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 sx - sy + cos 2u + txy sin 2u sx + sy + sx - sy - 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 ksi ksi ksi 60Њ 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 ¢Fx¿ + 2¢A sin 60° cos 60° + 5¢ A sin 60 sin 60 +QâFx = 0; + 2ÂA cos 60 sin 60° - 8¢A cos 60° cos 60° = ¢Fx¿ = -3.482 ¢A ¢Fy¿ + 2¢A sin 60° sin 60 - 5Â A sin 60 cos 60 +aâFy = 0; - 8¢A cos 60° sin 60° - 2¢A cos 60° cos 60° = ¢Fy¿ = 4.629 ¢A From the definition, sx¿ = lim¢A:0 ¢Fx¿ = -3.48 ksi ¢A tx¿y¿ = lim¢A:0 ¢Fy¿ Ans Ans = 4.63 ksi ¢A 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 60Њ 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 +RâFx = 0; ÂFx + 500 ¢A sin 60° sin 60° + 350¢A sin 60° cos 60° +350¢A cos 60° sin 60° = ¢Fx¿ = -678.11 ÂA +QâFy = 0; ÂFy + 350ÂA sin 60° sin 60° - 500¢A sin 60° cos 60° -350¢A cos 60° cos 60° = ¢Fy¿ = 41.51 ¢A From the definition sx¿ = lim¢A:0 tx¿y¿ = lim¢A:0 ¢Fx¿ = -6.78 psi ¢A ¢Fy¿ Ans Ans = 41.5 psi ¢A 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 A 650 psi ¢Fx¿ - 400(¢Acos 60°)cos 60° + 650(¢ A sin 60°)cos 30° = Q+ âFx = 400 psi 60 ÂFx = -387.5¢A ¢Fy¿ - 650(¢Asin 60°)sin 30° - 400(¢ A cos 60°)sin 60° = a+ ©Fy¿ = B ¢Fy¿ = 455 ¢A sx¿ = lim¢A:0 sx¿y¿ = lim¢A:0 ¢Fx¿ = -388 psi ¢A ¢Fy¿ Ans Ans = 455 psi ¢A The negative sign indicates that the sense of sx¿, is opposite to that shown on FBD •9–5 Solve Prob 9–4 using the stress-transformation equations developed in Sec 9.2 sy = 400 psi sx = -650 psi sx¿ = = sx + sy sx - sy + 2 txy = A 400 psi u = 30° 650 psi cos 2u + txy sin 2u 60Њ -650 + 400 -650 - 400 + cos 60° + = -388 psi 2 Ans B The negative sign indicates sx¿, is a compressive stress tx¿y¿ = = -a sx - sy sin 2u + txy cos 2u -650 - 400 bsin 60° = 455 psi 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 60Њ 30Њ R+ ©Fy¿ = 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 = ÂFy = -34.82ÂA b+ âFx = ¢Fx¿ - 50¢A sin 30° sin 30° + 35¢A sin 30° sin 60° -90¢A cos 30° cos 30° + 35¢A cos 30° cos 60° = ¢Fx¿ = 49.69 ¢A sx¿ = lim¢A:0 ¢Fx¿ = 49.7 MPa ¢A tx¿y¿ = lim¢A:0 ¢Fy¿ Ans Ans = -34.8 MPa ¢A 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 60Њ 30Њ sy = 50 MPa sx = 90 MPa sx¿ = = sx + sy sx - sy + 2 txy = -35 MPa u = -150° cos 2u + txy sin 2u 90 - 50 90 + 50 + cos(-300°) + (-35) sin ( -300°) 2 = 49.7 MPa tx¿y¿ = - sx - sy = -a Ans sin 2u + txy cos 2u 90 - 50 bsin(-300°) + ( -35) cos ( -300°) = -34.8 MPa 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 45Њ 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° = ¢Fx¿ = -5 A 106 B ÂA âFy = 0; ÂFy + 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° = ¢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¿ Ans Ans = 40 MPa ¢A 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 •9–9 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 45 MPa 80 MPa 45Њ Stress Transformation Equations: u = +135° (Fig a) sx = 80 MPa sy = txy = 45 MPa we obtain, sx¿ = = sx + sy sx - sy + 2 cos u + txysin 2u 80 - 80 + + cos 270 + 45 sin 270° 2 Ans = -5 MPa tx¿y¿ = - = - sx - sy B sinu + txy cos 2u 80 - sin 270° + 45 cos 270° = 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 ksi A ksi 30Њ ksi B Force Equllibrium: For the sectioned element, ¢Fy¿ - 3(¢A sin 30°) sin 60° + 4(Â A sin 30)sin 30 a+ âFy = 0; -2(¢A cos 30°) sin 30° - 4(¢A cos 30°) sin 60° = ¢Fy¿ = 4.165 ¢A ¢Fx¿ + 3(¢A sin 30°) cos 60° + 4(¢ A sin 30°)cos 30° Q+ âFx = 0; -2(ÂA cos 30) cos 30 + 4(¢A cos 30°) cos 60° = ¢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¿ Ans Ans = 4.17 ksi ¢A 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 ksi Normal and Shear Stress: In accordance with the established sign convention, u = +60° sx = -3 ksi sy = ksi A txy = -4 ksi ksi 30Њ Stress Transformation Equations: Applying Eqs 9-1 and 9-2 sx¿ = = sx + sy sx - sy + 2 B cos 2u + txy sin 2u -3 - -3 + + cos 120° + (-4 sin 120°) 2 Ans = -2.71 ksi tx¿y¿ = - = - sx - sy