02-06_chap_gere

02-06_chap_gere - 144 CHAPTER 2 Axially Loaded Numbers...

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Problem 2.7-9 A slightly tapered bar AB of rectangular cross section and length L is acted upon by a force P (see figure). The width of the bar varies uniformly from b 2 at end A to b 1 at end B . The thickness t is constant. (a) Determine the strain energy U of the bar. (b) Determine the elongation d of the bar by equating the strain energy to the work done by the force P . Solution 2.7-9 Tapered bar of rectangular cross section 144 CHAPTER 2 Axially Loaded Numbers b 2 b 1 L A B P (a) S TRAIN ENERGY OF THE BAR Apply this integration formula to Eq. (1): From Appendix C: # dx a 1 bx 5 1 b ln ( a 1 bx ) 5 # L 0 P 2 dx 2 Et b ( x ) 5 P 2 2 Et # L 0 dx b 2 2 ( b 2 2 b 1 ) x L Ê (1) U 5 # [ N ( x )] 2 dx 2 E A ( x ) Ê (Eq. 2-41) 5 t B b 2 2 ( b 2 2 b 1 ) x L R A ( x ) 5 tb ( x ) b ( x ) 5 b 2 2 ( b 2 2 b 1 ) x L (b) E LONGATION OF THE BAR (E Q . 2-42) N OTE : This result agrees with the formula derived in prob. 2.3-11. d 5 2 U P 5 PL Et ( b 2 2 b 1 ) ln b 2 b 1 U 5 P 2 L 2 Et ( b 2 2 b 1 ) ln b 2 b 1 5 P 2 2 Et B 2 L ( b 2 2 b 1 ) ln b 1 2 2 L ( b 2 2 b 1 ) ln b 2 R 2 ( b 2 2 b 1 ) x L RR 0 L U 5 P 2 2 Et B 1 2 ( b 2 2 b 1 )( 1 L ) ln B b 2 b 2 b 1 L A B P dx b(x) x Problem 2.7-10 A compressive load P is transmitted through a rigid plate to three magnesium-alloy bars that are identical except that initially the middle bar is slightly shorter than the other bars (see figure). The dimensions and properties of the assembly are as follows: length L 5 1.0 m, cross-sectional area of each bar A 5 3000 mm 2 , modulus of elasticity E 5 45 GPa, and the gap s 5 1.0 mm. (a) Calculate the load P 1 required to close the gap. (b) Calculate the downward displacement d of the rigid plate when P 5 400 kN. (c) Calculate the total strain energy U of the three bars when P 5 400 kN. (d) Explain why the strain energy U is not equal to P d /2. ( Hint: Draw a load-displacement diagram.) L P s
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Solution 2.7-10 Three bars in compression SECTION 2.7 Strain Energy 145 s 5 1.0 mm L 5 1.0 m For each bar: A 5 3000 mm 2 E 5 45 GPa (a) L OAD P 1 REQUIRED TO CLOSE THE GAP For two bars, we obtain: (b) D ISPLACEMENT d FOR P 5 400 kN 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 0.001 m) 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 5 400 kN Outer bars: d 5 1.321 mm Middle bar: d 5 1.321 mm 2 s 5 0.321 mm (d) L OAD - DISPLACEMENT DIAGRAM U 5 243 J 5 243 N . m The strain energy U is not equal to because the 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 k N )(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 00 .
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This note was uploaded on 04/30/2008 for the course CVEN 305 taught by Professor Gardoni during the Spring '08 term at Texas A&M.

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02-06_chap_gere - 144 CHAPTER 2 Axially Loaded Numbers...

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