© 2014 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. 9–57. Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress. Specify the orientation of the element in each case. 50 MPa 30 MPa A(0, -30) B(50, 30) C(25, 0) R = CA = 2(25 - 0)2 + 302 = 39.05 s1 = 25 + 39.05 = 64.1 MPa Ans. s2 = 25 - 39.05 = - 14.1 MPa Ans. tan 2up = 30 = 1.2 25 - 0 up2 = 25.1° Ans. savg = 25.0 MPa Ans. tmax = R = 39.1 MPa Ans. in-plane tan 2us = 25 - 0 = 0.8333 30 us = - 19.9° Ans. Ans: s1 = 64.1 MPa, s2 = - 14.1 MPa, up2 = 25.1° savg = 25.0 MPa, tmax = 39.1 MPa, in-plane us = - 19.9° 901 © 2014 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. 9–58. Determine the equivalent state of stress if an element is oriented 25° counterclockwise from the element shown. 550 MPa A(0, -550) B(0, 550) C(0, 0) R = CA = CB = 550 sx¿ = - 550 sin 50° = - 421 MPa Ans. tx¿y¿ = - 550 cos 50° = - 354 MPa Ans. sy¿ = 550 sin 50° = 421 MPa Ans. Ans: sx¿ = - 421 MPa, tx¿y¿ = - 354 MPa, sy¿ = 421 MPa 902 © 2014 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. 8 ksi 9–59. Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress. Specify the orientation of the element in each case. 4 ksi 12 ksi A(-12, 4) B( - 8, - 4) C( -10, 0) R = CA = CB = 222 + 42 = 4.472 a) s1 = - 10 + 4.472 = - 5.53 ksi Ans. s2 = - 10 - 4.472 = - 14.5 ksi Ans. tan 2up = 4 2 2up = 63.43° up = - 31.7° b) tmax = R = 4.47 ksi Ans. in-plane savg = - 10 ksi Ans. 2us = 90 - 2up us = 13.3° Ans. Ans: (a) s1 = - 5.53 ksi, s2 = - 14.5 ksi, up = - 31.7° (b) tmax = 4.47 ksi, savg = - 10 ksi, in-plane us = 13.3° 903 © 2014 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. 200 MPa *9–60. Determine the principal stresses, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. 500 MPa 350 MPa Construction of the Circle: In accordance with the sign convention, sx = 350 MPa, sy = - 200 MPa, and txy = 500 MPa. Hence, savg = sx + sy 2 = 350 + ( -200) = 75.0 MPa 2 Ans. The coordinates for reference point A and C are A(350, 500) C(75.0, 0) The radius of the circle is R = 2(350 - 75.0)2 + 5002 = 570.64 MPa a) In-Plane Principal Stresses: The coordinate of points B and D represent s1 and s2 respectively. s1 = 75.0 + 570.64 = 646 MPa Ans. s2 = 75.0 - 570.64 = - 496 MPa Ans. Orientaion of Principal Plane: From the circle tan 2uP1 = 500 = 1.82 350 - 75.0 uP1 = 30.6° (Counterclockwise) Ans. b) Maximum In-Plane Shear Stress: Represented by the coordinates of point E on the circle. t max in-plane = R = 571 MPa Ans. Orientation of the Plane for Maximum In-Plane Shear Stress: From the circle tan 2us = 350 - 75.0 = 0.55 500 us = 14.4° (Clockwise) Ans. 904 © 2014 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. 9–61. Draw Mohr’s circle that describes each of the following states of stress. 5 MPa 20 ksi 20 ksi 5 MPa (b) (a) 18 MPa (c) 905 © 2014 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. 9–62. The grains of wood in the board make an angle of 20° with the horizontal as shown. Using Mohr’s circle, determine the normal and shear stresses that act perpendicular and parallel to the grains if the board is subjected to an axial load of 250 N. sx = 300 mm 60 mm 250 N 250 N 20⬚ 25 mm 250 P = = 166.67 kPa A (0.06)(0.025) R = 83.33 Coordinates of point B: sx¿ = 83.33 - 83.33 cos 40° sx¿ = 19.5 kPa Ans. tx¿y¿ = - 83.33 sin 40° = - 53.6 kPa Ans. Ans: sx¿ = 19.5 kPa, tx¿y¿ = - 53.6 kPa 906 © 2014 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. z 9–63. The post has a square cross-sectional area. If it is fixed supported at its base and a horizontal force is applied at its end as shown, determine (a) the maximum in-plane shear stress developed at A and (b) the principal stresses at A. 3 in. 3 in. 60 lb y x 18 in. A 1 in. 1 I = (3)(33) = 6.75 in4 12 sA = tA = Myx I = 1080(0.5) = - 80 psi 6.75 = 60(3) = 8.889 psi 6.75(3) VyQA It A(-80, 8.889) tmax QA = (1)(1)(3) = 3 in3 B(0, - 8.889) C( - 40, 0) = R = 2402 + 8.8892 = 41.0 psi Ans. in-plane s1 = - 40 + 40.9757 = 0.976 psi Ans. s2 = - 40 - 40.9757 = - 81.0 psi Ans. Ans: tmax in-plane = 41.0 psi , s1 = 0.976 psi , s2 = - 81.0 psi 907 © 2014 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. *9–64. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. 20 MPa 80 MPa 30 MPa In accordance to the established sign convention, sx = 30 MPa, sy = - 20 MPa and txy = 80 MPa. Thus, savg = sx + sy 30 + ( -20) = 5 MPa 2 = 2 Then, the coordinates of reference point A and the center C of the circle is A(30, 80) C(5, 0) Thus, the radius of circle is given by R = CA = 2(30 - 5)2 + (80 - 0)2 = 83.815 MPa Using these results, the circle shown in Fig. a, can be constructed. The coordinates of points B and D represent s1 and s2 respectively. Thus s1 = 5 + 83.815 = 88.8 MPa Ans. s2 = 5 - 83.815 = - 78.8 MPa Ans. Referring to the geometry of the circle, Fig. a tan 2(uP)1 = 80 = 3.20 30 - 5 uP = 36.3° (Counterclockwise) Ans. The state of maximum in-plane shear stress is represented by the coordinate of point E. Thus tmax = R = 83.8 MPa Ans. in-plane From the geometry of the circle, Fig. a, tan 2us = 30 - 5 = 0.3125 80 us = 8.68° (Clockwise) Ans. The state of maximum in-plane shear stress is represented by the element in Fig. c. 908 © 2014 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. 9–64. Continued 909 © 2014 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. 9–65. The thin-walled pipe has an inner diameter of 0.5 in. and a thickness of 0.025 in. If it is subjected to an internal pressure of 500 psi and the axial tension and torsional loadings shown, determine the principal stress at a point on the surface of the pipe. 200 lb 200 lb 20 lb⭈ft 20 lb⭈ft Section Properties: A = p A 0.2752 - 0.252 B = 0.013125p in2 J = p A 0.2754 - 0.254 B = 2.84768 A 10 - 3 B in4 2 Normal Stress: Since 0.25 r = = 10, thin wall analysis is valid. t 0.025 slong = pr 500(0.25) N 200 + = + = 7.350 ksi A 2t 0.013125p 2(0.025) shoop = pr 500(0.25) = = 5.00 ksi t 0.025 Shear Stress: Applying the torsion formula, t = 20(12)(0.275) Tc = 23.18 ksi = J 2.84768(10 - 3) Construction of the Circle: In accordance with the sign convention sx = 7.350 ksi, sy = 5.00 ksi, and txy = - 23.18 ksi. Hence, savg = sx + sy 2 = 7.350 + 5.00 = 6.175 ksi 2 The coordinates for reference points A and C are A(7.350, -23.18) C(6.175, 0) The radius of the circle is R = 2(7.350 - 6.175)2 + 23.182 = 23.2065 ksi In-Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 6.175 + 23.2065 = 29.4 ksi Ans. s2 = 6.175 - 23.2065 = - 17.0 ksi Ans. Ans: s1 = 29.4 ksi , s2 = - 17.0 ksi 910 © 2014 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. 9–66. Determine the principal stress and maximum in-plane shear stress at point A on the cross section of the pipe at section a–a. a 300 mm a 300 mm b 30 mm A B 200 mm 300 N Internal Loadings: Considering the equilibrium of the free-body diagram of the assembly’s segment, Fig. a, ©Fx = 0; N - 450 = 0 N = 450 N ©Fy = 0; Vy = 0 ©Fz = 0; Vz + 300 = 0 Vz = - 300 N ©Mx = 0; T + 300(0.2) = 0 T = - 60 N # m ©My = 0; My - 450(0.3) + 300(0.3) = 0 My = 45 N # m ©Mz = 0; Mz + 450(0.2) = 0 Mz = - 90 N # m Section Properties: The cross-sectional area, the moment of inertia about the y and z axes, and the polar moment of inertia of the pipe’s cross section are A = p(0.032 - 0.022) = 0.5p(10-3) m2 Iy = Iz = J = p (0.034 - 0.024) = 0.1625p(10-6) m4 4 p (0.034 - 0.024) = 0.325p(10-6) m4 2 Referring to Fig. b, (Qy)A = 0 (Qz)A = 4(0.03) p 4(0.02) p c (0.03)2 d c (0.022) d = 12.667(10-6) m3 3p 2 3p 2 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. sA = MzyA ( - 90)(- 0.03) N 450 = -3 A Iz 0.5p(10 ) 0.1625p(10-6) = - 5.002 MPa Since Vy = 0, [(txy)V]A = 0. However, the shear stress is the combination of torsional and transverse shear stress. Thus, (txz)A = [(txz)T]A - [(txz)V]A = Vz(Qz)A 60(0.03) 300[12.667(10-6)] Tc = 1.391 MPa = -6 J Iy t 0.325p(10 ) 0.1625p(10-6)(0.02) The state of stress at point A is represented by the element shown in Fig. c. 911 450 N 20 mm Section a – a © 2014 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. 9–66. Continued Construction of the Circle: sx = - 5.002 MPa, sz = 0, and txz = 1.391 MPa. Thus, savg = sx + sz 2 = - 5.002 + 0 = - 2.501 MPa 2 The coordinates of reference point A and the center C of the circle are A(- 5.002, 1.391) C(- 2.501, 0) Thus, the radius of the circle is R = CA = 2[- 5.002 - ( -2.501)]2 + 1.3912 = 2.862 MPa Using these results, the circle is shown in Fig. d. In-Plane Principal Stresses: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = - 2.501 + 2.862 = 0.361 MPa Ans. s2 = - 2.501 - 2.862 = - 5.36 MPa Ans. In-Plane Maximum Shear Stress: The coordinates of point E represent the state of maximum shear stress. Thus, tmax = |R| = 2.86 MPa Ans. in-plane Ans: s1 = 0.361 MPa , s2 = - 5.36 MPa, tmax = 2.86 MPa in-plane 912 © 2014 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. 9–67. Determine the principal stress and maximum in-plane shear stress at point B on the cross section of the pipe at section a–a. a 300 mm a 300 mm b 30 mm A B 200 mm 300 N Internal Loadings: Considering the equilibrium of the free-body diagram of the assembly’s cut segment, Fig. a, ©Fx = 0; N - 450 = 0 ©Fy = 0; Vy = 0 ©Fz = 0; Vz + 300 = 0 Vz = - 300 N ©Mx = 0; T + 300(0.2) = 0 T = - 60 N # m ©My = 0; My - 450(0.3) + 300(0.3) = 0 My = 45 N # m ©Mz = 0; Mz + 450(0.2) = 0 Mz = - 90 N # m N = 450 N Section Properties: The cross-sectional area, the moment of inertia about the y and z axes, and the polar moment of inertia of the pipe’s cross section are A = p (0.032 - 0.022) = 0.5p (10 - 3) m2 Iy = Iz = J = p (0.034 - 0.024) = 0.1625p(10 - 6) m4 4 p (0.034 - 0.024) = 0.325p(10 - 6) m4 2 Referring to Fig. b, (Qz)B = 0 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. sB = MyzB 45( -0.03) N 450 + = + -3 A ly 0.5p(10 ) 0.1625p(10 - 6) = - 2.358 MPa Since (Qz)B = 0, (txy)B = 0. Also Vy = 0. Then the shear stress along the y axis is contributed by torsional shear stress only. (txy)B = [(txy)T]B = 60(0.03) Tc = 1.763 MPa = J 0.325p(10 - 6) The state of stress at point B is represented on the two-dimensional element shown in Fig. c. 913 450 N 20 mm Section a – a © 2014 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. 9–67. Continued Construction of the Circle: sx = - 2.358 MPa, sy = 0, and txy = - 1.763 MPa . Thus, sx + sy -2.358 + 0 = = - 1.179 MPa savg = 2 2 The coordinates of reference point A and the center C of the circle are A( - 2.358, - 1.763) C( -1.179, 0) Thus, the radius of the circle is R = CA = 2[ -2.358 - ( -1.179)]2 + (- 1.763)2 = 2.121 MPa Using these results, the circle is shown in Fig. d. In-Plane Principal Stresses: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = - 1.179 + 2.121 = 0.942 MPa Ans. s2 = - 1.179 - 2.121 = - 3.30 MPa Ans. Maximum In-Plane Shear Stress: The coordinates of point E represent the state of maximum in-plane shear stress. Thus, tmax = |R| = 2.12 MPa Ans. in-plane Ans: s1 = 0.942 MPa, s2 = - 3.30 MPa, tmax = 2.12 MPa in-plane 914 © 2014 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. *9–68. The rotor shaft of the helicopter is subjected to the tensile force and torque shown when the rotor blades provide the lifting force to suspend the helicopter at midair. If the shaft has a diameter of 6 in., determine the principal stress and maximum in-plane shear stress at a point located on the surface of the shaft. 50 kip 10 kip⭈ft Internal Loadings: Considering the equilibrium of the free-body diagram of the rotor shaft’s upper segment, Fig. a, ©Fy = 0; N - 50 = 0 N = 50 kip ©My = 0; T - 10 = 0 T = 10 kip # ft Section Properties: The cross-sectional area and the polar moment of inertia of the rotor shaft’s cross section are A = p(32) = 9p m2 J = p 4 (3 ) = 40.5p in4 2 Normal and Shear Stress: The normal stress is contributed by axial stress only. sA = 50 N = = 1.768 ksi A 9p The shear stress is contributed by the torsional shear stress only. tA = 10(12)(3) Tc = = 2.829 ksi J 40.5p The state of stress at point A is represented by the element shown in Fig. b. 915 © 2014 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. 9–68. Continued Construction of the Circle: sx = 0, sy = 1.768 ksi, and txy = 2.829 ksi. Thus, savg = sx + sy 2 = 0 + 1.768 = 0.8842 ksi 2 The coordinates of reference point A and the center C of the circle are A(0, 2.829) C(0.8842, 0) Thus, the radius of the circle is R = CA = 2(0 - 0.8842)2 + 2.8292 = 2.964 ksi Using these results, the circle is shown in Fig. c. In-Plane Principal Stress: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = 0.8842 + 2.964 = 3.85 ksi Ans. s2 = 0.8842 - 2.964 = - 2.08 ksi Ans. Maximum In-Plane Shear Stress: The state of maximum shear stress is represented by the coordinates of point E, Fig. a. tmax = R = 2.96 ksi Ans. in-plane 916 © 2014 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. *9–69. The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not rotate while subjected to a force of 75 lb, determine the principal stress in the material on the cross section at point C. 75 lb B 3 in. A 4 in. C 0.4 in. 0.4 in. 0.2 in. 0.3 in. Internal Forces and Moment: As shown on FBD Section Properties: I = 1 (0.3) A 0.83 B = 0.0128 in3 12 QC = y¿A¿ = 0.3(0.2)(0.3) = 0.0180 in3 Normal Stress: Applying the flexure formula. sC = - - 300(0.2) My = = 4687.5 psi = 4.6875 ksi I 0.0128 Shear Stress: Applying the shear formula. tC = VQC 75.0(0.0180) = = 351.6 psi = 0.3516 ksi It 0.0128(0.3) Construction of the Circle: In accordance with the sign convention, sx = 4.6875 ksi, sy = 0, and txy = 0.3516 ksi. Hence, savg = sx + sy 2 = 4.6875 + 0 = 2.34375 ksi 2 The coordinates for reference points A and C are A(4.6875, 0.3516) C(2.34375, 0) The radius of the circle is R = 2(4.6875 - 2.34375)2 + 0.35162 = 2.370 ksi In-Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 2.34375 + 2.370 = 4.71 ksi Ans. s2 = 2.34375 - 2.370 = - 0.0262 ksi Ans. Ans: s1 = 4.71 ksi, s2 = - 0.0262 ksi 917 © 2014 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. 9–70. A spherical pressure vessel has an inner radius of 5 ft and a wall thickness of 0.5 in. Draw Mohr’s circle for the state of stress at a point on the vessel and explain the significance of the result. The vessel is subjected to an internal pressure of 80 psi. 45⬚ 1.25 m Normal Stress: s1 = s2 = pr 80(5)(12) = = 4.80 ksi 2t 2(0.5) Mohr’s circle: A(4.80, 0) B(4.80, 0) C(4.80, 0) Regardless of the orientation of the element, the shear stress is zero and the state of stress is represented by the same two normal stress components. Ans. Ans: Regardless of the orientation of the element, the shear stress is zero and the state of stress is represented by the same two normal stress components. 918 © 2014 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. 9–71. The cylindrical pressure vessel has an inner radius of 1.25 m and a wall thickness of 15 mm. It is made from steel plates that are welded along the 45° seam. Determine the normal and shear stress components along this seam if the vessel is subjected to an internal pressure of 8 MPa. sx = 45⬚ 1.25 m pr 8(1.25) = = 333.33 MPa 2t 2(0.015) sy = 2sx = 666.67 MPa A(333.33, 0) B(666.67, 0) C(500, 0) 333.33 + 666.67 = 500 MPa 2 Ans. tx¿y¿ = - R = 500 - 666.67 = - 167 MPa Ans. sx¿ = Ans: sx¿ = 500 MPa, tx¿y¿ = - 167 MPa 919 © 2014 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. *9–72. Determine the normal and shear stresses at point D that act perpendicular and parallel, respectively, to the grains. The grains at this point make an angle of 30° with the horizontal as shown. Point D is located just to the left of the 10-kN force. 10 kN A 100 mm D B 30⬚ 1m 100 mm D 100 mm Using the method of section and consider the FBD of the left cut segment, Fig. a + c ©Fy = 0; 5 - V = 0 a + ©MC = 0; V = 5 kN M = 5 kN # m M - 5(1) = 0 The moment of inertia of the rectangular cross - section about the neutral axis is I = 1 (0.1)(0.33) = 0.225(10 - 3) m4 12 Referring to Fig. b, QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3 The normal stress developed is contributed by bending stress only. For point D, y = 0.05 m. Then s = My 5(103)(0.05) = 1.111 MPa (T) = I 0.225(10 - 3) The shear stress is contributed by the transverse shear stress only. Thus, t = 5(103)(0.001) VQD = 0.2222 MPa = It 0.225(10 - 3)(0.1) The state of stress at point D can be represented by the element shown in Fig. c In accordance with the established sign convention, sx = 1.111 MPa, sy = 0 and txy = - 0.2222 MPa, Thus. savg = sx + sy 2 = 1.111 + 0 = 0.5556 MPa 2 Then, the coordinate of reference point A and the center C of the circle are A(1.111, -0.2222) C(0.5556, 0) Thus, the radius of the circle is given by R = 2(1.111 - 0.5556)2 + (- 0.2222)2 = 0.5984 MPa Using these results, the circle shown in Fig. d can be constructed. Referring to the geometry of the circle, Fig. d, a = tan - 1 a 0.2222 b = 21.80° 1.111 - 0.5556 b = 180° - (120° - 21.80°) = 81.80° 920 1m 300 mm 2m C © 2014 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. 9–72. Continued Then sx¿ = 0.5556 - 0.5984 cos 81.80° = 0.4702 MPa = 470 kPa Ans. tx¿y¿ = 0.5984 sin 81.80° = 0.5922 MPa = 592 kPa Ans. 921 © 2014 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. 9–73. Determine the principal stress at point D, which is located just to the left of the 10-kN force. 10 kN A 100 mm D B 30⬚ 1m 100 mm D 100 mm Using the method of section and consider the FBD of the left cut segment, Fig. a, + c ©Fy = 0; 5 - V = 0 a + ©MC = 0; V = 5 kN M = 5 kN # m M - 5(1) = 0 I = 1 (0.1)(0.33) = 0.225(10 - 3) m4 12 Referring to Fig. b, QD = y¿A¿ = 0.1(0.1)(0.1) = 0.001 m3 The normal stress developed is contributed by bending stress only. For point D, y = 0.05 m s = My 5(103)(0.05) = 1.111 MPa (T) = I 0.225(10 - 3) The shear stress is contributed by the transverse shear stress only. Thus, t = 5(103)(0.001) VQD = 0.2222 MPa = It 0.225(10 - 3)(0.1) The state of stress at point D can be represented by the element shown in Fig. c. In accordance with the established sign convention, sx = 1.111 MPa, sy = 0, and txy = - 0.2222 MPa. Thus, savg = sx + sy 2 = 1.111 + 0 = 0.5556 MPa 2 Then, the coordinate of reference point A and center C of the circle are A(1.111, - 0.2222) C(0.5556, 0) Thus, the radius of the circle is R = CA = 2(1.111 - 0.5556)2 + (- 0.2222)2 = 0.5984 MPa Using these results, the circle shown in Fig. d can be constructed. In-Plane Principal Stresses. The coordinates of points B and D represent s1 and s2, respectively. Thus, s1 = 0.5556 + 0.5984 = 1.15 MPa Ans. s2 = 0.5556 - 0.5984 = - 0.0428 MPa Ans. 922 1m 300 mm 2m C © 2014 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. 