587_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 587_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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SECTION 7.2 Plane Stress 581 For * cos (150°) + t xy sin(150°) 4800 psi ± 2200 psi + s y 2 + 2200 psi ² s y 2 s x 1 ± 4800 psi u ± 75°: (2) Solve Eqs. (1) and (2): s y ± 3805 psi t xy ± 2205 psi ; or 0.93301 s y + 0.50000 t xy ± 4652.6 psi Problem 7.2-18 The surface of an airplane wing is subjected to plane stress with normal stresses s x and s y and shear stress t xy , as shown in the figure. At a counterclockwise angle u ± 32˚ from the x axis, the normal stress is 37 MPa tension, and at an angle u ± 48˚, it is 12 MPa compression. If the stress s x equals 110 MPa tension, what are the stresses s y and t xy ? y x O t xy s y s x = 110 MPa Solution 7.2-18 Find and For * cos (64°) + t xy sin (64°) 37 MPa ± 110 MPa + s y 2 + 110 MPa
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Unformatted text preview: ² s y 2 s x 1 ± 37 MPa u ± 32° s x 1 ± s x + s y 2 + s x ² s y 2 cos(2 u ) + t xy sin(2 u ) t xy s y At u ± 48°, s x 1 ±² 12 MPa (compression) At u ± 32°, s x 1 ± 37 MPa (tension) s x ± 110 MPa s y unknown s xy unknown (1) For (2) Solve Eqs. (1) and (2): s y ± ² 60.7 MPa t xy ± ² 27.9 MPa ; or 0.55226 s y + 0.99452 t xy ± ² 61.25093 MPa * cos (96°) + t xy sin (96°) ² 12 MPa ± 110 MPa + s y 2 + 110 MPa ² s y 2 s x 1 ±² 12 MPa u ± 48°: or 0.28081 s y + 0.89879 t xy ± ² 42.11041 MPa 07Ch07.qxd 9/27/08 1:18 PM Page 581...
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