176_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

176_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.5.qxd 9/25/08 170 3:00 PM Page 170 CHAPTER 2 Axially Loaded Members STRESS IN THE PLASTIC CYLINDER sp Pp 2np Es As Ep Ap EpAp D L(Ep Ap + 2Es As) N Es As Ep 54.0 2 21.6 2np N ab LD sp 106 N 2(1)(1.0 mm) N ab 200 mm D ; 25.0 MPa SUBSTITUTE NUMERICAL VALUES: 15 2EsAs 2 10 N /m Problem 2.5-21 Solve the preceding problem if the data for the assembly are as follows: length L 10 in., pitch of the bolt threads p 0.058 in., modulus of elasticity for steel Es 30 106 psi, modulus of elasticity for the plastic Ep 500 ksi, cross-sectional area of one bolt As 0.06 in.2, and crosssectional area of the plastic cylinder Ap 1.5 in.2 Steel bolt L Solution 2.5-21 Plastic cylinder and two steel bolts COMPATIBILITY EQUATION L 10 in. p 0.058 in. s Es 30 106 psi elongation of steel bolt p shortening of plastic cylinder s As (Eq. 2) 500 ksi Ap np 0.06 in.2 (for one bolt) Ep p 1.5 in.2 n 1 (see Eq. 2-22) FORCE-DISPLACEMENT RELATIONS EQUILIBRIUM EQUATION Ps tensile force in one steel bolt Pp compressive force in plastic cylinder Pp 2Ps ds (Eq. 1) Ps L Es As dp Pp L Ep Ap (Eq. 3, Eq. 4) SOLUTION OF EQUATIONS Substitute (3) and (4) into Eq. (2): Pp L Ps L + Es As Ep Ap np (Eq. 5) ...
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