Normal Stress: s = sA = Myz Mzy N + A Iz

09 Solutions 46060 6/8/10 3:13 PM Page 619 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–1. Prove that the sum of the normal stresses sx + sy = sx¿ + sy¿ is constant. See Figs. 9–2a and 9–2b. Stress Transformation Equations: Applying Eqs. 9-1 and 9-3 of the text. sx¿ + sy¿ = sx + sy 2 + + sx - sy 2 sx + sy 2 cos 2u + txy sin 2u - sx - sy 2 cos 2u - txy sin 2u sx¿ + sy¿ = sx + sy (Q.E.D.) 619 09 Solutions 46060 6/8/10 3:13 PM Page 620 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–2. 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. A 8 ksi 2 ksi 5 ksi 60⬚ B Referring to Fig a, if we assume that the areas of the inclined plane AB is ¢A, then the area of the horizontal and vertical of the triangular element are ¢A cos 60° and ¢A sin 60° respectively. The forces act acting on these two faces indicated on the FBD of the triangular element, Fig. b. +Q©Fx¿ = 0; ¢Fx¿ + 2¢A sin 60° cos 60° + 5¢ A sin 60° sin 60° + 2¢A cos 60° sin 60° - 8¢A cos 60° cos 60° = 0 ¢Fx¿ = -3.482 ¢A +a©Fy¿ = 0; ¢Fy¿ + 2¢A sin 60° sin 60° - 5¢ A sin 60° cos 60° - 8¢A cos 60° sin 60° - 2¢A cos 60° cos 60° = 0 ¢Fy¿ = 4.629 ¢A From the definition, sx¿ = lim¢A:0 ¢Fx¿ = -3.48 ksi ¢A tx¿y¿ = lim¢A:0 ¢Fy¿ ¢A Ans. = 4.63 ksi Ans. The negative sign indicates that sx¿, is a compressive stress. 620 09 Solutions 46060 6/8/10 3:13 PM Page 621 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–3. 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. 500 psi B 60⬚ A Referring to Fig. a, if we assume that the area of the inclined plane AB is ¢A, then the areas of the horizontal and vertical surfaces of the triangular element are ¢A sin 60° and ¢A cos 60° respectively. The force acting on these two faces are indicated on the FBD of the triangular element, Fig. b +R©Fx¿ = 0; ¢Fx¿ + 500 ¢A sin 60° sin 60° + 350¢A sin 60° cos 60° +350¢A cos 60° sin 60° = 0 ¢Fx¿ = -678.11 ¢A +Q©Fy¿ = 0; ¢Fy¿ + 350¢A sin 60° sin 60° - 500¢A sin 60° cos 60° -350¢A cos 60° cos 60° = 0 ¢Fy¿ = 41.51 ¢A From the definition sx¿ = lim¢A:0 tx¿y¿ = lim¢A:0 ¢Fx¿ = -6.78 psi ¢A ¢Fy¿ ¢A Ans. = 41.5 psi Ans. The negative sign indicates that sx¿, is a compressive stress. 621 350 psi 09 Solutions 46060 6/8/10 3:13 PM Page 622 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–4. 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. Q+ ©Fx¿ = 0 A 400 psi 650 psi ¢Fx¿ - 400(¢Acos 60°)cos 60° + 650(¢ A sin 60°)cos 30° = 0 60⬚ ¢Fx¿ = -387.5¢A a+ ©Fy¿ = 0 ¢Fy¿ - 650(¢Asin 60°)sin 30° - 400(¢ A cos 60°)sin 60° = 0 B ¢Fy¿ = 455 ¢A sx¿ = lim¢A:0 sx¿y¿ = lim¢A:0 ¢Fx¿ = -388 psi ¢A ¢Fy¿ ¢A Ans. = 455 psi Ans. The negative sign indicates that the sense of sx¿, is opposite to that shown on FBD. Solve Prob. 9–4 using the stress-transformation equations developed in Sec. 9.2. •9–5. sy = 400 psi sx = -650 psi sx¿ = = sx + sy 2 + sx - sy 2 txy = 0 A 400 psi u = 30° 650 psi cos 2u + txy sin 2u 60⬚ -650 - 400 -650 + 400 + cos 60° + 0 = -388 psi 2 2 Ans. B The negative sign indicates sx¿, is a compressive stress. tx¿y¿ = = -a sx - sy 2 sin 2u + txy cos 2u -650 - 400 bsin 60° = 455 psi 2 Ans. 622 09 Solutions 46060 6/8/10 3:13 PM Page 623 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–6. 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. 90 MPa A 35 MPa 60⬚ 30⬚ R+ ©Fy¿ = 0 B 50 MPa ¢Fy¿ - 50¢A sin 30° cos 30° - 35¢A sin 30° cos 60° + 90¢A cos 30° sin 30° + 35¢A cos 30° sin 60° = 0 ¢Fy¿ = -34.82¢A ¢Fx¿ - 50¢A sin 30° sin 30° + 35¢A sin 30° sin 60° b+ ©Fx¿ = 0 -90¢A cos 30° cos 30° + 35¢A cos 30° cos 60° = 0 ¢Fx¿ = 49.69 ¢A sx¿ = lim¢A:0 ¢Fx¿ = 49.7 MPa ¢A tx¿y¿ = lim¢A:0 ¢Fy¿ ¢A Ans. = -34.8 MPa Ans. The negative signs indicate that the sense of sx¿, and tx¿y¿ are opposite to the shown on FBD. 9–7. Solve Prob. 9–6 using the stress-transformation equations developed in Sec. 9.2. Show the result on a sketch. 90 MPa A 35 MPa 60⬚ 30⬚ sy = 50 MPa sx = 90 MPa sx¿ = = sx + sy 2 + sx - sy 2 txy = -35 MPa u = -150° cos 2u + txy sin 2u 90 - 50 90 + 50 + cos(-300°) + (-35) sin ( -300°) 2 2 = 49.7 MPa tx¿y¿ = - sx - sy = -a 2 Ans. sin 2u + txy cos 2u 90 - 50 bsin(-300°) + ( -35) cos ( -300°) = -34.8 MPa 2 The negative sign indicates tx¿y¿ acts in -y¿ direction. 623 Ans. B 50 MPa 09 Solutions 46060 6/8/10 3:13 PM Page 624 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–8. Determine the normal stress and shear stress acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1. 45 MPa B 80 MPa 45⬚ A Force Equllibrium: Referring to Fig. a, if we assume that the area of the inclined plane AB is ¢A, then the area of the vertical and horizontal faces of the triangular sectioned element are ¢A sin 45° and ¢A cos 45°, respectively. The forces acting on the free-body diagram of the triangular sectioned element, Fig. b, are ©Fx¿ = 0; ¢Fx¿ + c45 A 106 B ¢A sin 45° dcos 45° + c45 A 106 B ¢A cos 45° dsin 45° - c80 A 106 B ¢A sin 45° dcos 45° = 0 ¢Fx¿ = -5 A 106 B ¢A ©Fy¿ = 0; ¢Fy¿ + c45 A 106 B ¢A cos 45° dcos 45° - c45 A 106 B ¢A sin 45° dsin 45° - c80 A 106 B ¢ A sin 45° dsin 45° = 0 ¢Fy¿ = 40 A 106 B ¢A Normal and Shear Stress: From the definition of normal and shear stress, sx¿ = lim¢A:0 ¢Fx¿ = -5 MPa ¢A tx¿y¿ = lim¢A:0 ¢Fy¿ ¢A Ans. = 40 MPa Ans. The negative sign indicates that sx¿ is a compressive stress. 624 09 Solutions 46060 6/8/10 3:13 PM Page 625 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Determine the normal stress and shear stress acting on the inclined plane AB. Solve the problem using the stress transformation equations. Show the result on the sectioned element. •9–9. 45 MPa 80 MPa 45⬚ Stress Transformation Equations: u = +135° (Fig. a) sx = 80 MPa sy = 0 txy = 45 MPa we obtain, sx¿ = = sx + sy 2 + sx - sy 2 cos u + txysin 2u 80 - 0 80 + 0 + cos 270 + 45 sin 270° 2 2 = -5 MPa tx¿y¿ = = - sx - sy 2 B Ans. sinu + txy cos 2u 80 - 0 sin 270° + 45 cos 270° 2 = 40 MPa Ans. The negative sign indicates that sx¿ is a compressive stress. These results are indicated on the triangular element shown in Fig. b. 625 A 09 Solutions 46060 6/8/10 3:13 PM Page 626 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–10. 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. 2 ksi A 3 ksi 30⬚ B Force Equllibrium: For the sectioned element, a+ ©Fy¿ = 0; 4 ksi ¢Fy¿ - 3(¢A sin 30°) sin 60° + 4(¢ A sin 30°)sin 30° -2(¢A cos 30°) sin 30° - 4(¢A cos 30°) sin 60° = 0 ¢Fy¿ = 4.165 ¢A Q+ ©Fx¿ = 0; ¢Fx¿ + 3(¢A sin 30°) cos 60° + 4(¢ A sin 30°)cos 30° -2(¢A cos 30°) cos 30° + 4(¢A cos 30°) cos 60° = 0 ¢Fx¿ = -2.714 ¢A Normal and Shear Stress: For the inclined plane. sx = lim¢A:0 tx¿y¿ = lim¢A:0 ¢Fx¿ = -2.71 ksi ¢A ¢Fy¿ ¢A Ans. = 4.17 ksi Ans. Negative sign indicates that the sense of sx¿, is opposite to that shown on FBD. 9–11. Solve Prob. 9–10 using the stress-transformation equations developed in Sec. 9.2. Show the result on a sketch. 2 ksi Normal and Shear Stress: In accordance with the established sign convention, u = +60° sx = -3 ksi sy = 2 ksi A txy = -4 ksi 3 ksi 30⬚ Stress Transformation Equations: Applying Eqs. 9-1 and 9-2. sx¿ = = sx + sy 2 + sx - sy 2 B cos 2u + txy sin 2u -3 - 2 -3 + 2 + cos 120° + (-4 sin 120°) 2 2 = -2.71 ksi tx¿y¿ = = - Ans. sx - sy 2 4 ksi sin 2u + txy cos 2u -3 - 2 sin 120° + (-4 cos 120°) 2 = 4.17 ksi Ans. Negative sign indicates sx¿, is a compressive stress 626 09 Solutions 46060 6/8/10 3:13 PM Page 627 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–12. Determine the equivalent state of stress on an element if it is oriented 50° counterclockwise from the element shown. Use the stress-transformation equations. 10 ksi 16 ksi sx = -10 ksi sy = 0 txy = -16 ksi u = +50° sx¿ = = sx + sy 2 = -a = sx - sy 2 cos 2u + txy sin 2u -10 + 0 -10 - 0 + cos 100° + (-16)sin 100° = -19.9 ksi 2 2 tx¿y¿ = - a sy¿ = + sx - sy 2 b sin 2u + txy cos 2u -10 - 0 b sin 100° + (-16)cos 100° = 7.70 ksi 2 sx + sy 2 - Ans. sx - sy 2 Ans. cos 2u - txy sin 2u -10 + 0 -10 - 0 - a bcos 100° - (-16)sin 100° = 9.89 ksi 2 2 627 Ans. 09 Solutions 46060 6/8/10 3:13 PM Page 628 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Determine the equivalent state of stress on an element if the element is oriented 60° clockwise from the element shown. Show the result on a sketch. •9–13. 350 psi 75 psi 200 psi In accordance to the established sign covention, u = -60° (Fig. a) sx = 200 psi sy = -350 psi txy = 75 psi Applying Eqs 9-1, 9-2 and 9-3, sx¿ = = sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u 200 - ( -350) 200 + (-350) + cos (-120°) + 75 sin (-120°) 2 2 = -277.45 psi = -277 psi sy¿ = = sx + sy 2 - sx - sy 2 Ans. cos 2u - txy sin 2u 200 - ( -350) 200 + ( -350) cos (-120°) - 75 sin ( -120°) 2 2 = 127.45 psi = 127 psi tx¿y¿ = = - sx - sy 2 Ans. sin 2u + txy cos 2u 200 - (-350) sin (-120°) + 75 cos (-120°) 2 = 200.66 psi = 201 psi Ans. Negative sign indicates that sx¿ is a compressive stress. These result, can be represented by the element shown in Fig. b. 628 09 Solutions 46060 6/8/10 3:13 PM Page 629 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–14. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Show the results on each element. 30 ksi 12 ksi sx = -30 ksi sy = 0 txy = -12 ksi a) sx + sy s1, 2 = 2 ; C a sx - sy 2 2 b + txy 2 = -30 + 0 -30 - 0 2 ; a b + (-12)2 2 C 2 s1 = 4.21 ksi Ans. s2 = -34.2 ksi Ans. Orientation of principal stress: txy tan 2uP = (sx - sy)>2 -12 = 0.8 (-30-0)>2 = -70.67° uP = 19.33° and Use Eq. 9-1 to determine the principal plane of s1 and s2. sx + sy sx¿ = 2 + sx - sy 2 cos 2u + txy sin 2u u = 19.33° sx¿ = -30 - 0 -30 + 0 + cos 2(19.33°) + (-12)sin 2(19.33°) = -34.2 ksi 2 2 Therefore uP2 = 19.3° Ans. and uP1 = -70.7° Ans. b) tmaxin-plane = savg = C a sx - sy 2 sx + sy 2 = 2 b + txy 2 = -30 - 0 2 b + (-12)2 = 19.2 ksi C 2 a -30 + 0 = -15 ksi 2 Ans. Ans. Orientation of max, in - plane shear stress: tan 2uP = -(sx - sy)>2 txy uP = -25.2° and = -(-30 - 0)>2 = -1.25 -12 64.3° Ans. By observation, in order to preserve equllibrium along AB, tmax has to act in the direction shown in the figure. 629 09 Solutions 46060 6/8/10 3:13 PM Page 630 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–15. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Show the results on each element. 80 MPa 50 MPa 60 MPa In accordance to the established sign convention, sx = -60 MPa s1, 2 = = sx + sy 2 sy = -80 MPa ; C a sx - sy 2 txy = 50 MPa 2 b + txy 2 -60 + (-80) -60 - (-80) 2 ; c d + 502 2 C 2 = -70 ; 22600 s1 = -19.0 MPa s2 = -121 MPa txy tan 2uP = = (sx - sy)>2 uP = 39.34° Ans. 50 = 5 [-60 - (-80)]>2 -50.65° and Substitute u = 39.34° into Eq. 9-1, sx¿ = = sx + sy sx - sy + 2 2 cos 2u + txy sin 2u -60 - ( -80) -60 + (-80) + cos 78.69° + 50 sin 78.69° 2 2 = -19.0 MPa = s1 Thus, (uP)1 = 39.3° (uP)2 = -50.7° Ans. The element that represents the state of principal stress is shown in Fig. a. t max in-plane = C a sx - sy 2 tan 2uS = 2 b + txy 2 = -(sx - sy)>2 txy = -60 - ( -80) 2 d + 502 = 51.0 MPa C 2 c -[-60 - (-80)]>2 = -0.2 50 uS = -5.65° and 84.3° By Inspection, t max Ans. Ans. has to act in the sense shown in Fig. b to maintain in-plane equilibrium. savg = sx + sy 2 = -60 + (-80) = -70 MPa 2 The element that represents the state of maximum in - plane shear stress is shown in Fig. c. 630 09 Solutions 46060 6/8/10 3:13 PM Page 631 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–15. Continued 631 09 Solutions 46060 6/8/10 3:13 PM Page 632 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–16. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Sketch the results on each element. sx = 45 MPa sy = -60 MPa 60 MPa 30 MPa 45 MPa txy = 30 MPa a) s1, 2 = = sx + sy 2 ; C a sx - sy 2 2 b + txy 2 45 - (-60) 2 45 - 60 a ; b + (30)2 2 C 2 s1 = 53.0 MPa Ans. s2 = -68.0 MPa Ans. Orientation of principal stress: tan 2uP = txy (sx - sy)>2 uP = 14.87, = 30 = 0.5714 (45 - (-60))>2 -75.13 Use Eq. 9-1 to determine the principal plane of s1 and s2: sx¿ = = sx + sy + 2 sx - sy 2 cos 2u + txy sin 2u, where u = 14.87° 45 + (-60) 45 - (-60) + cos 29.74° + 30 sin 29.74° = 53.0 MPa 2 2 Therefore uP1 = 14.9° Ans. and uP2 = -75.1° Ans. b) tmaxin-plane = savg = C a sx - sy sx - sy 2 2 = 2 b + txy 2 = 45 - (-60) 2 b + 302 = 60.5 MPa C 2 a 45 + (-60) = -7.50 MPa 2 Ans. Ans. Orientation of maximum in - plane shear stress: tan 2uS = -(sx - sy)>2 txy = -(45 - ( -60))>2 = -1.75 30 uS = -30.1° Ans. uS = 59.9° Ans. and By observation, in order to preserve equilibrium along AB, tmax has to act in the direction shown. 