Mechanics assignment (please check if my answers are correct)

[AnkleBreakeer]

Homework Statement

As an engineer you are tasked to design a 2m long shaft which is to be manufactured either as a solid circular bar or circular tube. The shaft is required to transmit a torque of 1000Nm without exceeding the allowable strength of the material.
Address the following:
a) Material Selection and mechanical properties of material
b) Factor of safety used in the design
c) Required diameter of the solid shaft
d) Required outer diameter if the thickness of the tube is specified as one-eighth of the outer diameter
e) Angle of twist for the hollow shaft
After the shaft is manufactured, you are informed by the client that the shaft is subjected to an additional thrust force of 5000N:
f) Buckling load of the shaft
g) Provide a revised factor of safety to the client supported with appropriate calculations

Homework Equations

τ/r=T/J=(Gθ)/L
where τ is the shear stress
J=((pi)D^4)/32
Pcr=((pi)^2EI)/L^2
where Pcr is the buckling load

The Attempt at a Solution

a) I chose A36 steel
b) I chose standard factor safety 4
c) Diameter of the solid shaft=52mm
d) Outer Diameter of the hollow shaft=59mm
e) Angle of twist for the hollow shaft=0.0319 rad or 1.83 degrees
f) Buckling load for the hollow shaft=200.65kN
g) I wasn't able to find an answer for this sum. Hoping for some help to complete this question

I hope someone checks my answers to see if they are correct and if they are not, please tell my where I have gone wrong.

Thank you
(Sorry for my bad English)

 

[BvU]

Hello Breaker and welcome to PF. I notice you haven't had any help yet and maybe I can help you to increase your chances a little. Remember the majority of the PF population are simple physicists like me. Not dumb, but non-engineers. When I read your stuff my fingers itch to do the desired checking, but I lack the knowledge of what all these symbols stand for (because they are used for many purposes all over the place in physics). So I need help to help you.
For example: all variables and given/known data ? Maybe that's all, but then I'll come back to that under 3.

2. Homework Equations . A lot of variables that I can't find under 1. Like r, J, G, θ, D, E, I. I could guess but the chance of getting them all correct are small. And any of them missed means wasting our time. Dimensions are also a good thing to have at hand. When checking calculations, dimensions are extremely valuable in finding omissions/mistakes etc.

3. a) good choice. But do you think I can telepathically check this answer ? Let alone check that you arrived at this answer via the correct path and not because it's in the back of the book?
b) good choice too. 3.95 would probably not be enough. Furthermore see under a).
c) Should follow from a, b and some of the relevant equations. You've already done it, so show your work.

etc. etc. I turn in now, but perhaps someone else will jump on this if you help him/her a little. Engineers might even be able to do so without all these dumb questions. Physicists are curious folks, eager to learn and generally very good with math and mechanics. A few nudges and you have them at your fingertips! Good luck!

 

[PhanthomJay]

I frankly am not doing the math when you are using USA A36 steel but then calculating using SI which is frowned upon. But most of your equations look ok it is plug and chug once you calculate the allowable shear stress in A36 steel with your chosen safety factor. I don't see your calc for the tube steel however, you don't show how you arrived at J for a tube.
For buckling you have chosen k = 1, but end conditions are not given.
Safety factor for combined shear and axial stress will be reduced...some calculate combined stress using sq rt sum of squares , but there are other formulas dictated sometimes by Code.

 

[BvU]

This is fun. An engineer frowning on SI units. Don't listen to him/her when he/she talks about units: the scientific world and the civilized world are much better off with a decimal system and SI units.
One thing in his/her favour: he (she?) speaks the lingo and knows what he/sho is talking about. I would be interested, but I'm afraid this post won't make me anything the wiser.

By the way: once I have grandchildren I won't be able to change units any more either, so I don't blame him/her ;-) All in good spirit.

 

[PhanthomJay]

Would be nice if we could all use one system of measure but the USA is essentially the last holdout and like it or not, conversion might happen some day but not in our lifetime old timer. A36 steel has a minimum tensile yield strength of 36,000 psi. Perhaps that's about 200 000 000 Pascals. It's E modulus is 29 000 000 psi , or maybe 250 000 000 000 Pascals. Too many zeros , I might be off by a factor of 1000. That would be disastrous. We have a 'feel' for psi, and no sense for what a pascal is. And although I'm too old to convert, even the youngins fresh out of the university in civil structural engineering are taught SI only in passing...or else they'd never get a job in the States. That's the way it is. And will be.

