07-02ChapGere.0008

# 07-02ChapGere.0008 - s y(b Determine the other principal...

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Solution 7.3-19 Plane stress s x 5 6500 psi t xy 52 2800 psi s y 5 ? One principal stress 5 7300 psi (tension) (a) S TRESS s y Because s x is smaller than the given principal stress, we know that the given stress is the larger principal stress. s 1 5 7300 psi Substitute numerical values and solve for s y : s y 52 2500 psi s 1 5 s x 1 s y 2 1 B ¢ s x 2 s y 2 2 1 t xy 2 (b) P RINCIPAL STRESSES 2 u p 5 2 31.891 8 and u p 5 2 15.945 8 2 u p 5 148.109 8 and u p 5 74.053 8 For 2 u p 5 2 31.891 8 : For 2 u p 5 148.109 8 : Therefore, s 1 5 7300 psi and s 2 52 3300 psi and u p 2 5 74.05 8 u p 1 52 15.95 8 s x 1 52 3300 psi s x 1 5 7300 psi s x 1 5 s x 1 s y 2 1 s x 2 s y 2 cos 2 u 1 t xy sin 2 u tan 2 u p 5 2 t xy s x 2 s y 52 0.62222 446 CHAPTER 7 Analysis of Stress and Strain x O u p 2 5 74.05 8 7300 psi 3300 psi y Problem 7.3-20 An element in plane stress is subjected to stresses s x 52 68.5 MPa and t xy 5 39.2 MPa (see figure). It is known that one of the principal stresses equals 26.3 MPa in tension. (a) Determine the stress
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Unformatted text preview: s y . (b) Determine the other principal stress and the orientation of the principal planes; then show the principal stresses on a sketch of a properly oriented element. Solution 7.3-20 Plane stress y x O 39.2 MPa s y 68.5 MPa } s x 5 2 68.5 MPa t xy 5 39.2 MPa s y 5 ? One principal stress 5 26.3 MPa (tension) (a) S TRESS s y Because s x is smaller than the given principal stress, we know that the given stress is the larger principal stress. s 1 5 26.3 MPa Substitute numerical values and solve for s y : s y 5 10.1 MPa s 1 5 s x 1 s y 2 1 B ¢ s x 2 s y 2 ≤ 2 1 t xy 2 A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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## This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Tech.

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