02-01ChapGere.0015

02-01ChapGere.0015 - L 4 — L 2 — b t P P Solution 2.3-4...

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Problem 2.3-4 A rectangular bar of length L has a slot in the middle half of its length (see figure). The bar has width b , thickness t , and modulus of elasticity E . The slot has width b /4. (a) Obtain a formula for the elongation d of the bar due to the axial loads P . (b) Calculate the elongation of the bar if the material is high-strength steel, the axial stress in the middle region is 160 MPa, the length is 750 mm, and the modulus of elasticity is 210 GPa. SECTION 2.3 Changes in Lengths under Nonuniform Conditions 77 (b) I NCREASE IN P 3 FOR NO CHANGE IN LENGTH P 5 increase in force P 3 The force P must produce a shortening equal to 0.0131 in. in order to have no change in length. P 5 1310 lb 5 P (120 in.) (30 3 10 6 psi)(0.40 in. 2 ) 0.0131 in. 5 d 5 PL EA b 4 L 4
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Unformatted text preview: L 4 — L 2 — b t P P Solution 2.3-4 Bar with a slot b 4 — b P P L 2 — L 4 — L 4 — t 5 thickness L 5 length of bar (a) E LONGATION OF BAR 5 PL Ebt ¢ 1 4 1 4 6 1 1 4 ≤ 5 7 PL 6 Ebt d 5 a N i L i EA i 5 P ( L / 4) E ( bt ) 1 P ( L / 2) E ( 3 4 bt ) 1 P ( L / 4) E ( bt ) S TRESS IN MIDDLE REGION Substitute into the equation for d : (b) S UBSTITUTE NUMERICAL VALUES : d 5 7(160 MPa)(750 mm) 8 (210 GPa) 5 0.500 mm s 5 160 MPa Ê L 5 750 mm Ê E 5 210 GPa 5 7 s L 8 E d 5 7 PL 6 Ebt 5 7 L 6 E ¢ P bt ≤ 5 7 L 6 E ¢ 3 s 4 ≤ s 5 P A 5 P ( 3 4 bt ) 5 4 P 3 bt Ê or P bt 5 3 s 4 P 120 in. A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Tech.

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