632 09 Solutions 46060 6/8/10 3:13 PM Page 633 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Determine the equivalent state of stress on an element at the same point which represents (a) the principal stress, 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. Sketch the results on each element. •9–17. 75 MPa 125 MPa 50 MPa Normal and Shear Stress: sx = 125 MPa sy = -75 MPa txy = -50 MPa In - Plane Principal Stresses: s1,2 = = sx - sy ; 2 B a sx - sy 2 2 b + txy 2 125 + (-75) 125 - (-75) 2 a ; b + (-50)2 2 2 B = 25 ; 212500 s2 = -86.8 MPa s1 = 137 MPa Ans. Orientation of Principal Plane: tan 2uP = txy A sx - sy B >2 = -50 A 125 -(-75) B >2 = -0.5 up = -13.28° and 76.72° Substitute u = -13.28° into sx¿ = = sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u 125 - (-75) 125 + (-75) + cos(-26.57°)+(-50) sin(-26.57°) 2 2 = 137 MPa = s1 Thus, A up B 1 = -13.3° and A up B 2 = 76.7° Ans. 125 - (-75)>(-50) The element that represents the state of principal stress is shown in Fig. a. Maximum In - Plane Shear Stress: t max in-plane = C ¢ sx - sy 2 ≤ + txy 2 = 2 -100 - 0 2 b + 252 = 112 MPa 2 B a Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = - A sx - sy B >2 txy = - A 125 - ( -75) B >2 -50 = 2 us = 31.7° and 122° 633 Ans. 09 Solutions 46060 6/8/10 3:13 PM Page 634 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–17. Continued By inspection, t max has to act in the same sense shown in Fig. b to maintain in-plane equilibrium. Average Normal Stress: savg = sx + sy 2 = 125 + (-75) = 25 MPa 2 Ans. The element that represents the state of maximum in - plane shear stress is shown in Fig. c. 634 09 Solutions 46060 6/8/10 3:13 PM Page 635 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–18. A point on a thin plate is subjected to the two successive states of stress shown. Determine the resultant state of stress represented on the element oriented as shown on the right. sy = sx¿ + sy¿ 2 + sx¿ - sy¿ 2 ⴙ 60⬚ Stress Transformation Equations: Applying Eqs. 9-1, 9-2, and 9-3 u = -30°, sx¿ = -200 MPa, to element (a) with sy¿ = -350 MPa and tx¿y¿ = 0. (sx)a = 350 MPa cos 2u + tx¿y¿ sin 2u -200 - (-350) -200 + (-350) + cos (-60°) + 0 2 2 = -237.5 MPa A sy B a = = sx¿ + sy¿ 2 - sx¿ - sy¿ 2 cos 2u - tx¿y¿ sin 2u -200 - (-350) -200 + (-350) cos (-60°) - 0 2 2 = -312.5 MPa A txy B a = = - sx¿ - sy¿ 2 sin 2u + tx¿y¿ cos 2u -200 - (-350) sin (-60°) + 0 2 = 64.95 MPa For element (b), u = 25°, sx¿ = sy¿ = 0 and sx¿y¿ = 58 MPa. (sx)b = sx¿ + sy¿ 2 + sx¿ - sy¿ 2 cos 2u + tx¿y¿ sin 2u = 0 + 0 + 58 sin 50° = 44.43 MPa A sy B b = sx¿ + sy¿ 2 - sx¿ - sy¿ 2 cos 2u - tx¿y¿ sin 2u = 0 - 0 - 58 sin 50° = -44.43 MPa A txy B b = - sx¿ - sy¿ 2 58 MPa 200 MPa sin 2u + tx¿y¿ cos 2u = -0 + 58 cos 50° = 37.28 MPa Combining the stress components of two elements yields ss = (sx)a + (sx)b = -237.5 + 44.43 = -193 MPa Ans. sy = A sy B a + A sy B b = -312.5 - 44.43 = -357 MPa Ans. txy = A txy B a + A txy B b = 64.95 + 37.28 = 102 MPa Ans. 635 25⬚ ⴝ txy sx 09 Solutions 46060 6/8/10 3:13 PM Page 636 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–19. The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. Sketch the results on each element. 160 MPa 120 MPa In accordance to the established sign Convention, sx = 0 sy = 160 MPa s1, 2 = = sx + sy 2 ; B a txy = -120 MPa sx - sy 2 2 b + t2xy 0 + 160 0 - 160 2 ; a b + (-120)2 2 2 B = 80 ; 220800 s2 = -64.2 MPa s1 = 224 MPa tan 2up = txy (sx - sy)>2 = Ans. -120 = 1.5 (0 - 160)>2 and -61.85° up = 28.15° Substitute u = 28.15° into Eq. 9-1, sx¿ = = sx + sy + 2 sx - sy 2 cos 2u + txy sin 2u 0 - 160 0 + 160 + cos 56.31° + (-120) sin 56.31° 2 2 = -64.22 = s2 Thus, (up)1 = -61.8° (up)2 = 28.2° Ans. The element that represents the state of principal stress is shown in Fig. a tmax in-plane = B a sx - sy 2 tan 2us = -(sx - sy)>2 us = -16.8° By inspection, equilibrium. 2 b + t2xy = tmax in-plane savg = txy 0 - 160 2 b + (-120)2 = 144 MPa 2 B a Ans. -(0 - 160)>2 = -0.6667 -120 = and 73.2° Ans. has to act in the sense shown in Fig. b to maintain sx + sy 2 = 0 + 160 = 80 MPa 2 Ans. The element that represents the state of Maximum in - plane shear stress is shown in Fig. (c) 636 09 Solutions 46060 6/8/10 3:13 PM Page 637 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–19. Continued 637 09 Solutions 46060 6/8/10 3:13 PM Page 638 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–20. The stress acting on two planes at a point is indicated. Determine the normal stress sb and the principal stresses at the point. a 4 ksi 60⬚ 45⬚ b 2 ksi sb a Stress Transformation Equations: Applying Eqs. 9-2 and 9-1 with u = -135°, sy = 3.464 ksi, txy = 2.00 ksi, tx¿y¿ = -2 ksi, and sx¿ = sb¿., tx¿y¿ = -2 = - sx - sy 2 sin 2u + txy cos 2u sx - 3.464 sin (-270°) + 2cos ( -270°) 2 sx = 7.464 ksi sx¿ = sy = sx - sy 2 + sx - sy 2 cos 2u + txy sin 2u 7.464 - 3.464 7.464 + 3.464 + cos (-270°) + 2sin (-270°) 2 2 = 7.46 ksi Ans. In - Plane Principal Stress: Applying Eq. 9-5. s1, 2 = = sx + sy 2 ; B a sx - sy 2 2 b + t2xy 7.464 - 3.464 2 7.464 + 3.464 ; a b + 22 2 2 B = 5.464 ; 2.828 s1 = 8.29 ksi s2 = 2.64 ksi Ans. 638 b 09 Solutions 46060 6/8/10 3:13 PM Page 639 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •9–21. The stress acting on two planes at a point is indicated. Determine the shear stress on plane a–a and the principal stresses at the point. b a ta 45⬚ 60 ksi 60⬚ txy = 60 cos 60° = 30 ksi sa = 80 = sx + sy 2 sx - sy + 2 51.962 + sy 2 + cos 2u + txy sin 2u 51.962 - sy 2 cos (90°) + 30 sin (90°) sy = 48.038 ksi ta = - a = -a sx - sy 2 b sin 2u + txy cos u 51.962 - 48.038 bsin (90°) + 30 cos (90°) 2 ta = -1.96 ksi s1, 2 = = sx + sy 2 ; Ans. C a sx - sy 2 2 b + t2xy 51.962 - 48.038 2 51.962 + 48.038 ; a b + (30)2 2 C 2 s1 = 80.1 ksi Ans. s2 = 19.9 ksi Ans. 639 90⬚ a b sx = 60 sin 60° = 51.962 ksi 80 ksi 09 Solutions 46060 6/8/10 3:13 PM Page 640 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–22. The T-beam is subjected to the distributed loading that is applied along its centerline. Determine the principal stress at point A and show the results on an element located at this point. 100 kN/m A 1m 0.5 m 200 mm 75 mm ' ©yA 0.1(0.2)(0.02) + 0.21(0.02)(0.2) = = 0.155 m ©A 0.2(0.02) + 0.02(0.2) 1 (0.02)(0.2 3) + 0.02(0.2)(0.155 - 0.1)2 12 I = + 1 (0.2)(0.023) + 0.2(0.02)(0.21 - 0.155)2 12 = 37.6667(10 - 6) m4 Referring to Fig. b, QA = y¿A¿ = 0.1175(0.075)(0.02) = 0.17625(10 - 3) m3 Using the method of sections and considering the FBD of the left cut segment of the beam, Fig. c, + c ©Fy = 0; V - 100(1) = 0 a + ©MC = 0; 100(1)(0.5) - M = 0 M = 50 kN # m V = 100 kN The normal stress developed is contributed by bending stress only. For point A, y = 0.155 - 0.075 = 0.08 m. Thus s = My 50(103) (0.08) = 106 MPa = I 37.6667(10 - 6) The shear stress is contributed by the transverse shear stress only. Thus, t = 100(103)[0.17625(10 - 3)] VQA = 23.40(106)Pa = 23.40 MPa = It 37.6667(10 - 6) (0.02) The state of stress of point A can be represented by the element shown in Fig. c. Here, sx = -106.19 MPa, sy = 0 and txy = 23.40 MPa. s1, 2 = = sx + sy 2 ; B a sx - sy 2 2 b + txy 2 -106.19 + 0 -106.19 - 0 2 ; b + 23.402 a 2 2 B = -53.10 ; 58.02 s1 = 4.93 MPa 20 mm 200 mm 20 mm The location of the centroid c of the T cross-section, Fig. a, is y = A s2 = -111 MPa Ans. 640 09 Solutions 46060 6/8/10 3:13 PM Page 641 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–22. Continued tan 2up = txy (sx - sy)>2 up = -11.89° = ans 23.40 = -0.4406 ( -106.19 - 0)>2 78.11° Substitute u = -11.89°, sx¿ = = sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u -106.19 - 0 -106.19 + 0 + cos (-23.78°) + 23.40 5m (-23.78°) 2 2 = -111.12 MPa = s2 Thus, (up)1 = 78.1° (up)2 = -11.9° Ans. The state of principal stress can be represented by the element shown in Fig. e. 641 09 Solutions 46060 6/8/10 3:13 PM Page 642 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The wood beam is subjected to a load of 12 kN. If a grain of wood in the beam at point A makes an angle of 25° with the horizontal as shown, determine the normal and shear stress that act perpendicular and parallel to the grain due to the loading. •9–23. I = 12 kN 1m 2m A 25⬚ 300 mm 75 mm 1 (0.2)(0.3)3 = 0.45(10 - 3) m4 12 QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3 sA = MyA 13.714(103)(0.075) = 2.2857 MPa (T) = I 0.45(10 - 3) tA = 6.875(103)(1.6875)(10 - 3) VQA = 0.1286 MPa = It 0.45(10 - 3)(0.2) sx = 2.2857 MPa sx¿ = sx¿ = sx + sy 2 + sy = 0 sx - sy 2 txy = -0.1286 MPa u = 115° cos 2u + txy sin 2u 2.2857 - 0 2.2857 + 0 + cos 230° + (-0.1286)sin 230° 2 2 = 0.507 MPa tx¿y¿ = - sx - sy = -a 2 Ans. sin 2u + txy cos 2u 2.2857 - 0 b sin 230° + (-0.1286)cos 230° 2 = 0.958 MPa Ans. 642 4m 200 mm 09 Solutions 46060 6/8/10 3:13 PM Page 643 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–24. The wood beam is subjected to a load of 12 kN. Determine the principal stress at point A and specify the orientation of the element. 12 kN 1m 2m A 25⬚ I = 300 mm 75 mm 1 (0.2)(0.3)3 = 0.45(10 - 3) m4 12 QA = yA¿ = 0.1125(0.2)(0.075) = 1.6875(10 - 3) m3 sA = MyA 13.714(103)(0.075) = 2.2857 MPa (T) = I 0.45(10 - 3) tA = 6.875(103)(1.6875)(10 - 3) VQA = 0.1286 MPa = It 0.45(10 - 3)(0.2) sx = 2.2857 MPa s1, 2 = = sx + sy 2 ; sy = 0 C a txy = -0.1286 MPa sx - sy 2 2 b + t2xy 2.2857 + 0 2.2857 - 0 2 ; a b + (-0.1286)2 2 C 2 s1 = 2.29 MPa Ans. s2 = -7.20 kPa Ans. tan 2up = txy (sx - sy)>2 = -0.1286 (2.2857 - 0)>2 up = -3.21° Check direction of principal stress: sx¿ = = sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u 2.2857 - 0 2.2857 + 0 + cos (-6.42°) - 0.1285 sin (-6.42) 2 2 = 2.29 MPa 643 4m 200 mm 09 Solutions 46060 6/8/10 3:13 PM Page 644 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The bent rod has a diameter of 20 mm and is subjected to the force of 400 N. Determine the principal stress and the maximum in-plane shear stress that is developed at point A. Show the results on a properly oriented element located at this point. •9–25. 100 mm 150 mm 400 N 400 N 250 mm A Using the method of sections and consider the FBD of the rod’s left cut segment, Fig. a. + ©F = 0; : x N - 400 = 0 N = 400 N a + ©MC = 0; 400(0.25) - M = 0 M = 100 N # m A = p(0.012) = 0.1(10 - 3) p m2 p (0.014) = 2.5(10 - 9)p m4 4 I = The normal stress developed is the combination of axial and bending stress. Thus, My N ; A I s = For point A, y = C = 0.01 m. s = 100(0.01) 400 -3 0.1(10 )p 2.5(10 - 9)p = -126.05 (106)Pa = 126.05 MPa (C) Since no torque and transverse shear acting on the cross - section, t = 0 The state of stress at point A can be represented by the element shown in Fig. b Here, sx = -126.05 MPa, sy = 0 and txy = 0. Since no shear stress acting on the element s1 = sy = 0 s2 = sx = -126 MPa Ans. Thus, the state of principal stress can also be represented by the element shown in Fig. b. tmax in-plane = B a sx - sy 2 tan 2us = - 2 b + t2xy = (sx - sy)>2 txy us = 45° tx¿y¿ = - -126.05 - 0 2 b + 02 = 63.0 MPa 2 B a = - (-126.05 - 0)>2 = q 0 and -45° sx - sy 2 = - = 63.0 = sin 2u + txy cos 2u -126.05 - 0 sin 90° + 0 cos 90° 2 tmax in-plane 644 Ans. 09 Solutions 46060 6/8/10 3:13 PM Page 645 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–25. Continued tmax This indicates that in-plane acts toward the positive sense of y¿ axis at the face of element defined by us = 45° savg = sx + sy 2 = -126.05 + 0 = -63.0 MPa 2 Ans. The state of maximum In - plane shear stress can be represented by the element shown in Fig. c 645 09 Solutions 46060 6/8/10 3:13 PM Page 646 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–26. The bracket is subjected to the force of 3 kip. Determine the principal stress and maximum in-plane shear stress at point A on the cross section at section a–a. Specify the orientation of this state of stress and show the results on elements. 