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A rigid horizontal bar of negligible mass is connected to two rods as shown. If the system is initially stress-free. Calculate the temperature change that will cause a tensile stress of 90 MPa in the brass rod Assume that both rods are subjected to the change in temperature Brass L=2m A = 1200 mm² E = 100 GPa a = 18.7 µm/(m °C) 3m Copper L=3m A-1500 mm E- 120 GPa a = 16.8 µm/(m*C)

A rigid horizontal bar of negligible mass is connected to two rods as shown. If the system is initially stress-free. Calculate the temperature change that will cause a tensile stress of 90 MPa in the brass rod Assume that both rods are subjected to the change in temperature Brass L=2m A = 1200 mm² E = 100 GPa a = 18.7 µm/(m °C) 3m Copper L=3m A-1500 mm E- 120 GPa a = 16.8 µm/(m*C)

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
Problem 1.1MA
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this is about thermal stress
5:05M. docs.google.com/draw Thermal Stress 4 Saved to Drive File Edit View Tools Help Last edit was made 3 days ago by Noel Amatong 6 View only. A rigid horizontal bar of negligible mass is connected to two rods as shown. If the system is initially stress-free. Calculate the temperature change that will cause a tensile stress of 90 MPa in the brass rod. Assume that both rods are subjected to the change in temperature Brass A = 1200 mm² E = 100 GPa a 18.7 µm/(m °C) 3 m ||| { | | 44% - Share Copper L=3m A 1500 mm² E-120 GPa a = 16.8 µm/(m*C) D Ⓒ e 9
Transcribed Image Text:5:05M. docs.google.com/draw Thermal Stress 4 Saved to Drive File Edit View Tools Help Last edit was made 3 days ago by Noel Amatong 6 View only. A rigid horizontal bar of negligible mass is connected to two rods as shown. If the system is initially stress-free. Calculate the temperature change that will cause a tensile stress of 90 MPa in the brass rod. Assume that both rods are subjected to the change in temperature Brass A = 1200 mm² E = 100 GPa a 18.7 µm/(m °C) 3 m ||| { | | 44% - Share Copper L=3m A 1500 mm² E-120 GPa a = 16.8 µm/(m*C) D Ⓒ e 9
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