What are the Material Properties of a Twisted Shaft and Tensile Test Results?

In summary: After calculating G and E, you can calculate v using the relation v= lateral strain/longditudinal strain.
  • #1
Malawaki
13
0

Homework Statement


A shaft of 0.3 metres long and 45mm diameter twists 2 degrees under a torque of 7kNm.
On a tensile test the same material extends 0.015mm on a length of 110mm. If the tensile force producing this extension was 500N and the diameter of the test piece was 5mm determine the values of :
G, E, v, K

Any help/ guidance would be greatly appreciated

Kindest Regards
 
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  • #2
Malawaki said:
Any help/ guidance would be greatly appreciated

Any attempts/elaborations/efforts from your side first would be appreciated even more. :wink:
 
  • #3
My first thoughts on the question were to figure out v then with v= lateral strain/longditudinal strain i can only figure out one of these strains i then got stuck as v is in the other formula. Apoint in the right direction would be helpful

Regards
 
  • #4
If the applied load in the first case causes torsion, then you can use the relation [tex]\phi = \frac{M L}{G I}[/tex], where I is the polar moment of inertia and I = [tex]\frac{d^4 \Pi}{32}[/tex].
 
  • #5
i take it that i re-arrange the formula to get G am I correct

Regards
 
  • #6
Malawaki said:
i take it that i re-arrange the formula to get G am I correct

Regards

Yes, you can calculate G, and I hope it works out.
 
  • #7
Am i correct in saying that M is the bending moment and in this case it is = to 500 Newtons
 
  • #8
Malawaki said:
Am i correct in saying that M is the bending moment and in this case it is = to 500 Newtons

No, it seems to equal 7 kNm.
 
  • #9
The answer i get is very small 4.22 x10-7

Is my working out right

7 x 0.32/ (0.045to the power 4 x pi / 32) x 2
 
  • #10
Malawaki said:
The answer i get is very small 4.22 x10-7

Is my working out right

7 x 0.32/ (0.045to the power 4 x pi / 32) x 2

Looks right, except the length of the shaft is 0.3 m
 
  • #11
what is 4.22 x 10 -7 measured in and what forumula do i use next

Regards
 
  • #12
Malawaki said:
what is 4.22 x 10 -7 measured in and what forumula do i use next

Regards

Well, how is the shear modulus G defined?

Next, you can calculate the module of elasticity E. Further on, there is a neat relation between G, E and v, do you know it?
 
  • #13
yes i do know what it is but the formual has two unknown values in it poisson's ratio and E. I also know that i have to calculate poissons ratio next and i know this is equal to lateral strain / longditudinal strain. But i don't know how to get these values from the question

Regards
 
  • #14
Malawaki said:
yes i do know what it is but the formual has two unknown values in it poisson's ratio and E. I also know that i have to calculate poissons ratio next and i know this is equal to lateral strain / longditudinal strain. But i don't know how to get these values from the question

Regards

You already calculated G. Now you can calculate E, since you know the force which caused the given extension at the tensile test. Then you can easily calculate v from the mentioned relation.
 
  • #15
I think i am doing something wrong i calculated G which i feel is right with other collegues of mine but am i right in saying E = stress / strain then i get a value of V = .765 which can not be right
 
  • #16
Did you turn the torque from [kNm] to [Nm]?
 
  • #17
I have solved that problem now calculating poissons ratio having difficulty transposing the formula to find v
 
  • #18
Thanks for all the help i solved it in the end

Many thanks
 

FAQ: What are the Material Properties of a Twisted Shaft and Tensile Test Results?

What is physics stress and why is it important?

Physics stress refers to the physical strain or pressure placed on a material or object. In physics, stress is a crucial concept as it helps us understand how forces and objects interact with each other. By studying stress, we can determine the strength, stability, and durability of various materials, which is essential in creating safe and efficient structures and machines.

How is stress different from strain?

While stress and strain are often used interchangeably, they are different concepts. Stress is the force applied to an object, while strain is the resulting deformation or change in shape of the object. In simpler terms, stress is the cause, and strain is the effect. They are related through the material's elasticity, which describes how much it can deform under stress before breaking.

What are the three types of stress?

The three types of stress are tensile, compressive, and shear stress. Tensile stress occurs when a material is stretched, compressive stress when it is compressed, and shear stress when it is twisted or sheared. These types of stress can act on an object simultaneously, and understanding their effects is crucial in designing and analyzing structures and machines.

How is stress calculated and measured?

Stress is calculated by dividing the applied force by the cross-sectional area of the object. It is measured in units of force per unit area, such as newtons per square meter (N/m²) or pounds per square inch (psi). The amount of stress a material can withstand before breaking is known as its ultimate strength. Stress can also be measured experimentally using instruments such as strain gauges and load cells.

How does stress affect the behavior of materials?

The behavior of materials under stress depends on their properties and the type of stress they experience. Some materials, such as rubber, can stretch significantly before breaking, while others, like ceramics, are more brittle and will break under very little stress. Stress can also cause materials to undergo permanent deformation, known as plastic deformation, or it can cause them to return to their original shape when the stress is removed, known as elastic deformation.

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