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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 41 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •1–57. The specimen failed in a tension test at an angle of 52° when the axial load was 19.80 kip. If the diameter of the specimen is 0.5 in., determine the average normal and average shear stress acting on the area of the inclined failure plane. Also, what is the average normal stress acting on the cross section when failure occurs? 52 0.5 in. + b © F x = 0; V - 19.80 cos 52° = 0 V = 12.19 kip +a © F y = 0; N - 19.80 sin 52° = 0 N = 15.603 kip Inclined plane: s¿ = P A t œ avg = V A ; tœ avg = 12.19 p(0.25) 2 Cross section: ; s¿ =15.603 p(0.25) 2 sin 52° s = P A ; s = 19.80 = 101 ksi 2 p(0.25) t avg = V A ; t avg = 0 sin 52° = 62.6 ksi = 48.9 ksi Ans. Ans. Ans. Ans. 1–58. The anchor bolt was pulled out of the concrete wall and the failure surface formed part of a frustum and cylinder. This indicates a shear failure occurred along the cylinder BC and tension failure along the frustum AB. If the shear and normal stresses along these surfaces have the magnitudes shown, determine the force P that must have been applied to the bolt. A 45 P 45 Average Normal Stress: For the frustum, A = 2pxL = 2p(0.025 + 0.025)A 2 0.05 2 + 0.05 2 B = 0.02221 m 2 50 mm 3 MPa 3 MPa B 30 mm 4.5 MPa s = P A ; 3A106 B = F 1 = 66.64 kN F 1 0.02221 C 25 mm 25 mm Average Shear Stress: For the cylinder, A = p(0.05)(0.03) = 0.004712 m 2 t avg = V A ; F 2 4.5A106 B = 0.004712 Equation of Equilibrium: F 2 = 21.21 kN + c ©F y = 0; P - 21.21 - 66.64 sin 45° = 0 P = 68.3 kN Ans. 41

01 Solutions 46060 5/6/10 2:43 PM Page 42 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–59. The open square butt joint is used to transmit a force of 50 kip from one plate to the other. Determine the average normal and average shear stress components that this loading creates on the face of the weld, section AB. 50 kip 30 30 Equations of Equilibrium: a + © F y = 0; N - 50 cos 30° = 0 N = 43.30 kip + Q© F x = 0; -V + 50 sin 30° = 0 V = 25.0 kip A B 6 in. 2 in. 50 kip Average Normal and Shear Stress: 2 A¿ =a b(6) = 13.86 in2 sin 60° s = N = 43.30 = 3.125 ksi A¿ 13.86 t avg = V = 25.0 = 1.80 ksi A¿ 13.86 Ans. Ans. *1–60. If P = 20 kN, determine the average shear stress developed in the pins at A and C. The pins are subjected to double shear as shown, and each has a diameter of 18 mm. C Referring to the FBD of member AB,Fig.a a+©M A = 0; F BC sin 30° (6) - 20(2) - 20(4) = 0 F BC = 40 kN : + ©F x = 0; A x - 40 cos 30° = 0 A x = 34.64 kN A 2 m 2 m 30 2 m B + c ©F y = 0; A y - 20 - 20 + 40 sin 30° A y = 20 kN Thus, the force acting on pin A is F A = 2 A x 2 + A y 2 = 2 34.64 2 + 20 2 = 40 kN Pins A and C are subjected to double shear. Referring to their FBDs in Figs. b and c, V A = F A 2 = 40 2 = 20 kN V C = F BC = 40 = 20 kN 2 2 = 81(10 - 6 )p m 2 . Thus The cross-sectional area of Pins A and C are A A = A C = p 4 (0.0182 ) t A = V A A A = 20(103 ) 81(10 - 6 )p = 78.59(106 ) Pa = 78.6 MPa t C = V C A C = 20(103 ) 81(10 - 6 )p = 78.59(106 ) Pa = 78.6 MPa Ans. Ans. P P 42

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