586_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 586_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

This preview shows page 1. Sign up to view the full content.

580 CHAPTER 7 Analysis of Stress and Strain Problem 7.2-16 Solve the preceding problem for the element shown in the figure. 24.3 MPa 62.5 MPa 24.0 MPa y x O u = 55 ° Solution 7.2-16 Transform from s x 1 ± 56.5 MPa ; s x 1 ± s x + s y 2 + s x ² s y 2 cos (2 u ) + t xy sin (2 u ) u ±² 55° t xy ±² 24 MPa s x ±² 24.3 MPa s y ± 62.5 MPa u ± 55° to u ± s y 1 ±² 18.3 MPa ; s y 1 ± s x + s y ² s x 1 t x 1 y 1 ±² 32.6 MPa ; t x 1 y 1 ±² s x ² s y 2 sin (2 u ) + t xy cos (2 u ) Problem 7.2-17 A plate in plane stress is subjected to normal stresses s x and s y and shear stress t xy , as shown in the figure. At counterclockwise angles u ± 35˚ and u ± 75˚ from the x axis, the normal stress is 4800 psi tension. If the stress
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s x equals 2200 psi tension, what are the stresses s y and t xy ? y x O t xy s y s x = 2200 psi Solution 7.2-17 Find and 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 ± 35° and u ± 75°, s x 1 ± 4800 psi s x ± 2200 psi s y unknown t xy unknown For (1) or 0.32899 s y + 0.93969 t xy ± 3323.8 psi * cos (70°) + t xy sin(70°) 4800 psi ± 2200 psi + s y 2 + 2200 psi ² s y 2 s x 1 ± 4800 psi u ± 35° 07Ch07.qxd 9/27/08 1:18 PM Page 580...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online