3 kip 3 kip a 3 in. a A 0.25 in. 2 in. 0.25 in. Internal Loadings: Consider the equilibrium of the free - body diagram from the bracket’s left cut segment, Fig. a. + ©F = 0; : x N - 3 = 0 M = 12 kip # in Section a – a Normal and Shear Stresses: The normal stress is the combination of axial and bending stress. Thus, s = My N A I The cross - sectional area and the moment of inertia about the z axis of the bracket’s cross section is A = 1(2) - 0.75(1.5) = 0.875 in2 I = 1 1 (1) A 23 B (0.75) A 1.53 B = 0.45573 in4 12 12 For point A, y = 1 in. Then sA = (-12)(1) 3 = 29.76 ksi 0.875 0.45573 Since no shear force is acting on the section, tA = 0 The state of stress at point A can be represented on the element shown in Fig. b. In - Plane Principal Stress: sx = 29.76 ksi, sy = 0, and txy = 0. Since no shear stress acts on the element, s1 = sx = 29.8 ksi s2 = sy = 0 Ans. The state of principal stresses can also be represented by the elements shown in Fig. b Maximum In - Plane Shear Stress: in-plane = C ¢ sx - sy 2 ≤ + txy 2 = 2 29.76 - 0 2 b + 02 = 14.9 ksi 2 B a Ans. Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = - A sx - sy B >2 txy 0.25 in. N = 3 kip ©MO = 0; 3(4) - M = 0 t max B 1 in. = - (29.76 - 0)>2 = -q 0 us = -45° and 45° Ans. 646 09 Solutions 46060 6/8/10 3:13 PM Page 647 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–26. Continued Substituting u = -45° into tx¿y¿ = = - sx - sy 2 sin 2u + txy cos 2u 29.76 - 0 sin(-90°) + 0 2 = 14.9 ksi = t max in-plane This indicates that t max is directed in the positive sense of the y¿ axes on the ace in-plane of the element defined by us = -45°. Average Normal Stress: savg = sx + sy 2 = 29.76 + 0 = 14.9 ksi 2 Ans. The state of maximum in - plane shear stress is represented by the element shown in Fig. c. 647 09 Solutions 46060 6/8/10 3:13 PM Page 648 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–27. The bracket is subjected to the force of 3 kip. Determine the principal stress and maximum in-plane shear stress at point B on the cross section at section a–a. Specify the orientation of this state of stress and show the results on elements. 3 kip 3 kip a 3 in. a A 0.25 in. 2 in. 0.25 in. Internal Loadings: Consider the equilibrium of the free - body diagram of the bracket’s left cut segment, Fig. a. + ©F = 0; : x N - 3 = 0 M = 12 kip # in Section a – a Normal and Shear Stresses: The normal stress is the combination of axial and bending stress. Thus, s = My N A I The cross - sectional area and the moment of inertia about the z axis of the bracket’s cross section is A = 1(2) - 0.75(1.5) = 0.875 in2 I = 1 1 (1) A 23 B (0.75) A 1.53 B = 0.45573 in4 12 12 For point B, y = -1 in. Then sB = (-12)(-1) 3 = -22.90 ksi 0.875 0.45573 Since no shear force is acting on the section, tB = 0 The state of stress at point A can be represented on the element shown in Fig. b. In - Plane Principal Stress: sx = -22.90 ksi, sy = 0, and txy = 0. Since no shear stress acts on the element, s1 = sy = 0 s2 = sx = -22.90 ksi Ans. The state of principal stresses can also be represented by the elements shown in Fig. b. Maximum In - Plane Shear Stress: in-plane = C ¢ sx - sy 2 ≤ + txy 2 = 2 -22.90 - 0 2 b + 02 = 11.5 ksi 2 B a Ans. Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = - A sx - sy B >2 txy 0.25 in. N = 3 kip ©MO = 0; 3(4) - M = 0 t max B 1 in. = - (-22.9 - 0)>2 = -q 0 us = 45° and 135° Ans. 648 09 Solutions 46060 6/8/10 3:13 PM Page 649 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–27. Continued Substituting u = 45° into tx¿y¿ = = - sx - sy 2 sin 2u + txy cos 2u -22.9 - 0 sin 90° + 0 2 = 11.5 ksi = t max in-plane This indicates that t max is directed in the positive sense of the y¿ axes on the in-plane element defined by us = 45°. Average Normal Stress: savg = sx + sy 2 = -22.9 + 0 = -11.5 ksi 2 Ans. The state of maximum in - plane shear stress is represented by the element shown in Fig. c. 649 09 Solutions 46060 6/8/10 3:13 PM Page 650 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–28. The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam at point A and at point B. These points are located at the top and bottom of the web, respectively. Although it is not very accurate, use the shear formula to determine the shear stress. 8 kN/m A B 1m A 10 mm B Internal Forces and Moment: As shown on FBD(a). 200 mm Section Properties: A = 0.2(0.22) - 0.19(0.2) = 6.00 A 10 - 3 B m2 1 1 (0.2) A 0.223 B (0.19) A 0.22 B = 50.8 A 10 - 6 B m4 12 12 I = QA = QB = y¿A¿ = 0.105(0.01)(0.2) = 0.210 A 10 - 3 B m3 Normal Stress: s = = My N ; A I 21.65(103) 73.5(103)(0.1) ; 6.00(10 - 3) 50.8(10 - 6) sA = 3.608 + 144.685 = 148.3 MPa sB = 3.608 - 144.685 = -141.1 MPa VQ . It Shear Stress: Applying the shear formula t = tA = tB = 36.5(103) C 0.210(10 - 3) D 50.8(10 - 6)(0.01) = 15.09 MPa In - Plane Principal Stress: sx = 148.3 MPa, sy = 0, and txy = -15.09 MPa for point A. Applying Eq. 9-5. s1, 2 = = sx + sy 2 ; C sx - sy a 2 2 b + t2xy 148.3 + 0 148.3 - 0 2 ; a b + (-15.09)2 2 C 2 = 74.147 ; 75.666 s2 = -1.52 MPa s1 = 150 MPa Ans. sx = -141.1 MPa, sy = 0, and txy = -15.09 MPa for point B. Applying Eq. 9-5. s1, 2 = = sx + sy 2 ; C a sx - sy 2 2 b + t2xy ( -141.1) - 0 2 -141.1 + 0 ; a b + (-15.09)2 2 C 2 = -70.538 ; 72.134 s1 = 1.60 MPa s2 = -143 MPa Ans. 650 3m 110 mm 30⬚ 25 kN 10 mm 200 mm 10 mm 09 Solutions 46060 6/8/10 3:13 PM Page 651 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam at point A, which is located at the top of the web. Although it is not very accurate, use the shear formula to determine the shear stress. Show the result on an element located at this point. •9–29. 120 kN/m 30 kN A 0.3 m 0.9 m Using the method of sections and consider the FBD of the left cut segment of the bean, Fig. a + c ©Fy = 0; V 1 2 a + ©MC = 0; 1 2 (90)(0.9) - 30 = 0 V = 70.5 kN (90)(0.9)(0.3) + 30(0.9) - M = 0 M = 39.15 kN # m 150 mm The moment of inertia of the cross - section about the bending axis is I = 1 1 (0.15)(0.193) (0.13)(0.153) = 49.175(10 - 6) m4 12 12 Referring to Fig. b, QA = y¿A¿ = 0.085 (0.02)(0.15) = 0.255 (10 - 3) m3 The normal stress developed is contributed by bending stress only. For point A, y = 0.075 m. Thus, s = My 39.15(103)(0.075) = 59.71(106)Pa = 59.71 MPa (T) = I 49.175(10 - 6) The shear stress is contributed by the transverse shear stress only. Thus t = 70.5(103) C 0.255(10 - 3) D VQA = 18.28(106)Pa = 18.28 MPa = It 49.175(10 - 6) (0.02) Here, sx = 59.71 MPa, sy = 0 and txy = 18.28 MPa. s1, 2 = = sx + sy 2 ; C a sx - sy 2 2 b + txy 59.71 + 0 59.71 - 0 2 ; a b + 18.282 2 C 2 = 29.86 ; 35.01 s2 = -5.15 MPa s1 = 64.9 MPa tan 2uP = txy (sx - sy)>2 uP = 15.74° = and Ans. 18.28 = 0.6122 (59.71 - 0)>2 -74.26° Substitute u = 15.74°, sx¿ = = sx + sy 2 + sx - sy 2 A 20 mm cos 2u + txy sin 2u 59.71 - 0 59.71 + 0 + cos 31.48° + 18.28 sin 31.48° 2 2 = 64.9 MPa = s1 651 20 mm 150 mm 20 mm 09 Solutions 46060 6/8/10 3:13 PM Page 652 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–29. Continued Thus, A uP B 1 = 15.7° A uP B 2 = -74.3° Ans. The state of principal stress can be represented by the element shown in Fig. d 652 09 Solutions 46060 6/8/10 3:13 PM Page 653 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–30. The cantilevered rectangular bar is subjected to the force of 5 kip. Determine the principal stress at points A and B. 1 I = (3)(63) = 54 in4 12 1.5 in. A 1.5 in. A = (6)(3) = 18 in2 QA = 2.25(1.5)(3) = 10.125 in3 1.5 in. 1.5 in. B 1 in. 3 in. QB = 2(2)(3) = 12 in3 1 in. Point A: 15 in. 3 in. 3 5 4 5 kip sA 45(1.5) Mxz P 4 = + = + = 1.472 ksi A I 18 54 tA = Vz QA It = 3(10.125) = 0.1875 ksi 54(3) sx = 1.472 ksi s1, 2 = = sy = 0 sx + sy 2 ; C a sx - sy 2 txy = 0.1875 ksi 2 b + txy 2 1.472 - 0 2 1.472 + 0 ; a b + 0.18752 2 C 2 s1 = 1.50 ksi Ans. s2 = -0.0235 ksi Ans. Point B: sB = tB = 45(1) Mxz P 4 = = -0.6111 ksi A I 18 54 Vz QB It = 3(12) = 0.2222 ksi 54(3) sx = -0.6111 ksi s1, 2 = = sx + sy 2 ; sy = 0 C a sx - sy 2 txy = 0.2222 ksi 2 b + txy 2 -0.6111 - 0 2 -0.611 + 0 ; a b + 0.2222 2 C 2 s1 = 0.0723 ksi Ans. s2 = -0.683 ksi Ans. 653 09 Solutions 46060 6/8/10 3:13 PM Page 654 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–31. Determine the principal stress at point A on the cross section of the arm at section a–a. Specify the orientation of this state of stress and indicate the results on an element at the point. 7.5 mm A 50 mm 7.5 mm Support Reactions: Referring to the free - body diagram of the entire arm shown in Fig. a, 7.5 mm ©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0 FCD = 2166.67 N + ©F = 0; : x Bx - 2166.67 cos 30° = 0 Bx = 1876.39 N Section a – a + c ©Fy = 0; 2166.67 sin 30° - 500 - By = 0 By = 583.33 N D 20 mm Internal Loadings: Consider the equilibrium of the free - body diagram of the arm’s left segment, Fig. b. + ©F = 0; : x 1876.39 - N = 0 N = 1876.39 N + c ©Fy = 0; V - 583.33 = 0 V = 583.33 N 583.33(0.15) - M = 0 M = 87.5N # m + ©MO = 0; 0.15 m 1 1 (0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4 12 12 Referring to Fig. b, QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. Thus, = -1876.39 0.5625 A 10 -3 B + MyA N + A I 87.5(0.0175) 0.16367 A 10 - 6 B = 6.020 MPa The shear stress is caused by transverse shear stress. tA = 583.33 C 3.1875 A 10 - 6 B D VQA = = 1.515 MPa It 0.16367 A 10 - 6 B (0.0075) The share of stress at point A can be represented on the element shown in Fig. d. In - Plane Principal Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. We have s1,2 = = sx + sy 2 ; C ¢ sx - sy 2 ≤ + txy 2 2 6.020 - 0 2 6.020 + 0 ; a b + 1.5152 2 C 2 s1 = 6.38 MPa s2 = -0.360 MPa Ans. 654 C a A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2 sA = a B Section Properties: The cross - sectional area and the moment of inertia about the z axis of the arm’s cross section are I = 60⬚ 0.15 m 0.35 m 500 N 09 Solutions 46060 6/8/10 3:13 PM Page 655 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–31. Continued Orientation of the Principal Plane: tan 2uP = txy A sx - sy B >2 = 1.515 = 0.5032 (6.020 - 0)>2 up = 13.36° and 26.71° Substituting u = 13.36° into sx¿ = = sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u 6.020 + 0 6.020 - 0 + cos 26.71° + 1.515 sin 26.71° 2 2 = 6.38 MPa = s1 Thus, A uP B 1 = 13.4 and A uP B 2 = 26.71° Ans. The state of principal stresses is represented by the element shown in Fig. e. 655 09 Solutions 46060 6/8/10 3:13 PM Page 656 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–32. Determine the maximum in-plane shear stress developed at point A on the cross section of the arm at section a–a. Specify the orientation of this state of stress and indicate the results on an element at the point. 