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For the bar shown in Fig., calculate the reaction produced by the lower support on the bar. Take E = 200 GN/m2. Find also the stresses in the bars. |
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Answer» Let R1 = reaction at the upper support; R2 = reaction at the lower support when the bar touches it. If the bar MN finally rests on the lower support, we have R1 + R2 = 55kN = 55000 N For bar LM, the total force = R1 = 55000 – R2 (tensile) For bar MN, the total force = R2 (compressive) δL1 =extension of LM = [(55000 – R2) × 1.2]/[(110 × 10 – 6) × 200 × 109] δL2 = contraction of MN = [R2 × 2.4]/[(220 × 10–6) × 200 × 109] In order that N rests on the lower support, we have from compatibility equation δL1 – δL2 = 1.2/1000 = 0.0012 m Or, [(55000 – R2) × 1.2]/[(110 × 10–6) × 200 × 109] – [R2 × 2.4]/[(220 × 10–6) × 200 × 109] = 0.0012 on solving; R2 = 16500N or, 16.5 KN R1 = 55-16.5 = 38.5 KN Stress in LM = R1/A1 = 38.5/110 × 10–6 = 0.350 × 106 kN/m2 = 350 MN/m2 Stress in MN = R2/A2 = 16.5/220 × 10–6 = 0.075 × 106 kN/m2 = 75 MN/m2 |
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