PROBLEM 7.1 For the given state of stress, determine the normal and shearing stresses exerted on the oblique face of the shaded triangular element shown. Use a method of analysis based on the equilibrium of that element, as was done in the derivations of Sec. 7.2. SOLUTION Stresses Areas Forces F = 0: σ A − 15 A sin 30° cos 30° − 15A cos 30° sin 30° + 10A cos 30 cos 30° = 0 σ = 30 sin 30° cos 30° − 10 cos 2 30° σ = 5.49 ksi ΣF = 0: τ A + 15 A sin 30° sin 30° − 15 A cos 30° cos 30° − 10 A cos 30° sin 30° = 0 τ = 15(cos 2 30° − sin 2 30°) + 10 cos 30° sin 30° τ = 11.83 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.4 For the given state of stress, determine the normal and shearing stresses exerted on the oblique face of the shaded triangular element shown. Use a method of analysis based on the equilibrium of that element, as was done in the derivations of Sec. 7.2. SOLUTION Stresses Areas Forces ΣF = 0: σ A + 18 A cos 15° sin 15° + 45 A cos 15° cos 15° − 27 A sin 15° sin 15° + 18A sin 15° cos 15° = 0 σ = −18 cos 15° sin 15° − 45 cos 2 15° + 27sin 2 15° − 18 sin 15° cos 15° σ = −49.2 MPa ΣF = 0: τ A + 18 A cos 15° cos 15° − 45 A cos 15° sin 15° − 27 A sin 15° cos 15° − 18 A sin 15° sin 15° = 0 τ = −18(cos 2 15° − sin 2 15°) + (45 + 27) cos 15° sin 15° τ = 2.41 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.5 For the given state of stress, determine (a) the principal planes, (b) the principal stresses. SOLUTION σ x = −60 MPa σ y = −40 MPa τ xy = 35 MPa (a) tan 2θ p = 2τ xy σx −σ y = (2) (35) = −3.50 −60 + 40 2θ p = −74.05° (b) σ max, min = = σx +σy 2 θ p = −37.0°, 53.0° ± −60 − 40 ± 2 σx −σ y 2 2 −60 + 40 2 2 + τ xy 2 + (35) 2 = −50 ± 36.4 MPa σ max = −13.60 MPa σ min = −86.4 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.8 For the given state of stress, determine (a) the principal planes, (b) the principal stresses. SOLUTION σ x = −8 ksi σ y = 12 ksi τ xy = 5 ksi (a) tan 2θ p = 2τ xy σx − σy 2(5) = −0.5 −8 − 12 = 2θ p = −26.5651° (b) σ max, min = = σx + σy 2 ± −8 + 12 ± 2 = 2 ± 11.1803 θ p = −13.3°, 76.7° 2 σx − σy 2 +τ xy 2 −8 − 12 2 2 + (5)2 σ max = 13.18 ksi σ min = −9.18 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.19 A steel pipe of 12-in. outer diameter is fabricated from 14 -in. -thick plate by welding along a helix that forms an angle of 22.5° with a plane perpendicular to the axis of the pipe. Knowing that a 40-kip axial force P and an 80-kip ⋅ in. torque T, each directed as shown, are applied to the pipe, determine σ and τ in directions, respectively, normal and tangential to the weld. SOLUTION 1 d 2 = 6 in., t = 0.25 in. 2 c1 = c2 − t = 5.75 in. d 2 = 12 in., c2 = ( π J = (c 2 ) π − c ) = (6 2 A = π c22 − c12 = π (62 − 5.752 ) = 9.2284 in 2 4 2 4 1 4 − 5.754 ) = 318.67 in 4 Stresses: σ =− P A 40 = −4.3344 ksi 9.2284 Tc τ= 2 J (80)(6) = = 1.5063 ksi 318.67 σ x = 0, σ y = −4.3344 ksi, τ xy = 1.5063 ksi =− Choose the x′ and y ′ axes, respectively, tangential and normal to the weld. Then σ w = σ y′ and τ w = τ x′y′ θ = 22.5° σ y′ = σx +σy σx −σ y cos 2θ − τ xy sin 2θ 2 2 (−4.3344) [ −(−4.3344)] = − cos 45° − 1.5063 sin 45° 2 2 = −4.76 ksi τ x′y′ = − − σ w = −4.76 ksi σx −σy sin 2θ + τ xy cos 2θ 2 [ −(−4.3344)] =− sin 45° + 1.5063 cos 45° 2 = −0.467 ksi τ w = −0.467 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 7.22 Two steel plates of uniform cross section 10 × 80 mm are welded together as shown. Knowing that centric 100-kN forces are applied to the welded plates and that the in-plane shearing stress parallel to the weld is 30 MPa, determine (a) the angle β, (b) the corresponding normal stress perpendicular to the weld. SOLUTION Area of weld: Aw = = (a) Fs = 0: Fs − 100sin β = 0 τw = Fs Aw sin β cos β = (b) 30 × 106 = (10 × 10−3 )(80 × 10−3 ) cos β 800 × 10−6 2 m cos β Fs = 100sin β kN = 100 × 103 sin β N 100 × 103 sin β = 125 × 106 sin β cos β −6 800 × 10 / cos β 1 30 × 106 sin 2β = = 0.240 2 125 × 106 Fn = 0: Fn − 100cos β = 0 β = 14.34° Fn = 100cos14.34° = 96.88 kN Aw = 800 × 10−6 = 825.74 × 10−6 m 2 cos14.34 σ = Fn 96.88 × 103 = = 117.3 × 106 Pa Aw 825.74 × 10−6 σ = 117.3 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.