7.5 mm A 50 mm 7.5 mm 20 mm 7.5 mm Section a – a Support Reactions: Referring to the free - body diagram of the entire arm shown in Fig. a, ©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0 FCD = 2166.67 N + ©F = 0; : x Bx - 2166.67 cos 30° = 0 Bx = 1876.39 N + c ©Fy = 0; 2166.67 sin 30° - 500 - By = 0 By = 583.33 N 60⬚ 1876.39 - N = 0 N = 1876.39 N + c ©Fy = 0; V - 583.33 = 0 V = 583.33 N + ©MO = 0; 583.33(0.15) - M = 0 M = 87.5 N # m 0.15 m A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2 1 1 (0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4 12 12 Referring to Fig. b, QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. Thus, sA = = MyA N + A I -1876.39 0.5625 A 10 -3 B + 87.5(0.0175) 0.16367 A 10 - 6 B = 6.020 MPa The shear stress is contributed only by transverse shear stress. tA = 583.33 C 3.1875 A 10 - 6 B D VQA = = 1.515 MPa It 0.16367 A 10 - 6 B (0.0075) Maximum In - Plane Shear Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. tmax in-plane = C ¢ sx - sy 2 ≤ + txy 2 = 2 6.020 - 0 2 b + 1.5152 = 3.37 MPa B 2 a 656 C a Section Properties: The cross - sectional area and the moment of inertia about the z axis of the arm’s cross section are I = a B Internal Loadings: Considering the equilibrium of the free - body diagram of the arm’s left cut segment, Fig. b, + ©F = 0; : x D Ans. 0.15 m 0.35 m 500 N 09 Solutions 46060 6/8/10 3:13 PM Page 657 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–32. Continued Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = - A sx - sy B >2 txy = - (6.020 - 0)>2 = -1.9871 1.515 us = -31.6° and 58.4° Ans. Substituting u = -31.6° into tx¿y¿ = = - sx - sy 2 sin 2u + txy cos 2u 6.020 - 0 sin(-63.29°) + 1.515 cos(-63.29°) 2 = 3.37 MPa = t max in-plane This indicates that t max is directed in the positive sense of the y¿ axis on the face in-plane of the element defined by us = -31.6°. Average Normal Stress: savg = sx + sy 2 = 6.020 + 0 = 3.01 MPa 2 Ans. The state of maximum in - plane shear stress is represented on the element shown in Fig. e. 657 09 Solutions 46060 6/8/10 3:13 PM Page 658 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The clamp bears down on the smooth surface at E by tightening the bolt. If the tensile force in the bolt is 40 kN, determine the principal stress at points A and B and show the results on elements located at each of these points. The cross-sectional area at A and B is shown in the adjacent figure. •9–33. 300 mm 50 mm 30 mm 100 mm B A Support Reactions: As shown on FBD(a). E Internal Forces and Moment: As shown on FBD(b). Section Properties: I = 1 (0.03) A 0.053 B = 0.3125 A 10 - 6 B m4 12 QA = 0 QB = y¿A¿ = 0.0125(0.025)(0.03) = 9.375 A 10 - 6 B m3 Normal Stress: Applying the flexure formula s = - sA = sB = - 2.40(103)(0.025) = -192 MPa 0.3125(10 - 6) 2.40(103)(0) 0.3125(10 - 6) = 0 Shear Stress: Applying the shear formula t = tA = tB = My . I 24.0(103)(0) 0.3125(10 - 6)(0.03) VQ It = 0 24.0(103) C 9.375(10 - 6) D 0.3125(10 - 6)(0.03) = 24.0 MPa In - Plane Principal Stresses: sx = 0, sy = -192 MPa, and txy = 0 for point A. Since no shear stress acts on the element. s1 = sx = 0 Ans. s2 = sy = -192 MPa Ans. sx = sy = 0 and txy = -24.0 MPa for point B. Applying Eq. 9-5 s1,2 = sx + sy 2 ; C a sx - sy 2 2 b + t2xy = 0 ; 20 + (-24.0)2 = 0 ; 24.0 s1 = 24.0 s2 = -24.0 MPa Ans. 658 B A 25 mm 100 mm 50 mm 09 Solutions 46060 6/8/10 3:13 PM Page 659 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–33. Continued Orientation of Principal Plane: Applying Eq. 9-4 for point B. tan 2up = txy A sx - sy B >2 up = -45.0° = and -24.0 = -q 0 45.0° Subsututing the results into Eq. 9-1 with u = -45.0° yields sx¿ = sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u = 0 + 0 + [-24.0 sin (-90.0°)] = 24.0 MPa = s1 Hence, up1 = -45.0° up2 = 45.0° Ans. 659 09 Solutions 46060 6/8/10 3:13 PM Page 660 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–34. Determine the principal stress and the maximum inplane shear stress that are developed at point A in the 2-in.-diameter shaft. Show the results on an element located at this point. The bearings only support vertical reactions. 300 lb Using the method of sections and consider the FBD of shaft’s left cut segment, Fig. a, + ©F = 0; : x N - 3000 = 0 + c ©Fy = 0; 75 - V = 0 a + ©MC = 0; M - 75(24) = 0 N = 3000 lb V = 75 lb M = 1800 lb # in A = p(12) = p in2 I = p 4 p (1 ) = in4 4 4 Also, QA = 0 The normal stress developed is the combination of axial and bending stress. Thus My N ; A I s = For point A, y = C = 1 in. Then s = 1800(1) 3000 p p>4 = -1.337 (103) psi = 1.337 ksi (c) The shear stress developed is due to transverse shear force. Thus, t = VQA = 0 It The state of stress at point A, can be represented by the element shown in Fig. b. Here, sx = -1.337 ksi, sy = 0 is txy = 0. Since no shear stress acting on the element, s1 = sy = 0 s2 = sx = -1.34 ksi Ans. Thus, the state of principal stress can also be represented by the element shown in Fig. b. t max in-plane = C a sx - sy 2 tan 2us = us = 45° 2 b + t2xy = -1.337 - 0 2 b + 02 = 0.668 ksi - 668 psi Ans. C 2 a (sx - sy)>2 txy and = - (-1.337 - 0)>2 = q 0 -45° Ans. Substitute u = 45°, tx¿y¿ = = - sx - sy 2 sin 2u + txy cos 2u -1.337 - 0 sin 90° + 0 2 = 0.668 ksi = 668 psi = A 3000 lb tmax in-plane 660 24 in. 3000 lb 12 in. 12 in. 09 Solutions 46060 6/8/10 3:13 PM Page 661 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–34. Continued tmax This indicates that in-plane acts toward the positive sense of y¿ axis at the face of the element defined by us = 45°. Average Normal Stress. The state of maximum in - plane shear stress can be represented by the element shown in Fig. c. 661 09 Solutions 46060 6/8/10 3:13 PM Page 662 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–35. The square steel plate has a thickness of 10 mm and is subjected to the edge loading shown. Determine the maximum in-plane shear stress and the average normal stress developed in the steel. sx = 5 kPa sy = -5 kPa tmax sx - sy in-plane = = savg = C a 2 50 N/m txy = 0 200 mm b + t2xy 5 + 5 2 b + 0 = 5 kPa C 2 a sx + sy 3 50 N/m 2 Ans. 200 mm 5 - 5 = = 0 2 Ans. Note: tan 2us = tan 2us = -(sx - sy)>2 txy -(5 + 5)>2 = q 0 us = 45° *9–36. The square steel plate has a thickness of 0.5 in. and is subjected to the edge loading shown. Determine the principal stresses developed in the steel. sx = 0 s1,2 = sy = 0 sx + sy 2 ; 16 lb/in. txy = 32 psi C a sx - sy 2 2 b + t2xy = 0 ; 20 + 322 4 in. s1 = 32 psi Ans. s2 = -32 psi Ans. Note: tan 2up = 16 lb/in. 4 in. txy (sx - sy)>2 = 32 = q 0 up = 45° 662 09 Solutions 46060 6/8/10 3:13 PM Page 663 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stress and the maximum in-plane shear stress that is developed at point A. The bearings only support vertical reactions. •9–37. P F F A L 2 Support Reactions: As shown on FBD(a). Internal Forces and Moment: As shown on FBD(b). Section Properties: A = p 2 d 4 p d 4 p 4 a b = d 4 2 64 I = QA = 0 Normal Stress: Mc N ; A I s = -F ; d2 = p 4 sA = A B pL d 4 2 p 4 d 64 2PL 4 a - Fb 2 d pd Shear Stress: Since QA = 0, tA = 0 In - Plane Principal Stress: sx = 4 2PL a - Fb. pd2 d sy = 0 and txy = 0 for point A. Since no shear stress acts on the element, s1 = sx = 2PL 4 a - Fb d pd2 Ans. s2 = sy = 0 Ans. Maximum In - Plane Shear Stress: Applying Eq. 9-7 for point A, t max in-plane = = = Q £ B a 4 2 pd sx - sy 2 2 b + t2xy A 2PL d - FB - 0 2 ≥ + 0 2 2PL 2 a - Fb d pd2 Ans. 663 L 2 09 Solutions 46060 6/8/10 3:13 PM Page 664 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–38. A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown. Determine the shear stress acting along the seam, which is at 30° from the vertical, when the tube is subjected to an axial force of 10 N. The paper is 1 mm thick and the tube has an outer diameter of 30 mm. P = A s = p 4 = - 10 N 10 N 30 mm 10 = 109.76 kPa (0.032 - 0.0282) sx = 109.76 kPa tx¿y¿ = - 30⬚ sx - sy 2 sy = 0 txy = 0 u = 30° sin 2u + txy cos 2u 106.76 - 0 sin 60° + 0 = -47.5 kPa 2 Ans. 9–39. Solve Prob. 9–38 for the normal stress acting perpendicular to the seam. 30⬚ 10 N 10 N 30 mm s = sn = = P = A p 4 10 = 109.76 kPa (0.032 - 0.0282) sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u 109.76 + 0 109.76 - 0 + cos (60°) + 0 = 82.3 kPa 2 2 Ans. 664 09 Solutions 46060 6/8/10 3:13 PM Page 665 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–40. Determine the principal stresses acting at point A of the supporting frame. Show the results on a properly oriented element located at this point. 800 mm B A 300 mm 12 mm 150 mm 5 B 15 mm 130 mm A y = ©yA 0.065(0.13)(0.015) + 0.136(0.15)(0.012) = = 0.0991 m ©A 0.13(0.015) + 0.15(0.012) I = 1 (0.015)(0.133) + 0.015(0.13)(0.0991 - 0.065)2 12 1 (0.15)(0.012 3) + 0.15(0.012)(0.136 - 0.0991)2 = 7.4862(10 - 6) m4 12 + QA = 0 A = 0.13(0.015) + 0.15(0.012) = 3.75(10 - 3) m2 Normal stress: s = Mc P + A I sA = -3.6(103) -3 3.75(10 ) - 5.2767(103)(0.0991) 7.4862(10 - 6) = -70.80 MPa Shear stress: tA = 0 Principal stress: s1 = 0 Ans. s2 = -70.8 MPa Ans. 665 4 3 6 kN 09 Solutions 46060 6/8/10 3:13 PM Page 666 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Determine the principal stress acting at point B, which is located just on the web, below the horizontal segment on the cross section. Show the results on a properly oriented element located at this point. Although it is not very accurate, use the shear formula to calculate the shear stress. •9–41. 800 mm B A 300 mm 12 mm y = ©yA 0.065(0.13)(0.015) + 0.136(0.15)(0.012) = = 0.0991 m ©A 0.13(0.015) + 0.15(0.012) I = 1 (0.015)(0.133) + 0.015(0.13)(0.0991 - 0.065)2 12 + A 1 (0.15)(0.0123) + 0.15(0.012)(0.136 - 0.0991)2 = 7.4862(10 - 6) m4 12 Normal stress: Mc P + A I sB = - 3.6(103) 3.75(10 - 3) + 5.2767(103)(0.130 - 0.0991) 7.4862(10 - 6) = 20.834 MPa Shear stress: tB = VQ -4.8(103)(0.0369)(0.15)(0.012) = -2.84 MPa = It 7.4862(10 - 6)(0.015) Principal stress: s1,2 = a 20.834 + 0 20.834 - 0 2 b ; a b + (-2.84)2 2 C 2 s1 = 21.2 MPa Ans. s2 = -0.380 MPa Ans. tan 2up = A -2.84 20.834 - 0 2 B up = -7.63° Ans. 666 5 B 15 mm 130 mm A = 0.13(0.015) + 0.15(0.012) = 3.75(10 - 3) m2 s = 150 mm 4 3 6 kN 09 Solutions 46060 6/8/10 3:13 PM Page 667 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–42. The drill pipe has an outer diameter of 3 in., a wall thickness of 0.25 in., and a weight of 50 lb>ft. If it is subjected to a torque and axial load as shown, determine (a) the principal stress and (b) the maximum in-plane shear stress at a point on its surface at section a. 1500 lb 800 lb⭈ft 20 ft a 20 ft Internal Forces and Torque: As shown on FBD(a). Section Properties: A = p 2 A 3 - 2.52 B = 0.6875p in2 4 J = p A 1.54 - 1.254 B = 4.1172 in4 2 s = N -2500 = = -1157.5 psi A 0.6875p Normal Stress: Shear Stress: Applying the torsion formula. t = 800(12)(1.5) Tc = = 3497.5 psi J 4.1172 a) In - Plane Principal Stresses: sx = 0, sy = -1157.5 psi and txy = 3497.5 psi 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 0 + (-1157.5) 0 - (-1157.5) 2 ; a b + (3497.5)2 2 C 2 = -578.75 ; 3545.08 s1 = 2966 psi = 2.97 ksi Ans. s2 = -4124 psi = -4.12 ksi Ans. b) Maximum In - Plane Shear Stress: Applying Eq. 9-7 t max in-plane = = C a sx - sy 2 2 b + t2xy ¢ 0 - (-1157.5) 2 ≤ + (3497.5)2 C 2 = 3545 psi = 3.55 ksi Ans. 667 09 Solutions 46060 6/8/10 3:13 PM Page 668 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–43. Determine the principal stress in the beam at point A. 60 kN 50 mm 150 kN A A 60 mm 0.5 m Using the method of sections and consider the FBD of the beam’s left cut segment, Fig. a, + ©F = 0; : x 150 - N = 0 N = 150 kN + c ©Fy = 0; V - 60 = 0 V = 60 kN a + ©MC = 0; M = 30 kN # m 60(0.5) - M = 0 A = 0.06(0.15) = 0.009 m2 1 (0.06)(0.153) = 16.875(10 - 6) m4 12 I = Referring to Fig. b, QA = y¿A¿ = 0.05 (0.05)(0.06) = 0.15(10 - 3) m3 The normal stress developed is the combination of axial and bending stress. Thus My N ; A I s = For point A, y = 0.075 - 0.05 = 0.025 m. Then s = 30(103)(0.025) -150(103) 0.009 16.875(10 - 6) = -61.11(106) Pa = 61.11 MPa (c) The shear stress developed is due to the transverse shear, Thus, t = 60(103) C 0.15(10 - 3) D VQA = 8.889 MPa = It 16.875(10 - 6) (0.06) Here, sx = -61.11 MPa, sy = 0 and txy = 8.889 MPa, s1, 2 = = sx + sy 2 ; C a sx - sy 2 2 b + t2xy -61.11 + 0 -61.11 - 0 2 ; a b + 8.8892 2 C 2 = -30.56 ; 31.82 s2 = -62.4 MPa s1 = 1.27 MPa tan 2uP = txy (sx - sy)>2 uP = -8.11° = and Ans. 8.889 = -0.2909 (-61.11 - 0)>2 81.89° 668 0.25 m 150 mm 09 Solutions 46060 6/8/10 3:13 PM Page 669 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–43. Continued Substitute u = -8.11°, sx¿ = = sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u -61.11 - 0 -61.11 + 0 + cos (-16.22°) + 8.889 sin (-16.22°) 2 2 = -62.4 MPa = s2 Thus, (uP)1 = 81.9° (uP)2 = -8.11° The state of principal stresses can be represented by the elements shown in Fig. (c) 669 09 Solutions 46060 6/8/10 3:13 PM Page 670 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–44. Determine the principal stress at point A which is located at the bottom of the web. Show the results on an element located at this point. 