9–73. Continued Referring to the geometry of the circle, Fig. d, tan (2uP)1 = 0.2222 = 0.4 1.111 - 0.5556 (uP)1 = 10.9° (Clockwise) Ans. The state of principal stresses is represented by the element shown in Fig. e. Ans: s1 = 1.15 MPa, s2 = - 0.0428 MPa 923 © 2014 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. 9–74. If the box wrench is subjected to the 50 lb force, determine the principal stress and maximum in-plane shear stress at point A on the cross section of the wrench at section a–a. Specify the orientation of these states of stress and indicate the results on elements at the point. 12 in. 50 lb 0.5 in. 2 in. a a A B Section a – a Internal Loadings: Considering the equilibrium of the free-body diagram of the wrench’s segment, Fig. a, ©Fy = 0; Vy + 50 = 0 Vy = - 50 lb ©Mx = 0; T + 50(12) = 0 T = - 600 lb # in ©Mz = 0; Mz - 50(2) = 0 Mz = 100 lb # in Section Properties: The moment of inertia about the z axis and the polar moment of inertia of the wrench’s cross section are Iz = p (0.54) = 0.015625p in4 4 J = p (0.54) = 0.03125p in4 2 Referring to Fig. b, (Qy)A = y¿A¿ = 4(0.5) p c (0.52) d = 0.08333 in3 3p 2 Normal and Shear Stress: The shear stress of point A along the z axis is (txz)A = 0. However, the shear stress along the y axis is a combination of torsional and transverse shear stress. (txy)A = [(txy)T]A - [(txy)V]A = Vy(Qy)A Tc + J lzt = 600(0.5) -50(0.08333) + = 2.971 ksi 0.03125p 0.015625p(l) The state of stress at point A is represented by the two-dimensional element shown in Fig. c. 924 © 2014 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. 9–74. Continued Construction of the Circle: sx = sy = 0, and txy = 2.971 ksi. Thus, savg = sx + sy 2 = 0 + 0 = 0 2 The coordinates of reference point A and the center C of the circle are A(0, 2.971) C(0, 0) Thus, the radius of the circle is R = CA = 2(0 - 0)2 + 2.9712 = 2.971 ksi Using these results, the circle is shown in Fig. d. In-Plane Principal Stress: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = 0 + 2.971 = 2.97 ksi Ans. s2 = 0 - 2.971 = - 2.97 ksi Ans. Maximum In-Plane Shear Stress: Since there is no normal stress acting on the element, tmax in-plane = (txy)A = 2.97 ksi Ans. Ans: s1 = 2.97 ksi, s2 = - 2.97 ksi, up1 = 45.0°, up2 = - 45.0°, tmax = 2.97 ksi, us = 0° in-plane 925 © 2014 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. 9–75. If the box wrench is subjected to the 50 lb force, determine the principal stress and maximum in-plane shear stress at point B on the cross section of the wrench at section a–a. Specify the orientation of these states of stress and indicate the results on elements at the point. 12 in. 50 lb 0.5 in. 2 in. a a A B Section a – a Internal Loadings: Considering the equilibrium of the free-body diagram of the wrench’s cut segment, Fig. a, ©Fy = 0; Vy + 50 = 0 Vy = - 50 lb ©Mx = 0; T + 50(12) = 0 T = - 600 lb # in ©Mz = 0; Mz - 50(2) = 0 Mz = 100 lb # in Section Properties: The moment of inertia about the z axis and the polar moment of inertia of the wrench’s cross section are Iz = p (0.54) = 0.015625p in4 4 J = p (0.54) = 0.03125p in4 2 Referring to Fig. b, (Qy)B = 0 Normal and Shear Stress: The normal stress is caused by the bending stress due to Mz. (sx)B = - MzyB Iz = - 100(0.5) = - 1.019 ksi 0.015625p The shear stress at point B along the y axis is (txy)B = 0 since (Qy)B. However, the shear stress along the z axis is caused by torsion. (txz)B = 600(0.5) Tc = = 3.056 ksi J 0.03125p The state of stress at point B is represented by the two-dimensional element shown in Fig. c. 926 © 2014 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. 9–75. Continued Construction of the Circle: sx = - 1.019 ksi, sz = 0, and txz = - 3.056 ksi . Thus, savg = sx + sy 2 = - 1.019 + 0 = - 0.5093 ksi 2 The coordinates of reference point A and the center C of the circle are A( - 1.019, - 3.056) C( - 0.5093, 0) Thus, the radius of the circle is R = CA = 2 [- 1.019 - ( -0.5093)]2 + ( -3.056)2 = 3.0979 ksi Using these results, the circle is shown in Fig. d. In-Plane Principal Stress: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = - 0.5093 + 3.0979 = 2.59 ksi Ans. s2 = - 0.5093 - 3.0979 = - 3.61 ksi Ans. Maximum In-Plane Shear Stress: The coordinates of point E represent the maximum in-plane stress, Fig. a. tmax = R = 3.10 ksi Ans. in-plane Ans: s1 = 2.59 ksi, s2 = - 3.61 ksi, up1 = - 40.3°, up2 = 49.7, tmax = 3.10 ksi , us = 4.73° in-plane 927 © 2014 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. *9–76. The ladder is supported on the rough surface at A and by a smooth wall at B. If a man weighing 150 lb stands upright at C, determine the principal stresses in one of the legs at point D. Each leg is made from a 1-in.-thick board having a rectangular cross section. Assume that the total weight of the man is exerted vertically on the rung at C and is shared equally by each of the ladder’s two legs. Neglect the weight of the ladder and the forces developed by the man’s arms. B C 12 ft 3 in. 1 in. D 3 in. 1 in. 1 in. 5 ft D 4 ft A A = 3(1) = 3 in2 I = 1 (1)(33) = 2.25 in4 12 QD = y¿A¿ = (1)(1)(1) = 1 in3 sD = My 35.52(12)(0.5) -P - 77.55 = = - 120.570 psi A I 3 2.25 tD = 8.88(1) VQD = = 3.947 psi It 2.25(1) A( -120.57, - 3.947) B(0, 3.947) C( -60.285, 0) R = 2(60.285)2 + (3.947)2 = 60.412 s1 = - 60.285 + 60.4125 = 0.129 psi Ans. s2 = - 60.285 - 60.4125 = - 121 psi Ans. 928 5 ft © 2014 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. 9–77. Draw the three Mohr’s circles that describe each of the following states of stress. 5 ksi (a) Here, smin = 0, sint = 3 ksi and smax = 5 ksi. The three Mohr’s circles of this state of stress are shown in Fig. a 3 ksi (b) Here, smin = 0, sint = 140 MPa and smax = 180 MPa. The three Mohr’s circles of this state of stress are shown in Fig. b (a) 929 180 MPa 140 MPa (b) © 2014 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. 9–78. Draw the three Mohr’s circles that describe the following state of stress. 300 psi Here, smin = - 300 psi, sint = 0 and smax = 400 psi. The three Mohr’s circles for this state of stress are shown in Fig. a. 400 psi 930 © 2014 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. z 9–79. The stress at a point is shown on the element. Determine the principal stresses and the absolute maximum shear stress. x y 90 MPa 20 MPa 60 MPa For y – z plane: The center of the cricle is at sAvg = sy + sz 2 = 60 + 90 = 75 MPa 2 R = 2(75 - 60)2 + (- 20)2 = 25 MPa s1 = 75 + 25 = 100 MPa s2 = 75 - 25 = 50 MPa Thus, tabs max = s1 = 100 Ans. s2 = 50 MPa Ans. s3 = 0 MPa Ans. smax - smin 100 - 0 = = 50 MPa 2 2 Ans. Ans: s1 = 100 MPa, s2 = 50 MPa, s3 = 0, tabs = 50 MPa max 931 © 2014 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. z *9–80. The stress at a point is shown on the element. Determine the principal stresses and the absolute maximum shear stress. x Mohr’s circle for the element in the y–z plane, Fig. a, will be drawn first. In accordance with the established sign convention, sy = 30 psi, sz = 120 psi and tyz = 70 psi. Thus savg = sy + sz 2 = 30 + 120 = 75 psi 2 C(75, 0) Thus, the radius of the circle is R = CA = 2(75 - 30)2 + 702 = 83.217 psi Using these results, the circle shown in Fig. b. The coordinates of point B and D represent the principal stresses From the results, smax = 158 psi smin = - 8.22 psi sint = 0 psi Ans. Using these results, the three Mohr’s circles are shown in Fig. c, From the geometry of the three circles, tabs max = 120 psi 70 psi 30 psi Thus the coordinates of reference point A and the center C of the circle are A(30, 70) y 158.22 - ( -8.22) smax - smin = = 83.2 psi 2 2 932 Ans. © 2014 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. z *9–81. The stress at a point is shown on the element. Determine the principal stresses and the absolute maximum shear stress. x y 6 ksi Mohr’s circle for the element in x–z plane, Fig. a, will be drawn first. In accordance with the established sign convention, sx = - 1 ksi , sz = 0 and txz = 6 ksi . Thus savg = sx + sz = 2 -1 - 6 = - 3.5 ksi 2 1 ksi 1 ksi Thus, the coordinates of reference point A and the center C of the circle are A(- 1, 1) C( - 3.5, 0) Thus, the radius of the circle is R = CA = 2[- 1 - ( - 3.5)]2 + 12 = 2.69 ksi Using these results, the circle is shown in Fig. b, The coordinates of points B and D represent s2 and s3, respectively. s2 = - (3.5 - 2.69) = - 0.807 ksi s3 = - (3.5 + 2.69) = - 6.19 ksi From the results obtained, s2 = - 0.807 ksi tabs max = s3 = - 6.19 ksi s1 = 0 ksi Ans. smax - smin - 6.19 - 0 = = -3.10 ksi 2 2 Ans. Ans: s2 = - 0.807 ksi, s3 = - 6.19 ksi, s1 = 0, tabs = - 3.10 ksi max 933 © 2014 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. z 9–82. The stress at a point is shown on the element. Determine the principal stresses and the absolute maximum shear stress. y x 150 MPa 120 MPa For x – z plane: R = CA = 2(120 - 60)2 + 1502 = 161.55 s1 = 60 + 161.55 = 221.55 MPa s2 = 60 - 161.55 = - 101.55 MPa s1 = 222 MPa tabs max = s2 = 0 MPa s3 = - 102 MPa Ans. 221.55 - ( - 101.55) smax - smin = = 162 MPa 2 2 Ans. Ans: s1 = 222 MPa, s2 = 0 MPa, s3 = - 102 MPa, tabs = 162 MPa max 934 © 2014 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. z 9–83. The state of stress at a point is shown on the element. Determine the principal stresses and the absolute maximum shear stress. x For y - z plane: A(5, - 4) B( - 2.5, 4) y 2.5 ksi C(1.25, 0) 4 ksi R = 23.752 + 42 = 5.483 s1 = 1.25 + 5.483 = 6.733 ksi 5 ksi s2 = 1.25 - 5.483 = - 4.233 ksi Thus, savg = tabs max = s1 = 6.73 ksi Ans. s2 = 0 Ans. s3 = - 4.23 ksi Ans. 6.73 + (- 4.23) = 1.25 ksi 2 6.73 - ( -4.23) smax - smin = = 5.48 ksi 2 2 Ans. Ans: s1 = 6.73 ksi, s2 = 0, s3 = - 4.23 ksi, tabs = 5.48 ksi max 935 © 2014 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. 9–85. The solid cylinder having a radius r is placed in a sealed container and subjected to a pressure p. Determine the stress components acting at point A located on the center line of the cylinder. Draw Mohr’s circles for the element at this point. A r p -s(dz)(2r) = L0 p(r du) dz sin u u - 2s = p p L0 sin u du = p( - cos u)|0 s = -p The stress in every direction is s1 = s2 = s3 = - p Ans. Ans: The stress in every direction is s1 = s2 = s3 = - p 936 © 2014 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. 9–86. The plate is subjected to a tensile force P = 5 kip. If it has the dimensions shown, determine the principal stresses and the absolute maximum shear stress. If the material is ductile it will fail in shear. Make a sketch of the plate showing how this failure would appear. If the material is brittle the plate will fail due to the principal stresses. Show how this failure occurs. s = P ⫽ 5 kip 2 in. 2 in. P ⫽ 5 kip 12 in. 0.5 in. P 5000 = = 2500 psi = 2.50 ksi A (4)(0.5) s1 = 2.50 ksi Ans. s2 = s3 = 0 Ans. tabs = max s1 = 1.25 ksi 2 Ans. Failure by shear: Failure by principal stress: Ans: s1 = 2.50 ksi, s2 = s3 = 0, tabs = 1.25 ksi max 937 © 2014 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. 9–87. Determine the principal stresses and absolute maximum shear stress developed at point A on the cross section of the bracket at section a–a. 12 in. 6 in. 5 3 a 4 a 0.5 in. B 0.25 in. A 0.25 in. 0.25 in. 1.5 in.1.5 in. Section a – a Internal Loadings: Considering the equilibrium of the free-body diagram of the bracket’s upper cut segment, Fig. a, + c ©Fy = 0; 3 N - 500 a b = 0 5 N = 300 lb + ©F = 0; ; x 4 V - 500 a b = 0 5 V = 400 lb 3 4 ©MO = 0; M - 500 a b (12) - 500 a b (6) = 0 5 5 M = 6000 lb # in Section Properties: The cross-sectional area and the moment of inertia of the bracket’s cross section are A = 0.5(3) - 0.25(2.5) = 0.875 in2 I = 1 1 (0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4 12 12 Referring to Fig. b. QA = x1œ A1œ + x2œ A2œ = 0.625(1.25)(0.25) + 1.375(0.25)(0.5) = 0.3672 in3 Normal and Shear Stress: The normal stress is sA = 300 N = = - 342.86 psi A 0.875 The shear stress is contributed by the transverse shear stress. tA = VQA 400(0.3672) = = 734.85 psi It 0.79948(0.25) The state of stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 0, sy = - 342.86 psi, and txy = 734.85. Thus, savg = sx + sy 2 = 0 + ( - 342.86) = - 171.43 psi 2 The coordinates of reference point A and the center C of the circle are A(0, 734.85) C( - 171.43, 0) Thus, the radius of the circle is R = CA = 2[0 - ( -171.43)]2 + 734.852 = 754.58 psi 938 500 lb © 2014 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. 9–87. Continued Using these results, the circle is shown in Fig. d. In-Plane Principal Stresses: The coordinates of reference point B and D represent s1 and s2, respectively. s1 = - 171.43 + 754.58 = 583.2 psi s2 = - 171.43 - 754.58 = - 926.0 psi Three Mohr’s Circles: Using these results, smax = 583 psi sint = 0 smin = - 926 psi Ans. Absolute Maximum Shear Stress: tabs max = 583.2 - ( - 926.0) smax - smin = = 755 psi 2 2 Ans. Ans: s1 = 583 psi, s2 = 0, s3 = - 926 psi, tabs = 755 psi max 939 © 2014 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. *9–88. Determine the principal stresses and absolute maximum shear stress developed at point B on the cross section of the bracket at section a–a. 12 in. Internal Loadings: Considering the equilibrium of the free-body diagram of the 6 in. bracket’s upper cut segment, Fig. a, + c ©Fy = 0; 3 N - 500 a b = 0 5 + ©F = 0; ; x 4 V - 500 a b = 0 5 B Section Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the bracket’s cross section are A = 0.5(3) - 0.25(2.5) = 0.875 in2 1 1 (0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4 12 12 Referring to Fig. b, QB = 0 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. sB = 6000(1.5) MxB N 300 + = + = 10.9 ksi A I 0.875 0.79948 Since QB = 0, tB = 0. The state of stress at point B is represented on the element shown in Fig. c. In-Plane Principal Stresses: Since no shear stress acts on the element, s1 = 10.91 ksi s2 = 0 Three Mohr’s Circles: Using these results, smax = 10.91 ksi sint = smin = 0 Ans. Absolute Maximum Shear Stress: tabs max = 0.25 in. 1.5 in.1.5 in. Section a – a M = 6000 lb # in smax - smin 10.91 - 0 = = 5.46 ksi 2 2 Ans. 940 500 lb 0.25 in. A 0.25 in. V = 400 lb 4 3 ©MO = 0; M - 500 a b (12) - 500 a b (6) = 0 5 5 I = 4 a 0.5 in. N = 300 lb 5 3 a © 2014 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. 9–89. The solid propeller shaft on a ship extends outward from the hull. During operation it turns at v = 15 rad>s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft. If the shaft has an outer diameter of 250 mm, determine the principal stresses at any point located on the surface of the shaft. 0.75 m A T F Power Transmission: Using the formula developed in Chapter 5, P = 900 kW = 0.900 A 106 B N # m>s 0.900(106) P = = 60.0 A 103 B N # m v 15 T0 = Internal Torque and Force: As shown on FBD. Section Properties: A = p A 0.252 B = 0.015625p m2 4 J = p A 0.1254 B = 0.3835 A 10 - 3 B m4 2 Normal Stress: s = - 1.23(106) N = = - 25.06 MPa A 0.015625p Shear Stress: Applying the torsion formula, t = 60.0(103) (0.125) Tc = 19.56 MPa = J 0.3835(10 - 3) In-Plane Principal Stresses: sx = - 25.06 MPa, sy = 0 and txy = 19.56 MPa for any point on the shaft’s surface. Applying Eq. 9-5, s1,2 = = sx + sy 2 ; C a sx - sy 2 2 b + t2xy - 25.06 - 0 2 - 25.06 + 0 ; a b + (19.56)2 2 C 2 = - 12.53 ; 23.23 s1 = 10.7 MPa s2 = - 35.8 MPa Ans. Ans: s1 = 10.7 MPa, s2 = - 35.8 MPa 941 © 2014 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. 9–90. The solid propeller shaft on a ship extends outward from the hull. During operation it turns at v = 15 rad>s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft. If the shaft has a diameter of 250 mm, determine the maximum in-plane shear stress at any point located on the surface of the shaft. 0.75 m A T F Power Transmission: Using the formula developed in Chapter 5, P = 900 kW = 0.900 A 106 B N # m>s T0 = 0.900(106) P = = 60.0 A 103 B N # m v 15 Internal Torque and Force: As shown on FBD. Section Properties: A = p A 0.252 B = 0.015625p m2 4 J = p A 0.1254 B = 0.3835 A 10 - 3 B m4 2 Normal Stress: s = - 1.23(106) N = = - 25.06 MPa A 0.015625p Shear Stress: Applying the torsion formula. t = 60.0(103) (0.125) Tc = 19.56 MPa = J 0.3835 (10 - 3) Maximum In-Plane Principal Shear Stress: sx = - 25.06 MPa, sy = 0, and txy = 19.56 MPa for any point on the shaft’s surface. Applying Eq. 9-7, t max in-plane a sx - sy 2 b + t2xy = C = - 25.06 - 0 2 b + (19.56)2 C 2 2 a = 23.2 MPa Ans. Ans: tmax in-plane 942 = 23.2 MPa © 2014 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. 