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Stress Question

  1. Jun 13, 2008 #1
    1. The problem statement, all variables and given/known data
    The length of a unstressed girder is found to be 1.2m at a temp of 20 oC. A tensile load is then applied and some time later the temp is 50 oC and the length of the girder 1.202m evaluate:
    (a) the tensile stress due to the applied load in the girder.

    (b) the factor of safety that is in operation if the UTS for the girder is 550MN/m2

    (c) the strain and hence the change in length of the girder, if the temp remains at 20 oC

    E = 200 GN/m2 coefficient of linear expansion = 12 x 10 - 6

    2. Relevant equations

    X = length x coefficient of linear expansion x change in temp

    strain = coefficent of linear expansion x change in temp

    stress = modulus of elasticity x coefficient of linear expansion x change in temp

    stress = U.T.S/ F.S

    3. The attempt at a solution
    (a) stress = 200 x 10 9 x 12 x 10 - 6 x 30 oC ( this is from 50 oC- 20 oC = 30 oC)
    = 72 MN/m2

    (b) f.s = u.t.s/stress
    550 x 10 6 / 72 x 10 6 = 7.63

    (c) strain = coefficient of linear expansion x change in temp
    12 x 10 -6 x 20 oC = 0.24 x 10 -3

    if someone would be so kind to tell me if this correct it will be much appreciated.
     
  2. jcsd
  3. Jun 13, 2008 #2

    Mapes

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    This isn't quite right. The strain from the load and the strain from thermal expansion are additive. Try subtracting the thermal strain from the total strain to get the strain from the load only and using that to calculate the load.
     
  4. Jun 13, 2008 #3
    is this for part c or all of the question because i have taken away 50 oC - 20 oC = 30 oC
    thanks for your reply.
     
  5. Jun 13, 2008 #4

    Mapes

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    The problem is with part (a). What you have calculated is the stress that would exist in the girder if it were fixed at both ends and underwent a temperature increase of 30°C. But that's not what the question is looking for.
     
  6. Jun 14, 2008 #5
    This is what i did originally

    stress/modulus of elasticity = coefficient of linear expansion x temp change - total strain
    where modulus of elasticity = change in length / original length

    = 0.02/1200 = 0.01666 x 10 -3 mm

    therefore stress = E ( COEFFICIENT OF LINEAR EXPANSION X CHANGE IN TEMP - TOTAL STRAIN

    200x10 9 x ( 12x10-6 x 30 - 0.01666 x 10 -3) = 68.668 x 10 6

    i got told it was transposed wrong but dont no where just pulling my hair out now
     
  7. Jun 14, 2008 #6

    Mapes

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    This is closer. The total strain is the thermal strain plus the strain from the load. You just need to express this correctly in equation form. Also, you've used a couple different length values (is the change in length 0.002m or 0.02mm?), so make sure to check that.
     
  8. Jun 14, 2008 #7
    so is this correct

    stress = E ( coeffficient of linear expansion x change in temp + total strain

    so stress = 200x10 9 x 12 x 10-6 x 30 + 1.666x10-6 = 72 x 10 6
     
  9. Jun 14, 2008 #8

    Mapes

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    No. Read my posts again.
     
  10. Jun 14, 2008 #9
    try again

    stress = E x coefficient of linear expansion x total strain + change in temp

    transposition arhh
     
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