Calculating change in length under max load

In summary, a student is designing a suspension pull rod for a car and needs to determine the minimum tube diameter that can be used to withstand a maximum load of 2500N. The maximum allowed normal stress in the tube is 100 MPa and the available tube diameters are 10mm, 8mm, and 6mm. They also need to calculate the change in length under max load conditions and determine the minimum bolt size for the wishbone rocker. Equations for stress, strain, and change in length can be used to solve these problems.
  • #1
felixj500
5
0
Please help, i can't answer part B, but think i am using the right equation! Thanks

A student is designing a suspension pull rod for a car. The rod consists of a central tube
section that has a wall thickness of 1.0 mm. The following tube diameters are available: 10mm,
8 mm and 6 mm and the maxmimum allowed normal stress in the tube is 100 MPa. If the pull rod
is designed to withstand a maximum load of 2500N what is the minimum tube diameter that
can be used?

2b) Calculate the change in length for the chosen tube and length 400 mm under max load
conditions, E = 210 GPa.

2c) Each end of the pull rod is to be mounted on a wishbone rocker by using a steel bolt in double
shear. Using the design load of 2500 N and assuming a factor of safety of 1.5, what is the
minimum bolt size that can be used if the shear stress allowed in the bolt is 150 MPa.
Assume bolt sizes available are 4 mm, 5 mm, 6 mm and 8 mm.
 
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  • #2
If you know the max stress is 100Mpa (100x10^6 N /m^2) and you know the load is
2500N then you can work out the min cross sectional area of material required.
With the given diameters and wall thicknesses it should be possible to select what will do the job. I have not calculated anything yet but I will have a go.
The change in length should be a straight forward calculation because you are given Young's modulus = 210 GPa
Youngs modulus = stress/strain and strain = extension/original length
Hope this helps
 
  • #3
Are you using these equations for part b

E=σ/ε where σ is your stress and ε is your strain. You would use this equation to calculate your strain.

For change in length you would use this equation

ε=ΔL/L where ΔL is the change in length and L is the original length.

These equations should solve your problem.
 

1. What is the purpose of calculating change in length under max load?

The purpose of calculating change in length under max load is to determine the amount of deformation or elongation that a material experiences when it is subjected to a maximum load or stress. This information is important in understanding the behavior and strength of the material.

2. How is change in length under max load calculated?

Change in length under max load is calculated by subtracting the original length of the material from its final length after being subjected to the maximum load. This difference is then divided by the original length and multiplied by 100 to get the percentage change.

3. What factors can affect the change in length under max load?

The change in length under max load can be affected by various factors such as the type of material, its composition and structure, the magnitude and duration of the load, and the environmental conditions (e.g. temperature, humidity).

4. Why is it important to calculate change in length under max load?

Calculating change in length under max load is important because it provides valuable information about the strength and behavior of a material. It can help in selecting the right material for a specific application, designing structures and machines, and ensuring safety and reliability in various industries.

5. Is there a limit to the amount of change in length a material can experience under max load?

Yes, there is a limit to the amount of change in length a material can experience under max load. This is known as the elastic limit or yield point, beyond which the material will undergo permanent deformation or failure. It is important to stay within this limit to ensure the material's structural integrity and prevent potential hazards or accidents.

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