_______________________________________________________________________ PROBLEM (*9.66) Loads P and R are applied at the free end of the post having a diameter d shown in Fig. P9.66. Calculate the principal stresses and the maximum shearing stress: (a) At point A. (b) At point B. Indicate the results on properly oriented elements. R = 800π ⋅ N, d = 50 mm Given: P = 400π ⋅ N, *SOLUTION x T=0.4P R C My=0.4R d = 50 mm P = 400π N R = 800π N B A Mz=0.6P P y z 16T 16(0.4 × 400π ) = 20.48 MPa = πd3 π (0.05)3 800π 32(0.6 × 400π ) R M y σx = + z = − − 2 A I π (0.025) π (0.05)3 = −1.28 − 61.44 = −62.7 MPa (a) τ t = 62.7 62.7 2 ) + (20.48) 2 ± (− 2 2 = −31.35 ± 37.45 σ 1 = 6.1 MPa σ 2 = −68.8 MPa τ max = 37.5 MPa 2τ 2(20.48) = θ p ' = −16.6o tan 2θ p ' = , σx −62.7 σ 1,2 = − x A τt σx Continued on next slide Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. τ (MPa) 68.8 MPa x σ2 2θ ' p 16.6o y σ1 C (−62.7, −20.48 ) σ (MPa) A 6.1 MPa 37.5 MPa (b) x σx = R Myz + A I B τ = τt +τ d VQ 4 V 4P = = (Example 9.4) Ib 3 A 3π r 2 4 400π = = 0.85 MPa 3 π (0.025) 2 and τ = τ t + τ d = 20.48 + 0.85 = 21.33 MPa τd = Similarly, 32(0.4 × 800π ) π (0.05)3 = −1.28 − 81.92 = −83.2 MPa σ x = −1.28 − Continued on next slide Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 48.4 MPa Thus, 83.2 83.2 2 ) + (21.33) 2 m (− 2 2 = −41.6 m 46.75 σ 1 = 5.15 MPa σ 2 = −88.4 MPa σ 1,2 = − 13.6o B 5.15 MPa 46.8 MPa τ max = 46.8 MPa 2τ 2(21.33) = , tan 2θ p ' = σx −83.2 θ p ' = −13.6o Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.