# Homework Help: Stress and Strain Coursework Question

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1. Oct 13, 2014

### 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 – 0.32

Calculate :

(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

#### Attached Files:

• ###### Fig1.jpg
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2. Oct 13, 2014

### billy_joule

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

3. Oct 13, 2014

### 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

Pete

4. Oct 13, 2014

### Staff: Mentor

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

5. Nov 14, 2014

### Dean Cockerill

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

6. Nov 14, 2014

### SteamKing

Staff Emeritus
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.

7. Nov 20, 2014

### 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

#### Attached Files:

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• ###### Eng2.jpg
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8. Feb 24, 2016

### 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.

9. Jun 8, 2017

### Coffee_Guru

10. Jun 9, 2017

### Staff: Mentor

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