624_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 624_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 07Ch07.qxd 9/27/08 618 1:22 PM Page 618 CHAPTER 7 Analysis of Stress and Strain Problem 7.6-5 An element of aluminum in triaxial stress (see figure) is 4750 psi subjected to stresses sx 5200 psi (tension), sy 3090 psi (com(compression), and sz pression). It is also known that the normal strains in the x and y directions 502.3 * 10 6 (shortare Px 7138.8 * 10 6 (elongation) and Py sx ening). What is the bulk modulus K for the aluminum? y sy sz sx O x sz sy z Probs. 7.6-5 and 7.6-6 Solution 7.6-5 Triaxial stress (bulk modulus) sx 5200 psi sy sz ây Substitute numerical values and rearrange: 4750 psi 3090 psi âx 713.8 * 10 6 6 502.3 * 10 (713.8 * 10 6 )E 5200 + 7840 (1) 6 4750 (2) ( 502.3 * 10 Find K. )E 2110 Units: E = psi Eq. (7-53 a): âx Eq. (7-53 b): ây sx E sy E E E Solve simultaneously Eqs. (1) and (2): (sy + sz) E (sx + sy) 10.801 * 106 psi 0.3202 E Eq. (7-16): K 3(1 2) 10.0 * 10 6 psi ; Problem 7.6-6 sx Py Solve the preceding problem if the material is nylon subjected to compressive stresses 4.5 MPa, sy 3.6 MPa , and sz 2.1 MPa, and the normal strains are Px 740 * 10 6 and 320 * 10 6 (shortenings). Solution 7.6-6 Triaxial stress (bulk modulus) sx sz ây 4.5 MPa sy 740 * 10 6 Find K. Eq. (7-53 a): âx Eq. (7-53 b): ây 6 ( 740 * 10 6 )E 4.5 + 5.7 (1) ( 320 * 10 2.1 MPa âx 320 * 10 Substitute numerical values and rearrange: 3.6 MPa 6 )E 3.6 + 6.6 (2) Units: E = MPa sx E sy E E E (sy + sz) Solve simultaneously Eqs. (1) and (2): E (sz + sx) 3,000 MPa Eq. (7-16): K 3.0 GPa E 3(1 2) 0.40 5.0 GPa ; ...
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