01-01ChapGere.0019

01-01ChapGere.0019 - e ∴ Initial slope 5 18,000 ksi At e...

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Solution 1.4-5 Wire stretched by forces P SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio 19 L 5 4ft 5 48 in. d 5 0.125 in. P 5 600 lb C OPPER ALLOY (Eq. 1) (a) S TRESS - STRAIN DIAGRAM (From Eq. 1) 0 # e # 0.03 ( s 5 ksi) s 5 18,000 e 1 1 300 e A LTERNATIVE FORM OF THE STRESS - STRAIN RELATIONSHIP Solve Eq. (1) for e in terms of s : (Eq. 2) This equation may also be used when plotting the stress-strain diagram. (b) E LONGATION d OF THE WIRE From Eq. (2) or from the stress-strain diagram: e 5 0.0147 d 5 e L 5 (0.0147)(48 in.) 5 0.71 in. S TRESS AND STRAIN AT POINT B (see diagram) s B 5 48.9 ksi e B 5 0.0147 E LASTIC RECOVERY e E R ESIDUAL STRAIN e R e R 5 e B 2 e E 5 0.0147 2 0.0027 5 0.0120 (c) Permanent set 5 e R L 5 (0.0120)(48 in.) 5 0.58 in. (d) Proportional limit when reloaded 5 s B s B 5 49 ksi e E 5 s B Slope 5 48.9 ksi 18,000 ksi 5 0.00272 s 5 P A 5 600 lb p 4 (0.125 in.) 2 5 48,900 psi 5 48.9 ksi 0 # s # 54 ksi Ê ( s 5 ksi) e 5 s 18,000 2 300 s 60 40 20 0 0.01 0.02 0.03 e e R e B e E = e B e R s = 54 ksi s B s (ksi) B I NITIAL SLOPE OF STRESS - STRAIN CURVE Take the derivative of s with respect to
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Unformatted text preview: e : ∴ Initial slope 5 18,000 ksi At e 5 0, d s d e 5 18,000 ksi 5 18,000 (1 1 300 e ) 2 d s d e 5 (1 1 300 e )(18,000) 2 (18,000 e )(300) (1 1 300 e ) 2 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio When solving the problems for Section 1.5, assume that the material behaves linearly elastically. Problem 1.5-1 A high-strength steel bar used in a large crane has diameter d 5 2.00 in. (see figure). The steel has modulus of elasticity E 5 29 3 10 6 psi and Poisson’s ratio n 5 0.29. Because of clearance requirements, the diameter of the bar is limited to 2.001 in. when it is compressed by axial forces. What is the largest compressive load P max that is permitted? P d P A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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