07-03ChapGere.0002 - Show all results on sketches of...

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Solution 7.4-8 Pure shear SECTION 7.4 Mohr’s Circle for Plane Stress 453 s x 5 0 s y 5 0 t xy 52 16 MPa (a) E LEMENT AT u 5 20 8 (All stresses in MPa) 2 u 5 40 8 u 5 20 8 R 5 16 MPa Origin O is at center of circle. Point D : Point D 9 : t x 1 y 1 5 R cos 2 u 5 12.26 MPa s x 1 5 R sin 2 u 5 10.28 MPa t x 1 y 1 52 R cos 2 u 52 12.26 MPa s x 1 52 R sin 2 u 52 10.28 MPa (b) P RINCIPAL STRESSES Point P 1 : s 1 5 R 5 16 MPa Point P 2 : s 2 52 R 52 16 MPa u p 2 5 45 8 2 u p 2 5 90 8 u p 1 5 135 8 2 u p 1 5 270 8 D' D A ( u 5 0) 2 u 2 u p 2 2 u p 1 P 1 P 2 R R O s x 1 ( u 5 20 8 ) B ( u 5 90 8 ) t x 1 y 1 16 x O 12.3 MPa u 5 20 8 10.3 MPa 10.3 MPa y D D' x O u p 2 5 45 8 16 MPa 16 MPa y P 2 P 1 Problem 7.4-9 An element in pure shear is subjected to stresses t xy 5 4000 psi, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a slope of 3 on 4 (see figure) and (b) the principal stresses.
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Unformatted text preview: Show all results on sketches of properly oriented elements. O 3 4 y x 4000 psi Solution 7.4-9 Pure shear s x 5 s y 5 t xy 5 4000 psi (a) E LEMENT AT A SLOPE OF 3 ON 4 (All stresses in psi) 2 u 5 73.740 8 u 5 36.870 8 R 5 4000 psi Origin O is at center of circle. u 5 arctan 3 4 5 36.870 8 u 3 4 D' D A ( u 5 0) 2 u 2 u p 1 P 1 P 2 R R O s x 1 B ( u 5 90 8 ) t x 1 y 1 4000 16.260 8 R 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 Institute of Technology.

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