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 shon. Use a method of analysis based on the equilibrium of that element, as as done in the derivations of Sec. 7.. Stresses Areas Forces F = : σ A 15 A sin 3 cos 3 15 A cos 3 sin 3 + 1 A cos 3 cos 3 = σ = 3 sin 3 cos 3 1 cos 3 Σ F = : τ A+ 15 A sin 3 sin 3 15 A cos 3 cos 3 1 A cos 3 sin 3 = τ = 15(cos 3 sin 3 ) + 1 cos 3 sin 3 σ = 5.49 ksi τ = 11.83 ksi PROPRIETARY MATERIAL. 1 The McGra-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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 shon. Use a method of analysis based on the equilibrium of that element, as as done in the derivations of Sec. 7.. Stresses Areas Forces Σ F = : σ A+ 18 A cos 15 sin 15 + 45 A cos 15 cos 15 7 A sin 15 sin 15 + 18 A sin 15 cos 15 = σ = 18 cos 15 sin 15 45 cos 15 + 7sin 15 18 sin 15 cos 15 σ = 49. MPa Σ F = : τ A+ 18 A cos 15 cos 15 45 A cos 15 sin 15 7 A sin 15 cos 15 18 A sin 15 sin 15 = τ = 18(cos 15 sin 15 ) + (45 + 7) cos 15 sin 15 τ =.41 MPa PROPRIETARY MATERIAL. 1 The McGra-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROBLEM 7.5 For the given state of stress, determine (a) the principal planes, (b) the principal stresses. σ = MPa σ = 4 MPa τ = 35 MPa x y xy (a) τ xy ()(35) tan θ p = 3.5 σ σ = + 4 = x y θ p = 74.5 θ p = 37., 53. (b) σ max, min σx + σy σx σy = ± + τ xy 4 + 4 = ± + (35) = 5 ± 3.4 MPa σ = 13. MPa max σ = 8.4 MPa min PROPRIETARY MATERIAL. 1 The McGra-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROBLEM 7.8 For the given state of stress, determine (a) the principal planes, (b) the principal stresses. σ = 8ksi σ = 1ksi τ = 5ksi x y xy (a) τ xy (5) tan θ p = =.5 σ σ 8 1 = x y θ p =.551 θ p = 13.3, 7.7 (b) σ max, min σx + σy σx σ y = ± + τ xy 8 + 1 8 1 = ± + (5) = ± 11.183 σ max = 13.18 ksi σ min = 9.18 ksi PROPRIETARY MATERIAL. 1 The McGra-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROBLEM 7.19 A steel pipe of 1-in. outer diameter is fabricated from 1 -in.-thick plate by 4 elding along a helix that forms an angle of.5 ith a plane perpendicular to the axis of the pipe. Knoing that a 4-kip axial force P and an 8-kip in. torque T, each directed as shon, are applied to the pipe, determine σ and τ in directions, respectively, normal and tangential to the eld. Stresses: 1 d = 1 in., c = d = in., t =.5 in. c = c t = 5.75 in. 1 ( 1 ) ( 1 ) A= π c c = π( 5.75 ) = 9.84 in π π J = c c = ( 5.75 ) = 318.7 in 4 4 4 4 4 P σ = A 4 = = 4.3344 ksi 9.84 Tc τ = J (8)() = = 1.53 ksi 318.7 σ =, σ = 4.3344 ksi, τ = 1.53 ksi x y xy Choose the x and y axes, respectively, tangential and normal to the eld. Then σ = σ and τ = τ θ =.5 y xy σx + σy σx σ y σy = cos θ τxysin θ ( 4.3344) [ ( 4.3344)] = cos 45 1.53 sin 45 = 4.7 ksi σ = 4.7 ksi σx σy τxy = sin θ + τxycos θ [ ( 4.3344)] = sin 45 + 1.53 cos45 =.47 ksi τ =.47 ksi PROPRIETARY MATERIAL. 1 The McGra-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROBLEM 7. To steel plates of uniform cross section 1 8 mm are elded together as shon. Knoing that centric 1-kN forces are applied to the elded plates and that the in-plane shearing stress parallel to the eld is 3 MPa, determine (a) the angle β, (b) the corresponding normal stress perpendicular to the eld. Area of eld: A = = 3 3 (1 1 )(8 1 ) cos β 8 1 cos β m (a) (b) F = : F 1sin β = F = 1sin β kn = 1 1 sin β N s s s 3 1 1 sin β Fs τ = 3 1 = = 15 1 sin βcos β A 8 1 / cos β 1 3 1 sin βcos β = sin β = =.4 15 1 F = : F 1cosβ = F = 1cos14.34 = 9.88 kn A n n n 8 1 = = 85.74 1 m cos14.34 3 Fn 9.88 1 σ = = = 117.3 1 Pa A 85.74 1 3 β = 14.34 σ = 117.3 MPa PROPRIETARY MATERIAL. 1 The McGra-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,