9–91. The steel pipe has an inner diameter of 2.75 in. and an outer diameter of 3 in. If it is fixed at C and subjected to the horizontal 20-lb force acting on the handle of the pipe wrench at its end, determine the principal stresses in the pipe at point A, which is located on the surface of the pipe. 20 lb 12 in. 10 in. Internal Forces, Torque and Moment: As shown on FBD. A Section Properties: B p I = A 1.54 - 1.3754 B = 1.1687 in4 4 J = C p A 1.54 - 1.3754 B = 2.3374 in4 2 y z x (QA)z = ©y¿A¿ = 4(1.5) 1 4(1.375) 1 c p A 1.52 B d c p A 1.3752 B d 3p 2 3p 2 = 0.51693 in3 Normal Stress: Applying the flexure formula s = sA = My z Iy , 200(0) = 0 1.1687 Shear Stress: The transverse shear stress in the z direction and the torsional shear VQ stress can be obtained using shear formula and torsion formula, tv = and It Tr ttwist = , respectively. J tA = (tv)z - ttwist = 20.0(0.51693) 240(1.5) 1.1687(2)(0.125) 2.3374 = - 118.6 psi In-Plane Principal Stress: sx = 0, sz = 0 and txz = - 118.6 psi for point A. Applying Eq. 9-5 s1,2 = sx + sz 2 ; C a sx - sz 2 2 b + t2xz = 0 ; 20 + ( - 118.6)2 s1 = 119 psi s2 = - 119 psi Ans. Ans: s1 = 119 psi, s2 = - 119 psi 943 © 2014 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. *9–92. Solve Prob. 9–91 for point B, which is located on the surface of the pipe. 20 lb 12 in. 10 in. A B Internal Forces, Torque and Moment: As shown on FBD. C y Section Properties: z I = p A 1.54 - 1.3754 B = 1.1687 in4 4 J = p A 1.54 - 1.3754 B = 2.3374 in4 2 x (QB)z = 0 Normal Stress: Applying the flexure formula s = sB = My z Iv , 200(1.5) = 256.7 psi 1.1687 Shear Stress: Torsional shear stress can be obtained using torsion formula, Tr . ttwist = J tB = ttwist = 240(1.5) = 154.0 psi 2.3374 In - Plane Prinicipal Stress: sx = 256.7 psi, sy = 0, and txy = - 154.0 psi for point B. Applying Eq. 9-5 s1,2 = = sx + sy 2 ; C sx - sy a 2 2 b + t2xy 256.7 - 0 2 256.7 + 0 a ; b + ( -154.0)2 2 C 2 = 128.35 ; 200.49 s1 = 329 psi s2 = - 72.1 psi Ans. 944 © 2014 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. 9–93. Determine the equivalent state of stress if an element is oriented 40° clockwise from the element shown. Use Mohr’s circle. 10 ksi 6 ksi A(6, 0) B(- 10, 0) C( - 2, 0) R = CA = CB = 8 sx¿ = - 2 + 8 cos 80° = - 0.611 ksi Ans. tx¿y¿ = 8 sin 80° = 7.88 ksi Ans. sy¿ = - 2 - 8 cos 80° = - 3.39 ksi Ans. Ans: sx¿ = - 0.611 ksi, tx¿y¿ = 7.88 ksi, sy¿ = - 3.39 ksi 945 © 2014 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. 9–94. The crane is used to support the 350-lb load. Determine the principal stresses acting in the boom at points A and B. The cross section is rectangular and has a width of 6 in. and a thickness of 3 in. Use Mohr’s circle. 5 ft 3 in. 5 ft B A 45° A = 6(3) = 18 in2 I = 45° 1 (3)(63) = 54 in4 12 QB = (1.5)(3)(3) = 13.5 in3 QA = 0 For point A: sA = - My 1750(12)(3) P 597.49 = = - 1200 psi A I 18 54 tA = 0 s1 = 0 Ans. s2 = - 1200 psi = - 1.20 ksi Ans. For point B: sB = - tB = P 597.49 = = - 33.19 psi A 18 VQB 247.49(13.5) = = 20.62 psi It 54(3) A(-33.19, - 20.62) B(0, 20.62) C( - 16.60, 0) R = 216.602 + 20.622 = 26.47 s1 = - 16.60 + 26.47 = 9.88 psi Ans. s2 = - 16.60 - 26.47 = - 43.1 psi Ans. Ans: Point A: s1 = 0, s2 = - 1.20 ksi, Point B: s1 = 9.88 psi, s2 = - 43.1 psi 946 © 2014 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. 9–95. Determine the equivalent state of stress on an element at the same point which represents (a) the principal stresses, and (b) the maximum in-plane shear stress and the associated average normal stress. Also, for each case, determine the corresponding orientation of the element with respect to the element shown and sketch the results on the element. 30 ksi 10 ksi Normal and Shear Stress: sx = 0 sy = - 30 ksi txy = - 10 ksi In-Plane Principal Stresses: s1, 2 = = sx + sy ; 2 A a sx - sy 2 2 2 b + txy 0 + ( -30) 0 - ( - 30)2 2 ; b + (- 10) 2 Aa 2 = - 15 ; 2325 s1 = 3.03 ksi s2 = - 33.0 ksi Ans. Orientation of Principal Plane: tan 2up = txy (sx - sy)>2 = - 10 = - 0.6667 [0 - (- 30)]>2 up = - 16.845° and 73.15° Substituting u = - 16.845° into sx + sy sx¿ = = sx - sy + 2 2 cos 2u + txy sin 2u 0 + ( - 30) 0 - ( -30) + cos ( - 33.69°) - 10 sin ( -33.69°) 2 2 = 3.03 ksi = s1 Thus, (up)1 = - 16.8° and (up)2 = 73.2° Ans. The element that represents the state of principal stress is shown in Fig. a. Maximum In-Plane Shear Stress: tmax in-plane = a B sx - sy 2 2 b + txy2 = 0 - ( - 30) 2 b + ( -10)2 = 18.0 ksi 2 B a 947 Ans. © 2014 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. 9–95. Continued Orientation of the Plane of Maximum In-Plane Shear Stress: tan 2us = - (sx - sy)>2 txy = [0 - (- 30)]>2 = 1.5 - 10 us = 28.2° and 118° By inspection, tmax has to act in the same sense shown in Fig. b to maintain in-plane equilibrium. Average Normal Stress: savg = sx + sy 2 = 0 + ( - 30) = - 15 ksi 2 Ans. The element that represents the state of maximum in-plane shear stress is shown in Fig. c. Ans: s1 = 3.03 ksi, s2 = - 33.0 ksi, up1 = - 16.8° and up2 = 73.2°, tmax = 18.0 ksi , savg = - 15 ksi, us = 28.2° in-plane 948 © 2014 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. *9–96. The propeller shaft of the tugboat is subjected to the compressive force and torque shown. If the shaft has an inner diameter of 100 mm and an outer diameter of 150 mm, determine the principal stress at a point A located on the outer surface. 10 kN A 2 kN·m Internal Loadings: Considering the equilibrium of the free-body diagram of the propeller shaft’s right segment, Fig. a, ©Fx = 0; 10 - N = 0 N = 10 kN ©Mx = 0; T - 2 = 0 T = 2 kN # m Section Properties: The cross - sectional area and the polar moment of inertia of the propeller shaft’s cross section are A = p A 0.0752 - 0.052 B = 3.125p A 10 - 3 B m2 J = p A 0.0754 - 0.054 B = 12.6953125p A 10 - 6 B m4 2 Normal and Shear Stress: The normal stress is a contributed by axial stress only. sA = 10 A 103 B N = = - 1.019 MPa A 3.125p A 10 - 3 B The shear stress is contributed by the torsional shear stress only. tA = 2 A 103 B (0.075) Tc = = 3.761 MPa J 12.6953125p A 10 - 6 B The state of stress at point A is represented by the element shown in Fig. b. Construction of the Circle: sx = - 1.019 MPa, sy = 0, and txy = - 3.761 MPa. Thus, savg = sx + sy 2 = -1.019 + 0 = - 0.5093 MPa 2 The coordinates of reference point A and the center C of the circle are A(-1.019, - 3.761) C( - 0.5093, 0) Thus, the radius of the circle is R = CA = 2[- 1.019 - ( - 0.5093)]2 + (- 3.761)2 = 3.795 MPa Using these results, the circle is shown in Fig. c. In-Plane Principal Stress: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = - 0.5093 + 3.795 = 3.29 MPa Ans. s2 = - 0.5093 - 3.795 = - 4.30 MPa Ans. 949 © 2014 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. 9–96. Continued Orientation of the Principal Plane: Referring to the geometry of the circle, Fig. c, tan 2 A up B 2 = 3.761 = 7.3846 1.019 - 0.5093 A up B 2 = 41.1° (clockwise) Ans. The state of principal stresses is represented on the element shown in Fig. d. 950 © 2014 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. *9–97. The box beam is subjected to the loading shown. Determine the principal stress in the beam at points A and B. 1200 lb 800 lb 6 in. A 6 in. B 8 in. 8 in. A B 3 ft 2.5 ft 2.5 ft 5 ft Support Reactions: As shown on FBD(a). Internal Forces and Moment: As shown on FBD(b). Section Properties: I = 1 1 (8) A 83 B (6) A 63 B = 233.33 in4 12 12 QA = QB = 0 Normal Stress: Applying the flexure formula. s = - My I sA = - - 300(12)(4) = 61.71 psi 233.33 sB = - - 300(12)(- 3) = - 46.29 psi 233.33 Shear Stress: Since QA = QB = 0, then tA = tB = 0. In-Plane Principal Stress: sx = 61.71 psi, sy = 0, and txy = 0 for point A. Since no shear stress acts on the element, s1 = sx = 61.7 psi Ans. s2 = sy = 0 Ans. sx = - 46.29 psi, sy = 0, and txy = 0 for point B. Since no shear stress acts on the element, s1 = sy = 0 Ans. s2 = sx = - 46.3 psi Ans. Ans: Point A: s1 = 61.7 psi, s2 = 0 Point B: s1 = 0, s2 = - 46.3 psi 951 © 2014 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. 9–98. The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. 60 MPa 30 MPa 45 MPa a) sx = 45 MPa s1, 2 = = txy = 30 MPa sy = - 60 MPa sx + sy Aa ; 2 sx - sy 2 2 b + txy 2 45 - 60 45 - ( -60) 2 2 ; a b + 30 2 A 2 tan 2up = Ans. s2 = - 68.0 MPa s1 = 53.0 MPa txy sx - sy 2 30 = 45 - ( - 60) 2 = 0.5714 up = 14.87° and - 75.13° Use Eq. 9–1 to determine the principal plane of s1 and s2 sx + sy sx ¿ = sx - sy + 2 2 cos 2u + txy sin 2u u = uy = 14.87° sx¿ = 45 - ( -60) 45 + ( -60) + cos 29.74° + 30 sin 29.74° = 53.0 MPa 2 2 Therefore, up1 = 14.9° ; b) tmax = A in-plane savg = a sx - sy 2 sx + sy 2 tan 2us = us = - 30.1° = 2 = b + txy 2 Aa Ans. 45 - ( -60) 2 2 b + 30 = 60.5 MPa 2 45 + ( -60) = - 7.50 MPa 2 (sx - sy) 2 txy up2 = - 75.1° = - [45 - ( - 60)] 2 Ans. 30 Ans. Ans. = -1.75 and 59.9° Ans. By observation, in order to preserve equilibrium, tmax = 60.5 MPa has to act in the direction shown in the figure. Ans: s1 = 53.0 MPa, s2 = - 68.0 MPa, up1 = 14.9°, up2 = - 75.1°, tmax = 60.5 MPa, in-plane savg = - 7.50 MPa, us = - 30.1° and 59.9° 952 © 2014 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. 9–99. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1. 14 ksi A 20 ksi 50⬚ B + Q©F x¿ = 0; sx¿ ¢A + 14 ¢A sin 50° cos 40° + 20 ¢A cos 50° cos 50° = 0 sx¿ = - 16.5 ksi a + ©Fy¿ = 0; Ans. tx¿y¿ ¢A + 14 ¢A sin 50° sin 40° - 20 ¢A cos 50° sin 50° = 0 tx¿y¿ = 2.95 ksi Ans. Ans: sx¿ = - 16.5 ksi, tx¿y¿ = 2.95 ksi 953 © 2014 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. 10–1. Prove that the sum of the normal strains in perpendicular directions is constant. Px + Py Px¿ = 2 Px - Py + Px + Py Py¿ = 2 2 Px - Py - 2 cos 2u + cos 2u - gxy 2 gxy 2 sin 2u (1) sin 2u (2) Adding Eq. (1) and Eq. (2) yields: Px¿ + Py¿ = Px + Py = constant (Q.E.D.) 954 © 2014 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. 10–2. The state of strain at the point has components of Px = 200 110-62, Py = - 300 110-62, and gxy = 400(10-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of 30° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane. y x In accordance with the established sign convention, Px = 200(10 - 6), Px + Py Px¿ = Px - Py + 2 = c Py = - 300(10 - 6) 2 cos 2u + gxy 2 gxy = 400(10 - 6) u = 30° sin 2u 200 + ( -300) 200 - ( -300) 400 + cos 60° + sin 60° d(10 - 6) 2 2 2 = 248 (10 - 6) gx¿y¿ 2 = -a Ans. Px - Py 2 b sin 2u + gxy 2 cos 2u gx¿y¿ = e - C 200 - ( - 300) D sin 60° + 400 cos 60° f(10 - 6) = - 233(10 - 6) Px + Py Py¿ = = c 2 Ans. Px - Py - 2 cos 2u - gxy 2 sin 2u 200 - ( - 300) 200 + ( -300) 400 cos 60° sin 60° d(10 - 6) 2 2 2 = - 348(10 - 6) Ans. The deformed element of this equivalent state of strain is shown in Fig. a Ans: Px¿ = 248(10 - 6), gx¿y¿ = - 233(10 - 6), Py¿ = - 348(10 - 6) 955 © 2014 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. 10–3. The state of strain at a point on a wrench Px = 120110-62, has components Py = - 180110-62, gxy = 150(10-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within x–y plane. Px = 120(10 - 6) Py = - 180(10 - 6) ; Aa Px + Py a) P1, 2 = 2 = c Px - Py 2 2 b + a gxy = 150(10 - 6) gxy 2 b 2 120 + ( - 180) 120 - ( -180) 2 150 2 10 - 6 ; b + a b d a A 2 2 2 P1 = 138(10 - 6); P2 = - 198(10 - 6) Ans. Orientation of P1 and P2 tan 2up = gxy = Px - Py 150 = 0.5 [120 - (- 180)] up = 13.28° and - 76.72° Use Eq. 10–5 to determine the direction of P1 and P2 Px + Py Py¿ = 2 Px - Py + cos 2u + 2 gxy 2 sin 2u u = up = 13.28° Py¿ = J 120 + ( -180) 120 - ( - 180) 150 + cos (26.56°) + sin 26.56° d10 - 6 2 2 2 = 138(10 - 6) = P1 Therefore up1 = 13.3°; gmax b) in-plane 2 gmax in-plane = Aa = 2c Px - Py Aa 2 2 Ans. gxy 2 2 b + a 2 b 2 120 - ( -180) 2 + a 150 b d 10 - 6 = 335(10 - 6) b 2 2 Px + Py Pavg = up2 = - 76.7° = c 120 + (- 180) d 10 - 6 = - 30.0(10 - 6) 2 956 Ans. Ans. © 2014 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. 10–3. Continued Orientation of gmax tan 2us = - (Px - Py) gxy = us = - 31.7° and -[120 - ( - 180)] = - 2.0 150 Ans. 58.3° Use Eq. 10–11 to determine the sign of gmax in-plane gx¿y¿ 2 Px - Py = - 2 sin 2u + gxy 2 cos 2u u = us = - 31.7° gx¿y¿ = 2 c - 120 - (- 180) 150 sin ( - 63.4°) + cos ( -63.4°) d10 - 6 = 335(10 - 6) 2 2 Ans: P1 = 138(10 - 6), P2 = - 198(10 - 6), up1 = 13.3°, up2 = - 76.7°, gmax = 335(10 - 6), Pavg = - 30.0(10 - 6) in-plane us = - 31.7° and 58.3° 957 © 2014 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. y *10–4. The state of strain at the point on the gear tooth has components gxy = Px = 850110-62, Py = 480110-62, -6 650(10 2. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. Px = 850(10 - 6) Py = 480(10 - 6) Px + Py a) P1, 2 = 2 = c Aa ; Px - Py 2 gxy = 650(10 - 6) b + a 2 x gxy 2 b 2 2 2 850 + 480 ; a 850 - 480 b + a 650 b d(10 - 6) 2 A 2 2 P1 = 1039(10 - 6) P2 = 291(10 - 6) Ans. Ans. Orientation of P1 and P2: gxy tan 2uy = 650 850 # 480 = Px - Py uy = 30.18° and 120.18° Use Eq. 10–5 to determine the direction of P1 and P2: Px + Py Px¿ = Px - Py + 2 2 cos 2u + gxy 2 sin 2u u = uy = 30.18° Px¿ = c 850 + 480 850 - 480 650 + cos (60.35°) + sin (60.35°) d (10 - 6) 2 2 2 = 1039(10 - 6) up2 = 120° Ans. 850 - 480 2 + 650 2 d (10 - 6) = 748(10 - 6) a b b 2 2 Ans. Therefore, up1 = 30.2° Ans. b) gmax Aa in-plane 2 gmax in-plane = = 2c 2 2 Aa Px + Py Pavg = Px - Py = a 2 b + a gxy 2 b 2 850 + 480 b (10 - 6) = 665(10 - 6) 2 Ans. 958 © 2014 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. 10–4. Continued Orientation of gmax: tan 2ut = - (Px - Py) gxy = - (850 - 480) 650 ut = - 14.8° and 75.2° Ans. Use Eq. 10–6 to determine the sign of gmax : in-plane gx¿y¿ 2 Px - Py = - 2 sin 2u + gxy 2 cos 2u; u = ut = - 14.8° gx¿y¿ = [ -(850 - 480) sin ( - 29.65°) + 650 cos ( -29.65°)](10 - 6) = 748(10 - 6) 959 © 2014 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. y 10–5. The state of strain at the point on the gear tooth has the components Px = 520110-62, Py = - 760110-62, gxy = 750(10-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. Px = 520(10 - 6) Py = - 760(10 - 6) Px + Py a) P1, 2 = ; 2 = c Aa Px - Py 2 x gxy = - 750(10 - 6) 2 b + a gxy 2 b 2 2 2 520 + ( - 760) ; a 520 - ( -760) b + a - 750 b d10 - 6 2 A 2 2 P1 = 622(10 - 6); P2 = - 862(10 - 6) Ans. Orientation of P1 and P2 tan 2up = gxy = Px - Py -750 = - 0.5859; up = - 15.18° and up = 74.82° [520 - ( - 760)] Use Eq. 10–5 to determine the direction of P1 and P2. Px + Py Px¿ = Px - Py + 2 2 cos 2u + gxy 2 sin 2u u = up = - 15.18° Px¿ = J 520 + (- 760) 520 - ( - 760) -750 + cos ( - 30.36°) + sin ( -30.36°) d10 - 6 2 2 2 = 622 (10 - 6) = P1 Therefore, up1 = - 15.2° and up2 = 74.8° gmax b) in-plane = 2 gmax in-plane gxy 2 Px - Py 2 b + a 2 2 b A = 2c a 520 - ( -760) 2 - 750 2 d 10 - 6 = - 1484 (10 - 6) b + a A 2 2 b Px + Py Pavg = 2 Ans. a = c 520 + ( -760) d 10 - 6 = - 120 (10 - 6) 2 Ans. Ans. 960 © 2014 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. 10–5. Continued Orientation of gmax in-plane tan 2us = : - (Px - Py) = gxy -[520 - ( - 760)] = 1.7067 - 750 us = 29.8° and us = - 60.2° Ans. Use Eq. 10–6 to check the sign of gmax : in-plane gx¿y¿ 2 Px - Py = - gx¿y¿ = 2 c - 2 sin 2u + gxy 2 cos 2u; u = us = 29.8° 520 - (- 760) -750 sin (59.6°) + cos (59.6°) d10 - 6 = - 1484 (10 - 6) 2 2 Ans: P1 = 622(10 - 6), P2 = - 862(10-6), up1 = - 15.2° and up2 = 74.8°, gmax = - 1484(10 - 6), Pavg = - 120(10 - 6), in-plane us = 29.8° and -60.2° 961 © 2014 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. y 10–6. A differential element on the bracket is subjected to plane strain that has the following components: Px = 150110-62, Py = 200110-62, gxy = - 700(10-62. Use the strain-transformation equations and determine the equivalent in plane strains on an element oriented at an angle of u = 60° counterclockwise from the original position. Sketch the deformed element within the x–y plane due to these strains. Px = 150 (10 - 6) Px + Py Px¿ = Px - Py + 2 = c Py = 200 (10 - 6) 2 cos 2u + gxy 2 x gxy = - 700 (10 - 6) u = 60° sin 2u 150 - 200 - 700 150 + 200 + cos 120° + a b sin 120° d 10 - 6 2 2 2 = - 116 (10 - 6) Px + Py Py¿ = = c Px - Py - 2 Ans. 2 cos 2u - gxy 2 sin 2u 150 + 200 150 - 200 - 700 cos 120° - a b sin 120° d 10 - 6 2 2 2 = 466 (10 - 6) gx¿y¿ 2 Px - Py = - gx¿y¿ = 2 c - 2 Ans. sin 2u + gxy 2 cos 2u 150 - 200 -700 sin 120° + a b cos 120° d 10 - 6 = 393 (10 - 6) 2 2 Ans. Ans: Px¿ = - 116(10 - 6), Py¿ = 466(10 - 6), gx¿y¿ = 393(10 - 6) 962 © 2014 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. y 10–7. Solve Prob. 10–6 for an element oriented u = 30° clockwise. x Px = 150 (10 - 6) Px + Py Px¿ = Px - Py + 2 = c Py = 200 (10 - 6) 2 cos 2u + gxy 2 gxy = - 700 (10 - 6) u = - 30° sin 2u 150 - 200 - 700 150 + 200 + cos ( -60°) + a b sin ( -60°) d10 - 6 2 2 2 = 466 (10 - 6) Px + Py Py¿ = = c Ans. Px - Py - 2 2 cos 2u - gxy 2 sin 2u 150 - 200 - 700 150 + 200 cos ( -60°) - a b sin ( -60°) d10 - 6 2 2 2 = - 116 (10 - 6) gx¿y¿ 2 Px - Py = - gx¿y¿ = 2 c - 2 sin 2u + Ans. gxy 2 cos 2u - 700 150 - 200 sin ( - 60°) + cos ( - 60°) d10 - 6 2 2 = - 393(10 - 6) Ans. Ans: Px¿ = 466(10 - 6), Py¿ = - 116(10 - 6), gx¿y¿ = - 393(10 - 6) 963 © 2014 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. *10–8. The state of strain at the point on the bracket has components Px = - 200110-62, Py = - 650110-62, gxy ⫽ - 175110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 20° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane. Px = - 200(10 - 6) Px + Py Px¿ = Px - Py + 2 = c Py = - 650(10 - 6) 2 cos 2u + gxy 2 y x gxy = - 175(10 - 6) u = 20° sin 2u (- 200) - ( -650) ( -175) - 200 + (- 650) + cos (40°) + sin (40°) d(10 - 6) 2 2 2 = - 309(10 - 6) Px + Py Py¿ = Px - Py - 2 = c Ans. 2 cos 2u - gxy 2 sin 2u - 200 - ( - 650) ( -175) - 200 + (- 650) cos (40°) sin (40°) d(10 - 6) 2 2 2 = - 541(10 - 6) gx¿y¿ 2 Px - Py = - 2 Ans. sin 2u + gxy 2 cos 2u gx¿y¿ = [ -(- 200 - ( -650)) sin (40°) + ( - 175) cos (40°)](10 - 6) = - 423(10 - 6) Ans. 964 © 2014 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. 10–9. The state of strain at the point has components of Px = 180110-62, Py = - 120110-62, and gxy = - 100110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. y x a) In accordance with the established sign convention, Px = 180(10 - 6), Py = - 120(10 - 6) and gxy = - 100(10 - 6). Px + Py P1, 2 = ; 2 = b Aa Px - Py 2 2 b + a gxy 2 b 2 180 + ( - 120) 180 - ( -120) 2 -100 2 -6 ; c d + a b r (10 ) 2 A 2 2 = A 30 ; 158.11 B (10 - 6) P1 = 188(10 - 6) tan 2uP = P2 = - 128(10 - 6) gxy Ans. - 100(10 - 6) C 180 - ( -120) D (10 - 6) = Px - Py uP = - 9.217° and = - 0.3333 80.78° Substitute u = - 9.217°, Px + Py Px¿ = 2 = c Px - Py + 2 cos 2u + gxy 2 sin 2u 180 + (- 120) 180 - ( - 120) -100 + cos ( - 18.43°) + sin ( -18.43) d(10 - 6) 2 2 2 = 188(10 - 6) = P1 Thus, (uP)1 = - 9.22° (uP)2 = 80.8° Ans. The deformed element is shown in Fig (a). gmax Px - Py 2 gxy 2 in-plane = b + a b b) 2 Aa 2 2 gmax in-plane tan 2us = - a = b2 A Px - Py gxy c 180 - (- 120) 2 - 100 2 -6 -6 d + a b r (10 ) = 316 A 10 B 2 2 b = -c C 180 - ( - 120) D (10 - 6) us = 35.78° = 35.8° and - 100(10 - 6) Ans. s = 3 - 54.22° = - 54.2° Ans. 965 © 2014 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. 