150 kN/m Using the method of sections, consider the FBD of the bean’s left cut segment, Fig. a, + c ©Fy = 0; V - I = 1 (100)(0.6) = 0 2 M = 6 kN # m 10 mm A 1 1 (0.15)(0.223) (0.14)(0.23) = 39.7667(10 - 6) m4 12 12 150 mm Referring to Fig. b QA = y¿A¿ = 0.105 (0.01)(0.15) = 0.1575(10 - 3) m3 The normal stress developed is due to bending only. For point A, y = 0.1 m. Then s = My I = 6(103)(0.1) = 15.09(106)Pa = 15.09 MPa (c) 39.7667(10 - 6) The shear stress developed is due to the transverse shear. Thus, t = 30(103) C 0.1575(10 - 3) D VQA = 11.88(106)Pa = 11.88 MPa = It 39.7667(10 - 6)(0.01) Here, sx = -15.09 MPa, sy = 0 And txy = 11.88 MPa. s1, 2 = = sx + sy 2 ; C a sx - sy 2 2 b + t2xy -15.09 + 0 -15.09 - 0 2 ; a b + 11.882 2 C 2 = -7.544 ; 14.074 s2 = -21.6 MPa s1 = 6.53 MPa tan 2uP = txy (sx - sy)>2 uP = -28.79° = Ans. 11.88 = -1.575 (-15.09 - 0)>2 and 61.21° Substitute u = 61.21°, sx¿ = = sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u -15.09 + 0 -15.09 - 0 + cos 122.42° + 11.88 sin 122.42° 2 2 = 6.53 MPa = s1 Thus, (uP)1 = 61.2° 0.3 m V = 30 kN 1 (100)(0.6)(0.2) - M = 0 2 a + ©MC = 0; A 0.6 m (uP)2 = -28.8° Ans. The state of principal stresses can be represented by the element shown in Fig. d. 670 10 mm 200 mm 10 mm 09 Solutions 46060 6/8/10 3:13 PM Page 671 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–44. Continued 671 09 Solutions 46060 6/8/10 3:13 PM Page 672 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Determine the maximum in-plane shear stress in the box beam at point A. Show the results on an element located at this point. •9–45. 10 kip 4 kip A B Using the method of section, consider the FBD, of bean’s left cut segment, Fig. a, + c ©Fy = 0; 8 - 10 + V = 0 a + ©MC = 0; M + 10(1.5) - 8(3.5) = 0 4 in. A M = 13 kip # ft 4 in. The moment of inertia of the cross - section about the neutral axis is Referring to Fig. b, QA = 0 The normal stress developed is contributed by the bending stress only. For point A, y = C = 3 in. My = I 13(12)(3) = 5.40 ksi (c) 86.6667 The shear stress is contributed by the transverse shear stress only. Thus t = VQA = 0 It The state of stress at point A can be represented by the element shown in Fig. c Here, sx = -5.40 ksi, sy = 0 and txy = 0. tmax in-plane = C a sx - sy 2 tan 2us = - 2 b + txy 2 = (sx - sy)>2 txy us = 45° -5.40 - 0 2 b + 02 = 2.70 ksi C 2 = - a Ans. (-5.40 - 0)>2 = q 2 -45° and Substitute u = 45°, tx¿y¿ = - sx - sy = - 2 sin 2u + txy cos 2u -5.40 - 0 sin 90° + 0 2 = 2.70 ksi = tmax in-plane tmax This indicates that in-plane acts toward the positive sense of y¿ axis at the face of element defined by us = 45° savg = sx + sy 2 = -5.40 + 0 = -2.70 ksi 2 The state of maximum In - plane shear stress can be represented by the element shown in Fig. d. 672 B 6 in. 1 1 I = (6)(63) (4)(43) = 86.6667 in4 12 12 s = 1.5 ft 2 ft V = 2 kip 2 ft 0.5 ft 3 in. 3 in. 09 Solutions 46060 6/8/10 3:13 PM Page 673 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–45. Continued 673 09 Solutions 46060 6/8/10 3:13 PM Page 674 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–46. Determine the principal stress in the box beam at point B. Show the results on an element located at this point. 10 kip 4 kip A Using the method of sections, consider the FBD of bean’s left cut segment, Fig. a, + c ©Fy = 0; 8 - 10 + V = 0 a + ©MC = 0; M = 13 kip # ft M + 10(1.5) - 8(3.5) = 0 I = B V = 2 kip 4 in. A 1 1 (6)(63) (4)(43) = 86.6667 in4 12 12 4 in. Referring to Fig. b, QB = 2y1œ A1œ + 3 The normal stress developed is contributed by the bending stress only. For point B, y = 0. My s = = 0 I The shear stress is contributed by the transverse shear stress only. Thus VQB 2(103)(19) = = 219.23 psi It 86.6667(2) t = The state of stress at point B can be represented by the element shown in Fig. c Here, sx = sy = 0 and txy = 219.23 psi. s1, 2 = sx + sy ; 2 C a sx - sy 2 2 b + txy 2 = 0 ; 20 + 219.232 s2 = -219 psi s1 = 219 psi tan 2uP = txy (sx - sy)>2 uP = 45° = Ans. 219.23 = q 0 -45° and Substitute u = 45°, sx¿ = sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u = 0 + 0 + 219.23 sin 90° = 219 psi = s1 Thus, (uP)1 = 45° (uP)2 = -45° Ans. The state of principal stress can be represented by the element shown in Fig. d. 674 B 6 in. = 2 C 1(2)(1) D + 2.5(1)(6) = 19 in y2œ A2œ 1.5 ft 2 ft 2 ft 0.5 ft 3 in. 3 in. 09 Solutions 46060 6/8/10 3:13 PM Page 675 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–46. Continued 675 09 Solutions 46060 6/8/10 3:13 PM Page 676 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–47. The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine the principal stresses acting at point A. Ix = Iy = J = p (0.025)4 = 0.306796(10 - 6) m4 4 450 mm 300 N⭈m p (0.025)4 = 0.613592(10 - 6) m4 2 25 mm 45 N⭈m QA = 0 sA 800 N 60(0.025) Mx c = 4.889 MPa = = I 0.306796(10 - 6) tA = Ty c J = 45(0.025) 0.613592(10 - 6) sx = 4.889 MPa s1, 2 = = sx + sy 2 = 1.833 MPa sy = 0 ; C a txy = -1.833 MPa sx - sy 2 2 b + txy 2 4.889 - 0 2 4.889 + 0 ; b + (-1.833)2 a 2 C 2 s1 = 5.50 MPa Ans. s2 = -0.611 MPa Ans. Solve Prob. 9–47 for point B. *9–48. Ix = Iy = p (0.025)4 = 0.306796(10 - 6) m4 4 450 mm p J = (0.025)4 = 0.613592(10 - 6) m4 2 QB = yA¿ = 300 N⭈m 4(0.025) 1 a b p (0.0252) = 10.4167(10 - 6) m3 3p 2 800 N VzQB It - Ty c J -6 = 800(10.4167)(10 ) -6 0.306796(10 )(0.05) sx = 0 s1, 2 = sy = 0 sx + sy 2 ; C a sx - sy 2 - 45(0.025) 0.61359(10 - 6) = -1.290 MPa txy = -1.290 MPa 2 b + txy 2 = 0 ; 2(0)2 + ( -1.290)2 s1 = 1.29 MPa Ans. s2 = -1.29 MPa Ans. 676 A B 25 mm 45 N⭈m sB = 0 tB = A B 09 Solutions 46060 6/8/10 3:13 PM Page 677 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The internal loadings at a section of the beam are shown. Determine the principal stress at point A. Also compute the maximum in-plane shear stress at this point. •9–49. 50 mm A 200 mm 50 mm 50 mm y 200 mm Section Properties: z A = 0.2(0.3) - 0.15(0.2) = 0.030 m4 800 kN 1 1 Iz = (0.2) A 0.33 B (0.15) A 0.23 B = 0.350 A 10 - 3 B m4 12 12 Iy = 1 1 (0.1) A 0.23 B + (0.2) A 0.053 B = 68.75 A 10 - 6 B m4 12 12 (QA)y = 0 Normal Stress: s = sA = Myz Mzy N + A Iz Iy -30(103)(0.1) 40(103)(0.15) -500(103) + 3 0.030 0.350(10 ) 68.75(10 - 6) = -77.45 MPa Shear Stress: Since (QA)y = 0, tA = 0. In - Plane Principal Stresses: sx = -77.45 MPa. sy = 0. and txy = 0 for point A. Since no shear stress acts on the element. s1 = sy = 0 Ans. s2 = sz = -77.4 MPa Ans. Maximum In-Plane Shear Stress: Applying Eq. 9–7. t max in-plane = = C a sx - sy 2 2 b + t2xy -77.45 - 0 2 b + 0 C 2 a = 38.7 MPa Ans. 677 40 kN⭈m 30 kN⭈m 500 kN x 09 Solutions 46060 6/8/10 3:13 PM Page 678 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–50. The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800 N, and two moment components of 30 N # m and 40 N # m. Determine the principal stress at point A. Also calculate the maximum in-plane shear stress at this point. Ix = A 1 (0.1)(0.2)3 = 66.67(10 - 6) in4 12 40 N⭈m B C 50 mm QA = 0 sA 200 mm 50 mm 30(0.1) Mz 500 P = -20 kPa = = A Ix (0.1)(0.2) 66.67(10 - 6) 100 mm 30 N⭈m 500 N 800 N tA = 0 Here, the principal stresses are s1 = sy = 0 Ans. s2 = sx = -20 kPa Ans. t max in-plane= = 9–51. C a sx - sy 2 2 b + txy 2 -20 - 0 2 b + 0 = 10 kPa C 2 a Ans. Solve Prob. 9–4 using Mohr’s circle. A 400 psi 650 psi 60⬚ sx + sy 2 -650 + 400 = = -125 2 A(-650, 0) B(400, 0) C( -125, 0) B R = CA = = 650 - 125 = 525 sx¿ = -125 - 525 cos 60° = -388 psi Ans. tx¿y¿ = 525 sin 60° = 455 psi Ans. 678 09 Solutions 46060 6/8/10 3:13 PM Page 679 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–52. Solve Prob. 9–6 using Mohr’s circle. 90 MPa A 35 MPa 60⬚ 30⬚ sx = 90 MPa sx + sy 2 = sy = 50 MPa txy = -35 MPa A(90, -35) 90 + 50 = 70 2 R = 2(90 - 70)2 + (35)2 = 40.311 Coordinates of point B: f = tan - 1 a 35 b = 60.255° 20 c = 300° - 180° - 60.255° = 59.745° sx¿ = 70 - 40.311 cos 59.745° = 49.7 MPa Ans. tx¿ = -40.311 sin 59.745° = -34.8 MPa Ans. 679 B 50 MPa 09 Solutions 46060 6/8/10 3:13 PM Page 680 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Solve Prob. 9–14 using Mohr’s circle. •9–53. 30 ksi 12 ksi sx + sy 2 = -30 + 0 = -15 2 R = 2(30 - 15)2 + (12)2 = 19.21 ksi s1 = 19.21 - 15 = 4.21 ksi Ans. s2 = -19.21 - 15 = -34.2 ksi Ans. 2uP2 = tan - 1 tmax in-plane 12 ; (30 - 15) uP2 = 19.3° Ans. = R = 19.2 ksi Ans. savg = -15 ksi 2uP2 = tan - 1 12 + 90°; (30 - 15) Ans. us = 64.3° Ans. 680 09 Solutions 46060 6/8/10 3:13 PM Page 681 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–54. Solve Prob. 9–16 using Mohr’s circle. 350 psi 75 psi 200 psi sx + sy 2 = 45 - 60 = -7.5 MPa 2 R = 2(45 + 7.5)2 + (30)2 = 60.467 MPa s1 = 60.467 - 7.5 = 53.0 MPa Ans. s2 = -60.467 - 7.5 = -68.0 MPa Ans. 2uP1 = tan - 1 uP1 = 14.9° tmax 30 (45 + 7.5) counterclockwise Ans. = 60.5 MPa Ans. savg = -7.50 MPa Ans. in-plane 2uP1 = 90° - tan - 1 us1 = 30.1° 30 (45 + 7.5) clockwise Ans. 681 09 Solutions 46060 6/8/10 3:13 PM Page 682 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–55. Solve Prob. 9–12 using Mohr’s circle. sx + sy 2 = -10 + 0 = -5 ksi 2 10 ksi 16 ksi R = 2(10 - 5)2 + (16)2 = 16.763 ksi f = tan - 1 16 = 72.646° (10 - 5) a = 100 - 72.646 = 27.354° sx¿ = -5 - 16.763 cos 27.354° = -19.9 ksi Ans. tx¿y¿ = 16.763 sin 27.354° = 7.70 ksi Ans. sy¿ = 16.763 cos 27.354° - 5 = 9.89 ksi *9–56. Solve Prob. 9–11 using Mohr’s circle. 2 ksi Construction of the Circle: In accordance with the sign convention, sx = -3 ksi, sy = 2 ksi, and txy = -4 ksi. Hence, savg = sx + sy 2 = -3 + 2 = -0.500 ksi 2 3 ksi 30⬚ The coordinates for reference point A and C are A(-3, -4) A 4 ksi B C(-0.500, 0) The radius of circle is R = 2(3 - 0.5)2 + 42 = 4.717 ksi Stress on the Inclined Plane: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle. sx¿ = -0.500 - 4.717 cos 62.01° = -2.71 ksi Ans. tx¿y¿ = 4.717 sin 62.01° = 4.17 ksi Ans. 682 09 Solutions 46060 6/8/10 3:13 PM Page 683 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–57. Mohr’s circle for the state of stress in Fig. 9–15a is shown in Fig. 9–15b. Show that finding the coordinates of point P1sx¿ , tx¿y¿2 on the circle gives the same value as the stress-transformation Eqs. 9–1 and 9–2. R = sxœ = C Ca a B(sy, -txy) A(sx, txy) csx - a sx + sy sx + sy + 2 2 C a sx + sy 2 2 b d + t2xy = sx - sy 2 C a b, 0b sx - sy 2 2 b + t2xy 2 b + t2xy cos u¿ (1) u¿ = 2uP - 2u cos (2uP - 2u) = cos 2uP cos 2u + sin 2up sin 2u (2) From the circle: sx - cos 2uP = sin 2uP = 4A 4A sx + sy 2 sx - sy 2 txy sx - sy 2 B + 2 (3) t2xy (4) B 2 + t2xy Substitute Eq. (2), (3) and into Eq. (1) sx¿ = tx¿y¿ = sx + sy 2 C a + sx - sy 2 sx - sy 2 cos 2u + txy sin 2u QED 2 b + t2xy sin u¿ (5) sin u¿ = sin (2uP - 2u) = sin 2uP cos 2u - sin 2u cos 2uP (6) Substitute Eq. (3), (4), (6) into Eq. (5), tx¿y¿ = - sx - sy 2 sin 2u + txy cos 2u QED 683 09 Solutions 46060 6/8/10 3:13 PM Page 684 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. 9–59. Determine the equivalent state of stress if an element is oriented 20° clockwise from the element shown. 2 ksi Construction of the Circle: In accordance with the sign convention, sx = 3 ksi, sy = -2 ksi, and tx¿y¿ = -4 ksi. Hence, savg = sx + sy 2 = 3 + (-2) = 0.500 ksi 2 4 ksi The coordinates for reference points A and C are A(3, -4) 3 ksi C(0.500, 0) The radius of the circle is R = 2(3 - 0.500)2 + 42 = 4.717 ksi Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinate of point P on the circle, sy¿, can be determined by calculating the coordinates of point Q on the circle. sx¿ = 0.500 + 4.717 cos 17.99° = 4.99 ksi Ans. tx¿y¿ = -4.717 sin 17.99° = -1.46 ksi Ans. sy¿ = 0.500 - 4.717 cos 17.99° = -3.99 ksi Ans. 684 09 Solutions 46060 6/8/10 3:13 PM Page 685 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–60. Determine the equivalent state of stress if an element is oriented 30° clockwise from the element shown. Show the result on the element. 