# HW4 - 5.3 A 12-mm-diameter steel [E = 200 GPa] rod (2) is...

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5.3 A 12-mm-diameter steel [ E = 200 GPa] rod (2) is connected to a 30-mm-wide by 8-mm-thick rectangular aluminum [ E = 70 GPa] bar (1), as shown in Fig. P5.3. Determine the force P required to stretch the assembly 10.0 mm. Fig. P5.3 Solution The elongations in the two axial members are expressed by 1 1 2 2 1 2 1 1 2 2 and F L F L A E A E δ = = The total elongation of the assembly is thus 1 1 2 2 1 2 1 1 2 2 C F L F L u A E A E = + = + Since the internal forces F 1 and F 2 are equal to external load P , this expression can be simplified to 1 2 1 1 2 2 C L L u P A E A E = + For rectangular aluminum bar (1), the cross-sectional area is 2 1 (30 mm)(8 mm) 240 mm A = = and the cross-sectional area of steel rod (2) is 2 2 2 (12 mm) 113.0973 mm 4 A π = = The force P required to stretch the assembly 10.0 mm is thus 1 2 1 1 2 2 2 2 2 2 10.0 mm 450 mm 1,300 mm (240 mm )(70,000 N/mm ) (113.09739 mm )(200,000 N/mm ) 118,682.6 N = 118.7 kN C u P L L A E A E = + = + = Ans. Comment: P min = 118.7 kN

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5.7 Aluminum [ E = 70 GPa] member ABC supports a load of 28 kN, as shown in Fig. P5.7. Determine: (a) the value of load P such that the deflection of joint C is zero. (b) the corresponding deflection of joint B . Fig. P5.7 Solution Cut a FBD that exposes the internal axial force in member (2): 2 2 28 kN 0 28 kN y F F F Σ = = = Similarly, cut a FBD that exposes the internal axial force in member (1): 1 1 28 kN 0 28 kN y F P F F P Σ = = = The deflection at joint C , which must ultimately equal zero, can be expressed in terms of the member deformations δ 1 and 2 : 1 1 2 2 1 2 1 1 2 2 0 C F L F L u A E A E = + = + = or 1 1 2 2 1 1 2 2 2 2 2 2 (28,000 N )(1,000 mm) (28,000 N)(1,300 mm) 0 (50 mm) (70,000 N/mm ) (32 mm) (70,000 N/mm ) 4 4 C F L F L u A E A E P π = + = + = The only unknown in this equation is P , which can be computed as 80.5841 k 80. kN N 6 P = = Ans. The corresponding deflection at joint B can be found from the deformation in member (1): 1 1 1 2 2 1 1 (28,000 N 80,584 N)(1,300 mm) (50 mm) (70,000 N/m 0.497 m mm 0.497 mm ) 4 B F L u A E = = = = = Ans. Comment: (a) P = 80.6 kN; (b) u B = 0.497 mm downward.
5.13 Determine the extension, due to its own weight, of the conical bar shown in Fig. P5.13. The bar is made of aluminum alloy [ E = 10,600 ksi and γ = 0.100 lb/in. 3

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## This note was uploaded on 02/08/2011 for the course ME 650:291 taught by Professor Weng during the Fall '11 term at Rutgers.

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HW4 - 5.3 A 12-mm-diameter steel [E = 200 GPa] rod (2) is...

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