 

[AnkleBreakeer]

My working

a) Steel A36 should be used as our lecturer said so.
b) I researched a little and found that 4 is a pretty safe estimate for the safety factor
c)
Factor of safety= (shear stress)/(maximum allowable stress)
4=(145*〖10〗^6)/(maximum allowable stress)
maximum allowable stress=(145*〖10〗^6)/4
maximum allowable stress=36.25*〖10〗^6
T/J=τ/r
1000/(((πD^4)/32) )=((36.25*〖10〗^6 ))/((D/2) )
32000/((pi)D^4 )=(72.5*〖10〗^6)/D
32000/(pi)(72.5*〖10〗^6 ) =D^3
D=∛(32000/π(72.5*〖10〗^6 ) )
D=0.052m
D=52.0mm
d)
Do – Di = 2/8Do
Where Do & Di are outer diameter and inner diameter, respectively
Above equation can be simplified to:
Di = 6/8Do
T/J=τ/r
J=((pi)(Do^4-Di^4))/32
J=((pi)(Do^4-(6/8)^4 Do^4))/32
J=(2.15Do^4)/32
1000/(((2.15Do^4)/32))=(36.25*〖10〗^6)/(Do/2)
32000/((2.15(72.5*〖10〗^6 )))=(Do^4)/Do
0.059m=Do
Do=59mm
e)
T/J=Gθ/L
θ=TL/GJ
J=((pi)((59/1000)^4-(44.25/1000)^4))/32=8.13*〖10〗^(-7)
θ=(1000*2)/((77.2*〖10〗^9 )(8.13*〖10〗^(-7)))=0.0319 radians=1.83°
f)
I= (pi)((59/1000)^4-(44.25/1000)^4 )/64=4.066*〖10〗^(-7)
Pcr=((pi)^2 EI)/L^2 Where PCR is the buckling load
Pcr=((pi)^2 (200*〖10〗^9 )(4.066*〖10〗^(-7) ))/4=200.65*〖10〗^3 N=200.65kN
In the Buckling load equation, we are said to assume (L/k)>100 So the buckling load equation is valid.
g)
I seriously need help with this question please help

 

[BvU]

Poor Breakie: urgent questions and here are these old farts starting an age-old argument over units systems.
Sorry to have triggered that. I willl embark on trying to understand what's going on, do some reading up and get back to you (even if I can't do anything substantial).
You will understand that my two bits are no use in a) (first law of school constitution) and b) (where you adapt to whatever is customary in the application; I can imagine that a jet turbine shaft differs from a low rpm water pump,but that's it. So up to you).

You've done work showing details, I'll honour that by going through - albeit slowly.

 

[PhanthomJay]

Your calcs look pretty good, although I haven't checked the numbers too closely. For the last part, because the member is subject to both shear and axial compressive stresses, your safety factor is now less than 4. Theoretical analysis of such members can get a bit complex, considering principal stresses from Mohr's circle, and buckling to boot. Although I've done many bolt designs subject to tension and shear (analysis using Code empirical formulas), I'll be honest I don't think I have ever looked at a member subject to significant torsion (shear stresses) and compression (axial normal stresses), because in my practice shear loads (in members, not bolts) are typically small when compared to bending and axial stresses, and often ignored. So faced with this problem, I would be conservative and check that (f_s/F_s + f_a/F_a) is less than 1, and the safety factor would be the inverse of that result. Here f_s and F_s are the actual and ultimate shear stresses, respectively, and f_a and F_a are the actual and allowable compressive stresses, respectively, computed per P/A. In your case, f_s/F_s is 1/4, and you'll have to compute f_a/F_a, which if your number are correct, for the solid shaft, is only about 5/200 or .025, so rather not significant here. IMHO.

Say BvU I am not an old fart ...show some respect to your elders...me and and Jack Nicholson have earned it.:tongue:

 

[BvU]

c)
Factor of safety= 4 = (shear stress of material)/(maximum allowable stress in application)
$\tau$ shear stress, N/m²
$\tau_{\rm actual,\ max} = \tau_{\rm material,\ max}/4$
$\tau_{\rm material,\ max}=145$ MPa (found where ? wiki yield strength 250 MPa)

T = torque, Nm. Given: T = 1000 Nm.
r = radius
D = Diameter; r=D/2
J = second moment of area, m⁴.
Solid cylinder: $J =\pi r^4/2 = \pi D^4/32$
Hollow cylinder: $J =\pi (r_o^4- r_i^4) /2 = \pi (D_o^4-D_i^4)/32$

$\tau = T\ r/J$, hence $J/r = T/\tau$
$J/r = \pi D^3/16 \ \Rightarrow D = \left (16/\pi\ J/r\right )^{1\over 3} \Rightarrow$
$D = \left (16/\pi\ T/\left(\tau_{\rm material,\ max}/4\right) \right ) ^{1\over 3} = 0.052$ m

d)
$r = r_o$ (is this OK? I can't judge that) and $D_i = D_o - 2 D_i/8\Rightarrow D_i = 3/4 \ D_o \Rightarrow$
$J/r_o = \pi D_o^4 (1-({3\over 4})^4)/(32 r_o) \Rightarrow D = \left ({16 \over \pi\ \left (1-\left ({3\over 4}\right)^4\right)} \ T/\left(\tau_{\rm material,\ max}/4 \right)\right) ^{1\over 3} = 0.059$ m

So far, so good.

 

[nvn]

AnkleBreakeer: Excellent work. All of your answers for items (a) through (e) are correct.

I did not check items (f) and (g). I do not know what you mean by "k" in your last line of item (f) in post 6. Your last sentence of item (f) in post 6 might be incorrect.

I am hesitant to tell you how to work the problem in items (f) and (g), because this assignment seems like it could be worth a large portion of your grade, or could even be a take-home exam. Nonetheless, you did excellent work in items (a) through (e), except for one thing, as follows.

(1) Always leave a space between a numeric value and its following unit symbol. E.g., 59 mm, not 59mm. See the international standard for writing units (ISO 31-0).

 

[AnkleBreakeer]

g) Factor of Safety= (Buckling stress)/(working stress)
Cross sectional area=(pi)〖((59/2000)〗^2-(44.25/2000)^2)
Cross sectional area=1.196*〖10〗^(-3) m^2
Buckling stress=(200.65*〖10〗^3)/(1.196*〖10〗^(-3) )=167.8*〖10〗^6 Pa
Working stress=5000/(1.196*〖10〗^(-3) )=4.18*〖10〗^6 Pa
Factor of Safety= (167.8*〖10〗^6)/(4.18*〖10〗^6 )=40.13
Revised factor of safety was obtained as 40.13 as shown in the above calculations.

 

[nvn]

f) ... In the buckling load equation, we are said to assume (L/k) > 100. So the buckling load equation is valid.

AnkleBreakeer: (f) I currently disagree. Instead, for steel, I currently think the formula you listed is valid if L/k > 125.7, not 100. Therefore, I currently think the buckling critical stress for your round tubular shaft is 156.9 MPa, not 167.8 MPa.

(g) Buckling currently does not appear to govern. Hint 1: von Mises stress is sigma_vm = (sigma_axial^2 + 3*tau^2)^0.5. Compute new factor of safety.

 

[BvU]

e)

$T/J=G\theta/L$ G Shear modulus, 77.3 GPa
(I only have Wiki, 79.4 GPa, from $G = {E\over 2 (1+\nu)}$ , with

E Young modulus, 200 GPa
$\nu$ Poisson's ratio, 0.26 )​

$θ=TL/GJ$ with $T/J = \tau_{\rm material,\ max}/4\ / \ (D_0/2)$
$\$
$θ= {145 {\ \rm MPa}/4 \ 2 {\ \rm m} \over 77.3 {\ \rm GPa} \ D_0/2 {\ \rm m}}$ $= {145 \over 77.3 \ 59} = 0.0318$

f)

${\rm P_{cr}} = {\pi^2 E\ I \over L^2}$
P_cr buckling load, N
I Area moment of inertia, m⁴.
$I_x = I_y = {\pi \over 4} (r_o^4 - r_i^4) = J/2$, $I_z = 2I_x$
I take it you want $I_x$ because you use the D⁴/64 ?

${\rm P_{cr}} = 2.01\ {10}^5$ N.

I don't see the L/k > 100 (I use k=1) but that's probably for L in other units ?

So e) and f) are same as yours as well.

g)

I see you have good help with g), so I'll leave you to it and peek at the outcome later.