.49 8.49 Three forces are applied to the machine component ABD as shown. Knowing that the cross section containing point H is a 20 x 40-mm rectangle, determine the principal stresses and the maximum shearing stress at point H. Equivalent force-couple system at section containing point H: Fx = -3 kN, Fy = -0.5 kN, Fz = -2.5 kN Mx = 0, My = (0.150)(2500) = 375 N·m Mz = -(0.150)(500) = -75 N·m A = (20)(40) = 800 mm² = 800 x 10?? m² Iz = 1/12 (40)(20)³ = 26.667 x 10³ mm? = 26.667 x 10?? m? ?H = P/A - MyY/Iz = -3000/(800 x 10??) - (-75)(10 x 10?³)/(26.667 x 10??) = 24.375 MPa ?H = 3/2 |Vz|/A = 3/2 · 2500 / (800 x 10??) = 4.6875 MPa Use Mohr's circle. ?c = 1/2 ?H = 12.1875 MPa R = ?((24.375/2)² + (4.6875)²) = 13.0579 MPa ?a = ?c + R, ?a = 25.2 MPa ?b = ?c - R, ?b = -0.87 MPa tan 2?p = 2?H/?H = (2)(4.6875)/24.375 = 0.3846 ?a = 10.5°, ?b = 100.5° ?max = R, ?max = 13.06 MPa
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The question involves determining the principal stresses and the maximum shearing stress at point H in a machine component subjected to multiple forces. The formula for shear stress mentioned, Shear Stress = 3/2 V/A, seems to be a simplification or specific case Show more…
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Where did the formula Shear Stress = 3/2 V/A come from?
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