ksi sin 2u + txy cos 2u -3 - sin 120° + (-4 cos 120°) = 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 sy = sx = -10 ksi txy = -16 ksi u = +50° sx¿ = = sx + sy = -a = cos 2u + txy sin 2u -10 - -10 + + cos 100° + ( -16)sin 100° = -19.9 ksi 2 tx¿y¿ = - a sy¿ = sx - sy + sx - sy b sin 2u + txy cos 2u -10 - b sin 100° + (-16)cos 100° = 7.70 ksi sx + sy sx - sy - Ans Ans cos 2u - txy sin 2u -10 + -10 - - a bcos 100° - (-16)sin 100° = 9.89 ksi 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 •9–13 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 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 sx - sy + 2 cos 2u + txy sin 2u 200 - ( -350) 200 + (-350) + cos ( -120°) + 75 sin (-120°) 2 = -277.45 psi = -277 psi sy¿ = = sx + sy sx - sy - 2 Ans cos 2u - txy sin 2u 200 - ( -350) 200 + (-350) cos ( -120°) - 75 sin ( -120°) 2 = 127.45 psi = 127 psi tx¿y¿ = - = - sx - sy Ans sin 2u + txy cos 2u 200 - (-350) sin (-120°) + 75 cos (-120°) = 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 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 = smin = -926 psi Ans Absolute Maximum Shear Stress: tabs max = 583.2 - (-926.0) smax - smin = - 755 psi 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 in bracket’s upper cut segment, Fig a, a + c ©Fy = 0; + ©F = 0; ; x N - 500 a b = N = 300 lb V - 500 a b = V = 400 lb 1 (0.5) A 33 B (0.25) A 2.53 B = 0.79948 in4 12 12 Referring to Fig b, QB = 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 = 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 = Three Mohr’s Circles: Using these results, smax = 10.91 ksi sint = smin = Ans Absolute Maximum Shear Stress: tabs max = 0.25 in 1.5 in.1.5 in Section a – a M = 6000 lb # in smax - smin 10.91 - = = 5.46 ksi 2 Ans 724 500 lb 0.25 in A 0.25 in A = 0.5(3) - 0.25(2.5) = 0.875 in2 s1 = 10.91 ksi 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 sB = B ©MO = 0; M - 500 a b(12) - 500 a b(6) = 5 I = 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 J = p A 0.1254 B = 0.3835 A 10 - B m4 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 = and txy = 19.56 MPa for any point on the shaft’s surface Applying Eq 9-5, s1,2 = = sx + sy ; C a sx - sy 2 b + t2xy -25.06 - -25.06 + ; a b + (19.56)2 C = -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 J = p A 0.1254 B = 0.3835 A 10 - B m4 Normal Stress: s = -1.23(106) N = = -25.06 MPa A 0.015625p Shear Stress: Applying the torsion formula t = 60.0(103) (0.125) Tc = 19.56 MPa = J 0.3835 (10 - 3) Maximum In - Plane Principal Shear Stress: sx = -25.06 MPa, sy = 0, and txy = 19.56 MPa for any point on the shaft’s surface Applying Eq 9-7, t max in-plane a sx - sy b + t2xy = C = -25.06 - 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 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 J = p A 1.54 - 1.3754 B = 2.3374 in4 C y z (QA)z = ©y¿A¿ x 4(1.5) 4(1.375) = c p A 1.52 B d c p A 1.3752 B d 3p 3p = 0.51693 in3 Normal Stress: Applying the flexure formula s = sA = My z Iy , 200(0) = 1.1687 Shear Stress: The transverse shear stress in the z direction and the torsional shear VQ stress can be obtained using shear formula and torsion formula, tv = and It Tr ttwist = , respectively J tA = (tv)z - ttwist = 20.0(0.51693) 240(1.5) 1.1687(2)(0.125) 2.3374 = -118.6 psi In - Plane Principal Stress: sx = 0, sz = and txz = -118.6 psi for point A Applying Eq 9-5 s1,2 = sx + sz ; C a sx - sz 2 b + t2xz = ; 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 y z x p J = A 1.54 - 1.3754 B = 2.3374 in4 (QB)z = 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 ; C sx - sy a 2 b + t2xy 256.7 - 256.7 + ; a b + ( -154.0)2 C = 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 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) = FCD = 350 lb Ans 350 - 150 - FAB = FAB = 200 lb Ans B 1.5 in 1.5 in Internal Forces and Moment: As shown on FBD(b) Section Properties: I = (1) A 1.53 B = 0.28125 in4 12 QE = 