Working out maximum stress and Factor of safety

[donniemateno]

ladies and gents

I have been set a question for homework which is one where if you get one part wrong then the rest of your answers are in correct. It is a 3 questions in one question and i have answered 1 and 2 but can't seem to do part 3

the question is

A torque T = 75Nm is applied to a thin-walled steel tube having a rectengular cross-section of 12mm by 72mm . the tube has a constant wall thickness of 4mm . calculate shear stress at the wall and the factor of safety using tresca theory assuming the yield strength is 180 MPa. using the table below ( I've attached as a jpeg) work out the maximal stress and calculate its value.

i have got :

used the formula t(shear)=T/(2tAm)

so 75/2*(0.04)*(12*72) = 10.9 MPa or rounded up to nearest whole 11 Mpa

Tresca theory

first work out hoop stress which is 11mpa * (0.72*0.012/0.004) = 2376 n/m^2 ( seems really high) divide my 180 mpa by this i get a fos = 0.76 so a fail?

part 3 of the question

the formula to use is tmax = T/(k1bt^2)

where b is the longer side of the strip
t is the shorter side thickness of tube
k1 is the empircal constant depending on the ratio of b/t and is obtained using the table ( attached as jpeg)

so bt on mine is ( this is a guess) by multiplying the 72mm by the 4mm and using the number on the table closest to mine? i got 288 so i have used 4 = b/t and k1 = 0.282

putting the numbers into the equation 75/(0.282*4^2) = 16.62Mpa

 

[nvn]

donniemateno: Watch your units. Some of your units are wrong. Let me show you an example, using correct units.

Am = (12 - 4)(72 - 4) = 544 mm^2
tau = T/(2*Am*t) = (75 000 N*mm)/(2*544*4) = 17.23 MPa​

Try again. Also, what is the general formula for Tresca theory? I thought it was tensile yield strength divided by 2, if I recall correctly.

In part 3, why are you using a solid cross section formula for a rectangular tube? It does not make sense to me, so far.

Can you post the given question, so we can see if you are doing it right?

By the way, always leave a space between a numeric value and its following unit symbol. E.g., 72 mm, not 72mm. See the international standard for writing units[/color] (ISO 31-0[/color]).

 

[donniemateno]

Watch your units. Some of your units are wrong. Let me show you an example, using correct units.
Am = (12 - 4)(72 - 4) = 544 mm^2
tau = T/(2*Am*t) = (75 000 N*mm)/(2*544*4) = 17.23 MPa
Try again. Also, what is the general formula for Tresca theory? I thought it was tensile yield strength divided by 2, if I recall correctly.

In part 3, why are you using a solid cross section formula for a rectangular tube? It does not make sense to me, so far.

So using your method I would get for part a

Am = (12-4)(72-4) = 544mm^2 i would then use tau = T/(2*Am*t) to get (75 000 N*mm)/(2*544*4) = 17.23 MPa?

the first part where you take 12-4 and 72 - 4 is this taking the wall thickness from both sides? i would of thought my Am would be (12*72-4)

I have attached the official question via a jpeg by the tresca formula I have been taught is you would be given a yield stregth in my case 180mpa this would then be divided by your hoop stress.

 

[nvn]

donniemateno: In post 1, you did not tell us about one of the sentences in part a. The correct answer is now Am = 12*72 = 864 mm^2, and tau = T/(2*Am*t) = (75 000 N*mm)/(2*864*4) = 10.85 MPa.

Please reread my last paragraph in post 2. The unit symbol for megapascal should be written, e.g., 180 MPa, not 180mpa.

For part 3, the b/t value you computed is wrong, your units are wrong, and your calculation is wrong.

 

[donniemateno]

so now using your value if i use it to work out my hoop stress i get (0.72*0.04)*10.85 = 195.3N/m^2

i then take my 180 yield stress and divide it by 195.3 = 0.92 which is a fail

part 3 of the question i don't see how the formula is wrong?

the formula i used is tmax = T/(k1bt^2)

where b is the longer side of the strip
t is the shorter side thickness of tube
k1 is the empircal constant depending on the ratio of b/t and is obtained using the table ( attached as jpeg)

b in my formula is 72 mm and t is the thickness which is 12 mm so multiplying them i would have (72*12)^2 and to get k1 i would divide my b by t so 72 / 12 = 6 which on the graph supplied would make my k1 factor 0.299

so plugging the numbers in again i would get 75000 / ((0.299)*(72*12)^2) = 0.34 which if remember rightly should be 33.6 MPa

 

[nvn]

donniemateno: Sorry, I currently do not know what you are trying to compute in your first sentence of post 5. It does not make sense to me yet. And the units are completely wrong. Makes no sense.

... which, on the graph supplied, would make my k1 factor 0.299.

That is correct.

Your calculation in the last sentence of post 5 is wrong; your arithmetic is wrong, and is not what the formula says. Try again.

 

[donniemateno]

the first part of my previous post is to work out stress which i then use to work out the factor of safety. OK let's make this abit simpler how would you work out the Factor of safety with the data you have been given for part b?

and which part ? my formula is tmax = T/(k1bt^2) yes?

i have T which is 75000
k1 I have established with your help is 0.299 which you then times by bt squared or...looking at the formula again do i only square t?

 

[nvn]

donniemateno: Yes, square only t. I currently do not have a response on the first part of post 5, due to time constraints. I might look at it in the near future, if I get a chance.

 

[donniemateno]

so should my last part be :

tmax = T/(k1bt^2) yes?

= 75000 / ( 0.299 * 72 * ( 12^2)

=24.19 MPa or 24.2 MPa rounded

 

[donniemateno]

I have looked at the safety factor again and in my notes it states I need to get principal stresses; SF = yiled stress / difference between principle stresses

So I think the number i get to divide 180 MPa should be the different between the 72 mm measurement and 12 mm measurement

so 10.9 * (0.72/0.04) = 19.62
and 10.9 * (0.12/0.04) = 3.27

so the difference between 19.62 - 3.27 = 16.35

180 MPa / 16.35 = 11

 

[nvn]

donniemateno: Is this a take-home test?

 

[donniemateno]

yes and no. we get questions which come up in a test he following week which are the same as the ones given to take home but the numbers are different, it's to teach us the method. so if you can't remember the method then you can't do the question

 

[nvn]

donniemateno: Your answer in post 9 is correct.

 

[donniemateno]

Hey thanks for your help.

Any idea on the tresca theory question ? am a little stumped. I know you said you are busy but any help would be appreciated

 

[nvn]

Try SF = 0.5*(tensile yield strength)/tau_max.

 

[donniemateno]

so 10.9 * (0.72/0.04) = 19.62
and 10.9 * (0.12/0.04) = 3.27

so the difference between 19.62 - 3.27 = 16.35

so are you saying try 0.5 (180 MPa / 16.35) = 5.5?

 

[nvn]

Post 16 currently does not make sense, to my knowledge. Remember, you computed tau_max earlier, in two posts. Just use the tau_max value you obtained earlier.

 

[donniemateno]

so divide by my 10.9 / 11 MPa?

so I would get a result of 8.18 or 8.2 rounded

 

[nvn]

donniemateno: Close. But do not round before calculating. Therefore, use 10.85.

And, round the final answer to three significant digits, not two significant digits.