10–9. Continued gmax The algebraic sign for in-plane when u = 35.78°. gx¿y¿ Px - Py gxy = -a b sin 2u + cos 2u 2 2 2 gx¿y¿ = e - C 180 - (- 120) D sin 71.56° + ( -100) cos 71.56° f(10 - 6) Pavg = - 316(10 - 6) Px + Py 180 + ( -120) = c d (10 - 6) = 30(10 - 6) = 2 2 Ans. The deformed element for the state of maximum in-plane shear strain is shown in Fig. b Ans: P1 = 188(10 - 6), P2 = - 128(10 - 6), up1 = - 9.22°, up 2 = 80.8°, gmax = 316(10 - 6), us = 35.8° and - 54.2°, in-plane Pavg = 30(10 - 6) 966 © 2014 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. 10–10. The state of strain at the point on the support has components of Px = 350110-62, Py = 400110-62, gxy = - 675(10-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. P a) Px + Py P1, 2 = = ; 2 A a Px - Py 2 2 b + a gxy 2 b 2 350 + 400 - 675 2 350 - 400 2 ; a b + a b 2 A 2 2 P1 = 713 (10 - 6) tan 2up = P2 = 36.6 (10 - 6) Ans. gxy = Px - Py Ans. - 675 (350 - 400) up1 = 133° Ans. b) (gx¿y¿)max = 2 (gx¿y¿)max = 2 A a Px - Py 2 2 gxy 2 b + a 2 b - 675 2 350 - 400 2 b + a b A 2 2 a (gx¿y¿)max = 677(10 - 6) Px + Py Pavg = tan 2us = 2 = Ans. 350 + 400 = 375 (10 - 6) 2 -(Px - Py) gxy = Ans. 350 - 400 675 us = - 2.12° Ans. Ans: (a) P1 = 713(10-6), P2 = 36.6(10 - 6), up1 = 133° = 677(10 - 6), Pavg = 375(10 - 6), (b) gmax in-plane us = - 2.12° 967 © 2014 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. y 10–11. The state of strain on an element has components Px = - 150110-62, Py = 450110-62, gxy = 200110-62. Determine the equivalent state of strain on an element at the same point oriented 30° counterclockwise with respect to the original element. Sketch the results on this element. Pydy dy gxy 2 gxy 2 dx Pxdx x Strain Transformation Equations: Px = - 150(10 - 6) Py = 450(10 - 6) gxy = 200(10 - 6) u = 30° We obtain Px + Py Px¿ = 2 = c Px - Py + 2 gxy cos 2u + 2 sin 2u - 150 - 450 200 -150 + 450 + cos 60° + sin 60° d(10 - 6) 2 2 2 = 86. 6(10 - 6) gx¿y¿ 2 = -a Ans. Px - Py 2 b sin 2u + gxy 2 cos 2u gx¿y¿ = [ -( -150 - 450) sin 60° + 200 cos 60°](10 - 6) = 620(10 - 6) Px + Py Py¿ = = c 2 Ans. Px - Py - 2 cos 2u - gxy 2 sin 2u - 150 + 450 - 150 - 450 200 cos 60° sin 60° d(10 - 6) 2 2 2 = 213(10 - 6) Ans. The deformed element for this state of strain is shown in Fig. a. Ans: Px¿ = 86.6(10 - 6), gx¿y¿ = 620(10 - 6), Py¿ = 213(10 - 6) 968 © 2014 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. y *10–12. The state of strain on an element has components Px = - 400110-62, Py = 0, gxy = 150110-62. Determine the equivalent state of strain on an element at the same point oriented 30° clockwise with respect to the original element. Sketch the results on this element. gxy dy 2 x gxy 2 dx Strain Transformation Equations: Px = - 400(10 - 6) Py = 0 gxy = 150(10 - 6) u = - 30° We obtain Px + Py Px¿ = 2 = c Px - Py + 2 gxy cos 2u + 2 sin 2u - 400 - 0 150 - 400 + 0 + cos ( -60°) + sin (-60°) d(10 - 6) 2 2 2 = - 365(10 - 6) gx¿y¿ = -a 2 Ans. Px - Py 2 b sin 2u + gxy 2 cos 2u gx¿y¿ = [ - ( -400 - 0) sin ( -60°) + 150 cos ( -60°)](10 - 6) = - 271(10 - 6) Px + Py Py¿ = = c 2 Ans. Px - Py - 2 cos 2u - gxy 2 sin 2u - 400 + 0 - 400 - 0 150 cos ( -60°) sin ( -60°) d(10 - 6) 2 2 2 = - 35.0(10 - 6) Ans. The deformed element for this state of strain is shown in Fig. a. 969 Pxdx © 2014 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. y 10–13. The state of plane strain on an element is Px = - 300110-62, Py = 0, and gxy = 150110-62. Determine the equivalent state of strain which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding elements for these states of strain with respect to the original element. gxy dy 2 x In-Plane Principal Strains: Px = - 300 A 10 - 6 B , Py = 0, and gxy = 150 A 10 - 6 B . We obtain Px + Py C¢ ; P1, 2 = 2 = C Px - Py 2 gxy 2 ≤ + ¢ ≤ 2 2 -300 + 0 - 300 - 0 2 150 2 ; ¢ ≤ + ¢ ≤ S A 10 - 6 B 2 C 2 2 = ( -150 ; 167.71) A 10 - 6 B P1 = 17.7 A 10 - 6 B P2 = - 318 A 10 - 6 B Ans. Orientation of Principal Strain: tan 2up = gxy = Px - Py 150 A 10 - 6 B ( -300 - 0) A 10 - 6 B = - 0.5 uP = - 13.28° and 76.72° Substituting u = - 13.28° into Eq. 9-1, Px + Py Px¿ = = c Px - Py + 2 cos 2u + 2 gxy 2 sin 2u -300 + 0 -300 - 0 150 + cos ( - 26.57°) + sin ( -26.57°) d A 10 - 6 B 2 2 2 = - 318 A 10 - 6 B = P2 Thus, A uP B 1 = 76.7° and A uP B 2 = - 13.3° Ans. The deformed element of this state of strain is shown in Fig. a. Maximum In-Plane Shear Strain: gmax in-plane 2 gmax in-plane = C¢ Px - Py 2 2 ≤ + ¢ gxy 2 ≤ 2 - 300 - 0 2 150 2 -6 -6 b + a b R A 10 B = 335 A 10 B 2 2 A = B2 a Ans. Orientation of the Maximum In-Plane Shear Strain: tan 2us = - ¢ Px - Py gxy ≤ = -C ( - 300 - 0) A 10 - 6 B 150 A 10 - 6 B S = 2 us = 31.7° and 122° Ans. 970 gxy 2 dx Pxdx © 2014 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. 10–13. Continued The algebraic sign for gx¿y¿ 2 = -¢ Px - Py 2 gmax in-plane ≤ sin 2u + when u = us = 31.7° can be obtained using gxy 2 cos 2u gx¿y¿ = [ -(- 300 - 0) sin 63.43° + 150 cos 63.43°] A 10 - 6 B = 335 A 10 - 6 B Average Normal Strain: Px + Py Pavg = 2 = a - 300 + 0 b A 10 - 6 B = - 150 A 10 - 6 B 2 Ans. The deformed element for this state of strain is shown in Fig. b. Ans: P1 = 17.7(10 - 6), P2 = - 318(10 - 6), up1 = 76.7° and up2 = - 13.3°, gmax = 335(10 - 6), us = 31.7° and 122°, in-plane Pavg = - 150(10 - 6) 971 © 2014 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. 10–14. The state of strain at the point on a boom of an hydraulic engine crane has components of Px = 250110-62, Py = 300110-62, and gxy = - 180110-62. Use the straintransformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case, specify the orientation of the element and show how the strains deform the element within the x–y plane. y a) In-Plane Principal Strain: Applying Eq. 10–9, Px + Py P1, 2 = ; 2 = B Aa Px - Py 2 b + a 2 gxy 2 b 2 250 - 300 2 250 + 300 - 180 2 -6 ; a b + a b R A 10 B 2 A 2 2 = 275 ; 93.41 P1 = 368 A 10 - 6 B P2 = 182 A 10 - 6 B Ans. Orientation of Principal Strain: Applying Eq. 10–8, gxy tan 2uP = - 180(10 - 6) = Px - Py (250 - 300)(10 - 6) uP = 37.24° and = 3.600 - 52.76° Use Eq. 10–5 to determine which principal strain deforms the element in the x¿ direction with u = 37.24°. Px + Py Px¿ = = c 2 Px - Py + 2 cos 2u + gxy 2 sin 2u 250 + 300 250 - 300 - 180 + cos 74.48° + sin 74.48° d A 10 - 6 B 2 2 2 = 182 A 10 - 6 B = P2 Hence, uP1 = - 52.8° and uP2 = 37.2° Ans. b) Maximum In-Plane Shear Strain: Applying Eq. 10–11, g max Px - Py 2 gxy 2 in-plane = b + a b 2 Aa 2 2 g max in-plane = 2B -180 2 250 - 300 2 -6 b + a b R A 10 B A 2 2 a = 187 A 10 - 6 B Ans. 972 x © 2014 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. 10–14. Continued Orientation of the Maximum In-Plane Shear Strain: Applying Eq. 10–10, tan 2us = - Px - Py us = - 7.76° and The proper sign of gx¿y¿ 2 Px - Py = - = - gxy 2 g max in-plane 250 - 300 = - 0.2778 - 180 Ans. 82.2° can be determined by substituting u = - 7.76° into Eq. 10–6. sin 2u + gxy 2 cos 2u gx¿y¿ = { -[250 - 300] sin (- 15.52°) + ( -180) cos ( -15.52°)} A 10 - 6 B = - 187 A 10 - 6 B Normal Strain and Shear Strain: In accordance with the sign convention, Px = 250 A 10 - 6 B Py = 300 A 10 - 6 B gxy = - 180 A 10 - 6 B Average Normal Strain: Applying Eq. 10–12, Px + Py Pavg = 2 = c 250 + 300 d A 10 - 6 B = 275 A 10 - 6 B 2 Ans. Ans: P1 = 368(10 - 6), P2 = 182(10 - 6), up1 = - 52.8° and up2 = 37.2°, gmax = 187(10 - 6), us = -7.76° and 82.2°, in-plane Pavg = 275 (10 - 6) 973 © 2014 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. y *10–16. The state of strain on an element has components Px = -300 (10 - 6), Py = 100 (10 - 6), gxy = 150 (10-6). Determine the equivalent state of strain, which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding elements for these states of strain with respect to the original element. Pydy dy gxy 2 gxy 2 dx In-Plane Principal Strains: Px = - 300(10 - 6), Py = 100(10 - 6), and gxy = 150(10 - 6). We obtain Px + Py P1, 2 = = c ; 2 a A Px - Py 2 2 b + a gxy 2 b 2 -300 + 100 150 2 - 300 - 100 2 ; d(10 - 6) a b + a 2 A 2 2 b = (- 100 ; 213.60)(10 - 6) P1 = 114(10 - 6) P2 = - 314(10 - 6) Ans. Orientation of Principal Strains: tan 2up = gxy 150(10 - 6) = Px - Py ( -300 - 100)(10 - 6) = - 0.375 up = - 10.28° and 79.72° Substituting u = - 10.28° into Px + Py Px¿ = 2 = c Px - Py + 2 cos 2u + gxy 2 sin 2u - 300 + 100 - 300 - 100 150 + cos ( - 20.56°) + sin ( -20.56°) d(10 - 6) 2 2 2 = - 314(10 - 6) = P2 Thus, (up)1 = 79.7° and (up)2 = - 10.3° Ans. The deformed element for the state of principal strain is shown in Fig. a. Maximum In-Plane Shear Strain: gmax in-plane 2 gmax in-plane = A a = c2 Px - Py A 2 a 2 b + a gxy 2 b 2 150 2 - 300 - 100 2 d(10 - 6) = 427(10 - 6) b + a 2 2 b 974 Ans. Pxdx x © 2014 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. 10–16. Continued Orientation of Maximum In-Plane Shear Strain: tan 2us = - a Px - Py gxy b = -c (- 300 - 100)(10 - 6) 150(10 - 6) d(10 - 6) = 2.6667 us = 34.7° and 125° The algebraic sign for gmax Ans. in-plane gx¿y¿ 2 = -a Px - Py 2 when u = us = 34.7° can be determined using b sin 2u + gxy 2 cos 2u gx¿y¿ = [ - (- 300 - 100) sin 69.44° + 150 cos 69.44°](10 - 6) = 427(10 - 6) Average Normal Strain: Px + Py Pavg = 2 = a - 300 + 100 b(10 - 6) = - 100(10 - 6) 2 Ans. The deformed element for this state of maximum in-plane shear strain is shown in Fig. b. 975 © 2014 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. 10–17. Solve part (a) of Prob. 10–3 using Mohr’s circle. Px = 120(10 - 6) Py = - 180(10 - 6) gxy = 150(10 - 6) A (120, 75)(10 - 6) C (- 30, 0)(10 - 6) R = C 2[120 - (-30)]2 + (75)2 D (10 - 6) = 167.71 (10 - 6) P1 = (- 30 + 167.71)(10 - 6) = 138(10 - 6) Ans. P2 = (- 30 - 167.71)(10 - 6) = - 198(10 - 6) Ans. tan 2uP = a 75 b , uP = 13.3° 30 + 120 Ans. Ans: P1 = 138(10 - 6), P2 = - 198(10 - 6), up = 13.3° 976 © 2014 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. 10–18. Solve part (b) of Prob. 10–3 using Mohr’s circle. Px = 120(10 - 6) Py = - 180(10 - 6) A (120, 75)(10 - 6) C ( - 30, 0)(10 - 6) gxy = 150(10 - 6) R = C 2[120 - ( - 30)]2 + (75)2 D (10 - 6) = 167.71 (10 - 6) gxy 2 gmax = R = 167.7(10 - 6) in-plane = 335(10 - 6) Ans. Pavg = - 30 (10 - 6) tan 2us = 120 + 30 75 Ans. us = - 31.7° Ans. Ans: g max in-plane = 335(10 - 6), Pavg = - 30(10 - 6), us = - 31.7° 977 © 2014 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. 10–19. Solve Prob. 10–4 using Mohr’s circle. Px = 850(10 - 6) Py = 480(10 - 6) gxy = 650(10 - 6) gxy 2 = 325(10 - 6) A(850, 325)(10 - 6) C(665, 0)(10 - 6) R = [2(850 - 665)2 + 3252](10 - 6) = 373.97(10 - 6) P1 = (665 + 373.97)(10 - 6) = 1039(10 - 6) Ans. P2 = (665 - 373.97)(10 - 6) = 291(10 - 6) Ans. tan 2up = 325 850 - 665 (Mohr’s circle) 2up = 60.35° up = 30.2° Ans. (element) gmax in-plane 2 gmax in-plane = R = 2(373.97)(10 - 6) = 748(10 - 6) Ans. Pavg = 665(10 - 6) Ans. 2us = 90° - 2up = 29.65° us = - 14.8° Ans. (Mohr’s circle) (element) Ans: P1 = 1039(10 - 6), P2 = 291(10 - 6), up = 30.2°, gmax = 748(10 - 9), Pavg = 665(10 - 6), in-plane us = - 14.8° 978 © 2014 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. *10–20. Solve Prob. 10–5 using Mohr’s circle. a) Px = 520(10 - 6) Py = - 760(10 - 6) gxy = - 750(10 - 6) gxy 2 = - 375(10 - 6) A(520, -375); C(- 120, 0) R = 2(520 + 120)2 + 3752 = 741.77 P1 = 741.77 - 120 = 622(10 - 6) Ans. P2 = - 120 - 741.77 = - 862(10 - 6) Ans. tan 2up1 = 375 = 0.5859 (120 + 520) up1 = 15.2° b) gmax gmax in-plane in-plane Ans. = 2R = 2(741.77) = - 1484(10 - 6) Ans. Pavg = - 120(10 - 6) tan 2us = Ans. (120 + 520) = 1.7067 375 us = 29.8° Ans. 979 © 2014 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. 10–21. Solve Prob. 10–7 using Mohr’s circle. Px = 150(10 - 6) u = - 30° Py = 200(10 - 6) gxy = - 700(10 - 6) gxy 2 = - 350(10 - 6) 2u = - 60° A(150, - 350); C(175, 0) R = 2(175 - 150)2 + (- 350)2 = 350.89 Coordinates of point B: Px¿ = 350.89 cos 34.09° + 175 = 466(10 - 6) gx¿y¿ 2 Ans. = - 350.89 sin 34.09° gx¿y¿ = - 393(10 - 6) Ans. Coordinates of point D: Py¿ = 175 - 350.89 cos 34.09° = - 116(10 - 6) Ans. Ans: -6 Px¿ = 466(10 - 6), gx¿y¿ = - 393(10 ), -6 Py¿ = - 116(10 ) 980 © 2014 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. 10–22. The strain at point A on the bracket has components Px = 300110-62, Py = 550110-62, -6 gxy = - 650110 2, Pz = 0. Determine (a) the principal strains at A in the x – y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. Px = 300(10 - 6) A(300, - 325)10 - 6 Py = 550(10 - 6) gxy = - 650(10 - 6) y gxy 2 A = - 325(10 - 6) x C(425, 0)10 - 6 R = C 2(425 - 300)2 + ( - 325)2 D 10 - 6 = 348.2(10 - 6) a) P1 = (425 + 348.2)(10 - 6) = 773(10 - 6) Ans. P2 = (425 - 348.2)(10 - 6) = 76.8(10 - 6) Ans. b) g max in-plane = 2R = 2(348.2)(10 - 6) = 696(10 - 6) Ans. 773(10 - 6) ; 2 Ans. c) gabs max 2 = gabs max = 773(10 - 6) Ans: (a) P1 = 773(10 - 6), P2 = 76.8(10 - 6), (b) gmax = 696(10 - 6), in-plane (c) gabs = 773(10 - 6) max 981 © 2014 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. 10–23. The strain at point A on a beam has components Px = 450(10 - 6), Py = 825(10 - 6), gxy = 275(10 - 6), Pz = 0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. Px = 450(10 - 6) Py = 825(10 - 6) A(450, 137.5)10 - 6 gxy = 275(10 - 6) A gxy 2 = 137.5(10 - 6) C(637.5, 0)10 - 6 R = [2(637.5 - 450)2 + 137.52]10 - 6 = 232.51(10 - 6) a) P1 = (637.5 + 232.51)(10 - 6) = 870(10 - 6) Ans. P2 = (637.5 - 232.51)(10 - 6) = 405(10 - 6) Ans. gmax = 2R = 2(232.51)(10 - 6) = 465(10 - 6) Ans. 870(10 - 6) ; 2 Ans. b) in-plane c) gabs max 2 = gabs max = 870(10 - 6) Ans: (a) P1 = 870(10 - 6), P2 = 405(10 - 6), = 465(10 - 6), (b) gmax in-plane (c) gabs = 870(10 - 6) max 982 © 2014 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. *10–24. The steel bar is subjected to the tensile load of 500 lb. If it is 0.5 in. thick determine the three principal strains. E = 29 (103) ksi , n = 0.3. 2 in. 500 lb 500 lb 15 in. sx = 500 = 500 psi 2(0.5) Px = 1 1 (500) = 17.2414 (10 - 6) (s ) = E x 29(106) sy = 0 sz = 0 Py = Pz = - vPx = - 0.3(17.2414)(10 - 6) = - 5.1724(10 - 6) P1 = 17.2(10 - 6) P2, 3 = - 5.17(10 - 6) Ans. 983 © 2014 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. 10–25. The 45° strain rosette is mounted on a machine element. The following readings are obtained from each gauge: Pa = 650(10 - 6), Pb = - 300(10 - 6), Pc = 480(10 - 6). Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and associated average normal strain. In each case show the deformed element due to these strains. a b 45⬚ c Pa = 650(10 - 6) Pb = - 300(10 - 6) Pc = 480(10 - 6) ua = 90° ub = 135° uc = 180° 45⬚ Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua 650(10 - 6) = Px cos2 90° + Py sin2 90° + gxy sin 90° cos 90° Px = 650(10 - 6) Pc = Px cos2 uc + Py sin2 uc + gxy sin uc cos uc 480(10 - 6) = Px cos2 180° + Py sin2 180° + gxy sin 180° cos 180° Py = 480(10 - 6) Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub - 300(10 - 6) = 480(10 - 6) cos2 135° + 650(10 - 6) sin2 135° + gxy sin 135° cos 135° gxy = 1730(10 - 6) gxy 2 = 865(10 - 6) A(480, 865)10 - 6 C(565, 0)10 - 6 R = ( 2(565 - 480)2 + 8652)10 - 6 = 869.17(10 - 6) a) P1 = (565 + 869.17)10 - 6 = 1434(10 - 6) Ans. P2 = (565 - 869.17)10 - 6 = - 304(10 - 6) Ans. tan 2up = 865 565 - 480 2up = 84.39° (Mohr’s circle) up = - 42.19° (element) b) gmax in-plane = 2R = 2(869.17)(10 - 6) = 1738(10 - 6) Ans. Pavg = 565(10 - 6) Ans. 2us = 90° - 2up = 5.61° (Mohr’s circle) us = 2.81° (element) Ans: P1 = 1434(10 - 6), P2 = - 304(10 - 6), gmax = 1738(10 - 6), Pavg = 565(10 - 6) in-plane 984 © 2014 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. 10–26. The 60° strain rosette is attached to point A on the surface of the support. Due to the loading the strain gauges give a reading of Pa = 300(10 - 6), Pb = - 150 (10-6), and Pc = - 450 (10 - 6). Use Mohr’s circle and determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of each element that has these states of strain with respect to the x axis. Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua 300(10 - 6) = Px cos2 0° + Py sin2 0° + gxy sin 0° cos 0° Px = 300(10 - 6) Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub - 150(10 - 6) = 300(10 - 6) cos2 60° + Py sin2 60° + gxy sin 60° cos 60° 0.75Py + 0.43301gxy = - 225(10 - 6) (1) 2 Pc = Px cos uc + Py sin uc + gxy sin uc cos uc - 450(10 - 6) = 300(10 - 6) cos2 120° + Py sin2 120° + gxy sin 120° cos 120° 0.75Py - 0.43301gxy = - 525(10 - 6) (2) Solving Eqs. (1) and (2), gxy = 346.41(10 - 6) gxy Construction of the Circle: Px = 300(10 - 6), Py = - 500(10 - 6), and = 173.20(10 - 6). 2 Thus Px + Py Pavg = 2 = c 300 + ( - 500) d (10 - 6) = - 100(10 - 6) 2 Ans. The coordinates of reference point A and center of C of the circle are A(300, 173.20)(10 - 6) x a Normal and Shear Strain: With ua = 0°, ub = 60°, and uc = 120°, we have Py = - 500(10 - 6) 60⬚ 60⬚ A 2 c b C( - 100, 0)(10 - 6) Thus, the radius of the circle is R = CA = a2[300 - ( - 100)]2 + 173.202 b(10 - 6) = 435.89(10 - 6) Using these results, the circle is shown in Fig. a. In-Plane Principal Strains: The coordinates of reference points B and D represent P1 and P2, respectively. P1 = (- 100 + 435.89)(10 - 6) = 336(10 - 6) Ans. P2 = (- 100 - 435.89)(10 - 6) = - 536(10 - 6) Ans. 985 © 2014 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. 