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The assembly shown in Fig.3 consists of an aluminum tube AB having a cross- sectional area of 400 mm². A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod. Take E = 200 GPa, E = 70 GPa. al

The assembly shown in Fig.3 consists of an aluminum tube AB having a cross- sectional area of 400 mm². A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod. Take E = 200 GPa, E = 70 GPa. al

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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The assembly shown in Fig.3 consists of an aluminum tube AB having a cross- sectional area of 400 mm². A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod. Take E = 200 GPa, E = 70 GPa. al 400 mm 80 kN 600 mm Fig,3 Solutions • Find the displacement of end C with respect to end B. PL -80(10)(0.6) AE #(0.005)200(10°) +0.003056 m → Displacement of end B with respect to the fixed end A, -s0(10° (0.4) AE 400(10 70(10°)| PL :-0.001143 = 0.001143 m → Since both displacements are to the right, 8c = 8c +8c 3 =0.0042m=4.20 mm → 80 kN PA- 80 kN 80 kN Pac 80 kN (b)
Transcribed Image Text:The assembly shown in Fig.3 consists of an aluminum tube AB having a cross- sectional area of 400 mm². A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod. Take E = 200 GPa, E = 70 GPa. al 400 mm 80 kN 600 mm Fig,3 Solutions • Find the displacement of end C with respect to end B. PL -80(10)(0.6) AE #(0.005)200(10°) +0.003056 m → Displacement of end B with respect to the fixed end A, -s0(10° (0.4) AE 400(10 70(10°)| PL :-0.001143 = 0.001143 m → Since both displacements are to the right, 8c = 8c +8c 3 =0.0042m=4.20 mm → 80 kN PA- 80 kN 80 kN Pac 80 kN (b)
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