Torsion and the modulus of rigidity

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SUMMARY

The discussion centers on calculating the modulus of rigidity (G) for a material using a torsion testing machine. The specimen has a diameter of 12.5mm, a torque (T) of 4.5Nm, and an angle of twist of 4.0 degrees over a gauge length of 0.75m. The polar moment of inertia (J) is calculated using the formula J = π(d^4)/32, leading to a value of J = 3.8349 x 10^-8 m^4. The final modulus of rigidity is determined to be G = 126 x 10^7 Pa, after correcting the diameter conversion from 0.025m to 0.0125m.

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Homework Statement


A specimen tested in a simple torsion testing machine , the specimen was 12.5mm in diameter and the angles were read over a gauge length of 0.75m Determine the modulus of rigidity of the material taking the torque T as 4.5Nm and the angle of twist as 4.0 degrees

Homework Equations


T/J=τMax/R=Gθ\L

The Attempt at a Solution


J=∏(0.025)4/32
J=3.8349x10^-8

4.0degress=0.0698132

4.5/3.8349x10^-8=G(0.0698132)/0.75
=>G=126x10^7
Wondering if this looks correct ??
 
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Where does the factor (0.025)^4 come from?

Don't you want pi*r^4/2 for this?
 
Dr.D said:
Where does the factor (0.025)^4 come from?

Don't you want pi*r^4/2 for this?
So I have for the polar moment of inertia to be ∏d^4/32
I did just see however I wrote 0.025 instead of 0.0125 which is the 12.5mm in m
 

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