Axially loaded structural member problem

In summary, the conversation is about determining the maximum axial force that can be applied to a structural member made of solid round bar steel without causing the axial stress to exceed 175MPa or the total elongation to exceed 0.14% of its length. The correct answer is 343.5 KN, but the method to obtain it is uncertain. The topic of stress and pressure is also mentioned, as well as the need to multiply stress by area and consider the amount of steel stretch per Pa.
  • #1
Pepsi24chevy
65
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Hey guys, I need setting up this problem. Here it is:
A structural member is fabricated from a solid round bar of steel with a diameter D=50mm. If the member is 6.0m long, determine the maximum axial force that can be applied if the axial stress is not to exceed 175MPa and the total elongation is not to exceed 0.14& of its length.

I know the answer is 343.5 KN but i am unsure of how to get it. Thanks
 
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  • #2
Stress or Pressure [Pa] is Force per Area [N/m^2]
Did you try multiplying the stress by the Area?

Do you know how much the steel stretches per Pa?
 
  • #3


To solve this problem, we need to use the formula for axial stress, which is given by:

σ = F/A

Where σ is the axial stress, F is the applied force, and A is the cross-sectional area of the structural member.

First, we need to calculate the cross-sectional area of the steel bar using its diameter:

A = πr^2 = π(25mm)^2 = 1963.5 mm^2

Next, we need to determine the maximum force that can be applied without exceeding the given stress limit of 175 MPa. This can be done by rearranging the formula:

F = σA = (175 MPa)(1963.5 mm^2) = 343.5 kN

This is the maximum axial force that can be applied to the structural member without exceeding the given stress limit.

To ensure that the total elongation does not exceed 0.14% of the length, we can use the formula for total elongation:

ε = ΔL/L

Where ε is the total elongation, ΔL is the change in length, and L is the original length of the member.

In this case, we know that the total elongation should not exceed 0.14% of 6.0m, which is equal to 0.0084m. We can rearrange the formula to solve for the maximum change in length:

ΔL = εL = (0.0084m)(6.0m) = 0.0504m

Therefore, the maximum change in length that the member can experience without exceeding the given limit is 0.0504m.

To summarize, the maximum axial force that can be applied to the structural member is 343.5 kN, and the maximum change in length that the member can experience is 0.0504m. These values ensure that the axial stress does not exceed 175 MPa and the total elongation does not exceed 0.14% of the length.
 
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