10–26. Continued Orientation of Principal Strain: Referring to the geometry of the circle, tan 2(up)1 = (up)1 = 11.7° 173.20(10 - 6) (300 + 100)(10 - 6) = 0.43301 (counterclockwise) Ans. The deformed element for the state of principal strain is shown in Fig. b. Maximum In-Plane Shear Strain: The coordinates of point E represent Pavg and gmax in-plane . Thus 2 gmax in-plane 2 gmax in-plane = R = (435.89)(10 - 6) = 872(10 - 6) Ans. Orientation of Maximum In-Plane Shear Strain: Referring to the geometry of the circle, tan 2us = us = 33.3° 300 + 100 = 2.3094 173.20 (clockwise) Ans. The deformed element for the state of maximum in-plane shear strain is shown in Fig. c. Ans: P1 = 336(10 - 6), P2 = - 536(10 - 6), up1 = 11.7° (counterclockwise), gmax = 872(10 - 6), Pavg = - 100(10 - 6), in-plane us = 33.3° (clockwise) 986 © 2014 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. 10–27. The strain rosette is attached at the point on the surface of the pump. Due to the loading, the strain gauges give a reading of Pa = - 250(10 - 6), Pb = - 300 (10-6), and Pc = - 200 (10 - 6). Determine (a) the in-plane principal strains, and (b) the maximum in-plane shear strain. Specify the orientation of each element that has these states of strain with respect to the x axis. b 60⬚ xx A 60⬚ c Normal and Shear Strains: With ua = 0°, ub = 60°, and uc = - 60°, we have Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua -250(10 - 6) = Px cos2 0° + Py sin2 0° + gxy sin 0° cos 0° Px = - 250(10 - 6) Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub 300(10 - 6) = - 250(10 - 6) cos2 60° + Py sin2 60° + gxy sin 60° cos 60° 0.75Py + 0.43301gxy = 362.5(10 - 6) (1) Pc = Px cos2 uc + Py sin2 uc + gxy sin uc cos uc - 200(10 - 6) = - 250(10 - 6) cos2 ( -60°) + Py sin2 ( -60°) + gxy sin ( -60°) cos ( -60°) 0.75Py - 0.43301gxy = - 137.5(10 - 6) (2) Solving Eqs. (1) and (2), we obtain Py = 150(10 - 6) gxy = 577.35(10 - 6) gxy Construction of the Circle: Px = - 250(10 - 6), Py = 150(10 - 6), and = 288.68(10 - 6). 2 Thus Px + Py - 250 + 150 Ans. = a b(10 - 6) = - 50(10 - 6) Pavg = 2 2 The coordinates of reference point A and center of C of the circle are A( - 250, 288.68)(10 - 6) a C( -50, 0)(10 - 6) Thus, the radius of the circle is R = CA = a2[ -250 - ( - 50)]2 + 288.682 b(10 - 6) = 351.19(10 - 6) Using these results, the circle is shown in Fig. a. In-Plane Principal Strains: The coordinates of reference points B and D represent P1 and P2, respectively. P1 = (- 50 + 351.19)(10 - 6) = 301(10 - 6) Ans. P2 = (- 50 - 351.19)(10 - 6) = - 401(10 - 6) Ans. 987 © 2014 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. 10–27. Continued Orientation of Principal Strain: Referring to the geometry of the circle, tan 2(up)2 = 288.68 = 1.4434 250 - 50 (up)2 = 27.6° (clockwise) Ans. The deformed element for the state of principal strain is shown in Fig. b. Maximum In-Plane Shear Strain: The coordinates of point E represent Pavg and gmax in-plane . Thus 2 gmax in-plane 2 gmax in-plane = R = 351.19(10 - 6) = 702(10 - 6) Ans. Orientation of Maximum In-Plane Shear Strain: Referring to the geometry of the circle, tan 2us = 250 - 50 = 0.6928 288.68 us = 17.4° (counterclockwise) Ans. The deformed element for the state of maximum in-plane shear strain is shown in Fig. c. Ans: P1 = 301(10 - 6), P2 = - 401(10 - 6), up2 = 27.6° (clockwise), gmax = 702(10 - 6), Pavg = - 50(10 - 6), in-plane us = 17.4° (counterclockwise) 988 © 2014 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. *10–28. The 60° strain rosette is mounted on a beam. The following readings are obtained from each gauge: Pa = 250(10 - 6), Pb = - 400 (10-6), Pc = 280(10 - 6). Determine (a) the in-plane principal strains and their orientation, and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains. b a 60⬚ c Pa = 250(10 - 6) Pb = - 400(10 - 6) Pc = 280(10 - 6) ua = 60° ub = 120° uc = 180° Pc = Px cos2 uc + Pg sin2 uc + gxy sin uc cos uc 280(10 - 6) = Px cos2 180° + Py sin2 180° + gxy sin 180° cos 180° Px = 280(10 - 6) Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua 250(10 - 6) = Px cos2 60° + Py sin2 60° + gxy sin 60° cos 60° 250(10 - 6) = 0.25Px + 0.75Py + 0.433gxy (1) Pb = Px cos 2 ub + Py sin2 ub + gxy sin ub cos ub - 400(10 - 6) = Px cos2 120° + Py sin 2 120° + gxy sin 120° cos 120° - 400(10 - 6) = 0.25Px + 0.75Py - 0.433gxy (2) Subtract Eq. (2) from Eq. (1) 650(10 - 6) = 0.866gxy gxy = 750.56(10 - 6) Py = - 193.33(10 - 6) gxy 2 = 375.28(10 - 6) A(280, 375.28)10 - 6 C(43.34, 0)10 - 6 R = ( 2(280 - 43.34)2 + 375.282)10 - 6 = 443.67(10 - 6) a) P1 = (43.34 + 443.67)10 - 6 = 487(10 - 6) Ans. P2 = (43.34 - 443.67)10 - 6 = - 400(10 - 6) Ans. tan 2up = 375.28 280 - 43.34 2up = 57.76° (Mohr’s circle) up = 28.89° (element) 989 60⬚ 60⬚ © 2014 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. 10–28. Continued b) gmax in-plane = 2R = 2(443.67)(10 - 6) = 887(10 - 6) Ans. Pavg = 43.3(10 - 6) Ans. 2us = 90° - 2uy = 32.24° (Mohr’s circle) us = 16.12° (element) 990 © 2014 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. 10–30. For the case of plane stress, show that Hooke’s law can be written as sx = E E 1Px + nPy2, sy = 1Py + nPx2 11 - n22 11 - n22 Generalized Hooke’s Law: For plane stress, sz = 0. Applying Eq. 10–18, Px = 1 A s - n sy B E x nEPx = A sx - n sy B n nEPx = n sx - n2 sy Py = (1) 1 (s - n sx) E y E Py = - n sx + sy (2) Adding Eq. (1) and Eq.(2) yields. nE Px + E Py = sy - n2 sy sy = E A nPx + Py B 1 - n2 (Q.E.D.) Substituting sy into Eq. (2) E Py = - nsx + sx = E A n Px + Py B 1 - n2 E A n Px + Py B 2 n (1 - n ) EPy - n EnPx + EPy - EPy + EPy n2 = = n(1 - n2) E (Px + n Py) 1 - n2 (Q.E.D.) 991 © 2014 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. 10–31. Use Hooke’s law, Eq. 10–18, to develop the straintransformation equations, Eqs. 10–5 and 10–6, from the stress-transformation equations, Eqs. 9–1 and 9–2. Stress Transformation Equations: sx¿ = sx + sy 2 tx¿y¿ = - sy¿ = sx - sy + sx - sy 2 sx + sy 2 (1) sin 2u + txy cos 2u sx - sy - 2 cos 2u + txy sin 2u 2 (2) cos 2u - txy sin 2u (3) Hooke’s Law: Px = n sy sx E E (4) Py = sy - n sx + E E (5) txy = G gxy G = (6) E 2(1 + n) (7) From Eqs. (4) and (5) (1 - n)(sx + sy) Px + Py = (8) E (1 + n)(sx - sy) Px - Py = (9) E From Eqs. (6) and (7) txy = E g 2(1 + n) xy (10) From Eq. (4) Px¿ = n sy¿ sx¿ E E (11) Substitute Eqs. (1) and (3) into Eq. (11) (1 - n)(sx + sy) Px¿ = (1 + n)(sx - sy) + 2E 2E cos 2u + (1 + n)txy sin 2u E (12) By using Eqs. (8), (9) and (10) and substitute into Eq. (12), Px + Py Px¿ = 2 Px - Py + 2 cos 2u + gxy 2 (Q.E.D.) sin 2u 992 © 2014 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. 10–31. Continued From Eq. (6). tx¿y¿ = Ggx¿y¿ = E g 2(1 + n) x¿y¿ (13) Substitute Eqs. (13), (6) and (9) into Eq. (2), E(Px - Py) E E gx¿y¿ = sin 2u + g cos 2u 2(1 + n) 2(1 + n) 2(1 + n) xy gx¿y¿ 2 (Px - Py) = - 2 sin 2u + gxy 2 (Q.E.D.) cos 2u 993 © 2014 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. *10–32. The principal plane stresses and associated strains in a plane at a point are s1 = 36 ksi, s2 = 16 ksi, P1 = 1.02(10-3), P2 = 0.180(10-3). Determine the modulus of elasticity and Poisson’s ratio. s3 = 0 P1 = 1 [s - v(s2 + s3)] E 1 1.02(10 - 3) = 1 [36 - v(16)] E 1.02(10 - 3)E = 36 - 16v P2 = (1) 1 [s - v(s1 + s3)] E 2 0.180(10 - 3) = 1 [16 - v(36)] E 0.180(10 - 3)E = 16 - 36v (2) Solving Eqs. (1) and (2) yields: E = 30.7(103) ksi Ans. v = 0.291 Ans. 994 © 2014 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. 10–33. A rod has a radius of 10 mm. If it is subjected to an axial load of 15 N such that the axial strain in the rod is Px = 2.75(10-6), determine the modulus of elasticity E and the change in its diameter. n = 0.23. sx = 15 = 47.746 kPa, p(0.01)2 Px = 1 [s - v(sy + sz)] E x 2.75(10 - 6) = sy = 0, sz = 0 1 [47.746(103) - 0.23(0 + 0)] E E = 17.4 GPa Ans. Py = Pz = - vPx = - 0.23(2.75)(10 - 6) = - 0.632(10 - 6) ¢d = 20(- 0.632(10 - 6)) = - 12.6(10 - 6) mm Ans. Ans: E = 17.4 GPa , ¢d = - 12.6(10 - 6) mm 995 © 2014 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. 10–34. The polyvinyl chloride bar is subjected to an axial force of 900 lb. If it has the original dimensions shown determine the change in the angle u after the load is applied. Epvc = 800 (103) psi, npvc = 0.20. 900 lb 900 lb 3 in. u 6 in. sx = 900 = 300 psi 3(1) sy = 0 Px = = Py = = 1 in. sz = 0 1 [sx - v(sy + sz)] E 1 [300 - 0] = 0.375(10 - 3) 800(103) 1 [s - v(sx + sz)] E y 1 [0 - 0.2(300 + 0)] = - 75(10 - 6) 800(103) a¿ = 6 + 6(0.375)(10 - 3) = 6.00225 in. b¿ = 3 + 3( - 75)(10 - 6) = 2.999775 in. 3 u = tan - 1 a b = 26.56505118° 6 u¿ = tan - 1 a 2.999775 b = 26.55474088° 6.00225 ¢u = u¿ - u = 26.55474088° - 26.56505118° = - 0.0103° Ans. Ans: ¢u = - 0.0103° 996 © 2014 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. 10–35. The polyvinyl chloride bar is subjected to an axial force of 900 lb. If it has the original dimensions shown determine the value of Poisson’s ratio if the angle u decreases by ¢u = 0.01° after the load is applied. Epvc = 800(103) psi. sx = 900 = 300 psi 3(1) Px = 1 [s - vpvc (sy + sz)] E x = Py = = sy = 0 900 lb 900 lb 3 in. u 6 in. 1 in. sz = 0 1 [300 - 0] = 0.375(10 - 3) 800(103) 1 [s - vpvc (sx + sz)] E y 1 [0 - vpvc (300 + 0)] = - 0.375(10 - 3)vpvc 800(103) a¿ = 6 + 6(0.375)(10 - 3) = 6.00225 in. b¿ = 3 + 3( -0.375)(10 - 3)vpvc = 3 - 1.125(10 - 3)vpvc 3 u = tan - 1 a b = 26.56505118° 6 u¿ = 26.56505118° - 0.01° = 26.55505118° tan u¿ = 0.49978185 = 3 - 1.125(10 - 3)vpvc 6.00225 vpvc = 0.164 Ans. Ans: npvc = 0.164 997 © 2014 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. *10–36. The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Est = 200 GPa and nst = 0.3. 20 mm 1000 r = = 100 7 10, the thin wall analysis is valid to t 10 determine the normal stress in the wall of the spherical vessel. This is a plane stress Normal Stresses: Since problem where smin = 0 since there is no load acting on the outer surface of the wall. smax = slat = pr p(1000) = = 50.0p 2t 2(10) (1) Normal Strains: Applying the generalized Hooke’s Law with Pmax = Plat = 0.012 = 0.600 A 10 - 3 B mm>mm 20 Pmax = 1 C s - n (slat + smin) D E max 0.600 A 10 - 3 B = 1 [50.0p - 0.3 (50.0p + 0)] 200(109) p = 3.4286 MPa = 3.43 MPa Ans. From Eq. (1) smax = slat = 50.0(3.4286) = 171.43 MPa Maximum In-Plane Shear (Sphere’s Surface): Mohr’s circle is simply a dot. As the result, the state of stress is the same consisting of two normal stresses with zero shear stress regardless of the orientation of the element. tmax = 0 Ans. smax - smin 171.43 - 0 = = 85.7 MPa 2 2 Ans. in-plane Absolute Maximum Shear Stress: tabs max = 998 © 2014 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. 10–37. Determine the bulk modulus for each of the following materials: (a) rubber, Er = 0.4 ksi, nr = 0.48, and (b) glass, Eg = 811032 ksi, ng = 0.24. a) For rubber: kr = Er 0.4 = = 3.33 ksi 3 (1 - 2 vr) 3[1 - 2(0.48)] Ans. b) For glass: kg = Eg 3(1 - 2 vg) = 8(103) = 5.13 (103) ksi 3[1 - 2(0.24)] Ans. Ans: (a) kr = 3.33 ksi , (b) kg = 5.13(103) ksi 999 © 2014 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. 10–38. The strain gauge is placed on the surface of a thinwalled steel boiler as shown. If it is 0.5 in. long, determine the pressure in the boiler when the gauge elongates 0.2(10-3) in. The boiler has a thickness of 0.5 in. and inner diameter of 60 in. Also, determine the maximum x, y in-plane shear strain in the material. Est = 29(103) ksi, nst = 0.3. P2 = 0.2(10 - 3) = 400(10 - 6) 0.5 P2 = 1 [s - v(s1 + s3)] E 2 s2 = where, 1 s 2 1 400(10 - 6) = y x 60 in. 0.5 in. s3 = 0 1 1 c s1 - 0.3s1 d 29(103) 2 s1 = 58 ksi Thus, p = 58(0.5) s1 t = = 0.967 ksi r 30 P1 = 1 [s - v(s2 + s3)] E 1 s3 = 0 and s2 = where, P1 = 58 = 29 ksi 2 1 [58 - 0.3(29 + 0)] = 1700(10 - 6) 29(103) gmax P1 - P2 in-plane 2 gmax Ans. in-plane = 2 = (1700 - 400)(10 - 6) = 1.30(10 - 3) Ans. Ans: p = 0.967 ksi , gmax in-plane 1000 = 1.30(10 - 3) © 2014 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. y 10–39. The strain in the x direction at point A on the A-36 structural-steel beam is measured and found to be Px = 100(10-6). Determine the applied load P. What is the shear strain gxy at point A? P 2 in. x A B 3 ft 4 ft 2 in. 12 in. A 6 in. Section Properties: I = 1 (6)(123) = 864 in4 12 QA = y¿A¿ = 5(6)(2) = 60 in3 Normal Stress: s = E Px = 29(103)(100)(10 - 6) = 2.90 ksi s = My ; I 2.90(103) = 4P(12)(4) 864 P = 13050 lb = 13.0 kip txy = gxy = Ans. 13.05(103)(60) VQ = = 151.04 psi It 864(6) txy G 151.04 = 11.0(106) = - 13.7(10 - 6) Ans. Ans: P = 13.0 kip, gxy = - 13.7(10 - 6) 1001 © 2014 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. y *10–40. The strain in the x direction at point A on the A-36 structural-steel beam is measured and found to be Px = 200(10-6). Determine the applied load P. What is the shear strain gxy at point A? P 2 in. x A B 3 ft 4 ft 2 in. A 6 in. Section Properties: QA = y¿A¿ = 5(6)(2) = 60 in3 I = 1 (6)(123) = 864 in4 12 Normal Stress: s = E Px = 29(103)(200)(10 - 6) = 5.80 ksi s = My ; I 5.80(103) = 4P(12)(4) 864 P = 26.1 kip Ans. Shear Stress and Shear Strain: tA = gxy = VQ 26.1(60) = = 0.302 ksi It 864(6) txy G = 0.302 = - 27.5(10 - 6) rad 11.0(103) Ans. 1002 12 in. © 2014 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. y 10–41. If a load of P = 3 kip is applied to the A-36 structural-steel beam, determine the strain Px and gxy at point A. P 2 in. x A B 3 ft 4 ft 2 in. A Section Properties: 12 in. 6 in. 3 QA = y¿A¿ = 2(6)(5) = 60 in I = 1 (6)(123) = 864 in4 12 Normal Stress and Strain: sA = Px = My 12.0(103)(12)(4) = = 666.7 psi I 864 sx 666.7 = 23.0(10 - 6) = E 29(106) Ans. Shear Stress and Shear Strain: tA = gxy = 3(103)(60) VQ = = 34.72 psi It 864(6) txy G 34.72 = 11.0(106) = - 3.16(10 - 6) Ans. Ans: Px = 23.0(10 - 6), gxy = - 3.16(10 - 6) 1003 © 2014 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. 26 ksi 10–42. The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal = 1011032 ksi and nal = 0.33, determine the principal strains. Px = 1 1 (s - v(sy + sz)) = (10 - 0.33( - 15 - 26)) = 2.35(10 - 3) E x 10(103) Py = 1 1 (s - v(sx + sz)) = ( -15 - 0.33)(10 - 26)) = - 0.972(10 - 3) Ans. E y 10(103) Pz = 1 1 (s - v(sx + sy)) = ( -26 - 0.33(10 - 15)) = - 2.44(10 - 3) E z 10(103) Ans. 15 ksi 10 ksi Ans. Ans: Px = 2.35(10 - 3), Py = - 0.972(10 - 3), Pz = - 2.44(10 - 3) 1004 © 2014 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. 10–43. A strain gauge a is attached in the longitudinal direction (x axis) on the surface of the gas tank. When the tank is pressurized, the strain gauge gives a reading of Pa = 100(10-6). Determine the pressure p in the tank. The tank has an inner diameter of 1.5 m and wall thickness of 25 mm. It is made of steel having a modulus of elasticity E = 200 GPa and Poisson’s ratio n = 13. b 45⬚ a x Normal Strain: With ua = 0, we have Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua 100(10 - 6) = Px cos2 0°+ Py sin2 0°+ gxy sin 0° cos 0° Px = 100(10 - 6) 0.75 r = = 30 7 10, thin-wall analysis can be used. t 0.025 Normal Stress: Since sy = s1 = pr p(0.75) = = 30 p t 0.025 sx = s2 = pr p(0.75) = = 15 p 2t 2(0.025) Generalized Hooke’s Law: This is a case of plane stress. Thus, nz = 0. Px = 1 c s - n(sy + sz) d E x 100(10 - 6) = 1 1 c 15p - (30p + 0) d 3 200(109) p = 4(106)N>m2 = 4 MPa Ans. Ans: p = 4 MPa 1005 © 2014 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. 10–44. Strain gauge b is attached to the surface of the gas tank at an angle of 45° with x axis as shown.When the tank is pressurized, the strain gauge gives a reading of Pb = 250(10 - 6). Determine the pressure in the tank.The tank has an inner diameter of 1.5 m and wall thickness of 25 mm. It is made of steel having a modulus of elasticity E = 200 GPa and Poisson’s ratio n = 13. Normal Stress: Since b 0.75 r = = 30 7 10, thin-wall analysis can be used. t 0.025 sy = s1 = pr p(0.75) = = 30 p t 0.025 sx = s2 = pr p(0.75) = = 15 p 2t 2(0.025) Normal Strain: Since no shear force acts along the x and y axes, gxy = 0. With ub = 45, we have Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub 250(10 - 6) = Px cos2 45° + Py sin2 45° + 0 Px + Py = 500(10 - 6) (1) Generalized Hooke’s Law: This is a case of plane stress. Thus, nz = 0. We have Px = 1 c s - n(sy + sz) d E x Px = 1 1 c 15p - (30p + 0) d 3 200(109) Px = 200(109) Py = 1 c s - n(sx + sz) d E y Py = 1 1 c 30p - (15p + 0) d 9 3 200(10 ) 5p (2) 25p Py = (3) 200(109) Substituting Eqs. (2) and (3) into Eq. (1), we obtain 5p 200(109) 25p + 200(109) = 500(10 - 6) p = 3.333(106) N>m2 = 3.33 MPa Ans. 1006 45⬚ a x © 2014 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. y 10–45. A material is subjected to principal stresses sx and sy. Determine the orientation u of a strain gauge placed at the point so that its reading of normal strain responds only to sy and not sx. The material constants are E and n. u x sx = sx + sy sx - sy + 2 cos 2u + txy sin 2u 2 Since txy = 0, sn = sx + sy sx - sy + 2 2 cos 2u cos 2u = 2 cos2 u - 1 sn = sy sy sx sx + + (sx - sy ) cos2 u + 2 2 2 2 = sy (1 - cos2 u) + sx cos2 u = sx cos2 u + sy sin2 u sn + 90° = sx + sy sx - sy - 2 2 cos 2u = sx - sy sy sx + - a b (2 cos2 u - 1) 2 2 2 = sy sy sx sx + - (sx - sy ) cos2 u + 2 2 2 2 = sx (1 - cos2 u) + sy cos2 u = sx sin2 u + sy cos2 u Pn = = 1 (s - n sn + 90°) E n 1 (s cos2 u + sy sin2 u - n sx sin2 u - n sy cos2 u) E x If Pn is to be independent of sx, then cos2 u - n sin2 u = 0 u = tan - 1 a 1 2n or tan2 u = 1/n b Ans. Ans: u = tan - 1 a 1007 1 2n b © 2014 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. 10–46. The principal strains in a plane, measured experimentally at a point on the aluminum fuselage of a jet aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is a case of plane stress, determine the associated principal stresses at the point in the same plane. Eal = 10(103) ksi and nal = 0.33. Normal Stresses: For plane stress, s3 = 0. Normal Strains: Applying the generalized Hooke’s Law. P1 = 1 C s - v (s2 + s3) D E 1 630 A 10 - 6 B = 1 [s1 - 0.33(s2 + 0)] 10(103) 6.30 = s1 - 0.33s2 P2 = [1] 1 C s2 - v (s1 + s3) D E 350 A 10 - 6 B = 1 C s2 - 0.33(s1 + 0) D 10(103) 3.50 = s2 - 0.33s1 [2] Solving Eqs.[1] and [2] yields: s1 = 8.37 ksi s2 = 6.26 ksi Ans. Ans: s1 = 8.37 ksi, s2 = 6.26 ksi 1008 © 2014 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. 