9 ksi 4 ksi In accordance to the established sign convention, sx = -6 ksi, sy = 9 ksi and txy = 4 ksi. Thus, savg = sx + sy 2 = -6 + 9 = 1.50 ksi 2 Then, the coordinates of reference point A and C are A(-6, 4) C(1.5, 0) The radius of the circle is R = CA = 2(-6 - 1.5)2 + 42 = 8.50 ksi Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a, a = tan - 1 a 4 b = 28.07° 6 + 1.5 b = 60° - 28.07° = 31.93° Then, sx¿ = 1.5 - 8.50 cos 31.93° = -5.71 ksi Ans. tx¿y¿ = -8.5 sin 31.95° = -4.50 ksi sy¿ = 8.71 ksi Ans. The results are shown in Fig. b. 685 6 ksi 09 Solutions 46060 6/8/10 3:13 PM Page 686 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Determine the equivalent state of stress for an element oriented 60° counterclockwise from the element shown. Show the result on the element. •9–61. 250 MPa 400 MPa In accordance to the established sign convention, sx = -560 MPa, sy = 250 MPa and txy = -400 MPa. Thus, savg = sx + sy 2 = -560 + 250 = -155 MPa 2 Then, the coordinate of reference points A and C are A(-560, -400) C(-155, 0) The radius of the circle is R = CA = 3 C -560 - (-155) D 2 + (-400)2 = 569.23 MPa Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a a = tan - 1 a 400 b = 44.64° 560 - 155 b = 120° - 44.64° = 75.36° Then, sx¿ = -155 - 569.23 cos 75.36° = -299 MPa Ans. tx¿y¿ = 569.23 sin 75.36° = 551 MPa Ans. sy¿ = -155 + 569.23 cos 75.36° = -11.1 MPa Ans. The results are shown in Fig. b. 686 560 MPa 09 Solutions 46060 6/8/10 3:13 PM Page 687 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–62. Determine the equivalent state of stress for an element oriented 30° clockwise from the element shown. Show the result on the element. 5 ksi In accordance to the established sign convention, sx = 2 ksi, sy = -5 ksi and txy = 0. Thus, savg = sx + sy 2 = 2 + (-5) = -1.50 ksi 2 Then, the coordinate of reference points A and C are A(2, 0) C(-1.5, 0) The radius of the circle is R = CA = 3 C 2 - (-1.5) D 2 + 02 = 3.50 ksi Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a, b = 60° Then, sx¿ = -1.50 + 3.50 cos 60° - 0.250 ksi Ans. tx¿y¿ = 3.50 sin 60° = 3.03 ksi Ans. sy¿ = -3.25 ksi Ans. The results are shown in Fig b. 687 2 ksi 09 Solutions 46060 6/8/10 3:13 PM Page 688 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–63. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. 15 ksi 5 ksi Construction of the Circle: In accordance with the sign convention, sx = 15 ksi, sy = 0 and txy = -5 ksi. Hence, sx + sy savg = 2 = 15 + 0 = 7.50 ksi 2 Ans. The coordinates for reference point A and C are A(15, -5) C(7.50, 0) The radius of the circle is R = 2(15 - 7.50)2 + 52 = 9.014 ksi a) In - Plane Principal Stress: The coordinates of points B and D represent s1 and s2, respectively. s1 = 7.50 + 9.014 = 16.5 ksi Ans. s2 = 7.50 - 9.014 = -1.51 ksi Ans. Orientation of Principal Plane: From the circle tan 2uP1 = 5 = 0.6667 15 - 7.50 uP1 = 16.8° (Clockwise) Ans. b) Maximum In - Plane Shear Stress: Represented by the coordinates of point E on the circle. tmax in-plane = -R = -9.01 ksi Ans. Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle tan 2us = us = 28.2° 15 - 7.50 = 1.500 5 (Counterclockwise) Ans. 688 09 Solutions 46060 6/8/10 3:13 PM Page 689 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *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 in-plane = R = 83.8 MPa Ans. 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 689 09 Solutions 46060 6/8/10 3:13 PM Page 690 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–64. Continued 690 09 Solutions 46060 6/8/10 3:13 PM Page 691 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •9–65. Determine the principal stress, the maximum inplane shear stress, and average normal stress. Specify the orientation of the element in each case. 120 psi 300 psi A(300, 120) B(0, -120) C(150, 0) R = 2(300 - 150)2 + 1202 = 192.094 s1 = 150 + 192.094 = 342 psi Ans. s2 = 150 - 192.094 = -42.1 psi Ans. tan 2uP = 120 = 0.8 300 - 150 uP1 = 19.3° Counterclockwise Ans. savg = 150 psi Ans. tmax Ans. in-plane = 192 psi tan 2us = 300 - 150 = 1.25 120 us = -25.7° Ans. 9–66. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. A(45, -50) B(30, 50) 30 MPa C(37.5, 0) 45 MPa R = CA = CB = 27.52 + 502 = 50.56 50 MPa a) tan 2uP = 50 7.5 s1 = 37.5 + 50.56 = 88.1 MPa Ans. s2 = 37.5 - 50.56 = -13.1 MPa Ans. 2uP = 81.47° uP = -40.7° b) t max in-plane = R = 50.6 MPa Ans. savg = 37.5 MPa Ans. 2us = 90 - 2uP us = 4.27° Ans. 691 09 Solutions 46060 6/8/10 3:13 PM Page 692 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–67. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case. 200 MPa 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 = us = 14.4° 350 - 75.0 = 0.55 500 (Clockwise) Ans. 692 09 Solutions 46060 6/8/10 3:13 PM Page 693 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–68. Draw Mohr’s circle that describes each of the following states of stress. 700 psi 4 ksi 40 MPa 600 psi (a) a) Here, sx = 600 psi, sy = 700 psi and txy = 0. Thus, savg = sx + sy 2 = 600 + 700 = 650 psi 2 Thus, the coordinate of reference point A and center of circle are A(600, 0) C(650, 0) Then the radius of the circle is R = CA = 650 - 600 = 50 psi The Mohr’s circle represents this state of stress is shown in Fig. a. b) Here, sx = 0, sy = 4 ksi and txy = 0. Thus, savg = sx + sy 2 = 0 + 4 = 2 ksi 2 Thus, the coordinate of reference point A and center of circle are A(0, 0) C(2, 0) Then the radius of the circle is R = CA = 2 - 0 = 2 psi c) Here, sx = sy = 0 and txy = -40 MPa. Thus, savg = sx + sy 2 = 0 Thus, the coordinate of reference point A and the center of circle are A(0, -40) C(0, 0) Then, the radius of the circle is R = CA = 40 MPa The Mohr’s circle represents this state of stress shown in Fig. c 693 (b) (c) 09 Solutions 46060 6/8/10 3:13 PM Page 694 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–68. Continued 694 09 Solutions 46060 6/8/10 3:13 PM Page 695 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–69. The frame supports the distributed loading of 200 N兾m. Determine the normal and shear stresses at point D that act perpendicular and parallel, respectively, to the grain. The grain at this point makes an angle of 30° with the horizontal as shown. 200 N/ m B 30⬚ 1m 200 mm 75 mm D 1.5 m C 100 mm 4m 60⬚ E Support Reactions: As shown on FBD(a). 50 mm 30 mm Internal Forces and Moment: As shown on FBD(b). 1.5 m 100 mm Section Properties: I = A 1 (0.1) A 0.23 B = 66.667 A 10 - 6 B m4 12 QD = y¿A¿ = 0.0625(0.075)(0.1) = 0.46875 A 10 - 3 B m3 Normal Stress: Applying the flexure formula. sD = - My 150(-0.025) = 56.25 kPa = I 66.667(10 - 6) Shear Stress: Applying the shear formula. tD = 50.0 C 0.46875(10 - 3) D VQD = 3.516 kPa = It 66.667(10 - 6)(0.1) Construction of the Circle: In accordance to the established sign convention, sx = 56.25 kPa, sy = 0 and txy = -3.516 kPa. Hence. savg = sx + sy 2 = 56.25 + 0 = 28.125 kPa 2 The coordinates for reference point A and C are A(56.25, -3.516) C(28.125, 0) The radius of the circle is R = 2(56.25 - 28.125)2 + 3.5162 = 28.3439 kPa Stresses on The Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle. Here, u = 60°. sx¿ = 28.125 - 28.3439 cos 52.875° = 11.0 kPa Ans. tx¿y¿ = -28.3439 sin 52.875° = -22.6 kPa Ans. 695 09 Solutions 46060 6/8/10 3:13 PM Page 696 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–70. The frame supports the distributed loading of 200 N兾m. Determine the normal and shear stresses at point E that act perpendicular and parallel, respectively, to the grain. The grain at this point makes an angle of 60° with the horizontal as shown. 200 N/ m B 30⬚ 1m 200 mm 75 mm D 1.5 m C 100 mm 4m 60⬚ E Support Reactions: As shown on FBD(a). 50 mm 30 mm Internal Forces and Moment: As shown on FBD(b). 1.5 m 100 mm Section Properties: A A = 0.1(0.05) = 5.00 A 10 - 3 B m2 Normal Stress: sE = N -250 = -50.0 kPa = A 5.00(10 - 3) Construction of the Circle: In accordance with the sign convention. sx = 0, sy = -50.0 kPa, and txy = 0. Hence. savg = sx + sy 2 = 0 + (-50.0) = -25.0 kPa 2 The coordinates for reference points A and C are A(0, 0) C(-25.0, 0) The radius of circle is R = 25.0 - 0 = 25.0 kPa Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by coordinates of point P on the circle. Here, u = 150°. sx = -25.0 + 25.0 cos 60° = -12.5 kPa Ans. tx¿y¿ = 25.0 sin 60° = 21.7 kPa Ans. 696 09 Solutions 46060 6/8/10 3:13 PM Page 697 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–71. The stair tread of the escalator is supported on two of its sides by the moving pin at A and the roller at B. If a man having a weight of 300 lb stands in the center of the tread, determine the principal stresses developed in the supporting truck on the cross section at point C. The stairs move at constant velocity. 1.25 ft 30⬚ A C 1.5 ft 0.5 ft B 0.5 ft Internal Forces and Moment: As shown on FBD (b). Section Properties: A = 2(0.5) = 1.00 in2 1 (0.5) A 23 B = 0.3333 in4 12 QB = y¿A¿ = 0.5(1)(0.5) = 0.250 in3 Normal Stress: s = sC = My N ; A I 475.48(0) -137.26 + = -137.26 psi 1.00 0.3333 Shear Stress: Applying the shear formula t = tC = VQ . It 79.25(0.250) = 118.87 psi 0.3333(0.5) Construction of the Circle: In accordance with the sign convention, sx = 0, sy = -137.26 psi, and txy = 118.87 psi. Hence, savg = sx + sy 2 = 0 + (-137.26) = -68.63 psi 2 The coordinates for reference points A and C are A(0, 118.87) 1 in. C 2 in. Support Reactions: As shown on FBD (a). I = 30⬚ C(-68.63, 0) The radius of the circle is R = 2(68.63 - 0)2 + 118.872 = 137.26 psi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = -68.63 + 137.26 = 68.6 psi Ans. s2 = -68.63 - 137.26 = -206 psi Ans. 697 0.5 in. 09 Solutions 46060 6/8/10 3:13 PM Page 698 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–72. 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 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. 698 20 lb⭈ft 09 Solutions 46060 6/8/10 3:13 PM Page 699 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •9–73. The cantilevered rectangular bar is subjected to the force of 5 kip. Determine the principal stress at point A. 1.5 in. A 1.5 in. 1 in. B 1.5 in. 1.5 in. 3 in. 1 in. 15 in. 3 in. 3 5 4 5 kip Internal Forces and Moment: As shown on FBD. Section Properties: A = 3(6) = 18.0 in2 I = 1 (3) A 63 B = 54.0 in4 12 QA = y¿A¿ = 2.25(1.5)(3) = 10.125 in3 Normal Stress: s = sA = My N ; A I 45.0(1.5) 4.00 + = 1.4722 ksi 18.0 54.0 Shear Stress: Applying the shear formula t = tA = VQ . It 3.00(10.125) = 0.1875 ksi 54.0(3) Construction of the Circle: In accordance with the sign convention, sx = 1.4722 ksi, sy = 0, and txy = -0.1875 ksi. Hence, savg = sx + sy 2 = 1.472 + 0 = 0.7361 ksi 2 The coordinates for reference points A and C are A(1.4722, -0.1875) C(0.7361, 0) The radius of the circle is R = 2(1.4722 - 0.7361)2 + 0.18752 = 0.7596 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 0.7361 + 0.7596 = 1.50 ksi Ans. s2 = 0.7361 - 0.7596 = -0.0235 ksi Ans. 699 09 Solutions 46060 6/8/10 3:13 PM Page 700 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–74. Solve Prob. 9–73 for the principal stress at point B. 1.5 in. A 1.5 in. 1 in. B 1.5 in. 1.5 in. 3 in. 1 in. 15 in. 3 in. 3 5 4 5 kip Internal Forces and Moment: As shown on FBD. Section Properties: A = 3(6) = 18.0 in2 1 (3) A 63 B = 54.0 in4 12 I = QB = y¿A¿ = 2(2)(3) = 12.0 in3 Normal Stress: My N ; A I s = 45.0(1) 4.00 = -0.6111 ksi 18.0 54.0 sB = Shear Stress: Applying the shear formula t = tB = VQ . It 3.00(12.0) = 0.2222 ksi 54.0(3) Construction of the Circle: In accordance with the sign convention, sx = -0.6111 ksi, sy = 0, and txy = -0.2222 ksi. Hence. savg = sx + sy 2 = -0.6111 + 0 = -0.3055 ksi 2 The coordinates for reference points A and C are A(-0.6111, -0.2222) C(-0.3055, 0) The radius of the circle is R = 2(0.6111 - 0.3055)2 + 0.22222 = 0.3778 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = -0.3055 + 0.3778 = 0.0723 ksi Ans. s2 = -0.3055 - 0.3778 = -0.683 ksi Ans. 700 09 Solutions 46060 6/8/10 3:13 PM Page 701 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–75. The 2-in.