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The square bar shown (Figure 1) is 70 mm thick and 5.4 m long and is fixed supported at both ends. A load directed leftward applied at point C, as shown, L2 = 3.3 m from the left end. The modulus of elasticity is E = 100 GPa .lf point C moves & = 0.13 mm to the left, what is the applied force?

The square bar shown (Figure 1) is 70 mm thick and 5.4 m long and is fixed supported at both ends. A load directed leftward applied at point C, as shown, L2 = 3.3 m from the left end. The modulus of elasticity is E = 100 GPa .lf point C moves & = 0.13 mm to the left, what is the applied force?

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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Learning Goal: To solve for forces in statically indeterminate bars with axial loads. When the number of reaction forces is greater than the number of equilibrium equations, the system is statically indeterminate. Solving for the reactions requires some additional equations. These additional equations come from considering compatibility relationships (i.e., continuity of displacements and relationships between displacements and loads). For an axially loaded member, the compatibility relationship for the deflections can be written by setting the total relative axial displacement between the ends of the member to a known value. The load- NL gives another AE displacement relationship 8 =E equation for the deflections. Once the internal normal force of each segment is written in terms of the end reactions and applied loads, there is enough information to solve for the reactions. Figure 1 of 2 > A B F +
Transcribed Image Text:Learning Goal: To solve for forces in statically indeterminate bars with axial loads. When the number of reaction forces is greater than the number of equilibrium equations, the system is statically indeterminate. Solving for the reactions requires some additional equations. These additional equations come from considering compatibility relationships (i.e., continuity of displacements and relationships between displacements and loads). For an axially loaded member, the compatibility relationship for the deflections can be written by setting the total relative axial displacement between the ends of the member to a known value. The load- NL gives another AE displacement relationship 8 =E equation for the deflections. Once the internal normal force of each segment is written in terms of the end reactions and applied loads, there is enough information to solve for the reactions. Figure 1 of 2 > A B F +
The square bar shown (Figure 1) is 70 mm thick and 5.4 m long and is fixed supported at both ends. A load directed leftward is applied at point C, as shown, L2 = 3.3 m from the left end. The modulus of elasticity is E = 100 GPa lf point C moves 8 = 0.13 mm to the left, what is the applied force?
Transcribed Image Text:The square bar shown (Figure 1) is 70 mm thick and 5.4 m long and is fixed supported at both ends. A load directed leftward is applied at point C, as shown, L2 = 3.3 m from the left end. The modulus of elasticity is E = 100 GPa lf point C moves 8 = 0.13 mm to the left, what is the applied force?
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