Torsion, angle of twist, diameter, load, momentum and find G

In summary: Then, which how do you find θ at the right end of the shaft given that:dθ / dx = T(x) / [G * Ip]and θ = 0 at the fixed end?Does this look right now?You say that T(x) = M x / L, but you only use M x in your integral expressionAlso, G has units, which you have neglected to show.I added L to my formula but the G stays same, which is is 2,5 GPa.
  • #1
roflpask
6
0
http://www.upload.ee/image/4760492/DSC_0002.JPG

Homework Statement


I have given angle of twist, diameter, uniform load, and length of the beam. I need to find the G.The load gives me some trouble and i don't know if i should solve it by using integration or? :)

Homework Equations

The Attempt at a Solution

 
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  • #2
roflpask said:

Homework Statement


I have given angle of twist, diameter, uniform load, and length of the beam. I need to find the G.The load gives me some trouble and i don't know if i should solve it by using integration or? :)

Why do you need to integrate anything?

1. Is the shaft prismatic along its entire length?

2. Does the polar moment of inertia change w.r.t. the length of the shaft?

3. Are the material properties of the shaft the same along its length?

4. Is the torque distributed evenly w.r.t. the length of the shaft?

If you can answer "Yes" to questions 1-4, integration is not necessary. (BTW, you haven't indicated if this shaft is fixed at one end in your diagram.)

Why wouldn't you check the formula for the angular deflection of this shaft given the known quantities and see if a reasonable value of G is obtained?

Remember, you should always draw a FBD to start your analysis.
 
  • #3
Thank you for your response. 1,3,4 are yes (2 dunno). I did a new drawing and wrote all the details i was given. The shaft is fixed at the end. At first i used this formula Θ=TL/GIp but teacher told me it is wrong, because i have distributed torque. After that he asked if i even know how should a stress diagram look like and i drew the same diagram as it is on the first picture.
 

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  • #4
roflpask said:
Thank you for your response. 1,3,4 are yes (2 dunno).

Well, let's start with answering question 2. On what physical property of the shaft does the value of Ip depend? Is it the length? Is it the diameter? Is it something else?

Then ask yourself, does this physical property of the shaft change with length? If it doesn't, then the value of Ip stays constant.

I did a new drawing and wrote all the details i was given. The shaft is fixed at the end. At first i used this formula Θ=TL/GIp but teacher told me it is wrong, because i have distributed torque.

Let's examine your teacher's comment.

So, dθ / dx = T(x) / [G * Ip], and you want to calculate G. How do you calculate θ given dθ / dx ? Where would the value of θ = 20 mRad be obtained on this shaft?
 
  • #5
The value of θ = 20 mRad would be obtained on the right end of the shaft.
 
  • #6
roflpask said:
The value of θ = 20 mRad would be obtained on the right end of the shaft.

Then, which how do you find θ at the right end of the shaft given that:

dθ / dx = T(x) / [G * Ip]

and θ = 0 at the fixed end?
 
  • #7
Does this look right now?
 

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  • #8
You say that T(x) = M x / L, but you only use M x in your integral expression

Also, G has units, which you have neglected to show.
 
  • #9
I added L to my formula but the G stays same, which is is 2,5 GPa
 

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Related to Torsion, angle of twist, diameter, load, momentum and find G

What is torsion?

Torsion is the twisting or rotation of a material due to an applied torque or moment. This can cause the material to deform or break, depending on its properties.

What is the angle of twist?

The angle of twist is the amount of rotation or angular displacement that occurs in a material when subjected to torsion. It is measured in radians and can be calculated by dividing the applied torque by the product of the material's shear modulus and the polar moment of inertia.

How does diameter affect torsion?

The diameter of a material can significantly impact its torsional strength. A larger diameter results in a higher torsional strength, meaning the material can withstand more torque before deforming or breaking. This is because a larger diameter provides more material to resist the twisting force.

What is the relationship between load and torsion?

Load and torsion are directly related, as the amount of load or force applied to a material can cause it to twist or rotate. The higher the load, the greater the torsional stress on the material, which can lead to deformation or failure.

How do you find G (shear modulus) in relation to torsion?

To find G, the shear modulus, in relation to torsion, you can use the equation G = T/(θ*r), where T is the applied torque, θ is the angle of twist, and r is the radius of the material. This equation can be rearranged to solve for any of these variables if the others are known.

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