HW7+Solutions - HOMEWORK PROBLEMS Week 7: Due Friday 28...

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HOMEWORK PROBLEMS Week 7: Due Friday 28 October 1. The assembly shown in Figure 1 comprises a hollow cylinder CD (seen in section) connected to a solid cylinder AB by a rigid disk at BC . The two cylinders are both made of steel for which E = 210 GPa and the hollow cylinder rests on a rigid support at D . If a force of 10 kN is applied at A , determine the average normal stress in AB and in CD and indicate whether these stresses are tensile or compressive. Determine also the distance that the end A moves downwards under the influence of the applied force. A B C D 20 4 60 120 70 10 kN hollow cylinder rigid disk solid cylinder Figure 1 Solution Clearly the cylinder AB is in tension and the tensile stress is σ AB = 10000 π × 20 2 / 4 = 31 . 83 N/mm 2 (MPa) . Figure 1.1 shows a free-body diagram of the body cut in the hollow cylinder CD , from which we conclude that F CD + 10 = 0 or F CD = - 10 kN . 1
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In other words, the hollow cylinder CD is in compression. Its cross-sectional area is A CD = π (78 2 - 70 2 ) 4 = 930 mm 2 and the stress is therefore σ CD = - 10000 930 = - 10 . 75 MPa . A B C 20 4 120 70 10 kN rigid disk F CD Figure 1.1 Measuring u upwards, we have u C = σ CD L CD E = - 10 . 75 × 60 210 × 10 3 = - 0 . 00307 mm and u B - u A = σ AB L AB E = 31 . 83 × 120 210 × 10 3 = 0 . 01819 mm . Since u B = u C , it then follows that u A = - 0 . 01819 - 0 . 00307 = - 0 . 02126 mm i.e. 0.02126 mm downwards. 2
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2. In Figure 2, the two horizontal bars ABC and DEF are rigid and originally horizontal and the links CE and DG are 1/4 inch diameter steel rods. All the pin joints are frictionless. If a force of 5000 lbs is applied at B , find the vertical displacement of the pins at C,E and D . ( E steel = 30 × 10 3 ksi). A B C D E F G 2 2 4 4 5.4 1.4 5000 lb all dimensions in feet Figure 2 Solution The rigid bar ABC is simply supported at its ends and loaded in the middle, so clearly the two reactions are both 2500 lb and the bar CD is loaded by a tensile force of 2500 lb. D E F 5.4 1.4 2500 lb F DG R 1 R 2 all dimensions in feet Figure 2.1 3
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DEF . Summing moments about F , we obtain 2500 × 1 . 4 - F DG × 6 . 8 = 0 so F DG = 514 . 7 lb . The rod DG is also in tension and its extension, which is also the downward displacement of the point D is given by δ DG = F DG L DG EA DG = 514 . 7 × 24 30 × 10 6 × π × (1 / 4) 2 / 4 = 0 . 0084 in . Notice that we have converted all these quantities to lbs and inches to avoid confusion over units. Since the rigid link FED pivots about F , the downward displacement at E is therefore given by u E = 0 . 0084 × 1 . 4 6 . 8 = 0 . 00173 in . Finally, for link CE , we have u C - u E = F CE L CE EA CE = 2500 × 24 30 × 10 6 × π × (1 / 4) 2 / 4 = 0 . 0407 in . It follows that the downward displacement of
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HW7+Solutions - HOMEWORK PROBLEMS Week 7: Due Friday 28...

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