Mechanics of Materials - Torsion

In summary, torsion is a type of mechanical stress that occurs when an object is twisted or rotated around its longitudinal axis. This can cause deformation and failure in materials, especially when they are not designed to withstand torsional forces. The strength of a material's resistance to torsion is measured by its torsional modulus, which is influenced by factors such as the material's shape, size, and composition. Torsion can also be affected by external factors such as temperature and loading conditions. Understanding the mechanics of torsion is crucial in designing and engineering structures that can withstand these types of stresses.
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  • #2
FAB is the force acting at the radius of the shaft. You used its torque.

0.15 is in mm, SI formulae require metres.

Fix these and you'll get 56.6 MPa
 
  • #3
Your first relevant equation is not correct and your second seems to be an allowable shear stress which is not applicable in this problem.
For a circular member subject to torsion from 'twisting' moments about the long axis, you should look up the formula for max shear torsional stress. The formula you noted applies for average transverse shear streses parallel to the cross sectional area.
 
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  • #4
FAB is the force acting at the radius of the shaft. You used its torque.
Yes, and I believed I used the right value.
0.15 is in mm, SI formulae require metres.

Fix these and you'll get 56.6 MPa

I fixed it to "M" but I still get the same numbers just the decimal point shifting.

Your first relevant equation is not correct and your second seems to be an allowable shear stress which is not applicable in this problem.

Not correct by not being relevant or not correct by not correct? My teacher gave us those formulas.

For a circular member subject to torsion from 'twisting' moments about the long axis, you should look up the formula for max shear torsional stress. The formula you noted applies for average transverse shear streses parallel to the cross sectional area.

This formula?

http://img42.imageshack.us/img42/2406/etaz.jpg
 
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  • #5
Femme_physics said:
Not correct by not being relevant or not correct by not correct? My teacher gave us those formulas.
not relevant
This formula?

http://img42.imageshack.us/img42/2406/etaz.jpg
Yes...where Io is the polar moment of inertia about the longitudinal axis of the shaft.
 
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  • #6
Femme_physics said:
Yes, and I believed I used the right value.
You used the right value of torque, but you didn't convert it to its tangential force equivalent at the radial distance from the shaft's axis. The formula you have been given requires the tangential force applied at the shaft surface. Force = torque / radius
I fixed it to "M" but I still get the same numbers just the decimal point shifting.
I just rechecked, and I get book's answer.
Not correct by not being relevant or not correct by not correct? My teacher gave us those formulas.
I verified your teacher's formula before I replied the first time, to confirm it derives from the general formula. It's correct.
 
  • #7
Erm...
[tex]{T/r \over \pi r^2} = {300 Nm / 0.015 m \over \pi (0.015 m)^2} = 28.3 MPa \ne 56.6 MPa[/tex]
 
  • #8
I like Serena said:
Erm...
[tex]{T/r \over \pi r^2} = {300 Nm / 0.015 m \over \pi (0.015 m)^2} = 28.3 MPa \ne 56.6 MPa[/tex]
Ah, there's a 2 that has gone missing somewhere. The polar moment of a round shaft is Pi/2 times r^4. The teacher's formula seems to be out by a factor of two.

I guess this means no one (except me) will have got the right answer?
 
  • #9
NascentOxygen said:
Ah, there's a 2 that has gone missing somewhere. The polar moment of a round shaft is Pi/2 times r^4. The teacher's formula seems to be out by a factor of two.

I guess this means no one (except me) will have got the right answer?

I believe PhantomJay had it right when he said Fp needed another formula that included a polar moment, which she just gave.

The first formula in the OP is apparently for the shear tension, not the shear torsion.
 
  • #10
Determine maximum shear stress on shaft, [itex]\tau _{max}[/itex], from the general formula:

[itex]\frac T J = \frac {\tau_{max}} {r} \; \;\; where\;\; for\;\; a\;\; solid\;\; cylindrical\; shaft\;\;J = \frac {\mathrm{\pi}} {2}\, r^{4}[/itex]
 
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  • #11
I like Serena said:
Erm...
[tex]{T/r \over \pi r^2} = {300 Nm / 0.015 m \over \pi (0.015 m)^2} = 28.3 MPa \ne 56.6 MPa[/tex]

Is this the formula you need to use to solve it? Because I don't see it in my formula sheet.
 
  • #12
Femme_physics said:
Is this the formula you need to use to solve it? Because I don't see it in my formula sheet.

No, this is not the formula you need to solve it.

