02-06ChapGere.0002 - load-displacement relation is not...

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Solution 2.7-10 Three bars in compression SECTION 2.7 Strain Energy 145 s 5 1.0mm L 5 1.0m For each bar: A 5 3000mm 2 E 5 45GPa (a) L OAD P 1 REQUIRED TO CLOSE THE GAP For two bars, we obtain: (b) D ISPLACEMENT d FOR P 5 400kN Since P . P 1 , all three bars are compressed. The force P equals P 1 plus the additional force required to compress all three bars by the amount d 2 s . or 400 kN 5 270 kN 1 3(135 3 10 6 N/m)( d 2 0.001m) Solving, we get d 5 1.321 mm P 5 P 1 1 3 ¢ EA L ( d 2 s ) P 1 5 270 kN P 1 5 2 ¢ EAs L 5 2(135 3 10 6 N / m)(1.0 mm) In general, d 5 PL EA and P 5 EA d L EA L 5 135 3 10 6 N / m (c) S TRAIN ENERGY U FOR P 5 400kN Outer bars: d 5 1.321mm Middle bar: d 5 1.321mm 2 s 5 0.321mm (d) L OAD - DISPLACEMENT DIAGRAM U 5 243 J 5 243N . m The strain energy U is not equal to because the
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Unformatted text preview: load-displacement relation is not linear. U 5 area under line OAB . under a straight line from O to B , which is larger than U . P d 2 5 area P d 2 P d 2 5 1 2 (400 kN)(1.321 mm) 5 264 N ? m 5 243 N ? m 5 243 J 5 1 2 (135 3 10 6 N / m)(3.593 mm 2 ) U 5 EA 2 L [2(1.321 mm) 2 1 (0.321 mm) 2 ] U 5 a EA d 2 2 L 100 0.5 1.0 1.5 2.0 200 400 300 400 kN 270 kN A B d = 1.0 mm d = 1.321 mm Displacement d (mm) Load P (kN) L P s = 1.0 mm 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 Institute of Technology.

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