-diameter drive shaft AB on the helicopter is subjected to an axial tension of 10 000 lb and a torque of 300 lb # ft. Determine the principal stress and the maximum in-plane shear stress that act at a point on the surface of the shaft. s = 10 000 P = 3.183 ksi = A p(1)2 t = 300(12)(1) Tc = 2.292 ksi = p 4 J 2 (1) s1, 2 = = t sx + sy 2 ; A ( sx - sy 2 B A )2 + t2xy 3.183 - 0 2 3.183 + 0 ; ( ) + (2.292)2 2 A 2 s1 = 4.38 ksi Ans. s2 = -1.20 ksi Ans. max in-plane = = A ( A ( sx - sy 2 )2 + t2xy 3.183 - 0 2 ) + (2.292)2 2 = 2.79 ksi Ans. 701 09 Solutions 46060 6/8/10 3:13 PM Page 702 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–76. 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 = - My -300(0.2) = = 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 = 2.3670 ksi In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 2.34375 + 2.3670 = 4.71 ksi Ans. s2 = 2.34375 - 2.3670 = -0.0262 ksi Ans. 702 09 Solutions 46060 6/8/10 3:13 PM Page 703 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. •9–77. 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. 9–78. 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) sx¿ = B(666.67, 0) C(500, 0) 333.33 + 666.67 = 500 MPa 2 Ans. tx¿y¿ = R = 666.67 - 500 = 167 MPa Ans. 703 09 Solutions 46060 6/8/10 3:13 PM Page 704 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. •9–79. 10 kN A 100 mm D 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 = VQD 5(103)(0.001) = 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 to 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° 704 B 30⬚ 1m 300 mm 2m C 09 Solutions 46060 6/8/10 3:13 PM Page 705 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–79. 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. 705 09 Solutions 46060 6/8/10 3:13 PM Page 706 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–80. 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 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 to 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. 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. 706 B 30⬚ 1m 300 mm 2m C 09 Solutions 46060 6/8/10 3:13 PM Page 707 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–80. 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 show in Fig. e. 707 09 Solutions 46060 6/8/10 3:13 PM Page 708 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Determine the principal stress at point A on the cross section of the hanger at section a–a. Specify the orientation of this state of stress and indicate the result on an element at the point. •9–81. 0.75 m 0.75 m a 250 mm a 900 N Internal Loadings: Considering the equilibrium of the free - body diagram of the hanger’s left cut segment, Fig. a, + ©F = 0; : x 900 - N = 0 N = 900 N + c ©Fy = 0; V - 900 = 0 V = 900 N a + ©MO = 0; 900(1) - 900(0.25) - M = 0 M = 675 N # m b 250 mm Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the hanger’s cross section are A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2 1 1 (0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4 12 12 Referring to Fig. b, QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04) = 18.875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stresses. Thus, sA = 675(0.025) MyA 900 N + = + = 9.074 MPa -3 A I 1.4 A 10 B 1.7367 A 10 - 6 B The shear stress is caused by the transverse shear stress. tA = 900 C 18.875 A 10 - 6 B D VQA = = 0.9782 MPa It 1.7367 A 10 - 6 B (0.01) The state of stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 9.074 MPa, sy = 0, and txy = 0.9782 MPa. Thus, savg = sx + sy 2 = 9.074 + 0 = 4.537 MPa 2 The coordinates of reference points A and the center C of the circle are A(9.074, 0.9782) C(4.537, 0) Thus, the radius of the circle is R = CA = 2(9.074 - 4.537)2 + 0.97822 = 4.641 MPa Using these results, the circle is shown in Fig. d. 708 b 900 N 5 mm 25 mm A 100 mm 5 mm 50 mm I = 0.5 m 5 mm Sections a – a and b – b 09 Solutions 46060 6/8/10 3:13 PM Page 709 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–81. Continued In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 4.537 + 4.641 = 9.18 MPa Ans. s2 = 4.537 - 4.641 = -0.104 MPa Ans. Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d, tan 2 A uP B 1 = 0.9782 = 0.2156 9.074 - 4.537 A uP B 1 = 6.08° (counterclockwise) Ans. The state of principal stresses is represented on the element shown in Fig. e. 709 09 Solutions 46060 6/8/10 3:13 PM Page 710 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–82. Determine the principal stress at point A on the cross section of the hanger at section b–b. Specify the orientation of the state of stress and indicate the results on an element at the point. 0.75 m 0.75 m a 250 mm a 900 N Internal Loadings: Considering the equilibrium of the free - body diagram of the hanger’s left cut segment, Fig. a, + c ©Fy = 0; V - 900 - 900 = 0 a + ©MO = 0; 900(2.25) + 900(0.25) - M = 0 Referring to Fig. b. QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04) = 18.875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is contributed by the bending stress only. 2250(0.025) MyA = = 32.39 MPa I 1.7367 A 10 - 6 B The shear stress is contributed by the transverse shear stress only. 1800 C 18.875 A 10 - 6 B D VQA = = 1.956 MPa It 1.7367 A 10 - 6 B (0.01) The state stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 32.39 MPa, sy = 0, and txy = 1.956 MPa. Thus, savg = sx + sy 2 = 32.39 + 0 = 16.19 MPa 2 The coordinates of reference point A and the center C of the circle are A(32.39, 1.956) C(16.19, 0) Thus, the radius of the circle is R = CA = 2(32.39 - 16.19)2 + 1.9562 = 16.313 MPa Using these results, the cricle is shown in Fig. d. 710 b 5 mm 25 mm A 100 mm 5 mm 50 mm 5 mm Sections a – a and b – b 1 1 (0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4 12 12 tA = 250 mm M = 2250 N # m A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2 sA = b 900 N V = 1800 N Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the hanger’s cross section are I = 0.5 m 09 Solutions 46060 6/8/10 3:13 PM Page 711 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–82. Continued In - Plane Principal Stresses: The coordinates of reference point B and D represent s1 and s2, respectively. s1 = 16.19 + 16.313 = 32.5 MPa Ans. s2 = 16.19 - 16.313 = -0.118 MPa Ans. Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d, tan 2 A uP B 1 = A uP B 1 = 3.44° 1.956 = 0.1208 32.39 - 16.19 (counterclockwise) Ans. The state of principal stresses is represented on the element shown in Fig. e. 711 09 Solutions 46060 6/8/10 3:13 PM Page 712 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–83. Determine the principal stresses and the maximum in-plane shear stress that are developed at point A. Show the results on an element located at this point. The rod has a diameter of 40 mm. 450 N 150 mm Using the method of sections and consider the FBD of the member’s upper cut segment, Fig. a, + c ©Fy = 0; 450 - N = 0 a + ©MC = 0; 100 mm A 150 mm N = 450 N 450(0.1) - M = 0 B M = 45 N # m A = p(0.022) = 0.4(10 - 3)p m2 I = 450 N p (0.024) = 40(10 - 9)p m4 4 The normal stress is the combination of axial and bending stress. Thus, s = My N + A I For point A, y = C = 0.02 m. s = 45 (0.02) 450 + = 7.520 MPa 0.4(10 - 3)p 40(10 - 9)p Since no transverse shear and torque is acting on the cross - section t = 0 The state of stress at point A can be represented by the element shown in Fig. b. In accordance to the established sign convention sx = 0, sy = 7.520 MPa and txy = 0. Thus savg = sx + sy 2 = 0 + 7.520 = 3.760 MPa 2 Then, the coordinates of reference point A and the center C of the circle are A(0, 0) C(3.760, 0) Thus, the radius of the circle is R = CA = 3.760 MPa Using this results, the circle shown in Fig. c can be constructed. Since no shear stress acts on the element, s1 = sy = 7.52 MPa s2 = sx = 0 Ans. The state of principal stresses can also be represented by the element shown in Fig. b. The state of maximum in - plane shear stress is represented by point B on the circle, Fig. c. Thus. tmax in-plane = R = 3.76 MPa Ans. 712 09 Solutions 46060 6/8/10 3:13 PM Page 713 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–83. Continued From the circle, 2us = 90° us = 45° (counter clockwise) Ans. The state of maximum In - Plane shear stress can be represented by the element shown in Fig. d. 713 09 Solutions 46060 6/8/10 3:13 PM Page 714 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–84. 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 circle 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 circle of this state of stress are shown in Fig. b (a) Draw the three Mohr’s circles that describe the following state of stress. •9–85. 180 MPa 140 MPa (b) 300 psi Here, smin = -300 psi, sint = 0 and smax = 400 psi. The three Mohr’s circle for this state of stress is shown in Fig. a. 400 psi 714 09 Solutions 46060 6/8/10 3:13 PM Page 715 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–86. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. z y x 80 MPa For y – z plane: A(0, -80) B(90, 80) C(45, 0) R = 2452 + 802 = 91.79 s1 = 45 + 91.79 = 136.79 MPa s2 = 45 - 91.79 = -46.79 MPa Thus, tabs max = s1 = 0 Ans. s2 = 137 MPa Ans. s3 = -46.8 MPa Ans. 136.79 - (-46.79) smax - smin = = 91.8 MPa 2 2 Ans. 715 90 MPa 09 Solutions 46060 6/8/10 3:13 PM Page 716 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–87. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. z Mohr’s circle for the element in y - 7 plane, Fig. a, will be drawn first. In accordance to the established sign convention, sy = 30 psi, sz = 120 psi and tyz = 70 psi. Thus savg = sy + sz 2 = x y 120 psi 70 psi 30 + 120 = 75 psi 2 30 psi Thus the coordinates of reference point A and the center C of the circle are A(30, 70) 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 circle are shown in Fig. c, From the geometry of the three circles, tabs max = 158.22 - ( -8.22) smax - smin = = 83.22 psi 2 2 716 Ans. 09 Solutions 46060 6/8/10 3:13 PM Page 717 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–88. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. z Mohr’s circle for the element in x - z plane, Fig. a, will be drawn first. In accordance to the established sign convention, sx = -2 ksi, sz = 0 and txz = 8 ksi. Thus savg = sx + sz 2 = -2 + 0 = -1 ksi 2 2 ksi 8 ksi Thus, the coordinates of reference point A and the center C of the circle are A( -2, 8) C(-1, 0) Thus, the radius of the circle is R = CA = 2[-2 - (-1)]2 + 82 = 265 ksi Using these results, the circle in shown in Fig. b, The coordinates of points B and D represent s1 and s2, respectively. s = -1 + 265 = 7.062 ksi smax = 7.06 ksi sint = 0 smin = -9.06 ksi From the results obtained, sint = 0 ksi smax = 7.06 ksi smin = -9.06 ksi Ans. Using these results, the three Mohr’s circles are shown in Fig, c. From the geometry of the cricle, tabs max = y x 7.06 - (-9.06) smax - smin = = 8.06 ksi 2 2 Ans. 717 09 Solutions 46060 6/8/10 3:13 PM Page 718 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. •9–89. z y x For x – y plane: 150 MPa 120 MPa 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. 9–90. The state of stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress. z x For y - z plane: A(5, -4) B(-2.5, 4) 4 ksi s1 = 1.25 + 5.483 = 6.733 ksi 5 ksi s2 = 1.25 - 5.483 = -4.233 ksi Thus, tabs max = 2.5 ksi C(1.25, 0) R = 23.752 + 42 = 5.483 savg = y 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. 718 09 Solutions 46060 6/8/10 3:13 PM Page 719 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–92. The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine the principal stress acting at points A and B and the absolute maximum shear stress. 