10–47. The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal = 10(103) ksi and nal = 0.33, determine the principal strains. 3 ksi P1 = 1 1 e 8 - 0.33 C 3 + (- 4) D f = 833 (10 - 6) C s - v(s2 + s3) D = E 1 10(103) P2 = 1 1 e 3 - 0.33 C 8 + (- 4) D f = 168 (10 - 6) C s2 - v(s1 + s3) D = E 10(103) P3 = 1 1 C s - v(s1 + s2) D = C - 4 - 0.33(8 + 3) D = - 763 (10 - 6) E 3 10(103) 8 ksi 4 ksi Using these results, P1 = 833(10 - 6) P2 = 168(10 - 6) P3 = - 763(10 - 6) Ans. Ans: P1 = 833(10 - 6), P2 = 168(10 - 6), P3 = - 763(10 - 6) 1009 © 2014 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. *10–48. The 6061-T6 aluminum alloy plate fits snugly into the rigid constraint. Determine the normal stresses sx and sy developed in the plate if the temperature is increased by ¢T = 50°C . To solve, add the thermal strain a¢T to the equations for Hooke’s Law. y 400 mm 300 mm x Generalized Hooke’s Law: Since the sides of the aluminum plate are confined in the rigid constraint along the x and y directions, Px = Py = 0. However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0. With the additional thermal strain term, we have Px = 0 = 1 c s - v A sy + sz B d + a¢T E x 1 68.9 A 109 B c sx - 0.35 A sy + 0 B d + 24 a10 - 6 b(50) sx - 0.35sy = - 82.68 A 106 B Py = 0 = (1) 1 C s - v A sx + sz B D + a¢T E y 1 68.9 a 109 b C sy - 0.35(sx + 0) D + 24 A 10 - 6 B (50) sy - 0.35sx = - 82.68 A 106 B (2) Solving Eqs. (1) and (2), sx = sy = - 127.2 MPa = 127 MPa (C) Ans. Since sx = sy and sy 6 sY, the above results are valid. 1010 © 2014 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. 10–49. Initially, gaps between the A-36 steel plate and the rigid constraint are as shown. Determine the normal stresses sx and sy developed in the plate if the temperature is increased by ¢T = 100°F . To solve, add the thermal strain a¢T to the equations for Hooke’s Law. y 0.0015 in. 6 in. 8 in. 0.0025 in. x Generalized Hooke’s Law: Since there are gaps between the sides of the plate and the rigid constraint, the plate is allowed to expand before it comes in contact with the constraint. dy dx 0.0025 0.0015 = = = 0.3125 A 10 - 3 B and Py = = 0.25 A 10 - 3 B . Thus, Px = Lx 8 Ly 6 However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0. With the additional thermal strain term, we have Px = 1 c s - v A sy + sz B d + a¢T E x 0.3125a 10 - 3 b = 1 29.0 a 10 b 3 C sx - 0.32 A sy + 0 B D + 6.60 A 10 - 6 B (100) sx - 0.32sy = - 10.0775 Py = (1) 1 C s - v A sx + sz B D + a¢T E y 0.25 A 10 - 3 B = 1 29.0 A 103 B C sy - 0.32(sx + 0) D + 6.60 A 10 - 6 B (100) sy - 0.32sx = - 11.89 (2) Solving Eqs. (1) and (2), sx = - 15.5 ksi = 15.5 ksi (C) Ans. sy = - 16.8 ksi = 16.8 ksi (C) Ans. Since sx 6 sY and sy 6 sY, the above results are valid. Ans: sx = 15.5 ksi(C), sy = 16.8 ksi(C) 1011 © 2014 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. y 10–50. The steel shaft has a radius of 15 mm. Determine the torque T in the shaft if the two strain gauges, attached to the surface of the shaft, report strains of Px ¿ = - 80(10 - 6) and Py ¿ = 80(10 - 6). Also, compute the strains acting in the x and y directions. Est = 200 GPa, nst = 0.3. T y¿ x¿ 45⬚ x T Px¿ = - 80(10 - 6) Py¿ = 80(10 - 6) Pure shear Px = Py = 0 Ans. Px ¿ = Px cos2 u + Py sin2 u + gxy sin u cos u u = 45° -80(10 - 6) = 0 + 0 + gxy sin 45° cos 45° gxy = - 160(10 - 6) Ans. Also, u = 135° 80(10 - 6) = 0 + 0 + g sin 135° cos 135° gxy = - 160(10 - 6) G = 200(109) E = = 76.923(109) 2(1 + n) 2(1 + 0.3) t = Gg = 76.923(109)(160)(10 - 6) = 12.308(106) Pa p 12.308(106) a b (0.015)4 2 tJ = = 65.2 N # m T = c 0.015 Ans. Ans: Px = Py = 0, gxy = - 160(10 - 6), T = 65.2 N # m 1012 © 2014 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. y 10–51. The shaft has a radius of 15 mm and is made of L2 tool steel. Determine the strains in the x¿ and y¿ direction if a torque T = 2 kN # m is applied to the shaft. T y¿ x¿ 45⬚ x T t = 2(103)(0.015) Tc = 377.26 MPa = p 4 J 2 (0.015 ) Stress-Strain Relationship: gxy = txy G 377.26(106) = - 75.0(109) = - 5.030(10 - 3) rad This is a pure shear case, therefore, Px = Py = 0 Applying Eq. 10–15, Px ¿ = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua Here ua = 45° Px ¿ = 0 + 0 - 5.030(10 - 3) sin 45° cos 45° = - 2.52(10 - 3) Px ¿ = - 2.52(10 - 3) Ans. Py ¿ = 2.52(10 - 3) Ans. Ans: Px¿ = - 2.52(10 - 3), Py¿ = 2.52(10 - 3) 1013 © 2014 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. *10–52. The metal block is fitted between the fixed supports. If the glued joint can resist a maximum shear stress of tallow = 2 ksi, determine the temperature rise that will cause the joint to fail. Take E = 10(103) ksi, n = 0.2, and a = 6.0(10 - 6)>°F. Hint: Use Eq. 10–18 with an additional strain term of a¢T (Eq. 4–4). 40⬚ Normal Strain: Since the aluminum is confined in the y direction by the rigid supports, then Py = 0 and sx = sz = 0. Applying the generalized Hooke’s Law with the additional thermal strain, Py = 0 = 1 C s - v(sx + sz) D + a¢T E y 1 C sy - 0.2(0 + 0) D + 6.0 A 10 - 6 B (¢T) 10.0(103) sy = - 0.06¢T Construction of the Circle: In accordance with the sign convention. sx = 0, sy = - 0.06¢T and txy = 0. Hence. savg = sx + sy 2 = 0 + ( - 0.06¢T) = - 0.03¢T 2 The coordinates for reference points A and C are A (0, 0) and C(-0.03¢T, 0). The radius of the circle is R = 2(0 - 0.03¢T)2 + 0 = 0.03¢T Stress on the inclined plane: The shear stress components tx¿y¿, are represented by the coordinates of point P on the circle. tx¿y¿ = 0.03¢T sin 80° = 0.02954¢T Allowable Shear Stress: tallow = tx¿y¿ 2 = 0.02954¢T ¢T = 67.7 °F Ans. 1014 © 2014 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. 10–53. Air is pumped into the steel thin-walled pressure vessel at C. If the ends of the vessel are closed using two pistons connected by a rod AB, determine the increase in the diameter of the pressure vessel when the internal gauge pressure is 5 MPa. Also, what is the tensile stress in rod AB if it has a diameter of 100 mm? The inner radius of the vessel is 400 mm, and its thickness is 10 mm. Est = 200 GPa and nst = 0.3. C 400 mm A B Circumferential Stress: s = 5(400) pr = = 200 MPa t 10 Note: longitudinal and radial stresses are zero. Circumferential Strain: P = 200(106) s = 1.0(10 - 3) = E 200(109) ¢d = P d = 1.0(10 - 3)(800) = 0.800 mm Ans. For rod AB: + ; ©Fx = 0; p TAB - 5(106) a b (0.82 - 0.12) = 0 4 TAB = 2474 kN sAB = 2474(103) TAB = 315 MPa = p 2 AAB 4 (0.1 ) Ans. Ans: ¢d = 0.800 mm, sAB = 315 MPa 1015 © 2014 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. 10–54. Determine the increase in the diameter of the pressure vessel in Prob. 10–53 if the pistons are replaced by walls connected to the ends of the vessel. C 400 mm A B Principal Stress: s1 = 5(400) pr = = 200 MPa; t 10 s2 = 1 s = 100 MPa 2 1 s3 = 0 Circumferential Strain: P1 = 1 1 [200(106) - 0.3{100(106) + 0}] [s - n(s2 + s3)] = E 1 200(109) = 0.85(10 - 3) ¢d = P1 d = 0.85(10 - 3)(800) = 0.680 mm Ans. Ans: ¢d = 0.680 mm 1016 © 2014 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. 10–55. A thin-walled spherical pressure vessel having an inner radius r and thickness t is subjected to an internal pressure p. Show that the increase in the volume within the vessel is ¢V = 12ppr4>Et211 - n2. Use a small-strain analysis. pr 2t s1 = s2 = s3 = 0 P1 = P2 = 1 (s - vs2) E 1 P1 = P2 = pr (1 - v) 2t E P3 = 1 (- v(s1 + s2)) E P3 = - V = v pr tE 4pr3 3 V + ¢V = 4p 4pr3 ¢r 3 ) (r + ¢r)3 = (1 + r 3 3 where ¢V V V, ¢r V r V + ¢V = PVol = ¢r 4p r3 b a1 + 3 r 3 ¢r ¢V = 3a b V r Since P1 = P2 = PVol = 3P1 = 2p(r + ¢r) - 2p r ¢r = r 2p r 3pr (1 - v) 2t E ¢V = VPVol = 2pp r4 (1 - v) Et (Q.E.D.) 1017 © 2014 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. *10–56. The thin-walled cylindrical pressure vessel of inner radius r and thickness t is subjected to an internal pressure p. If the material constants are E and n, determine the strains in the circumferential and longitudinal directions. Using these results, compute the increase in both the diameter and the length of a steel pressure vessel filled with air and having an internal gauge pressure of 15 MPa. The vessel is 3 m long, and has an inner radius of 0.5 m and a thickness of 10 mm. Est = 200 GPa, nst = 0.3. 3m 0.5 m Normal Stress: s1 = pr t s2 = pr 2t s3 = 0 Normal Strain: Pcir = = 1 [s - n(s2 + s3)] E 1 npr pr 1 pr a b = (2 - n) E t 2t 2Et Plong = = Ans. 1 [s - n(s1 + s3)] E 2 npr pr 1 pr a b = (1 - 2n) E 2t t 2Et Ans. Numerical Substitution: 15(106)(0.5) Pcir = 2(200)(109)(0.01) (2 - 0.3) = 3.1875 (10 - 3) ¢d = Pcir d = 3.1875 (10 - 3)(1000) = 3.19 mm 15(106)(0.5) Plong = 2(200)(109)(0.01) Ans. (1 - 2(0.3)) = 0.75(10 - 3) ¢L = Plong L = 0.75 (10 - 3)(3000) = 2.25 mm Ans. 1018 © 2014 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. 10–57. Estimate the increase in volume of the tank in Prob.10–56. 3m 0.5 m By basic principles, ¢V = p(r + ¢r)2(L + ¢L) - p r2 L = p(r2 + ¢r2 + 2 r ¢r)(L + ¢L) - p r2 L = p(r2L + r2 ¢L + ¢r2L + ¢r2 ¢L + 2 r ¢r L + 2 r ¢r ¢L - r2 L) = p(r2 ¢L + ¢r2 L + ¢r2 ¢L + 2 r ¢r L + 2 r ¢r ¢L) Neglecting the second order terms, ¢V = p(r2 ¢L + 2 r ¢r L) From Prob. 10–56, ¢L = 0.00225 m ¢r = ¢d = 0.00159375 m 2 ¢V = p[(0.52)(0.00225) + 2(0.5)(0.00159375)(3)] = 0.0168 m3 Ans. Ans: ¢V = 0.0168 m3 1019 © 2014 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. z 10–58. A soft material is placed within the confines of a rigid cylinder which rests on a rigid support. Assuming that Px = 0 and Py = 0, determine the factor by which the modulus of elasticity will be increased when a load is applied if n = 0.3 for the material. P x y Normal Strain: Since the material is confined in a rigid cylinder. Px = Py = 0. Applying the generalized Hooke’s Law, Px = 1 C s - v(sy + sz) D E x 0 = sx - v(sy + sz) Py = [1] 1 C sy - v(sx + sz) D E 0 = sy - v(sx + sz) [2] Solving Eqs.[1] and [2] yields: sx = sy = v s 1 - v z Thus, Pz = = 1 C s - v(sx + sy) D E z v v 1 cs - va s + s bd E z 1 - v z 1 - v z sz = E c1 - 2v2 d 1 - v = sz 1 - v - 2v2 c d E 1 - v = sz (1 + v)(1 - 2v) c d E 1 - v Thus, when the material is not being confined and undergoes the same normal strain of Pz, then the requtred modulus of elasticity is E¿ = sz = Pz The increase factor is k = 1 - v E (1 - 2v)(1 + v) E¿ 1 - v = E (1 - 2v)(1 + v) = 1 - 0.3 [1 - 2(0.3)](1 + 0.3) Ans. = 1.35 Ans: k = 1.35 1020 © 2014 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. 10–59. A material is subjected to plane stress. Express the distortion-energy theory of failure in terms of sx , sy , and txy . Maximum distortion energy theory: (s21 - s1 s2 + s22) = s2Y s1,2 = sx + sy Let a = ; 2 sx + sy 2 s1 = a + b; A a (1) sx - sy and b = 2 A a 2 2 b + txy sx - sy 2 2 2 b + txy s2 = a - b s21 = a2 + b2 + 2 a b; s22 = a2 + b2 - 2 a b s1 s2 = a2 - b2 From Eq. (1) (a2 + b2 + 2 a b - a2 + b2 + a2 + b2 - 2 a b) = sY2 (a2 + 3 b2) = s2Y (sx + sy)2 4 + 3 (sx - sy)2 4 + 3 t2xy = s2Y s2x + s2y - sxsy + 3 t2xy = s2Y Ans. Ans: s2x + s2y - sxsy + 3t2xy = s2y 1021 © 2014 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. *10–60. A material is subjected to plane stress. Express the maximum-shear-stress theory of failure in terms of sx , sy , and txy . Assume that the principal stresses are of different algebraic signs. Maximum shear stress theory: |s1 - s2| = sY s1,2 = (1) sx + sy ; 2 ` s1 - s2 ` = 2 A a A a sx - sy 2 sx - sy 2 2 2 b + txy 2 b + txy 2 From Eq. (1) 4 ca sx - sy 2 2 b + t2xy d = s2Y 2 (sx - sy) + 4 t2xy = s2Y Ans. 1022 © 2014 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. 10–61. The yield stress for a zirconium-magnesium alloy is sY = 15.3 ksi. If a machine part is made of this material and a critical point in the material is subjected to in-plane principal stresses s1 and s2 = - 0.5s1, determine the magnitude of s1 that will cause yielding according to the maximum-shear-stress theory. sY = 15.3 ksi s1 - s2 = 15.3 s1 - (- 0.5 s1) = 15.3 s1 = 10.2 ksi Ans. Ans: s1 = 10.2 ksi 1023 © 2014 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. 10–62. Solve Prob. 10–61 using the maximum-distortion energy theory. s21 - s1 s2 + s22 = sY2 s21 - s1(- 0.5s1) + (- 0.5s1)2 = sY2 1.75 s21 = (15.3)2 s1 = 11.6 ksi Ans. Ans: s1 = 11.6 ksi 1024 © 2014 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. 10–63. An aluminum alloy is to be used for a drive shaft such that it transmits 25 hp at 1500 rev>min. Using a factor of safety of 2.5 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum-distortion-energy theory. sY = 3.5 ksi. 1500(2p) = 50p 60 T = P v T = 25(550)(12) 3300 = p 50p t = Tc , J t = 3300 p c p 4 2c s1 = v = J = = p 4 c 2 6600 p2c3 6600 p2c3 s2 = s21 - s1 s2 + s22 = a 3a -6600 p2c3 sY 2 b F.S. 3.5(103) 2 6600 2 b = a b 2.5 p2c3 c = 0.9388 in. d = 1.88 in. Ans. Ans: d = 1.88 in. 1025 © 2014 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. *10–64. If a shaft is made of a material for which sY = 50 ksi, determine the torsional shear stress required to cause yielding using the maximum-distortion-energy theory. s1 = t s2 = - t s12 - s1 s2 + s22 = sY2 3t2 = 502 t = 28.9 ksi Ans. 1026 © 2014 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. 10–65. Solve Prob. 10–64 using the maximum-shear-stress theory. s1 = t s2 = - t |s1 - s2| = sY t - (- t) = 50 t = 25 ksi Ans. Ans: t = 25 ksi 1027 © 2014 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. 10–66. Derive an expression for an equivalent torque Te that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T. t = Te c J Principal Stress: s1 = t s2 = - t ud = 1 + v 2 (s1 - s1 s2 + s22) 3E 1 + v 1 + v 3 T2e c2 ( 3 t2) = a b 3E 3E J2 (ud)1 = Bending Moment and Torsion: s = Mc ; I t = Tc J Principal Stress: s1, 2 = s1 = s + 0 s - 0 2 2 ; a b + t A 2 2 s s2 + + t2 ; A4 2 Let a = s 2 b = s2 = s s2 + t2 A4 2 s2 + t2 A4 s21 = a2 + b2 + 2 a b s1 s2 = a2 - b2 s22 = a2 + b2 - 2 a b s21 - s1 s2 + s22 = 3 b2 + a2 ud = 1 + v 2 (s1 - s1 s2 + s22) 3E (ud)2 = = 1 + v 1 + v 3 s2 s2 (3 b2 + a2) = a + 3t2 + b 3E 3E 4 4 c2(1 + v) M2 1 + v 2 3 T2 (s + 3 t2) = a 2 + b 3E 3E I J2 (ud)1 = (ud)2 c2(1 + v) 3 Te2 c2(1 + v) M2 3 T2 = + b a 3E 3E J2 I2 J2 1028 © 2014 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. 10–66. Continued For circular shaft J = I p 2 p 4 c4 c4 =2 Te = J2 M2 + T2 A I2 3 Te = 4 2 M + T2 A3 Ans. Ans: Te = 1029 4 2 M + T2 A3 © 2014 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. 10–67. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T. Principal Stresses: s1 = Me c ; I ud = 1 + v 2 (s1 - s1 s2 + s22) 3E s2 = 0 1 + v M2e c2 a 2 b 3E I (ud)1 = (1) Principal Stress: s + 0 s - 0 2 3 ; a b + t A 2 2 s1, 2 = s1 = s s2 + + t2; A4 2 s2 = s s2 + t2 A4 2 Distortion Energy: Let a = 2 s , b = s + t2 A4 2 s21 = a2 + b2 + 2 a b s1 s2 = a2 - b2 s22 = a2 + b2 - 2 a b s22 - s1 s2 + s22 = 3 b2 + a2 Apply s = (ud)2 = = Mc ; I t = Tc J 1 + v 1 + v s2 3s2 (3 b2 + a2) = a + + 3 t2 b 3E 3E 4 4 1 + v 2 1 + v M2 c2 3 T2 c2 (s + 3 t2) = a 2 + b 3E 3E I J2 (2) Equating Eq. (1) and (2) yields: (1 + v) Me c2 1 + v M2 c2 3T2 c2 a 2 b = a 2 + b 3E 3E I I J2 M2e I 2 = M2 3 T2 + 2 I J2 M2e = M2 + 3 T2 a I 2 b J 1030 © 2014 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. 10–67. Continued For circular shaft I = J p 4 p 2 c4 c4 = 1 2 1 2 Hence, M2e = M2 + 3 T2 a b 2 Me = A M2 + 3 2 T 4 Ans. Ans: Me = 1031 A M2 + 3 2 T 4 © 2014 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. *10–68. The principal plane stresses acting on a differential element are shown. If the material is machine steel having a yield stress of sY = 700 MPa, determine the factor of safety with respect to yielding if the maximum-shear-stress theory is considered 50 MPa 80 MPa smax = 80 MPa smin = - 50 MPa tabs = 80 - ( - 50) smax - smin = = 65 MPa 2 2 tmax = sY 700 = = 350 MPa 2 2 F.S. = tmax 350 = = 5.38 tabs 65 max Ans. max 1032 © 2014 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. 10–69. The short concrete cylinder having a diameter of 50 mm is subjected to a torque of 500 N # m and an axial compressive force of 2 kN. Determine if it fails according to the maximum-normal-stress theory. The ultimate stress of the concrete is sult = 28 MPa. A = p (0.05)2 = 1.9635(10 - 3) m2 4 J = p (0.025)4 = 0.61359(10 - 4) m4 2 2 kN 500 N⭈m 500 N⭈m 2 kN 3 s = 2(10 ) P = 1.019 MPa = A 1.9635(10 - 3) t = 500(0.025) Tc = 20.372 MPa = J 0.61359(10 - 6) sx = 0 s1, 2 = s1,2 = sy = - 1.019 MPa sx + sy 2 ; Aa sx - sy 2 txy = 20.372 MPa 2 b + txy 2 0 - 1.018 0 - ( -1.019) 2 2 ; a b + 20.372 A 2 2 s1 = 19.87 MPa s2 = - 20.89 MPa Failure criteria: |s1| 6 salt = 28 MPa OK |s2| 6 salt = 28 MPa OK No. Ans. Ans: No. 1033 © 2014 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. 10–70. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same maximum shear stress as the combination of an applied moment M and torque T. Assume that the principal stresses are of opposite algebraic signs. Bending and Torsion: Mc Mc 4M = p 4 = ; I p c3 4 c s = t = Tc Tc 2T = p 4 = J p c3 2c The principal stresses: s1, 2 = tabs max sx + sy 2 ; Aa sx - sy 2 2 2 b + txy = = 2M 2 ; 2M2 + T2 p c3 p c3 = s1 - s2 2 = 2M2 + T2 2 p c3 4M pc3 4M + 0 2 3 ; Q ¢pc - 0 2 2 ≤ + a b 3 pc 2T 2 (1) Pure bending: s1 = tabs max Me c 4 Me Mc = p 4 = ; I c p c3 4 = s2 = 0 s1 - s2 2 Me = 2 p c3 (2) Equating Eq. (1) and (2) yields: 2 Me 2 2M2 + T2 = p c3 p c3 Me = 2M2 + T2 Ans. Ans: Me = 2M2 + T2 1034 © 2014 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. sy ⫽ 0.5sx 10–71. The plate is made of hard copper, which yields at sY = 105 ksi . Using the maximum-shear-stress theory, determine the tensile stress sx that can be applied to the plate if a tensile stress sy = 0.5sx is also applied. sx s1 = sx s2 = 1 s 2 x |s1| = sY sx = 105 ksi Ans. Ans: sx = 105 ksi 1035 © 2014 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. sy ⫽ 0.5sx *10–72. Solve Prob. 10–71 using the maximum-distortion energy theory. sx s1 = sx s2 = sx 2 s21 - s1 s2 + s22 = sY2 sx2 - sx2 sx2 + = (105)2 2 4 sx = 121 ksi Ans. 1036 © 2014 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. 