02-07ChapGere.0008 - mm ABC s 5 P 2 A 5 60 kN 480 mm 2 5...

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SECTION 2.11 Nonlinear Behavior 167 (c) B OTH P 1 AND P 2 ARE ACTING e 0.008477 e 0.002853 (Note that the displacement when both loads act simultaneously is not equal to the sum of the displacements when the loads act separately.) C AB BC 11.88 mm BC e ¢ L 3 1.71 mm BC : s P 2 A 60 kN 480 mm 2 125 MPa AB e ¢ 2 L 3 10.17 mm AB : s P 1 P 2 A 90 kN 480 mm 2 187.5 MPa Problem 2.11-3 A circular bar of length L 32 in. and diameter d 0.75 in. is subjected to tension by forces P (see figure). The wire is made of a copper alloy having the following hyperbolic stress-strain relationship : 1 18,0 3 0 0 0 0 0 0.03 ( ksi) (a) Draw a stress-strain diagram for the material. (b) If the elongation of the wire is limited to 0.25 in. and the maximum stress is limited to 40 ksi, what is the allowable load P ? Solution 2.11-3 Copper bar in tension P P L d (a) P 1 ACTS ALONE e 0.001389
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Unformatted text preview: mm ABC : s 5 P 2 A 5 60 kN 480 mm 2 5 125 MPa d c 5 e ¢ 2 L 3 ≤ 5 1.67 mm AB : s 5 P 1 A 5 30 kN 480 mm 2 5 62.5 MPa L 5 32in. d 5 0.75in. (a) S TRESS-STRAIN DIAGRAM s 5 18,000 e 1 1 300 e Ê # e # 0.03 Ê ( s 5 ksi) A 5 p d 2 4 5 0.4418 in. 2 (b) A LLOWABLE LOAD P Max. elongation d max 5 0.25in. Max. stress s max 5 40ksi Based upon elongation: B ASED UPON STRESS : Stress governs. P 5 s max A 5 (40ksi)(0.4418in. 2 ) 5 17.7 k s max 5 40 ksi s max 5 18,000 e max 1 1 300 e max 5 42.06 ksi e max 5 d max L 5 0.25 in. 32 in. 5 0.007813 P P L d 20 40 60 Slope = 18,000 ksi Asymptote equals 60 ksi 0.01 0.02 0.03 s (ksi) e A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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