# Stress and Strain

1. Dec 14, 2014

### stackemup

1. The problem statement, all variables and given/known data

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

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

(f) If the same component were subjected to a shear force of 7 kN as shown in FIG 2, calculate the shear strain in :
(i) The circular section
(ii) The square section

2. Relevant equations
unsure

3. The attempt at a solution
I have not made any attempt as of yet as I have not done this in a long time and need to be pointed in the right direction. I nwould greatly appreciate anyone who has the time, to take me through each question and guide me in the right direction. I am not asking for answers but for guidance on formulas to use and a bit of an explaination why each formula needs to be used. Thanks in advance.

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2. Dec 14, 2014

### SteamKing

Staff Emeritus
Sorry, but PF is not a tutoring service. The rules state that you should at least make an attempt at a solution in order to receive help. You must have gotten this problem from some text or other source. Check that source for a discussion of basic concepts like stress and strain.

3. Dec 14, 2014

### stackemup

Thanks for the info ... I will attempt and answer and post it.

4. Dec 15, 2014

### stackemup

I have made an attempt on (a) but im assuming it needs revising and quite a bit of work putting into it, but I would appreciate it if someone could help.
(a) The Stress in:
Stress = F / A
(i) The circular section
Area of cylinder = 2TTr^2 + h(2TTr)
= 7068.59 x 10^-3 m2
Force on cylinder
= 5 x 10^3
Stress
= 5 x 10^3 / 7068.59
= 707.3546
(ii) The square section
Area of square section = 1600 + 1600 + 2400 + 2400 + 2400 + 2400
= 12800 x 10^-3 m2
Force in square section
= 5 x 10^3
Stress
= 5 x 10^3 / 12800
= 0.390625

5. Dec 15, 2014

### SteamKing

Staff Emeritus
The area used for calculating stress is not the surface area of the cylinder but the cross-sectional area.

The area used for calculating stress is not the surface area of the block but the cross-sectional area.

The force is not being applied over the entire surface of the items but only on one particular portion of the part.

6. Dec 15, 2014

### BvU

Are the dimensions in millimeters ? Check your area calculations !

7. Dec 15, 2014

### stackemup

Heres my revised attempt -

(i) The circular section
CSA = TTr^2
= 706.8583 x 10^-3 m2
Force on cylinder
= 5 x 10^3
Stress
= 5 x 10^3 / 706.8583
= 7.0736
(ii) The square section
CSA = 1600 x 10^-3 m2
Force on square section
= 5 x 10^3
Stress
= 5 x 10^3 / 1600
= 3.125

8. Dec 15, 2014

### SteamKing

Staff Emeritus
You're getting closer. You should always indicate the units on all calculation results.

The unit of stress in SI is the pascal. 1 pascal = 1 Newton / m2

You must be careful with the units for your area calculations. Although 1 meter = 1000 millimeters, 1 sq. meter ≠ 1000 sq. mm. Do you see why?

9. Dec 15, 2014

### stackemup

How does this look?

(a) The Stress in:
Stress = F / A
(i) The circular section
CSA = TTr^2
Radius = 15 x 10^-3 m
CSA = 0.000707 m2
Force on cylinder
= 5 x 10^3 N
Stress
= 5 x 10^3 / 0.000707
= - 7.073553 Pa
(ii) The square section
CSA = 1600 x 10^-3 m2
Force on square section
= 5 x 10^3 N
Stress
= 5 x 10^3 / 1.6 m2
= - 3.125 Pa

Am I correct in thinking that compressive stresses are defined as negative in sign?

Last edited: Dec 15, 2014
10. Dec 15, 2014

### SteamKing

Staff Emeritus

Check your arithmetic here. You are also using the wrong value for the CSA.

Yes, but that's a minor detail. It's more important to check your work to eliminate careless arithmetic mistakes or mistakes in copying numbers.

11. Dec 15, 2014

### stackemup

(a) The Stress in:
Stress = F / A
(i) The circular section
CSA = TTr^2
Radius = 15 x 10^-3 m
CSA = 0.000707 m2
Force on cylinder
= 5 x 10^3 N
Stress
= 5 x 10^3 / 0.000707
= - 7.073553 MPa
(ii) The square section
CSA = 1600 x 10^-6 m2
Force on square section
= 5 x 10^3 N
Stress
= 5 x 10^3 / 0.0016 m2
= - 3.125 MPa

I have also attempted part B below.

(b) The strain in:

ε = dl / lo
= σ / E
= Stress / Young’s Modulus
(i) The circular section
Strain = 7.073553 / 200GNm
Strain = 3.53678 x 10^-8
(ii) The square section
Strain = 3.125 / 200GNm
Strain = 1.56250 x 10^-8

Last edited: Dec 15, 2014
12. Dec 15, 2014

### SteamKing

Staff Emeritus
For Part a), the stresses look good.

For Part b), you've backslid. The stresses you calculated are in units of MPa, not Pa. Also, Young's modulus is in units of GPa or GN/m2, not GNm. You've got to use the correct values of the stress to calculate correct values of the strain. You've got to be vigilant about making these silly mistakes.

13. Dec 15, 2014

### stackemup

This should now be correct -

(b) The strain in:

ε = dl / lo
= σ / E
= Stress / Young’s Modulus
(i) The circular section
Strain = 7.073553 / 200GN/m2
Strain = 0.0353678
(ii) The square section
Strain = 3.125 / 200GN/m2
Strain = 0.0156250

14. Dec 16, 2014

### BvU

Not there yet! 7.07 MPa / 200 GPa is not 0.035. Same for (ii)

15. Dec 21, 2014

### stackemup

Thanks for the support.

I am struggling to calculate part (d) The change in diameter of the circular section.

Is the below formula used to calculate this?

Change in diameter = - original diameter x Poisons ratio x (change in length / original length)

16. Dec 21, 2014

### BvU

Think of something you expect to remain constant. Diameter x length would remain constant with your formula. Would that be logical ?

17. Dec 21, 2014

### stackemup

My equation now doesn't seem correct.

I'm struggling to make any progress with this question.

18. Dec 22, 2014

### stackemup

I believe the lateral strain multiplied by the diameter gives the increase in diameter.

The lateral strain is obtained from poissons ration, and the positive sign determines an increase in lateral dimensions, which is expected when an object is under compression.

Last edited: Dec 22, 2014
19. Dec 22, 2014

### BvU

Sorry, I completely misread your post #15. Says transverse strain = - Poisson's ratio times axial stress/Young modulus , which is completely correct. My post #16 is wrong and distracting.

20. Dec 22, 2014

### stackemup

I am a litte confused now.

Is my post 18 correct?