Mechanics of materials torsion doubt

In summary: So if you have two different materials with different strengths then the torque you can apply is limited by the weakest material.In summary, the larger the difference in the material's strength, the smaller the torque that can be applied.
  • #1
thegreengineer
54
3
Moved from another forum, so homework template missing
Good evening. I've got stuck in this question from my book of Mechanics of Materials by Beer and Johnston about torsion.
The allowable shearing stress is 15 ksi in the 1.5-in.-diameter steel rod AB and 8 ksi in the 1.8-in.-diameter brass rod BC. Neglecting the effect of stress concentrations, determine the largest torque that can be applied at A.
This is the diagram of the problem
https://scontent-lax3-1.xx.fbcdn.net/v/t1.0-9/13327492_1734753976809139_591495701010647559_n.jpg?oh=c90b0b5caea0dff2cfa2524a864a764f&oe=57D2EC5B
And according to the section: answers to selected problems of the same book the answer is: T=9.16 kip*in.

So this is my attempt for the solution yet I know I'm not correct since the answer is different to the indicated in the book.
First of all what I did is that I know the formula for maximum shear stress in torsion is:
[itex]\tau_{max}=\frac{TC}{J}[\itex]
But I realized that I'm working with two different materials and not a single one so when I performed the operation for the Steel I got like almost 10 kip*in so I need to know how to calculate a torque when I have a rod of two distinct materials.
 
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  • #2
Excuse me for the failed LATEX writting but the formula goes:
tmax=TC/J
where tmax is the maximum shear stress, T is the torque, C is the radius of the rod and J is the polar moment of inertia
 
  • #3
MarcusAu314 said:
But I realized that I'm working with two different materials and not a single one so when I performed the operation for the Steel I got like almost 10 kip*in so I need to know how to calculate a torque when I have a rod of two distinct materials.
Calculate for both then whichever is lower is the limiting torque value and cannot be exceeded.
This can be shown by doing a free body diagram of each section of the shaft - one for the steel part, one for the brass.

It's analogous to tying ropes or strings together - you can only apply as much tension as the weakest rope can take. eg a bit of cotton thread can only support a kg or two, it doesn't matter if it's tied to a bit of fishing line or some carbon nanotube, the cotton still breaks at it's normal load.
 

FAQ: Mechanics of materials torsion doubt

What is torsion in mechanics of materials?

Torsion is a type of mechanical stress that occurs when a material is twisted or under a twisting force. It causes a shearing stress within the material, resulting in a rotational movement.

How is torsion different from other types of stress?

Torsion is different from other types of stress, such as tension or compression, because it involves a twisting force rather than a pulling or pushing force. It also causes a different type of deformation within the material.

What factors affect the torsional strength of a material?

The torsional strength of a material is affected by its shape, size, and material properties, such as elasticity and shear modulus. The geometry of the object being twisted also plays a role in its torsional strength.

What is the equation for calculating torsional stress?

The equation for calculating torsional stress is τ = Tc/J, where τ is the shear stress, T is the applied torque, c is the distance from the center of the object to the point of interest, and J is the polar moment of inertia of the object.

How can torsion failure be prevented in materials?

Torsion failure can be prevented by using materials with high torsional strength, avoiding sharp edges or corners where stress can concentrate, and designing structures with appropriate dimensions and shapes to distribute the torsional forces evenly.

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