Successfully reported this slideshow.
Upcoming SlideShare
×

# 3. Stress, Strain, Tension Test

23,807 views

Published on

Published in: Business, Health & Medicine
• Full Name
Comment goes here.

Are you sure you want to Yes No
• good job

Are you sure you want to  Yes  No
• excellent work and excellent team

Are you sure you want to  Yes  No

### 3. Stress, Strain, Tension Test

1. 1. Stresses and Strains <ul><li>Solid materials are deformable, not rigid. </li></ul><ul><li>We will study the stresses and strains that forces produce in a body </li></ul>
2. 2. A.) Axial Tensile and Compressive Stresses <ul><li>Consider a 2” x 4” piece of wood with a force P applied at each end. </li></ul>800 lb 800 lb 2” 4” A B
3. 3. <ul><li>Anywhere you cut this bar across its section, in order to keep the board from moving, the 800 lb force must act on that section. </li></ul><ul><li> F x = 0 = - 800 lb + P A = 0 </li></ul><ul><li>P A = 800 lb </li></ul>P A 800 lb A B
4. 4. <ul><li>We assume that the force is distributed evenly throughout the section so that an equal portion of the 800 lb force acts on each square inch of the cross-section </li></ul>800 lb 2” 4” 1” 1”
5. 5. <ul><li>Since we have 8 square makes, the amount of force on each square inch is: </li></ul><ul><li>800 lb = 100 lb = 100 psi </li></ul><ul><li>8in 2 in 2 </li></ul>
6. 6. <ul><li>Which is the definition of stress : </li></ul><ul><li> = P </li></ul><ul><li>A </li></ul><ul><li> = stress = unit stress </li></ul><ul><li>= average stress </li></ul><ul><li>= engineering stress </li></ul><ul><li> P = applied force </li></ul><ul><li> A= cross-sectional area over which the stress develops </li></ul>
7. 7. <ul><li>  t = Tensile Stress (produced by </li></ul><ul><li>Tensile Forces) </li></ul><ul><li>  c = Compressive Stress (produced by </li></ul><ul><li>Compressive Forces) </li></ul>
8. 8. <ul><li>B. Examples of Tensile and Compressive Stresses </li></ul>
9. 9. <ul><li>C.) TENSILE AND COMPRESSIVE </li></ul><ul><li> STRAINS AND DEFORMATIONS </li></ul>
10. 10. <ul><li>Example: Dock with wooden ladder for </li></ul><ul><li> a footbridge. </li></ul><ul><li>This is an example of deformation or </li></ul><ul><li>deflection due to bending stress which </li></ul><ul><li>we will cover later. </li></ul>
11. 11. <ul><li>Similarly, when a steel rod is in Tension, </li></ul><ul><li>it will deform, but it is not as noticeable. </li></ul><ul><li> = deformation = the amount a body is </li></ul><ul><li>lengthened by a tensile force and </li></ul><ul><li>shortened by a compressive force. </li></ul>L  T T
12. 12. <ul><li>To permit comparison with acceptable </li></ul><ul><li>values, the deformation is usually </li></ul><ul><li>converted to a unit basis, which is the </li></ul><ul><li>strain . </li></ul><ul><li> =   </li></ul><ul><li> L </li></ul><ul><li> = strain (= unit strain) </li></ul><ul><li> = deformation that occurs over length L </li></ul><ul><li>L = original length of member </li></ul>
13. 13. <ul><li>Example: a 3/8” cable, 100’ long stretches 1” before freeing a truck which is stuck in the mud. </li></ul><ul><li> </li></ul><ul><li>Find the strain in the cable. </li></ul>100’
14. 15. <ul><li> =   </li></ul><ul><li> L </li></ul><ul><li> = 1” </li></ul><ul><li>L = 100’ (12”/1) = 1200” </li></ul><ul><li> = 1” = 0.0008333 in/in </li></ul><ul><li> 1200” </li></ul><ul><li>We’ll come back to see if this is will break the cable. </li></ul>
15. 16. <ul><li>Review of Stress and Strain </li></ul><ul><li>Axial Stress and Strain </li></ul><ul><li> = P </li></ul><ul><li> A </li></ul><ul><li> =  L  </li></ul><ul><li>Shear Stress </li></ul><ul><li> = V  </li></ul><ul><li> A </li></ul>
