# Torsional Strain

#### Jim_Earle

hey everyone I am stuck on a problem and would appreciate any guidance you could give me.

here is the question:

Much as in exercise 8.1, the end of a .75-in-diameter rod is fixed to an immovable plate. the length of the rod is 18.0-in. A 50-lb weight is attached to a 16.0-in lever at the opposite end of the rod so the weight is hanging perpendicular to the longitudinal axis of the rod.
Reference problem 8.1 is this:

The end of a .75-in-diameter rod is fixed to an immovable plate as shown. The length of the rod is 18.0 in. A 50-lb weight is attached to a 16.0-in lever at the opposite end of the rod so that the weight is hanging perpendicular to the longitudinal axis of the rod, For steel, E=29,000,000 lb/in² and μ=.303. The poplar moment of inertia for a solid circular shaft may be found by J=πD⁴/32, where D is the shaft diameter.

A. Determine the minimum diameter ro so that the maximum torsional stress does not exceed 8,000 lb/in². (I have solved this problem)

To solve this I did the following:
Tmax= 8,000 lb/in²
T=FL = 50 lb * 16.0 in = 800 in-lb
Tmax solid = TR/J where J=πD⁴ / 32 and R=D/2 =
T(D/2) / πD⁴ / 32 so Tmax= 16T/π³ I took Tmax to both sides and D³ to get my final formula of
D³=16T/ π (Tmax solid) =
D³= 16(800 in-lb) / π (8,000 lb-in²) = .5092958179 in then took the invert by ∛(.5092958179 in) =
.7985890849 in thus D= .7985890849
To check I used 16T/ πD³ = 16 (800 in-lb) / π(.7985890849 in³) = 8,000.000 lb-in²

B. Determine the minimum diameter rod so that the maximum torsional stress does not exceed 8,000 lb/in². (this is the one I am stuck on)
I think I need G which I used E/2(1+μ) = 29,000,000 lb/in² / 2(1 +.303) = 11,128,165.77 lb/in²

please note π is meant to represent pi

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#### rock.freak667

Homework Helper
From what I understand you have a shaft, and attached to its free end is an eccentric load of 50lb.

So what you have is a case of both bending and torsion.

You can replace the 50lb acting at the 16in by a single 50lb acting at the free end of the shaft+ a torque of (50*16)lb*in.

So you know that

shear stress τ =16T/πD³

Now you just need to get bending stress using σbend=My/I.

Then combine using the principle stress equation and the shear stress theory.

#### Jim_Earle

attached is a diagram.

It looks like I miss typed my problem for 'B' it should read this.

Determine the minimum diameter rod so that torsional strain does not exceed 1.00 degrees.

Sorry about that. for some reason a and b were the same. I solved a already but am stuck on B

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#### rock.freak667

Homework Helper
You should know the equation

T/J = τ/r = Gθ/L

But in your first part, you did not account for th bending.

Due to the load being at the distance of 16 inches, you will have both bending and torsion happening. You only accounted for a case of only torsion.

Last edited:

#### Jim_Earle

Ok that confuses me even more. The question only asks for the diameter size so torsional stress does not exceed 1 degree. My book does not say to calculate bending in any of the examples and says to find the angle I need to use TL / JG and the result is in radians then I have to multiply that by 57.30 to get my degrees.

Could you possible elaborate more?

#### rock.freak667

Homework Helper
Ok that confuses me even more. The question only asks for the diameter size so torsional stress does not exceed 1 degree. My book does not say to calculate bending in any of the examples and says to find the angle I need to use TL / JG and the result is in radians then I have to multiply that by 57.30 to get my degrees.

Could you possible elaborate more?
For the first part, the eccentric loading, will cause bending + torsion.

For the second part, the torsion alone will produce the rotation, which you do not want to exceed 1 degree.

In all cases, your torque is T=50lb*16in = 800 lb-in.

For the second one:

You have T/J = τ/r = Gθ/L

you found 'G' and you have 'L' and you want θ=1° = ___ radians.

so you want T/J = Gθ/L and you have J=πD4/32, so you can find 'D'.

For the first one: You showed that τ=16T/πD3. This stress occurring due to torsion.

Due to bending you will have σ=My/I

y=D/2 and you can get I in terms of 'D'.

The 50lb load will be acting at a distance of 18in, so you can get M.

Do you know the principle stress equation?

#### Jim_Earle

For the first part, the eccentric loading, will cause bending + torsion.

For the second part, the torsion alone will produce the rotation, which you do not want to exceed 1 degree.

In all cases, your torque is T=50lb*16in = 800 lb-in.

For the second one:

You have T/J = τ/r = Gθ/L

you found 'G' and you have 'L' and you want θ=1° = ___ radians.

so you want T/J = Gθ/L and you have J=πD4/32, so you can find 'D'.

