To identify situations that can be modeled as an axially loaded bar and calculat

Question

To identify situations that can be modeled as an axially loaded bar and calculate the corresponding average normal stress.

In many applications, a solid bar is used to support a force along its length. This type of bar is called an axially loaded bar. When the bar is prismatic, homogenous, and isotropic and the applied load acts through the centroid of the cross section, the regions of the bar that are not near the applied loads will deform uniformly. Many common engineering materials, such as steel, aluminum, and other metals can be treated as homogenous and isotropic. Other materials like wood and composites are not necessarily homogenous or isotropic. In those cases, the deformation may not be uniform and more detailed methods of analysis are required. A section is "near" an applied load when it is closer to the load than the largest dimension of the cross section of the member.

For the following questions, consider the metal axial member shown in the figure below. Assume that applied loads are such that they act axially at the centroid of the cross section (i.e., the total axial load acting at the centroid at section A would be 2F2 and the total axial load acting at the centroid at section C would be 2F3).

Part B - The internal reaction at B

Assume that the axial member is in equilibrium and the applied forces are an unknown load F1, F2 = 13 kN , F3 = 12 kN , and F4 = 8 kN . Calculate the internal resultant normal force at section B( in kN)

Answer is not 13, or -13

Part C - The internal reaction at D

Assume that the axial member is in equilibrium and the applied forces are an unknown load F1, F2 = 13 kN , F3 = 12 kN , and F4 = 8 kN . Calculate the internal resultant normal force at section D.

Answer is not 2, 5, 9 kN

Part D - The average normal stress

Suppose the axial member has a circular cross section with a diameter of dB = 12 cm at section B and a diameter of dD = 6 cm at section D. What is the average normal stress in the section with the maximum magnitude stress?

Answer is not 353.677 kN

Explanation / Answer

Part B -

The axial member to be at equilibrium

F1 - 2 x F2 + 2 F3 - F4 = 0

13-2x13+2x12-8=0 ......However, this equation is not satisfying. Hence the member is not in equilibrium.

Part C -

The axial member to be at equilibrium

F1 - 2 x F2 + 2 F3 - F4 = 0

13-2x13+2x12-8=0 ......However, this equation is not satisfying. Hence the member is not in equilibrium.

Part D -

The axial member to be at equilibrium

F1 - 2 x F2 + 2 F3 - F4 = 0

13-2x13+2x12-8=0 ......However, this equation is not satisfying. Hence the member is not in equilibrium.

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