It was derived from your formula ##\tau_s={F_{max} \over A_s}## and I showed it was wrong.
You cannot use that formula here, since it applies to linear shear (by a regular force) and not to torsional shear (by rotational torque).The formula you should use is the one given by NascentOxygen in his last post.
This is effectively the same as the formula you showed in your answer to PhantomJay, except that you did not specify what ##I_0## was.
 
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  • #13
I like Serena said:
This is effectively the same as the formula you showed in your answer to PhantomJay, except that you did not specify what ##I_0## was.


Moment of inertia of a non-hollow cylinder.

http://img444.imageshack.us/img444/9633/mm2vh.jpg


But I don't have the mass in the question, so am not sure how am I supposed to use it.

No, this is not the formula you need to solve it.

It was derived from your formula ##\tau_s={F_{max} \over A_s}## and I showed it was wrong.
You cannot use that formula here, since it applies to linear shear (by a regular force) and not to torsional shear (by rotational torque).


The formula you should use is the one given by NascentOxygen in his last post.

Still quite off

http://img404.imageshack.us/img404/2576/69366748.jpg
 
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  • #14
Femme_physics said:
Moment of inertia of a non-hollow cylinder.

But I don't have the mass in the question, so am not sure how am I supposed to use it.

Are you sure it is the moment of inertia?
Isn't it the "polar" moment of inertia?



Femme_physics said:
Still quite off

http://img404.imageshack.us/img404/2576/69366748.jpg

Suppose you moved the decimal point 6 places to the left? :wink:
 
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  • #15
Are you sure it is the moment of inertia?
Isn't it the "polar" moment of inertia?

Oh yes, it's polar. *smacks forehead*

Ow...been a while since I smacked my forehead.

http://img443.imageshack.us/img443/734/offufs.jpg

Still off...darn it
Suppose you moved the decimal point 6 places to the left?
I didn't move anything. That's the result.
 
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  • #16
Femme_physics said:
Oh yes, it's polar. *smacks forehead*

Ow...been a while since I smacked my forehead.

:smile:


Femme_physics said:
((WAITT! Redoing exercise))
Damn it! Still wrong. This one is wayyyyyyy off.

I didn't move anything. That's the result.

In your result you have the unit [MPa], but it should be [Pa].
You still need to convert to [MPa], that is, move the decimal point 6 places over.
 
  • #17
In your result you have the unit [MPa], but it should be [Pa].
You still need to convert to [MPa], that is, move the decimal point 6 places over.

How do I make sure to always get the result in MPa? Do I use mm instead of m? BTW I edited my post above.
 
  • #18
Femme_physics said:
How do I make sure to always get the result in MPa? Do I use mm instead of m? BTW I edited my post above.

Yep. That should do the trick.
In your post above, you should use mm.
 
  • #20
Femme_physics said:
Alright, did it! Kinda rushed the diagrams though...but just for the record

Yep. You've got it down!

http://paniagua.es/wp-content/uploads/2011/03/smileygafas-115x115.png
 
  • #21
a w00t w00t! Femme-Fizzics in the hizzle yo!

Thanks ILS :)
 
  • #22
a w00t w00t! Femme-Fizzics in the hizzle yo!
My hizzle is the shizzle, y'all!
Party up in my hizzle!
Wanna go see my pimping new hizzle?
 

Related to Mechanics of Materials - Torsion

1. What is torsion and how does it relate to mechanics of materials?

Torsion is a type of stress that occurs when a material is twisted. In mechanics of materials, torsion is important because it can cause structural components to fail or deform. It is typically studied in relation to the strength and behavior of materials under different levels of torque or twisting forces.

2. What are the factors that affect torsion in materials?

There are several factors that can affect torsion in materials, including the material's properties (such as elasticity and yield strength), the geometry of the material (such as its cross-sectional shape and size), and the applied torque or twisting force.

3. How is torsion measured and quantified in mechanics of materials?

Torsion is typically measured and quantified through the use of shear stress and shear strain. Shear stress is the force per unit area that occurs when a material is twisted, while shear strain is the amount of deformation or change in shape that occurs as a result of this twisting force. These factors are used to calculate the torsional stress and strain in a material.

4. What are some common applications of torsion in the real world?

Torsion is a common phenomenon in many structures and systems in the real world. Some examples include the twisting of helicopter rotor blades, the torsion of a golf club during a swing, and the torsion of a car's driveshaft. It is also an important consideration in the design of bridges, buildings, and other structural components.

5. How does torsion differ from other types of stress in mechanics of materials?

Torsion is different from other types of stress, such as tension and compression, because it involves a twisting or rotational force rather than a straight or linear force. This can result in different types of deformation and failure in materials. Additionally, torsion is typically studied in relation to circular or cylindrical objects, while other types of stress may be studied in relation to different shapes and geometries.

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