450 mm A B 300 N⭈m 25 mm 45 N⭈m 800 N Internal Forces and Moment: As shown on FBD. Section Properties: Iz = p A 0.0254 B = 0.306796 A 10 - 6 B m4 4 J = p A 0.0254 B = 0.613592 A 10 - 6 B m4 2 (QA)x = 0 (QB)y = y¿A¿ = 4(0.025) 1 c (p) A 0.0252 B d = 10.417 A 10 - 6 B m3 3p 2 Normal stress: Applying the flexure formula. s = - Mzy sA = sB = - Iz -60.0(0.025) 0.306796(10 - 6) -60.0(0) 0.306796(10 - 6) = 4.889 MPa = 0 Shear Stress: Applying the torsion formula for point A, tA = 45.0(0.025) Tc = 1.833 MPa = J 0.613592(10 - 6) The transverse shear stress in the y direction and the torsional shear stress can be Tr VQ obtained using shear formula and torsion formula. tv = and ttwist = , It J respectively. tB = (tv)y - ttwist = 800 C 10.417(10 - 6) D -6 0.306796(10 )(0.05) - 45.0(0.025) 0.613592(10 - 6) = -1.290 MPa Construction of the Circle: sx = 4.889 MPa, sz = 0, and txz = -1.833 MPa for point A. Hence, savg = sx + sz 2 = 4.889 + 0 = 2.445 MPa 2 The coordinates for reference points A and C are A (4.889, –1.833) and C(2.445, 0). 719 09 Solutions 46060 6/8/10 3:13 PM Page 720 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–92. Continued The radius of the circle is R = 2(4.889 - 2.445)2 + 1.8332 = 3.056 MPa sx = sy = 0 and txy = -1.290 MPa for point B. Hence, savg = sx + sz = 0 2 The coordinates for reference points A and C are A(0. ‚–1.290) and C(0,0). The radius of the circle is R = 1.290 MPa In - Plane Principal Stresses: The coordinates of point B and D represent s1 and s2, respectively. For point A s1 = 2.445 + 3.056 = 5.50 MPa s2 = 2.445 - 3.506 = -0.611 MPa For point B s1 = 0 + 1.290 = 1.29 MPa s2 = 0 - 1.290 = -1.290 MPa Three Mohr’s Circles: From the results obtaired above, the principal stresses for point A are smax = 5.50 MPa sint = 0 smin = -0.611 MPa Ans. sint = 0 smin = -1.29 MPa Ans. And for point B smax = 1.29 MPa Absolute Maximum Shear Stress: For point A, tabs max = 5.50 - (-0.611) smax - smin = = 3.06 MPa 2 2 Ans. 1.29 - (-1.29) smax - smin = = 1.29 MPa 2 2 Ans. For point B, tabs max = 720 09 Solutions 46060 6/8/10 3:13 PM Page 721 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The propane gas tank has an inner diameter of 1500 mm and wall thickness of 15 mm. If the tank is pressurized to 2 MPa, determine the absolute maximum shear stress in the wall of the tank. •9–93. Normal Stress: Since r 750 = = 50 7 10, thin - wall analysis can be used. We have t 15 s1 = 2(750) pr = = 100 MPa t 15 s2 = pr 2(750) = = 50 MPa 2t 2(15) The state of stress of any point on the wall of the tank can be represented on the element shown in Fig. a Construction of Three Mohr’s Circles: Referring to the element, smax = 100 MPa sint = 50 MPa smin = 0 Using these results, the three Mohr’s circles are shown in Fig. b. Absolute Maximum Shear Stress: From the geometry of three circles, tabs max = smax - smin 100 - 0 = = 50 MPa 2 2 Ans. 721 09 Solutions 46060 6/8/10 3:13 PM Page 722 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–94. Determine the principal stress 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 - 500a 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 = N 300 = = -342.86 psi A 0.875 The shear stress is contributed by the transverse shear stress. tA = 400(0.3672) VQA = = 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 722 500 lb 09 Solutions 46060 6/8/10 3:13 PM Page 723 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–94. Continued Using these results, the cricle 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. 723 09 Solutions 46060 6/8/10 3:13 PM Page 724 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–95. Determine the principal stress 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, a + c ©Fy = 0; + ©F = 0; ; x 3 N - 500 a b = 0 5 N = 300 lb 4 V - 500a b = 0 5 V = 400 lb M = 6000 lb # in 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. 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, s2 = 0 s1 = 10.91 ksi Three Mohr’s Circles: Using these results, smax = 10.91 ksi sint = smin = 0 Ans. Absolute Maximum Shear Stress: tabs max = smax - smin 10.91 - 0 = = 5.46 ksi 2 2 Ans. 724 500 lb 0.25 in. A 0.25 in. A = 0.5(3) - 0.25(2.5) = 0.875 in2 sB = a 0.5 in. Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the bracket’s cross section are I = 4 B 3 4 ©MO = 0; M - 500a b(12) - 500 a b(6) = 0 5 5 5 3 0.25 in. 1.5 in.1.5 in. Section a – a 09 Solutions 46060 6/8/10 3:13 PM Page 725 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–96. 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 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 -25.06 - 0 2 ; a b + (19.56)2 2 C 2 = -12.53 ; 23.23 s1 = 10.7 MPa s2 = -35.8 MPa Ans. 725 F 09 Solutions 46060 6/8/10 3:13 PM Page 726 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •9–97. 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 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 = = C a sx - sy 2 2 b + t2xy -25.06 - 0 2 b + (19.56)2 C 2 a = 23.2 MPa Ans. 726 F 09 Solutions 46060 6/8/10 3:13 PM Page 727 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–98. 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 I = p A 1.54 - 1.3754 B = 1.1687 in4 4 J = p A 1.54 - 1.3754 B = 2.3374 in4 2 C y z (QA)z = ©y¿A¿ x 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 , respectively. ttwist = 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. 727 09 Solutions 46060 6/8/10 3:13 PM Page 728 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–99. Solve Prob. 9–98 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. Section Properties: C I = p A 1.54 - 1.3754 B = 1.1687 in4 4 y z x p J = A 1.54 - 1.3754 B = 2.3374 in4 2 (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 + 0 256.7 - 0 2 ; a b + ( -154.0)2 2 C 2 = 128.35 ; 200.49 s1 = 329 psi s2 = -72.1 psi Ans. 728 09 Solutions 46060 6/8/10 3:13 PM Page 729 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–100. The clamp exerts a force of 150 lb on the boards at G. Determine the axial force in each screw, AB and CD, and then compute the principal stresses at points E and F. Show the results on properly oriented elements located at these points. The section through EF is rectangular and is 1 in. wide. A 150 lb C G 0.5 in. E Support Reactions: FBD(a). a + ©MB = 0; + c ©Fy = 0; F FCD(3) - 150(7) = 0 FCD = 350 lb Ans. 350 - 150 - FAB = 0 FAB = 200 lb Ans. B 1.5 in. 1.5 in. Internal Forces and Moment: As shown on FBD(b). Section Properties: I = 1 (1) A 1.53 B = 0.28125 in4 12 QE = 0 QF = y¿A¿ = 0.5(0.5)(1) = 0.250 in3 Normal Stress: Applying the flexure formula s = - My , I sE = - -300(0.75) = 800 psi 0.28125 sF = - -300(0.25) = 266.67 psi 0.28125 VQ , It Shear Stress: Applying the shear formula t = tE = 200(0) = 0 0.28125(1) tF = 200(0.250) = 177.78 psi 0.28125(1) In - Plane Principal Stress: sx = 800 psi, sy = 0 and txy = 0 for point E. Since no shear stress acts upon the element. s1 = sx = 800 psi Ans. s2 = sy = 0 Ans. sx = 266.67 psi, sy = 0, and txy = 177.78 psi for point F. Applying Eq. 9-5 s1,2 = = sx + sy 2 ; C sx - sy a 2 2 b + t2xy 266.67 + 0 266.67 - 0 2 ; a b + 177.782 2 C 2 = 133.33 ; 222.22 s1 = 356 psi s2 = -88.9 psi Ans. 729 150 lb D 4 in. 09 Solutions 46060 6/8/10 3:13 PM Page 730 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–100. Continued Orientation of Principal Plane: Applying Eq. 9-4 for point F, tan 2up = txy A sx - sy B >2 up = 26.57° = and 177.78 = 1.3333 (266.67 - 0)>2 -63.43° Substituting the results into Eq. 9-1 with u = 26.57° yields sx¿ = = sx + sy 2 + sx - sy 2 cos 2u + txy sin 2u 266.67 + 0 266.67 - 0 + cos 53.13° + 177.78 sin 53.13° 2 2 = 356 psi = s1 Hence, up1 = 26.6° up2 = -63.4° Ans. 730 09 Solutions 46060 6/8/10 3:13 PM Page 731 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–101. The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stress and the maximum in-plane shear stress that is developed anywhere on the surface of the shaft. F T0 F T0 Internal Forces and Torque: As shown on FBD(b). Section Properties: A = p 2 d 4 J = p d 4 p 4 a b = d 2 2 32 Normal Stress: N -F 4F = p 2 = - 2 A pd 4 d s = Shear Stress: Applying the shear torsion formula, t = T0 A d2 B 16T0 Tc = p 4 = J pd3 d 32 16T0 4F , sy = 0, and txy = for any point on pd2 pd3 the shaft’s surface. Applying Eq. 9-5, In - Plane Principal Stress: sx = - s1,2 = = = s1 = sx + sy ; 2 - 4F2 pd C a + 0 ; 2 D sx - sy 2 ¢ - 4F2 pd 2 b + t2xy - 0 2 ≤ + a2 16T0 3 pd b 2 64T20 2 -F ; F2 + ≤ 2 ¢ C pd d2 64T20 2 -F + F2 + ≤ 2 ¢ C pd d2 s2 = - Ans. 64T20 2 F + F2 + ≤ 2 ¢ C pd d2 Ans. Maximum In - Plane Shear Stress: Applying Eq. 9-7, t max in-plane = = = C a sx - sy ¢ - 4F2 pd D 2 2 2 b + t2xy - 0 ≤ + a2 16T0 pd3 b 2 64T20 2 2 F + pd2 C d2 Ans. 731 09 Solutions 46060 6/8/10 3:13 PM Page 732 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–102. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the plane AB. A 50 MPa 30⬚ 28 MPa 100 MPa B Construction of the Circle: In accordance with the sign convention, sx = -50 MPa, sy = -100 MPa, and txy = -28 MPa. Hence, savg = sx + sy 2 = -50 + (-100) = -75.0 MPa 2 The coordinates for reference points A and C are A(–50, –28) and C(–75.0, 0). The radius of the circle is R = 2(75.0 - 50)2 + 282 = 37.54 MPa. Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle sx¿ = -75.0 + 37.54 cos 71.76° = -63.3 MPa Ans. tx¿y¿ = 37.54 sin 71.76° = 35.7 MPa Ans. 732 09 Solutions 46060 6/8/10 3:13 PM Page 733 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–103. 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 kN # m T - 2 = 0 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 is 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. 733 09 Solutions 46060 6/8/10 3:13 PM Page 734 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–103. Continued Orientation of the Principal Plane: Referring to the geometry of the circle, Fig. d, 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. 734 09 Solutions 46060 6/8/10 3:13 PM Page 735 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *9–104. The box beam is subjected to the loading shown. Determine the principal stress in the beam at points A and B. 6 in. A 6 in. B 8 in. 8 in. 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 1200 lb 800 lb 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. 735 A B 3 ft 2.5 ft 2.5 ft 5 ft 09 Solutions 46060 6/8/10 3:13 PM Page 736 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The wooden strut is subjected to the loading shown. Determine the principal stresses that act at point C and specify the orientation of the element at this point. The strut is supported by a bolt (pin) at B and smooth support at A. •9–105. 50 N 50 N 60⬚ C 100 mm 40 N 40 N B A 25 mm 50 mm 200 mm 200 mm 200 mm 200 mm 100 mm 100 mm QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3 1 (0.025)(0.13) = 2.0833(10 - 6) m4 12 I = Normal stress: sC = 0 Shear stress: VQC 44(31.25)(10 - 6) = 26.4 kPa = It 2.0833(10 - 6)(0.025) t = Principal stress: sx = sy = 0; s1,2 = txy = -26.4 kPa sx + sy 2 ; C a sx - sy 2 2 b + t2 xy = 0 ; 20 + (26.4)2 s1 = 26.4 kPa s2 = -26.4 kPa ; Ans. Orientation of principal stress: tan 2up = txy (sx - sy) = - q 2 up = +45° and -45° Use sx¿ Eq. 9-1 to determine the principal sx - sy sx + sy + cos 2u + txy sin 2u = 2 2 plane of s1 and s2 u = up = -45° sx¿ = 0 + 0 + (-26.4) sin( -90°) = 26.4 kPa Therefore, up1 = -45°; up2 = 45° Ans. 736 09 Solutions 46060 6/8/10 3:13 PM Page 737 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9–106. The wooden strut is subjected to the loading shown. If grains of wood in the strut at point C make an angle of 60° with the horizontal as shown, determine the normal and shear stresses that act perpendicular and parallel to the grains, respectively, due to the loading. The strut is supported by a bolt (pin) at B and smooth support at A. 50 N 50 N 60⬚ C 100 mm 40 N 40 N B A 25 mm 50 mm 200 mm 200 mm 200 mm 200 mm 100 mm 100 mm QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3 I = 1 (0.025)(0.13) = 2.0833(10 - 6) m4 12 Normal stress: sC = 0 Shear stress: t = VQC 44(31.25)(10 - 6) = 26.4 kPa = It 2.0833(10 - 6)(0.025) Stress transformation: sx = sy = 0; sx¿ = sx + sy 2 + sx - sy 2 txy = -26.4 kPa; u = 30° cos 2u + txy sin 2u = 0 + 0 + (-26.4) sin 60° = -22.9 kPa tx¿y¿ = - sx - sy 2 Ans. sin 2u + txy cos 2u = -0 + (-26.4) cos 60° = -13.2 kPa Ans. 737