For the first one: You showed that τ=16T/πD3. This stress occurring due to torsion.

Due to bending you will have σ=My/I

y=D/2 and you can get I in terms of 'D'.

The 50lb load will be acting at a distance of 18in, so you can get M.

Do you know the principle stress equation?
No I don't know the principal stress equation?

I guess I am confused by what you mean 'first part' are you referring to part a? did I get that incorrect? When I back check it it comes out perfect.

whats confusing me is the addition to the concept of bending into the mix when the problem does not ask for it or contain it.

I know 1° = .0174532925 radians. so if I use J = TL/Gθ I get .0741415291

and If I back check that by using θ=TL/JG I get .0174532925

but with all this I am not sure how I am closer to getting what the minimum diameter is lol. I think my brain is getting fried. This has been ongoing since yesterday afternoon.

This is what I know as given thus far

T = 800 lb-in
L= 18 in
E = 29,000,000 lb/in²
G = 11,128,165.77 lb/in²
μ = .303
weight = 50lb attached to a 16 in handle

I think J= .0741415291

Am I even close?

#### rock.freak667

Homework Helper
No I don't know the principal stress equation?

I guess I am confused by what you mean 'first part' are you referring to part a? did I get that incorrect? When I back check it it comes out perfect.

whats confusing me is the addition to the concept of bending into the mix when the problem does not ask for it or contain it.
Well from your diagram, the free end is not supported, and how the shaft is configured, you would have bending and torsion. But your analysis seems to assume that you only have torsion. But since you said it is correct, I will assume it is correct.

I know 1° = .0174532925 radians. so if I use J = TL/Gθ I get .0741415291

and If I back check that by using θ=TL/JG I get .0174532925

but with all this I am not sure how I am closer to getting what the minimum diameter is lol. I think my brain is getting fried. This has been ongoing since yesterday afternoon.

This is what I know as given thus far

T = 800 lb-in
L= 18 in
E = 29,000,000 lb/in²
G = 11,128,165.77 lb/in²
μ = .303
weight = 50lb attached to a 16 in handle

I think J= .0741415291

Am I even close?
Right so J= 0.0741415291 in4 and you know that J= πD4/32. So you can get 'D'.

#### Jim_Earle

i see what your saying about the bending now and I do agree with you. I think this book just brushed it off though, I think at least lol.

so I am assuming I got J correct then hey.

so if J=πD⁴ / 32

my formula should be D⁴= πJ/32 right? I am not sure ow to do D⁴

#### rock.freak667

Homework Helper
i see what your saying about the bending now and I do agree with you. I think this book just brushed it off though, I think at least lol.

so I am assuming I got J correct then hey.

so if J=πD⁴ / 32

my formula should be D⁴= πJ/32 right? I am not sure ow to do D⁴
When you get D4= some number, just put D=(some number)0.25 on your calculator.

#### Jim_Earle

ok I am still having some trouble. nothing is checking out correctly.

if I use D⁴ = πJ / 32 I get .0072788276 then I take that to the power of .25 and get .2920891052

My issue is this. if I assume that is D and plug it into J=πD⁴ / 32 I should get my original J value of .0741415291 but I don't. I get 7.145972285 E -4

in my original problem the shaft diameter was .75 in and the degrees it turned was 2. something. Thus I know the shaft has to be bigger than .2920891052. This just doesn't make sense. Do you see where I am going wrong? or is my math wrong?

#### rock.freak667

Homework Helper
ok I am still having some trouble. nothing is checking out correctly.

if I use D⁴ = πJ / 32 I get .0072788276 then I take that to the power of .25 and get .2920891052

My issue is this. if I assume that is D and plug it into J=πD⁴ / 32 I should get my original J value of .0741415291 but I don't. I get 7.145972285 E -4

in my original problem the shaft diameter was .75 in and the degrees it turned was 2. something. Thus I know the shaft has to be bigger than .2920891052. This just doesn't make sense. Do you see where I am going wrong? or is my math wrong?
You have J=πD4/32 which means that D4 = 32J/π

#### Jim_Earle

θ=TL / JG, J = TL/Gθ = 800 in-lb * 18 in / .0741415291 in * 11,128165.77 lb/ in² =
.0174532925 rad, 1 rad = 57.30° = .0174532925 * 57.30 = 1.0°so it checks out

J = πD⁴ / 32, 32J = πD⁴ = 32(.071415291 in) = 2.285289312 in.

D⁴ = 2.285289312 in / π = .7274301808 in, D⁴ = .7274301808, .7274301808) ^(.25) or ∜(.7274301808) = .9235232405 in

Check ---- J = πD⁴ / 32 = π(.9235232405 in⁴ / 32 = .0741415291 in so that checks out also

Thus my diameter D=.9235232405 in

Thanks so much for all of your help I really appreciate it :)

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