10–73. If the 2-in.-diameter shaft is made from brittle material having an ultimate strength of sult = 50 ksi, for both tension and compression, determine if the shaft fails according to the maximum-normal-stress theory. Use a factor of safety of 1.5 against rupture. 30 kip 4 kip · ft Normal Stress and Shear Stresses. The cross-sectional area and polar moment of inertia of the shaft’s cross-section are A = p A 12 B = pin2 J = p 4 p A 1 B = in4 2 2 The normal stress is caused by axial stress. s = N 30 = = - 9.549 ksi p A The shear stress is contributed by torsional shear stress. t = 4(12)(1) Tc = = 30.56 ksi p J 2 The state of stress at the points on the surface of the shaft is represented on the element shown in Fig. a. In-Plane Principal Stress. sx = - 9.549 ksi, sy = 0 and txy = - 30.56 ksi. We have s1, 2 = = sx + sy 2 ; Aa sx - sy 2 2 b + txy 2 - 9.549 - 0 2 -9.549 + 0 2 ; b + (- 30.56) Aa 2 2 = ( -4.775 ; 30.929) ksi s1 = 26.15 ksi s2 = - 35.70 ksi Maximum Normal-Stress Theory. sallow = sult 50 = = 33.33 ksi F.S. 1.5 |s1| = 26.15 ksi 6 sallow = 33.33 ksi (O.K.) |s2| = 35.70 ksi 7 sallow = 33.33 ksi (N.G.) Based on these results, the material fails according to the maximum normal-stress theory. Ans: Yes. 1037 © 2014 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. 10–74. If the 2-in.-diameter shaft is made from cast iron having tensile and compressive ultimate strengths of 1sult2t = 50 ksi and 1sult2c = 75 ksi, respectively, determine if the shaft fails in accordance with Mohr’s failure criterion. Normal Stress and Shear Stresses. The cross-sectional area and polar moment of inertia of the shaft’s cross-section are A = p A 12 B = p in2 J = p 4 p A 1 B = in4 2 2 30 kip 4 kip · ft The normal stress is contributed by axial stress. s = N 30 = = - 9.549 ksi p A The shear stress is contributed by torsional shear stress. t = 4(12)(1) Tc = = 30.56 ksi p J 2 The state of stress at the points on the surface of the shaft is represented on the element shown in Fig. a. In-Plane Principal Stress. sx = - 9.549 ksi, sy = 0, and txy = - 30.56 ksi. We have s1, 2 = = sx + sy 2 ; Aa sx - sy 2 2 b + txy 2 - 9.549 - 0 2 - 9.549 + 0 2 ; a b + ( -30.56) A 2 2 = (- 4.775 ; 30.929) ksi s1 = 26.15 ksi s2 = - 35.70 ksi Mohr’s Failure Criteria. As shown in Fig. b, the coordinates of point A, which represent the principal stresses, are located inside the shaded region. Therefore, the material does not fail according to Mohr’s failure criteria. Ans: No. 1038 © 2014 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. 10–75. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximum-shear-stress theory. 60 MPa 40 MPa In accordance with the established sign convention, sx = 70 MPa, sy = - 60 MPa and txy = 40 MPa. s1, 2 = = sx + sy 2 ; Aa sx - sy 2 70 MPa 2 b + txy 2 70 + (- 60) 70 - ( - 60) 2 2 ; c d + 40 A 2 2 = 5 ; 25825 s1 = 81.32 MPa s2 = - 71.32 MPa In this case, s1 and s2 have opposite sign. Thus, |s1 - s2| = |81.32 - ( -71.32)| = 152.64 MPa 6 sy = 250 MPa Based on this result, the steel shell does not yield according to the maximum shear stress theory. Ans: No. 1039 © 2014 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. *10–76. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximum-distortion-energy theory. 60 MPa 40 MPa In accordance with the established sign convention, sx = 70 MPa, sy = - 60 MPa and txy = 40 MPa. s1, 2 = = sx + sy 2 ; Aa sx - sy 2 2 b + txy 2 70 + ( - 60) 70 - ( -60) 2 2 ; c d + 40 A 2 2 = 5 ; 25825 s1 = 81.32 MPa s2 = - 71.32 MPa s1 2 - s1 s2 + s2 2 = 81.322 - 81.32( - 71.32) + ( - 71.32)2 = 17,500 6 sy 2 = 62500 Based on this result, the steel shell does not yield according to the maximum distortion energy theory. 1040 70 MPa © 2014 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. 10–77. If the A-36 steel pipe has outer and inner diameters of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-shear-stress theory. 900 N 150 mm A 100 mm 200 mm Internal Loadings. Considering the equilibrium of the free-body diagram of the post’s right cut segment Fig. a, ©Fy = 0; Vy + 900 - 900 = 0 Vy = 0 T = - 360 N # m ©Mx = 0; T + 900(0.4) = 0 ©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m Section Properties. The moment of inertia about the z axis and the polar moment of inertia of the pipe’s cross section are Iz = p A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4 4 J = p A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4 2 Normal Stress and Shear Stress. The normal stress is contributed by bending stress. Thus, sY = - MyA 90(0.015) = = - 42.31MPa Iz 10.15625p A 10 - 9 B The shear stress is contributed by torsional shear stress. t = 360(0.015) Tc = = 84.62 MPa J 20.3125p A 10 - 9 B The state of stress at point A is represented by the two-dimensional element shown in Fig. b. In-Plane Principal Stress. sx = - 42.31 MPa, sz = 0 and txz = 84.62 MPa. We have s1, 2 = = sx + sz 2 ; Aa sx - sz 2 2 b + txz 2 - 42.31 - 0 2 - 42.31 + 0 2 ; a b + 84.62 A 2 2 = ( - 21.16 ; 87.23) MPa s1 = 66.07 MPa s2 = - 108.38 MPa 1041 200 mm 900 N © 2014 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. 10–77. Continued Maximum Shear Stress Theory. s1 and s2 have opposite signs. This requires |s1 - s2| = sallow 66.07 - (- 108.38) = sallow sallow = 174.45 MPa The factor of safety is F.S. = sY 250 = = 1.43 sallow 174.45 Ans. Ans: F.S. = 1.43 1042 © 2014 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. 10–78. If the A-36 steel pipe has an outer and inner diameter of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-distortion-energy theory. 900 N 150 mm A 100 mm 200 mm Internal Loadings: Considering the equilibrium of the free-body diagram of the pipe’s right cut segment Fig. a, ©Fy = 0; Vy + 900 - 900 = 0 Vy = 0 T = - 360 N # m ©Mx = 0; T + 900(0.4) = 0 ©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m Section Properties. The moment of inertia about the z axis and the polar moment of inertia of the pipe’s cross section are Iz = p A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4 4 J = p A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4 2 Normal Stress and Shear Stress. The normal stress is caused by bending stress. Thus, sY = - MyA 90(0.015) = = - 42.31MPa Iz 10.15625p A 10 - 9 B The shear stress is caused by torsional stress. t = 360(0.015) Tc = 84.62 MPa = J 20.3125p A 10 - 9 B The state of stress at point A is represented by the two-dimensional element shown in Fig. b. In-Plane Principal Stress. sx = - 42.31 MPa, sz = 0 and txz = 84.62 MPa. We have s1, 2 = = sx + sz 2 ; Aa sx - sz 2 2 b + txz 2 - 42.31 - 0 2 -42.31 + 0 2 ; a b + 84.62 A 2 2 = ( -21.16 ; 87.23) MPa s1 = 66.07 MPa s2 = - 108.38 MPa 1043 200 mm 900 N © 2014 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. 10–78. Continued Maximum Distortion Energy Theory. s1 2 - s1s2 + s2 2 = sallow 2 66.072 - 66.07( - 108.38) + ( - 108.38)2 = sallow 2 sallow = 152.55 MPa Thus, the factor of safety is F.S. = sY 250 = = 1.64 sallow 152.55 Ans. Ans: F.S. = 1.64 1044 © 2014 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. 10–79. The yield stress for heat-treated beryllium copper is sY = 130 ksi. If this material is subjected to plane stress and elastic failure occurs when one principal stress is 145 ksi, what is the smallest magnitude of the other principal stress? Use the maximum-distortion-energy theory. Maximum Distortion Energy Theory: With s1 = 145 ksi, s21 - s1 s2 + s22 = s2Y 1452 - 145s2 + s22 = 1302 s22 - 145s2 + 4125 = 0 s2 = -( - 145) ; 2( - 145)2 - 4(1)(4125) 2(1) = 72.5 ; 33.634 Choose the smaller root, s2 = 38.9 ksi Ans. Ans: s2 = 38.9 ksi 1045 © 2014 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. *10–80. The yield stress for a uranium alloy is sY = 160 MPa. If a machine part is made of this material and a critical point in the material is subjected to plane stress, such that the principal stresses are s1 and s2 = 0.25s1, determine the magnitude of s1 that will cause yielding according to the maximum-distortion energy theory. s12 - s1 s2 + s22 = sY2 s12 - (s1)(0.25s1) + (0.25s1)2 = sY2 0.8125s12 = sY2 0.8125s12 = (160)2 s1 = 178 MPa Ans. 1046 © 2014 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. 10–81. Solve Prob. 10–80 using the maximum-shear-stress theory. s1 2 tabs = tabs = tallow max max ` s1 ` = 80; 2 tallow = sY 160 = = 80 MPa 2 2 Ans. s1 = 160 MPa Ans: s1 = 160 MPa 1047 © 2014 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. 10–82. The state of stress acting at a critical point on the seat frame of an automobile during a crash is shown in the figure. Determine the smallest yield stress for a steel that can be selected for the member, based on the maximumshear-stress theory. Normal and Shear Stress: In accordance with the sign convention. sx = 80 ksi sy = 0 25 ksi txy = 25 ksi 80 ksi In-Plane Principal Stress: Applying Eq. 9-5. s1,2 = = sx + sy ; 2 A a sx - sy 2 2 b + txy 2 80 - 0 2 80 + 0 2 ; a b + 25 2 A 2 = 40 ; 47.170 s1 = 87.170 ksi s2 = - 7.170 ksi Maximum Shear Stress Theory: s1 and s2 have opposite signs so |s1 - s2| = sY |87.170 - ( - 7.170)| = sY sY = 94.3 ksi Ans. Ans: sY = 94.3 ksi 1048 © 2014 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. 10–83. Solve Prob. 10–82 using the maximum-distortionenergy theory. 25 ksi Normal and Shear Stress: In accordance with the sign convention. sx = 80 ksi sy = 0 80 ksi txy = 25 ksi In-Plane Principal Stress: Applying Eq. 9-5. s1,2 = = sx + sy ; 2 a sx - s 2 2 b + txy A 2 80 - 0 2 80 + 0 2 ; a b + 25 2 A 2 = 40 ; 47.170 s1 = 87.170 ksi s2 = - 7.170 ksi Maximum Distortion Energy Theory: s21 - s1s2 + s22 = s2Y 87.1702 - 87.170(- 7.170) + ( -7.170)2 = s2Y sY = 91.0 ksi Ans. Ans: sY = 91.0 ksi 1049 © 2014 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. *10–84. A bar with a circular cross-sectional area is made of SAE 1045 carbon steel having a yield stress of sY = 150 ksi. If the bar is subjected to a torque of 30 kip # in. and a bending moment of 56 kip # in., determine the required diameter of the bar according to the maximum-distortion-energy theory. Use a factor of safety of 2 with respect to yielding. Normal and Shear Stresses: Applying the flexure and torsion formulas. 56 A d2 B Mc 1792 = = p d 4 I pd3 A B s = 4 t = Tc = J 2 30 A d2 B A B d 4 2 p 2 = 480 pd3 The critical state of stress is shown in Fig. (a) or (b), where sx = 1792 pd3 sy = 0 txy = 480 pd3 In-Plane Principal Stresses: Applying Eq. 9-5, s1,2 = sx + sy 2 1792 3 pd = = s1 = ; A + 0 2 ; D a sx - sy 2 1792 3 pd ¢ - 0 2 2 b + txy 2 2 ≤ + a 480 2 b pd3 896 1016.47 ; pd3 pd3 1912.47 pd3 s2 = - 120.47 pd3 Maximum Distortion Energy Theory: s21 - s1s2 + s22 = s2allow a 1912.47 120.47 120.47 2 150 2 1912.47 2 b - a bab + ab = a b 3 3 3 3 2 pd pd pd pd Ans. d = 2.03 in. 1050 © 2014 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. 10–85. The state of stress acting at a critical point on a machine element is shown in the figure. Determine the smallest yield stress for a steel that might be selected for the part, based on the maximum-shear-stress theory. 10 ksi The Principal Stresses: 4 ksi s1,2 = = sx + sy 2 ; A a sx - sy 2 b + 2 t2xy 8 ksi 8 - ( -10) 2 8 - 10 2 ; a b + 4 2 A 2 s1 = 8.8489 ksi s2 = - 10.8489 ksi Maximum Shear Stress Theory: Both principal stresses have opposite sign. hence, |s1 - s2| = sY 8.8489 - ( - 10.8489) = sY sY = 19.7 ksi Ans. Ans: sY = 19.7 ksi 1051 © 2014 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. 10–86. The principal stresses acting at a point on a thinwalled cylindrical pressure vessel are s1 = pr>t, s2 = pr>2t, and s3 = 0. If the yield stress is sY, determine the maximum value of p based on (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory. a) Maximum Shear Stress Theory: s1 and s2 have the same signs, then |s2| = sY 2 pr 2 = sY 2t p = 2t s r Y |s1| = sY 2 pr 2 = sY t p = t s (Controls!) r Y Ans. b) Maximum Distortion Energy Theory: s21 - s1s2 + s22 = s2Y a pr 2 pr 2 pr pr b - a b a b + a b = s2Y t t 2t 2t p = 2t 23r sY Ans. Ans: (a) p = (b) p = 1052 t s , r y 2t 23r sy © 2014 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. 10–87. If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the maximum-shearstress theory the maximum allowable shear stress is tallow = (16>pd3) 2M2 + T2. Assume the principal stresses to be of opposite algebraic signs. T M Section properties: I = p d 4 pd4 a b = ; 4 2 64 J = p d 4 pd4 a b = 2 2 32 Thus, s = M(d2 ) Mc 32 M = = p d4 I pd3 64 t = Tc = J T (d2 ) p d4 32 = 16 T pd3 The principal stresses: s1,2 = = sx + sy 2 ; A a sx - sy 2 2 b + txy 2 16 M 16 M 2 16 T 2 16 M 16 ; ; 2M2 + T2 a b + a b = 3 3 A pd pd p d3 pd3 p d3 Assume s1 and s2 have opposite sign, hence, tallow = 2 C 163 2M2 + T2 D s1 - s2 16 pd = = 2M2 + T2 2 2 pd3 (Q.E.D.) 1053 T M © 2014 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. *10–88. If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the maximum-normalstress theory the maximum allowable principal stress is sallow = 116>pd321M + 2M2 + T22. T M Section properties: I = p d4 ; 64 p d4 32 J = Stress components: s = M (d2 ) Mc 32 M = p 4 = ; I d p d3 64 t = T(d2 ) Tc 16 T = p 4 = J d p d3 32 The principal stresses: s1,2 = = sx + sy 2 ; A a sx - sy 2 2 b + txy = 2 32 M 3 pd 32 M + 0 2 3 ; D ¢ pd - 0 2 2 ≤ + a 16 T 2 b p d3 16 M 16 ; 2M2 + T2 p d3 p d3 Maximum normal stress theory. Assume s1 7 s2 sallow = s1 = = 16 M 16 + 2M2 + T2 p d3 p d3 16 [M + 2M2 + T2] p d3 (Q.E.D.) 1054 T M © 2014 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. 10–89. The gas tank has an inner diameter of 1.50 m and a wall thickness of 25 mm. If it is made from A-36 steel and the tank is pressured to 5 MPa, determine the factor of safety against yielding using (a) the maximum-shear-stress theory, and (b) the maximum-distortion-energy theory. (a) Normal Stress. Since 0.75 r = = 30 7 10, thin-wall analysis can be used.We have t 0.025 s1 = sh = 5(0.75) pr = = 150 MPa t 0.025 s2 = slong = pr 5(0.75) = = 75 MPa 2t 2(0.025) Maximum Shear Stress Theory. s1 and s2 have the sign. Thus, |s1| = sallow sallow = 150 MPa The factor of safety is F.S. = sY 250 = 1.67 = sallow 150 Ans. (b) Maximum Distortion Energy Theory. s1 2 - s1s2 + s2 2 = sallow 2 1502 - 150(75) + 752 = sallow 2 sallow = 129.90 MPa The factor of safety is F.S. = sY 250 = = 1.92 sallow 129.90 Ans. Ans: (a) F.S. = 1.67, (b) F.S. = 1.92 1055 © 2014 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. 10–90. The gas tank is made from A-36 steel and has an inner diameter of 1.50 m. If the tank is designed to withstand a pressure of 5 MPa, determine the required minimum wall thickness to the nearest millimeter using (a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of 1.5 against yielding. (a) Normal Stress. Assuming that thin-wall analysis is valid, we have s1 = sh = 5 A 106 B (0.75) 3.75 A 106 B pr = = t t t s2 = slong = 5 A 106 B (0.75) 1.875 A 106 B pr = = 2t 2t t Maximum Shear Stress Theory. sallow = 250 A 106 B sY = = 166.67 A 106 B Pa FS. 1.5 s1 and s2 have the same sign. Thus, |s1| = sallow 3.75 A 106 B = 166.67 A 106 B t t = 0.0225 m = 22.5 mm Since Ans. 0.75 r = = 33.3 7 10, thin-wall analysis is valid. t 0.0225 (b) Maximum Distortion Energy Theory. sallow = 250 A 106 B sY = = 166.67 A 106 B Pa F.S. 1.5 Thus, s1 2 - s1s2 + s2 2 = sallow 2 C 3.75 A 106 B t 3.2476 A 106 B t 2 S - C 3.75 A 106 B t SC 1.875 A 106 B t S + C 1.875 A 106 B t 2 S = c166.67 A 106 B d = 166.67 A 106 B t = 0.01949 m = 19.5 mm Since 2 Ans. 0.75 r = = 38.5 7 10, thin-wall analysis is valid. t 0.01949 Ans: (a) t = 22.5 mm, (b) t = 19.5 mm 1056 © 2014 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. 10–91. The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb # ft, a bending moment of 1500 lb # ft, and an axial thrust of 2500 lb. If the yield points for tension and shear sY = 100 ksi and tY = 50 ksi, respectively, determine the required diameter of the shaft using the maximum-shearstress theory. 2300 lb⭈ft A = p c2 sA = I = p 4 c 4 J = p 4 c 2 1500 lb⭈ft 2500 lb 1500(12)(c) P Mc 2500 2500 72 000 + b = -a + b + = -a 2 pc4 A I pc p c2 p c3 4 tA = s1,2 = Tc = J 2300(12)(c) = p c4 2 sx + sy = -a 2 ; A a 55 200 p c3 sx - sy 2 2 b + txy 2 2500 c + 72 000 2500c + 72 000 2 55200 2 b ; a b + a b 3 3 A 2p c 2p c p c3 (1) Assume s1 and s2 have opposite signs: |s1 - s2| = sY 2500c + 72 000 2 55 200 2 3 b + a b = 100(10 ) 3 A 2p c p c3 2 a (2500c + 72000)2 + 1104002 = 10 000(106)p2 c6 6.25c2 + 360c + 17372.16 - 10 000p2 c6 = 0 By trial and error: c = 0.750 57 in. Substitute c into Eq. (1): s1 = 22 193 psi s2 = - 77 807 psi s1 and s2 are of opposite signs OK Therefore, d = 1.50 in. Ans. Ans: d = 1.50 in. 1057 © 2014 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. *10–92. The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum-shear-stress theory. Use a factor of safety of 1.5 against yielding. A B T C Shear Stress: This is a case of pure shear, and the shear stress is contributed by p torsion. For the hollow segment, Jh = A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus, 2 (tmax)h = T(0.05) Tch = = 8626.28T Jh 1.845p A 10 - 6 B For the solid segment, Js = (tmax)s = p A 0.044 B = 1.28p A 10 - 6 B m4. Thus, 2 T(0.04) Tcs = = 9947.18T Js 1.28p A 10 - 6 B By comparision, the points on the surface of the solid segment are critical and their state of stress is represented on the element shown in Fig. a. In-Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have s1,2 = = sx + sy 2 ; C ¢ sx - sy 2 2 ≤ + t2xy 0 - 0 2 0 + 0 ; ¢ ≤ + (9947.18T)2 2 C 2 s1 = 9947.18T s2 = - 9947.18T Maximum Shear Stress Theory. sallow = 80 mm sY 250 = = 166.67 MPa F.S. 1.5 Since s1 and s2 have opposite signs, |s1 - s2| = sallow 9947.18T - ( -9947.18T) = 166.67 A 106 B T = 8377.58 N # m = 8.38 kN # m Ans. 1058 80 mm 100 mm T © 2014 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. 