16. 17. <ul><li>E.) The Relationship Between Stress </li></ul><ul><li>and Strain </li></ul><ul><li>As you apply load to a material, the strain increases constantly (or proportionately) with stress. </li></ul>
17. 18. <ul><li>Example: In a tension test you apply a gradually increasing load to a sample. You can determine the amount of strain (  that occurs in a sample at any given stress level (  . </li></ul><ul><li> (ksi)  (in/in x 0.001) </li></ul><ul><li> 0 0 </li></ul><ul><li> 3 1 </li></ul><ul><li> 6 2 </li></ul><ul><li> 9 3 </li></ul><ul><li> 12 4 </li></ul>
18. 19. Stress ,  (ksi) Strain ,  in/in x 0.001)           
19. 20. <ul><li>Since the stress is proportional to the strain, ratio of stress to strain is constant . </li></ul><ul><li> /  </li></ul><ul><li> (ksi)  (in/in x 0.001)  (ksi x 1000) </li></ul><ul><li> 0 0 0 </li></ul><ul><li> 3 1 3 </li></ul><ul><li> 6 2 3 </li></ul><ul><li> 9 3 3 </li></ul><ul><li>12 4 3 </li></ul>
20. 21. <ul><li>This constant ratio of stress to strain is called the Modulus of Elasticity (E). </li></ul><ul><li>E =  /  </li></ul><ul><li>The Modulus of Elasticity is always the same for a given material. We call it a material constant . </li></ul>
21. 22. <ul><li>Knowing E for a given material and : </li></ul><ul><li>E =  /  </li></ul><ul><li>1.) We can find how much stress is in the material if we know the strain: </li></ul><ul><li>  = E  </li></ul><ul><li>2.) We can find how much strain is in the material if we know the stress: </li></ul><ul><li>  =  E </li></ul>
22. 23. <ul><li>CAUTION ! </li></ul><ul><li>If the tension test continues, the stress will reach a level called the Proportional Limit (  PL ). If the stress is increased above  PL , the strain will increase at a higher rate. </li></ul>
23. 24.  Stress  ), ksi Strain (  ), in/in  PL
24. 25. <ul><li>Ex. Given: Previous Truck cable strain </li></ul><ul><li> Find: Stress in the steel cable </li></ul><ul><li> = 1” </li></ul><ul><li>L = 1200” </li></ul><ul><li> = 1” = 0.0008333 in/in </li></ul><ul><li> 1200” </li></ul><ul><li>   E ( as long as  PL ) </li></ul><ul><li> </li></ul><ul><li>E= 30,000,000 psi (for steel) </li></ul>
25. 26. <ul><li> E  = 30,000,000 psi (.0008333 in/in) </li></ul><ul><li>= 24,990 psi (pretty high) </li></ul><ul><li>CHECK: is  <  PL ? </li></ul><ul><li> = 24,990 psi <  PL = 34,000 psi (OK) </li></ul>
26. 27. <ul><li>D.) Material Properties found using the Tension Test </li></ul> Stress  ), ksi Strain (  ), in/in  PL  Y  U E =  =slope
27. 28. <ul><li>D.) Material Properties found using the Tension Test </li></ul><ul><li>1.) Ultimate Strength (  U ) - The maximum stress a material will withstand before failing. </li></ul><ul><li>2.) Yield Strength (  Y ) - The maximum stress a material will withstand before deforming permanently. </li></ul><ul><li>3.) Proportional Limit (  PL ) - The maximum stress a material will withstand before stress-strain relationship becomes non-linear. </li></ul>
28. 29. <ul><li>D.) Material Properties found using the Tension Test </li></ul><ul><li>4.) Modulus of Elasticity - the ratio of stress over strain in the linear region of the stress-strain curve. </li></ul><ul><li>5. Percentage Elongation-the plastic deformation at failure, as a percentage of the original length = (L f – L o )/ L o x 100 </li></ul>
29. 30. <ul><li>5.) Percent Elongation: </li></ul><ul><ul><li>Ductile Material - will undergo plastic deformation before failing </li></ul></ul> Stress Strain Ductile Material
30. 31. <ul><li>Brittle Material - will fail without any plastic deformation (opposite of ductile) </li></ul> Stress Strain Brittle Material 5.) Percent Elongation: