Stress and Strain Coursework Question

[PetePetePete]

I am struggling with a question in my coursework, and would appreciate some guidance.

The question is:

The component shown in Fig 1 is made from a material with the following properties and is subjected to a compressive force of 5kN.

Material Properties :

Young’s Modulus of Elasticity – 200 GNm-2
Modulus of Rigidity – 90 GNm-2
Poisons ratio – 0.32Calculate :

(a) The stress in :
(i) the circular section
(ii) the square section
(b) The strain in :
(i) The circular section
(ii) The square section
(c) The change in length of the component
(d) The change in diameter of the circular section

(Figure 1 is attached)

I have worked out question 1 i) and ii) to be 7.07Pa and 3.125Pa respectively, but I do not know where to begin with the rest of this question.

Thanks,
Pete

 

[billy_joule]

Show your working and relevant equations. What strain equations are you familiar with?

 

[PetePetePete]

So for the first question, I used Stress = Force / Area.

Working:
706.86mm2 / 5000N = 7.07Pa
1600mm2 / 5000N = 3.125Pa

I think I have figured out the 2nd question using Strain = Stress / Youngs Modulus

Working:
7.07Pa / 200GN = 3.535 x 10^-11
3.125Pa / 200GN = 1.5625 x 10*-11

Question c) I have attempted as: (F x length) / E

For the cylindrical part of the component:
= (7.07 x 60) / 200
= 2.121 x 10^-9

For the cuboid component:
= (3.125 x 60) / 200
= 9.375 x 10^-10

Total change in length = 3.0575 x 10^-9

But I am really struggling to find anything relating to a change in diameter when under compression. I know it is something to do with Poissons ratio, but not sure how to apply it.

I forgot, there is another part to the question:

(e) The change in the 40mm dimension on the square section

Appreciate your help.

Pete

 

[Chestermiller]

In terms of Poisson's ratio, how is the extensional strain in the radial direction related to the compressional strain in the axial direction?

Chet

 

[Dean Cockerill]

So for the first question, I used Stress = Force / Area.

Working:
706.86mm2 / 5000N = 7.07Pa
1600mm2 / 5000N = 3.125Pa

I think I have figured out the 2nd question using Strain = Stress / Youngs Modulus

Working:
7.07Pa / 200GN = 3.535 x 10^-11
3.125Pa / 200GN = 1.5625 x 10*-11

Question c) I have attempted as: (F x length) / E

For the cylindrical part of the component:
= (7.07 x 60) / 200
= 2.121 x 10^-9

For the cuboid component:
= (3.125 x 60) / 200
= 9.375 x 10^-10

Total change in length = 3.0575 x 10^-9

But I am really struggling to find anything relating to a change in diameter when under compression. I know it is something to do with Poissons ratio, but not sure how to apply it.

I forgot, there is another part to the question:

(e) The change in the 40mm dimension on the square section

Appreciate your help.

Pete

Pete. Did you get a reply to the rest of this question.

 

[SteamKing]

So for the first question, I used Stress = Force / Area.

Working:
706.86mm2 / 5000N = 7.07Pa
1600mm2 / 5000N = 3.125Pa

These values for the stresses are wrong. They're not even written correctly according to the formulas you show.

1 Pascal ≠ 1 Newton / 1 mm²

You need to check the definition of the Pascal unit and adjust your stress calculations accordingly.

 

[PetePetePete]

I have had another stab at it, and I would really appreciate it someone could sense check my answers? I haven't shown all the working on here, and the questions are in the attached files.

Q1 a)
Stress = Force / Area
i) 5000N / 706.68mm2 = -7.07MPa
ii) 5000N / 1600mm2 -3.125MPa

b)
i) Strain = Stress / E = -7.07MPa / 200GNm-2 = -3.535 x 10^-5
ii) -3.125MPa / 200GNm-2 = -1.5625 x 10^-5

c)
Compression = Strain x Original Length
Cylinder = -3.535 x 10^-5 x 60 = -2.121 x 10^-3
Cuboid = -9.375 x 10^-5 x 60 = -9.375 x 10^-4
Total change in length = -3.0585 x 10^-3 mm

d) Using Poissons Ratio @ 0.32 = 3.3936 x 10^-4 mm increase in diameter when compressed

e) 2 x 10^-4 mm increase in dimension when compressed

f) Stress = Force / Area
i) 7000 / 706.86 = 9.9029MPa
Shear Strain = Shear Stress / Modulus of Rigidity = 9.0929MPa / 90 x 10^9 = 1.1003 x 10-4
ii) 7000 / 1600 = 4.375MPa
4.375MPa / 90 x 10^9 = 4.86 x 10^-5

Thanks,
Pete

 

[willham0112]

I would really like to see know how you got 3.3936 x 10^-4 for question d?
I got 4.8936x10^-4 when using Change in diameter = - original diameter x Poisons ratio x (change in length / original length)?
Ive been stuck on this question for some time and would appreciate some help.

 

[Coffee_Guru]

I would really like to see know how you got 3.3936 x 10^-4 for question d?
I got 4.8936x10^-4 when using Change in diameter = - original diameter x Poisons ratio x (change in length / original length)?
Ive been stuck on this question for some time and would appreciate some help.[/QUOTE

3.3936 x 10^-4 is wrong, that's using the whole length of both the square and circular section, if you use only the circular section length then your answer is correct, as the question does only ask for circular section.

 

[Chestermiller]

This thread is almost 3 years old, and I'm closing it.