Calculation of Displacements in A992 Steel Rod Under Load Assignment Solutions

The A992 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 80 mm2, determine the displacement of B and A. Neglect the size of the couplings at B and C. D 0.75 m C 45 45 6 kN 6 kN SOLUTION 1.50 m B Normal Force And Stress: For segments AB, BC and CD, referring to the FBDs of the lower segments shown in Fig. a, b and c respectively, + c ΣFy = 0; PAB - 10 = 0 + c ΣFy = 0; 4 PBC - 2 c5a b d - 10 = 0 5 + c ΣFy = 0; PCD - 2(6 sin 45°) - 2 5 5 4 3 5 kN PAB = 10.0 kN 1m 5 kN A PBC = 18.0 kN 4 c5a b d - 10 = 0 5 4 3 10 kN PCD = 26.485 k N Since the rod has a constant cross-section and segment CD is subjected to greatest normal force, this segment will develop the greatest average normal stress. smax = sCD = NCD 26.485( 103 ) = 331.07 ( 106 ) Pa = 331.07 MP a = -6 A 80 (10 ) Displacement: For A992 steel, sy = 345 MPa and E = 200 GPa . Since smax 6 sy, NL 1 (PBCLBC + PCDLCD) = dB = Σ = AE AE 1 3 [18.0 ( 10 ) (1.5) 80( 10-6 ) [200( 109 ) ] + 26.485 ( 103 ) (0.75)] = 0.002929 m = 2.93 mm T Ans. 10.0( 103 ) (1) PABLAB + dB = + 0.002929 dA = AE 80 (10 -6) [200( 10 9) ] = 0.003554 m = 3.55 mm T Ans. Ans: dB = 2.93 mm T , dA = 3.55 mm T 4–13. The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14 mm2 and is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D when the distributed load is applied. C 300 N/m 1.5 m A a+ ΣMA = 0; 2m B D 2m 1200(2) - TCB(0.6)(2) = 0 TCB = 2000 N dB >C = (2000)(2.5) PL = 0.005183 5 = AE 14(10- 6 )(68.9)(10 9) (2.5051835)2 = (1.5)2 + (2)2 - 2(1.5)(2) cos u u = 90.248° u = 90.248° - 90° = 0.2478° = 0.004324 rad dD = u r = 0.004324(4000) = 17.3 mm Ans. Ans: dD = 17.3 mm 4–41. P The 10-mm-diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20 mm, and its inner diameter is 10 mm. If the yield stress for the steel is (s Y)st = 640 MPa, and for the bronze ( s Y)br = 520 MPa, determine the magnitude of the largest elastic load P that can be applied to the assembly. Est = 200 GPa, Ebr = 100 GPa. 10 mm 20 mm + c ΣFy = 0; Pst + Pbr - P = 0 (1) Assume failure of bolt: Pst = (sY)st(A) = 640 p 4 P (106 ) a b ( 0.012 ) = 50265.5 N dst = dbr PstL p 4 = ( 0.012) (200) ( 109 ) Pbr L p 4 ( 0.022 - 0.01 2) (100) (10 9) Pst = 0.6667 Pbr 50 265.5 = 0.6667 Pbr Pbr = 75398.2 N From Eq. (1) P = 50265.5 + 75398.2 = 125663.7 N = 126 kN (controls) Ans. Assume failure of sleeve: Pbr = (sY)br(A) = 520 ( 106 )ap b ( 0.022 - 0.01 2 ) = 122 522.11 N 4 Pst = 0.6667 Pbr = 0.6667(122 522.11) = 81 681.4 N From Eq. (1), P = 122 522.11 + 81 681.4 = 204 203.52 N = 204 kN Ans: P = 126 kN The support consists of a solid red brass C83400 copper post surrounded by a 304 stainless steel tube. Before the load is applied the gap between these two parts is 1 mm. Given the dimensions shown, determine the greatest axial load that can be applied to the rigid cap A without causing yielding of any one of the materials. P A 1 mm 0.25 m SOLUTION 60 mm 80 mm Require, 10 mm dst = dbr + 0.001 p[(0.05) 2 Fst (0.25) Fbr (0.25) + 0.00 1 = 2 9 - (0.04) 2 ]193(10 9) p (0.03) (101)(10 ) 0.45813 Fst = 0.87544 Fbr + 106 + c ΣFy = 0; Fst + Fbr - P = 0 (1) (2) Assume brass yields, then (Fbr)max = s g Abr = 70(106)( p )(0.03)2 = 197 920.3 N (Pg)br = sg >E = 70.0(106) 101(109 ) = 0.6931(10 - 3) mm >mm dbr = (Pg)brL = 0.6931(10 - 3)(0.25) = 0.1733 mm 6 1 mm Thus only the brass is loaded. P = Fbr = 198 kN Ans. Ans: P = 198 kN 4–85. The cylinder CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistance. Also, the two end rods AB and EF are heated from T1 = 30°C to T2 = 50°C. At the lower temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the force in rods AB and EF caused by the increase in temperature. Rods AB and EF are made of steel, and each has a cross-sectional area of 125 mm2. CD is made of aluminum and has a crosssectional area of 375 mm2. Est = 200 GPa, Eal = 70 GPa, ast = 12(10-6) >°C, and aal = 23(10-6) >°C. 0.7 mm B F C – 240 mm D A 300 mm + E dst + (dT)st = (dT)al - dal - 0.0007 Fst (0.3) (125)(10 -6 )(200)(10 9) + 12(10 - 6)(50 - 30)(0.3 ) = 23(10 - 6)(180 - 30)(0.24) - Fal (0.24) 375(10 - 6 )(70)(10 9) - 0.000 7 12.0Fst + 9.14286Fal = 56000 (1) + c ΣFy = 0; (2) Fal - 2Fst = 0 Solving Eqs. (1) and (2) yields: FAB = FEF = Fst = 1.85 kN Ans. FCD = Fal = 3.70 kN Ans: FAB = FEF = 1.85 kN