_______________________________________________________________________ PROBLEM (9.41) For a beam loaded as shown in Fig. P9.41, by taking into account the beam's own weight of 500 N/m, calculate the maximum normal and the maximum shearing stresses at point D of the cross section at midspan. Show the results on a properly oriented element. SOLUTION ∑M ∑F y A = 0 : R B ( 2) − 0.5( 2)1 − 20(0.5) = 0 , = 0: R B = 5.5 kN R A = 15.5 kN y 20kN 0.5 kN/m 80 A 15.5 kN C 0.5 m z B 1.5 m 5.5 kN 160 D 50 80 V, kN 15.5 15.25 -4.75 1m Vm=5 Mm = M, kN ⋅ m -5.5 5.5+ 5 2 x Dimensions are in millimeters (1)=5.25 x 1 (80)(160)3 = 27.31×106 mm 4 12 Q = 80 × 50 × 55 = 220 × 10 3 mm I= τD = σD = VQ 5 ×103 (220 ×10−6 ) = = 0.504 MPa Ib 27.31× 10−6 (0.08) Mc 5.25 ×103 (0.03) = = 5.77 MPa I 27.31×10−6 τ (MPa) τD D σD 2.885 R = (2.885) 2 + (0.504) 2 = 2.93 MPa σ2 σ1 R O C 2θ ’ p σ (psi) (5.77, -0.504) 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 max = 2.93 MPa σ 1 = 2.93 + 2.885 = 5.82 MPa σ 2 = −2.93 + 2.885 = −0.05 MPa 0.05 MPa and tan 2θ p ' = θ p ' = 4.95 0.504 2.885 2.93 MPa D 5.82 MPa 4.95o x 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.