Calculate Maximum Torque

[febbie22]

A 2.4 m long structural aluminium pipe (series 6063-T6) is used as part of a roof structure. As part of the design process it is required to assess the suitability of different sizes of pipe.

Calculate the maximum torque that can be applied to the shaft if the maximum allowable yield stress and angle of twist are to be complied with.

Outside Diameter = 33.40mm Wall Thickness = 3.38mm Nominal Size = 25.40

Shear Modulus = 27GPA Max Allowed Angle 15 degrees Max Allowed Stress 68800000

Just wondering i need to calculate the Torque and i know the torque equation but do i just choose two out of the 3 sections ie T/J = Stress/r or do i need to use all three, as it says in the question the max stress and max angle which are in the other two sections of the equations.

when i use two i get 174.85NM which i thought was kind of low

Cheers

[Shaun_W]

Well obviously the values of the three sections should all equal each other, i.e T/J = τ/r = Gθ/L = C otherwise one of your values is wrong. If you know the angle, the length of the specimen and its shear modulus then T = JGθ/L.

[febbie22]

This is what i have done but the answer is still not the same

Od = 0.03340m
Id = 0.02664m
Shear modulus = 27GPa
yield stress = 172Mpa
Max allowable shear stress = 0.4 * yield stress
Max Angle 15 degrees
lentgh = 2.4m

T = (Stress/r)*J

(172MPa * 0.4 /0.0167)* (Pie/32*(0.03340^4-0.02664^4)

= 299.63 Nm


The other way using

T= ((G*angle)/L)*J

27GPa*(15*(Pie/180))/ 2.4 * 7.27x10^-8

= 214.12N,


whats going wrong i dont get it

[Shaun_W]

The maximum allowable stress is 0.4 x the yield strength. You're using the yield stress in your calculation rather than the maximum allowable stress.

[febbie22]

but it says in the handout that i got that the maximum allowabale shear stress is equal to

0.4 * the yield stress

ive included the handout in an attachment

thanks for the help by the way, but how do i find the shear stress then.

cheers

[Shaun_W]

but it says in the handout that i got that the maximum allowabale shear stress is equal to

0.4 * the yield stress

Yes, which is what I said.


ive included the handout in an attachment

thanks for the help by the way, but how do i find the shear stress then.

τ = r(T/J) = r(Gθ/L)


cheers

It appears we go to the same university.

[febbie22]

hey, thanks for replying

so do i even need to bother with the 0.4 x the yield

when i can just use the formula you gave to find the stress

[Shaun_W]

Obviously you need to bother with the 0.4*yield because that's the maximum allowable stress.