10–93. The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum-distortion-energy theory. Use a factor of safety of 1.5 against yielding. A B T C Shear Stress. This is a case of pure shear, and the shear stress is contributed by p torsion. For the hollow segment, Jh = A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus, 2 (tmax)h = 80 mm 100 mm T T(0.05) Tch = = 8626.28T Jh 1.845p A 10 - 6 B For the solid segment, Js = (tmax)s = 80 mm p A 0.044 B = 1.28p A 10 - 6 B m4. Thus, 2 T(0.04) Tcs = = 9947.18T Js 1.28p A 10 - 6 B By comparision, the points on the surface of the solid segment are critical and their state of stress is represented on the element shown in Fig. a. In-Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have s1,2 = = sx + sy 2 ; C ¢ sx - sy 2 2 ≤ + t2xy 0 - 0 2 0 + 0 ; ¢ ≤ + (9947.18T)2 2 C 2 s1 = 9947.18T s2 = - 9947.18T Maximum Distortion Energy Theory. sallow = sY 250 = = 166.67 MPa F.S. 1.5 Then, s1 2 - s1s2 + s2 2 = sallow 2 (9947.18T)2 - (9947.18T)( - 9947.18T) + ( -9947.18T)2 = C 166.67 A 106 B D 2 T = 9673.60 N # m = 9.67 kN # m Ans. Ans: T = 9.67 kN # m 1059 © 2014 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. 10–94. In the case of plane stress, where the in-plane principal strains are given by P1 and P2, show that the third principal strain can be obtained from P3 = - [n>(1 - n)](P1 + P2), where n is Poisson’s ratio for the material. Generalized Hooke’s Law: In the case of plane stress, s3 = 0. Thus, P1 = 1 (s - ns2) E 1 (1) P2 = 1 (s - ns1) E 2 (2) n (s + s2) E 1 (3) P3 = - Solving for s1 and s2 using Eqs. (1) and (2), we obtain s1 = E(P1 + nP2) 2 1-n s2 = E(P2 + nP1) 1 - n2 Substituting these results into Eq. (3), P3 = - E(P2 + nP1) n E(P1 + nP2) c + d E 1 - n2 1 - n2 P3 = - (P1 +P2) + n(P1 +P2) n c d 1 - n 1+n P3 = - (P1 +P2)(1 + n) n c d 1 - n 1+n P3 = - n (P +P ) 1 - n 1 2 (Q.E.D.) Ans. 1060 © 2014 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. 10–95. The plate is made of material having a modulus of elasticity E = 200 GPa and Poisson’s ratio n = 13 . Determine the change in width a, height b, and thickness t when it is subjected to the uniform distributed loading shown. a ⫽ 400 mm 2 MN/m t ⫽ 20 mm 3 MN/m y b ⫽ 300 mm Normal Stress: The normal stresses along the x, y, and z axes are z 3(106) sx = = 150 MPa 0.02 sy = - x 2(106) = - 100 MPa 0.02 sz = 0 Generalized Hooke’s Law: Px = = 1 c s - n(sy + sz) d E x 1 1 e 150(106) - c - 100(106) + 0 d f 9 3 200(10 ) = 0.9167(10 - 3) Py = = 1 c s - n(sx + sz) d E y 1 1 e -100(106) - c 150(106) + 0 d f 3 200(109) = - 0.75(10 - 3) Pz = = 1 c s - n(sx + sy) d E z 1 1 e 0 - c 150(106) + (- 100)(106) d f 3 200(109) = - 83.33(10 - 6) Thus, the changes in dimensions of the plate are da = Pxa = 0.9167(10 - 3)(400) = 0.367 mm Ans. db = Pyb = - 0.75(10 - 3)(300) = - 0.225 mm Ans. dt = Pzt = - 83.33(10 - 6)(20) = - 0.00167 mm Ans. The negative signs indicate that b and t contract. Ans: da = 0.367 mm, db = - 0.255 mm, dt = - 0.00167 mm 1061 © 2014 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. *10–96. The principal plane stresses acting at a point are shown in the figure. If the material is machine steel having a yield stress of sY = 500 MPa, determine the factor of safety with respect to yielding if the maximum-shear-stress theory is considered. 100 MPa 150 MPa Here, the in plane principal stresses are s1 = sy = 100 MPa s2 = sx = - 150 MPa Since s1 and s2 have the same sign, F.S = sy |s1 - s2| = 500 = 2 |100 - ( - 150)| Ans. 1062 © 2014 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. 10–97. The components of plane stress at a critical point on a thin steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximumdistortion-energy theory. The yield stress for the steel is sY = 650 MPa. 340 MPa 65 MPa 55 MPa sx = - 55 MPa s1, 2 = = sy = 340 MPa sx + sy 2 ; A a sx - sy 2 txy = 65 MPa 2 b + txy 2 - 55 - 340 2 - 55 + 340 2 ; a b + 65 2 A 2 s1 = 350.42 MPa s2 = - 65.42 MPa (s1 2 - s1s2 + s2 ) = [350.422 - 350.42(- 65.42) + ( -65.42)2] = 150 000 6 s2Y = 422 500 OK No. Ans. Ans: No. 1063 © 2014 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. 10–98. The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: Pa = 600110-62, Pb = - 700110-62, and Pc = 350110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains. a 60⬚ 60⬚ b 60⬚ Strain Rosettes (60º): Applying Eq. 10-15 with Px = 600 A 10 - 6 B , Pb = - 700 A 10 - 6 B , Pc = 350 A 10 - 6 B , ua = 150°, ub = - 150° and uc = - 90°, c 350 A 10 - 6 B = Px cos2 (- 90°) + Py sin2 ( -90°) + gxy sin (- 90°) cos (-90°) Py = 350 A 10 - 6 B 600 A 10 - 6 B = Px cos2 150° + 350 A 10 - 6 B sin2 150° + gxy sin 150° cos 150° 512.5 A 10 - 6 B = 0.75 Px - 0.4330 gxy [1] -787.5 A 10 - 6 B = 0.75Px + 0.4330 gxy [2] - 700 A 10 - 6 B = Px cos2 ( -150°) + 350 A 10 - 6 B sin2( -150°) + gxy sin ( -150°) cos (- 150°) Solving Eq. [1] and [2] yields Px = - 183.33 A 10 - 6 B gxy = - 1501.11 A 10 - 6 B Construction of the Circle: With Px = - 183.33 A 10 - 6 B , Py = 350 A 10 - 6 B , and gxy = - 750.56 A 10 - 6 B . 2 Px + Py Pavg = 2 = a - 183.33 + 350 b A 10 - 6 B = 83.3 A 10 - 6 B 2 Ans. The coordinates for reference points A and C are A(- 183.33, -750.56) A 10 - 6 B C(83.33, 0) A 10 - 6 B The radius of the circle is R = a 2(183.33 + 83.33)2 + 750.562 b A 10 - 6 B = 796.52 A 10 - 6 B a) In-Plane Principal Strain: The coordinates of points B and D represent P1 and P2, respectively. P1 = (83.33 + 796.52) A 10 - 6 B = 880 A 10 - 6 B Ans. P2 = (83.33 - 796.52) A 10 - 6 B = - 713 A 10 - 6 B Ans. Orientation of Principal Strain: From the circle, tan 2uP1 = 750.56 = 2.8145 183.33 + 83.33 2uP2 = 70.44° 2uP1 = 180° - 2uP2 uP = 180° - 70.44° = 54.8° (Clockwise) 2 Ans. 1064 © 2014 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. 10–98. Continued b) Maximum In-Plane Shear Strain: Represented by the coordinates of point E on the circle. g max in-plane 2 g max in-plane = - R = - 796.52 A 10 - 6 B = - 1593 A 10 - 6 B Ans. Orientation of Maximum In-Plane Shear Strain: From the circle. tan 2us = 183.33 + 83.33 = 0.3553 750.56 us = 9.78° (Clockwise) Ans. Ans: Pavg = 83.3(10 - 6), P1 = 880(10 - 6), P2 = - 713(10 - 6), up = 54.8° (clockwise), gmax = - 1593(10 - 6), in-plane us = 9.78° (clockwise) 1065 © 2014 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. y 10–99. The state of strain at the point on the bracket has components Px = 350(10 - 6), Py = - 860(10 - 6), gxy = 250(10 - 6). Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 45° clockwise from the original position. Sketch the deformed element within the x–y plane due to these strains. x Px = 350(10 - 6) Px + Py Px ¿ = = c Px - Py + 2 Px - Py Px + Py - 2 gxy 2 u = - 45° sin 2u 2 cos 2u - gxy 2 sin 2u 350 - ( -860) 350 - 860 250 cos ( -90°) sin ( -90°) d (10 - 6) = - 130(10 - 6) 2 2 2 gx¿y¿ 2 2 cos 2u + gxy = 250(10 - 6) 350 - (- 860) 350 - 860 250 + cos ( -90°) + sin ( -90°) d (10 - 6) = - 380(10 - 6) Ans. 2 2 2 Py¿ = = c Py = - 860(10 - 6) Px - Py = - gx¿y¿ = 2 c- a 2 sin 2u + Ans. g cos 2u 2 350 - (- 860) 250 b sin (- 90°) + cos (- 90°)d (10 - 6) = 1.21(10 - 3) Ans. 2 2 Ans: Px¿ = - 380(10 - 6), Py¿ = - 130(10 - 6), gx¿y¿ = 1.21(10 - 3) 1066 © 2014 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. *10–100. The A-36 steel post is subjected to the forces shown. If the strain gauges a and b at point A give readings of Pa = 300(10 - 6) and Pb = 175(10 - 6), determine the magnitudes of P1 and P2. P1 P2 a + ©F = 0; : x P2 - V = 0 V = P2 + c ©Fy = 0; N - P1 = 0 N = P1 a + ©MO = 0; M + P2(2) = 0 M = 2P2 Section Properties: The cross-sectional area and the moment of inertia about the bending axis of the post’s cross-section are A = 4(2) = 8 in2 I = 1 (2) A 43 B = 10.667 in4 12 Referring to Fig. b, A Qy B A = x¿A¿ = 1.5(1)(2) = 3 in3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. sA = 2P2(12)(1) MxA P1 N + = + = 2.25P2 - 0.125P1 A I 8 10.667 The shear stress is caused by transverse shear stress. tA = P2(3) VQA = = 0.140625P2 It 10.667(2) Thus, the state of stress at point A is represented on the element shown in Fig. c. Normal and Shear Strain: With ua = 90° and ub = 45°, we have Pa = Px cos2ua + Py sin2ua + gxy sin ua cos ua 300 A 10 - 6 B = Px cos2 90° + Py sin2 90° + gxy sin 90° cos 90° Py = 300 A 10 - 6 B Pb = Px cos2ub + Py sin2 ub + gxy sin ub cos ub 175 A 10 - 6 B = Px cos2 45° + 300 A 10 - 6 B sin2 45° + gxy sin 45°cos 45° Px + gxy = 50 A 10 - 6 B (1) 1067 A 1 in. b 45⬚ A c Internal Loadings: Considering the equilibrium of the free-body diagram of the post’s segment, Fig. a, 2 in. 2 ft A 1 in. 4 in. c Section c–c © 2014 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. 10–100. Continued Since sy = sz = 0, Px = - vPy = - 0.32(300) A 10 - 6 B = - 96 A 10 - 6 B Then Eq. (1) gives gxy = 146 A 10 - 6 B Stress and Strain Relation: Hooke’s Law for shear gives tx = Ggxy 0.140625P2 = 11.0 A 103 B C 146 A 10 - 6 B D P2 = 11.42 kip = 11.4 kip Ans. Since sy = sz = 0, Hooke’s Law gives sy = EPy 2.25(11.42) - 0.125P1 = 29.0 A 103 B C 300 A 10 - 6 B D P1 = 136 kip Ans. 1068 © 2014 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. 10–101. A differential element is subjected to plane strain that has the following components; Px = 950110-62, Py = 420110-62, gxy = - 325110-62. Use the straintransformation equations and determine (a) the principal strains and (b) the maximum in plane shear strain and the associated average strain. In each case specify the orientation of the element and show how the strains deform the element. Aa Px + Py P1, 2 = ; 2 = c Px - Py 2 2 b + gxy 2 950 - 420 2 -325 2 950 + 420 -6 ; a b + a b d(10 ) 2 A 2 2 P1 = 996(10 - 6) Ans. P2 = 374(10 - 6) Ans. Orientation of P1 and P2 : gxy tan 2uP = -325 950 - 420 = Px - Py uP = - 15.76°, 74.24° Use Eq. 10.5 to determine the direction of P1 and P2. Px + Py Px¿ = Px - Py + 2 2 cos 2u + gxy 2 sin 2u u = uP = - 15.76° Px¿ = b ( - 325) 950 - 420 950 + 420 + cos (- 31.52°) + sin ( - 31.52°) r (10 - 6) = 996(10 - 6) 2 2 2 uP1 = - 15.8° Ans. uP2 = 74.2° Ans. b) gmax in-plane 2 gmax in-plane = Aa = 2c Px - Py 2 2 2 gxy 2 b 2 950 - 420 2 -325 2 -6 -6 b + a b d (10 ) = 622(10 ) A 2 2 a Px + Py Pavg = b + a = a Ans. 950 + 420 b (10 - 6) = 685(10 - 6) 2 Ans. 1069 © 2014 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. 10–101. Continued Orientation of gmax: tan 2uP = - (Px - Py) = gxy -(950 - 420) - 325 uP = 29.2° and uP = 119° Ans. Use Eq. 10.6 to determine the sign of gx¿y¿ 2 Px - Py = - 2 sin 2u + gxy 2 gmax in-plane : cos 2u u = us = 29.2° gx¿y¿ = 2 c - (950 - 420) -325 sin (58.4°) + cos (58.4°) d(10 - 6) 2 2 gxy = - 622(10 - 6) Ans: P1 = 996(10 - 6), P2 = 374(10 - 6), up1 = - 15.8, up2 = 74.2, gmax = 622(10 - 6), in-plane Pavg = 685(10 - 6), us = 29.2° and 119° 1070 © 2014 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. 10–102. The state of strain at the point on the bracket Px = - 130110-62, Py = 280110-62, has components gxy = 75110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. y x Px = - 130(10 - 6) Py = 280(10 - 6) gxy = 75(10 - 6) a) Px + Py P1, 2 = 2 = c Aa ; Px - Py 2 2 b + a gxy 2 b 2 - 130 + 280 - 130 - 280 2 75 2 -6 ; a b + a b d(10 ) 2 A 2 2 P1 = 283(10 - 6) Ans. P2 = - 133(10 - 6) Ans. Orientation of P1 and P2: tan 2up = gxy = Px - Py up = -5.18° 75 - 130 - 280 and 84.82° Use Eq. 10–5 to determine the direction of P1 and P2: Px + Py Px¿ = 2 Px - Py + 2 cos 2u + gxy 2 sin 2u u = up = -5.18° Px¿ = c - 130 + 280 - 130 - 280 75 + cos ( -10.37°) + sin (- 10.37°) d(10 - 6) = - 133(10 - 6) 2 2 2 Therefore up1 = 84.8° Ans. up2 = - 5.18° Ans. 1071 © 2014 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. 10–102. Continued b) gmax in-plane = 2 gmax in-plane A = 2c a 2 2 b + a gxy 2 b 2 - 130 - 280 2 75 2 b + a b d (10 - 6) = 417(10 - 6) A 2 2 a Px + Py Pavg = Px - Py 2 = a Ans. - 130 + 280 b (10 - 6) = 75.0(10 - 6) 2 Ans. Orientation of gmax: tan 2us = - (Px - Py) gxy = - (- 130 - 280) 75 us = 39.8° and us = 130° Ans. Use Eq. 10–16 to determine the sign of gx¿y¿ 2 Px - Py = - 2 sin 2u + gxy 2 cos 2u; gmax in-plane : u = us = 39.8° gx¿y¿ = -(- 130 - 280) sin (79.6°) + (75) cos (79.6°) = 417(10 - 6) Ans: P1 = 283(10 - 6), P2 = -133(10 - 6), up1 = 84.8°, up2 = -5.18°, gmax = 417(10 - 6), in-plane Pavg = 75.0(10 - 6), us = 39.8° and 130° 1072 © 2014 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. y 10–103. The state of plain strain on an element is Px = 400110-62, Py = 200110-62, and gxy = - 300110-62. Determine the equivalent state of strain, which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding element at the point with respect to the original element. Sketch the results on the element. Pydy dy Construction of the Circle: Px = 400 A 10 - 6 B , Py = 200 A 10 - 6 B , and gxy 2 = - 150 A 10 - 6 B . Thus, Px + Py Pavg = 2 = a 400 + 200 b A 10 - 6 B = 300 A 10 - 6 B 2 Ans. The coordinates for reference points A and the center C of the circle are A(400, -150) A 10 - 6 B C(300, 0) A 10 - 6 B The radius of the circle is R = CA = 2(400 - 300)2 + (- 150)2 = 180.28 A 10 - 6 B Using these results, the circle is shown in Fig. a. In-Plane Principal Stresses: The coordinates of points B and D represent P1 and P2, respectively. Thus, P1 = (300 + 180.28) A 10 - 6 B = 480 A 10 - 6 B Ans. P2 = (300 - 180.28) A 10 - 6 B = 120 A 10 - 6 B Ans. Orientation of Principal Plane: Referring to the geometry of the circle, tan 2 A up B 1 = 150 = 1.5 400 - 300 A up B 1 = 28.2° (clockwise) Ans. The deformed element for the state of principal strains is shown in Fig. b. Maximum In-Plane Shear Stress: The coordinates of point E represent Pavg and gmax . Thus in-plane gmax = - R = - 180.28 A 10 - 6 B in-plane 2 gmax in-plane = - 361 A 10 - 6 B Ans. Orientation of the Plane of Maximum In-plane Shear Strain: Referring to the geometry of the circle, tan 2us = 400 - 300 = 0.6667 150 uS = 16.8° (counterclockwise) Ans. The deformed element for the state of maximum In-plane shear strain is shown in Fig. c. 1073 gxy 2 gxy 2 dx x Pxdx © 2014 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. 10–103. Continued Ans: P1 = 480(10 - 6), P2 = 120(10 - 6), up1 = - 28.2° (clockwise), gmax = - 361(10 - 6), in-plane us = 16.8° (counterclockwise), Pavg = 300(10 - 6) 1074 © 2014 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. 8 kN/m 11–1. The simply supported beam is made of timber that has an allowable bending stress of sallow = 6.5 MPa and an allowable shear stress of tallow = 500 kPa. Determine its dimensions if it is to be rectangular and have a height-towidth ratio of 1.25. 2m Ix = 4m 2m 1 (b)(1.25b)3 = 0.16276b4 12 Qmax = y¿A¿ = (0.3125b)(0.625b)(b) = 0.1953125b3 Assume bending moment controls: Mmax = 16 kN # m sallow = Mmax c I 6.5(106) = 16(103)(0.625b) 0.16276b4 b = 0.21143 m = 211 mm Ans. h = 1.25b = 264 mm Ans. Check shear: Qmax = 1.846159(10 - 3) m3 I = 0.325248(10 - 3) m4 tmax = 16(103)(1.846159)(10 - 3) VQmax = 429 kPa 6 500 kPa‚ OK = It 0.325248(10 - 3)(0.21143) Ans: b = 211 mm, h = 264 mm 1075 © 2014 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. 11–2. Determine the minimum width of the beam to the nearest 14 in. that will safely support the loading of P = 8 kip. The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 15 ksi. P 6 ft 6 ft 6 in. B A Beam design: Assume moment controls. sallow = Mc ; I 24 = 48.0(12)(3) 1 3 12 (b)(6 ) b = 4 in. Ans. Check shear: tmax = 8(1.5)(3)(4) VQ = 0.5 ksi 6 15 ksi OK = 1 3 It 12 (4)(6) (4) Ans: Use b = 4 in. 1076 © 2014 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. 11–3. Solve Prob. 11–2 if P = 10 kip. P 6 ft 6 ft 6 in. B A Beam design: Assume moment controls. sallow = Mc ; I 24 = 60(12)(3) 1 3 12 (b)(6 ) b = 5 in. Ans. Check shear: tmax = VQ 10(1.5)(3)(5) = 0.5 ksi 6 15 ksi OK = 1 3 It 12 (5)(6) (5) Ans: Use b = 5 in. 1077 © 2014 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. *11–4. The brick wall exerts a uniform distributed load of 1.20 kip>ft on the beam. If the allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi, select the lightest wide-flange section with the shortest depth from Appendix B that will safely support the load. If there are several choices of equal weight, choose the one with the shortest height. 1.20 kip/ft 4 ft Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft. Assuming bending controls the design and applying the flexure formula. Sreq d = = Mmax sallow 44.55 (12) = 24.3 in3 22 Two choices of wide flange section having the weight 22 lb>ft can be made. They are W12 * 22 and W14 * 22. However, W12 * 22 is the shortest. Select W12 * 22 A Sx = 25.4 in3, d = 12.31 in., tw = 0.260 in. B V for the W12 * 22 wide tw d = 6.60 kip. Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax tmax = = Vmax tw d 6.60 0.260(12.31) = 2.06 ksi 6 tallow = 12 ksi (O.K!) Hence, Use Ans. W12 * 22 1078 10 ft 6 ft © 2014 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. 11–5. Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the machine loading shown. The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 14 ksi. 5 kip 2 ft Bending Stress: From the moment diagram, Mmax = 30.0 kip # ft. Assume bending controls the design. Applying the flexure formula. Sreq¿d = = Select W12 * 16 5 kip 2 ft 5 kip 2 ft 5 kip 2 ft 2 ft Mmax sallow 30.0(12) = 15.0 in3 24 A Sx = 17.1 in3, d = 11.99 in., tw = 0.220 in. B V for the W12 * 16 wide tw d = 10.0 kip Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax tmax = = Vmax tw d 10.0 0.220(11.99) = 3.79 ksi 6 tallow = 14 ksi (O.K!) Hence, Use Ans. W12 * 16 Ans: Use W12 * 16 1079 © 2014 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. 11–6. The spreader beam AB is used to slowly lift the 3000-lb pipe that is centrally located on the straps at C and D. If the beam is a W12 * 45, determine if it can safely support the load. The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi. 3000 lb 5 ft A 12 in. B 3 ft 6 ft C h = 1500 = 2700 lb tan 29.055° s = M ; S s = 5850(12) = 1.21 ksi 6 22 ksi 58.1 t = V ; A web t = 1500 = 371 psi 6 12 ksi OK (12.06)(0.335) 6 ft 3 ft D OK Yes. Ans. Ans: Yes. 1080