196_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

196_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.6.qxd 190 9/25/08 11:40 AM Page 190 CHAPTER 2 Axially Loaded Members The shear stress on the joint has a negative sign. Its numerical value cannot exceed allow 2.25 MPa. Therefore, 2.25 MPa 0.4592 (4.9 MPa)(sin )(cos ) or sin | (c) cos Therefore: sin 2 0.9184 66.69° or 90° Since If | t0 ` s0 x 113.31° or x sin cos | x cos2 2 cos 2 xcos2 or tan 63.43° 33.34° | 2 cos sin sin 56.66° 2? Numerical values only: 26.6° must be between 10° and 40°, we select NOTE: If | or 56.66° 33.3° | 2(0.4592) if WHAT IS ` 1 sin 2u 2 33.34° 2.25 MPa. | From trigonometry: sin u cos u Solving: 2 | 2 90° ; 26.6° and 63.4°, we find NOTE: For 0.98 MPa and 1.96 MPa. ; is between 10° and 33.3°, Thus, ` 2.25 MPa. is between 33.3° and 40°, Problem 2.6-15 Acting on the sides of a stress element cut from a bar in t0 ` s0 2 as required. 5000 psi uniaxial stress are tensile stresses of 10,000 psi and 5,000 psi, as shown in the figure. (a) Determine the angle and the shear stress sketch of the element. su = 10,000 psi u and show all stresses on a (b) Determine the maximum normal stress stress max in the material. tu tu and the maximum shear max 10,000 psi tu tu 5000 psi ...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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