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) = 0.28125(1) tF = 200(0.250) = 177.78 psi 0.28125(1) In - Plane Principal Stress: sx = 800 psi, sy = and txy = for point E Since no shear stress acts upon the element s1 = sx = 800 psi Ans s2 = sy = Ans sx = 266.67 psi, sy = 0, and txy = 177.78 psi for point F Applying Eq 9-5 s1,2 = = sx + sy ; C sx - sy a 2 b + t2xy 266.67 - 266.67 + ; a b + 177.782 C = 133.33 ; 222.22 s1 = 356 psi s2 = -88.9 psi Ans 729 150 lb D 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 sx - sy + cos 2u + txy sin 2u 266.67 - 266.67 + + cos 53.13° + 177.78 sin 53.13° 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 d J = p d p a b = d 2 32 Normal Stress: N -F 4F = p = - A pd d s = Shear Stress: Applying the shear torsion formula, t = T0 A d2 B 16T0 Tc = p = J d pd3 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 = sx + sy ; - 4F2 pd = C a + ; D sx - sy ¢ - 4F2 pd b + t2xy - 2 ≤ + a- 16T0 pd b = 64T20 -F ; F2 + ≤ ¢ C pd d2 s1 = 64T20 -F + F2 + ≤ ¢ C pd d2 Ans 64T20 F + F2 + ≤ ¢ C pd d2 Ans s2 = - Maximum In - Plane Shear Stress: Applying Eq 9-7, t max in-plane = = = C a D ¢ sx - sy - 4F2 pd 2 b + t2xy - ≤ + a- 16T0 pd3 b 64T20 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 30Њ 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 = -50 + (-100) = -75.0 MPa 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 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 = N = 10 kN ©Mx = 0; T - = T = kN # m Section Properties: The cross - sectional area and the polar moment of inertia of the propeller shaft’s cross section are A = p A 0.0752 - 0.052 B = 3.125p A 10 - B m2 J = p A 0.0754 - 0.054 B = 12.6953125p A 10 - B m4 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 - B The shear stress is contributed by the torsional shear stress only tA = A 103 B (0.075) Tc = = 3.761 MPa J 12.6953125p A 10 - 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 = -1.019 + = -0.5093 MPa 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 A up B = 3.761 = 7.3846 1.019 - 0.5093 A up B = 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 in A in B in in Support Reactions: As shown on FBD(a) Internal Forces and Moment: As shown on FBD(b) Section Properties: I = 1 (8) A 83 B (6) A 63 B = 233.33 in4 12 12 QA = QB = 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 = In - Plane Principal Stress: sx = 61.71 psi, sy = 0, and txy = for point A Since no shear stress acts on the element, s1 = sx = 61.7 psi Ans s2 = sy = Ans sx = -46.29 psi, sy = 0, and txy = for point B Since no shear stress acts on the element, s1 = sy = Ans s2 = sx = -46.3 psi Ans 735 A B ft 2.5 ft 2.5 ft 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 •9–105 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 50 N 50 N 60Њ 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 (0.025)(0.13) = 2.0833(10 - 6) m4 12 I = Normal stress: sC = 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 ; C a sx - sy 2 b + t2 xy = ; 20 + (26.4)2 s1 = 26.4 kPa s2 = -26.4 kPa ; Ans Orientation of principal stress: tan 2up = txy (sx - sy) = - q up = +45° and -45° Use sx¿ Eq 9-1 to determine the principal sx + sy sx - sy = + cos 2u + txy sin 2u 2 plane of s1 and s2 u = up = -45° sx¿ = + + (-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 60Њ 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 = (0.025)(0.13) = 2.0833(10 - 6) m4 12 Normal stress: sC = 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 sx - sy + 2 txy = -26.4 kPa; u = 30° cos 2u + txy sin 2u = + + (-26.4) sin 60° = -22.9 kPa tx¿y¿ = - sx - sy Ans sin 2u + txy cos 2u = -0 + (-26.4) cos 60° = -13.2 kPa Ans 737 .. .09 Solutions 46060 6/8/10 3:13 PM Page 620 © 2010 Pearson Education, Inc., Upper Saddle River, NJ... lim¢A:0 ¢Fy¿ Ans Ans = 4.63 ksi ¢A 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... Ans Ans = 41.5 psi ¢A 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