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Working out maximum stress and Factor of safety

  1. Mar 12, 2013 #1
    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 cant 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 ( ive 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) devide 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
     

    Attached Files:

  2. jcsd
  3. Mar 12, 2013 #2

    nvn

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    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 (ISO 31-0).
     
  4. Mar 13, 2013 #3
    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.
     

    Attached Files:

  5. Mar 13, 2013 #4

    nvn

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    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.
     
    Last edited: Mar 13, 2013
  6. Mar 13, 2013 #5
    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 dont 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
     
    Last edited: Mar 13, 2013
  7. Mar 13, 2013 #6

    nvn

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    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.
    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.
     
  8. Mar 13, 2013 #7
    the first part of my previous post is to work out stress which i then use to work out the factor of safety. OK lets 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?
     
  9. Mar 13, 2013 #8

    nvn

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    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.
     
  10. Mar 13, 2013 #9
    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
     
  11. Mar 13, 2013 #10
    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
     
    Last edited: Mar 13, 2013
  12. Mar 13, 2013 #11

    nvn

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    donniemateno: Is this a take-home test?
     
  13. Mar 14, 2013 #12
    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 cant remember the method then you cant do the question
     
  14. Mar 14, 2013 #13

    nvn

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    donniemateno: Your answer in post 9 is correct.
     
  15. Mar 14, 2013 #14
    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
     
  16. Mar 14, 2013 #15

    nvn

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    Try SF = 0.5*(tensile yield strength)/tau_max.
     
  17. Mar 16, 2013 #16
    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?
     
  18. Mar 16, 2013 #17

    nvn

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    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.
     
  19. Mar 17, 2013 #18
    so divide by my 10.9 / 11 MPa?

    so I would get a result of 8.18 or 8.2 rounded
     
  20. Mar 17, 2013 #19

    nvn

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    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.
     
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