• Support PF! Buy your school textbooks, materials and every day products Here!

Stress Question

  • Thread starter series111
  • Start date
  • #1
54
0

Homework Statement


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

Homework 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

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.
 

Answers and Replies

  • #2
Mapes
Science Advisor
Homework Helper
Gold Member
2,593
17
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.
 
  • #3
54
0
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.
 
  • #4
Mapes
Science Advisor
Homework Helper
Gold Member
2,593
17
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.
 
  • #5
54
0
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
 
  • #6
Mapes
Science Advisor
Homework Helper
Gold Member
2,593
17
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.
 
  • #7
54
0
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
 
  • #8
Mapes
Science Advisor
Homework Helper
Gold Member
2,593
17
No. Read my posts again.
 
  • #9
54
0
try again

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

transposition arhh
 

Related Threads for: Stress Question

  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
9K
  • Last Post
Replies
10
Views
2K
Replies
10
Views
2K
  • Last Post
Replies
3
Views
504
  • Last Post
Replies
5
Views
2K
Top