1. Elasticity by Ibrhim AlMohimeed
Chapter 4
Elasticity
by Ibrhim AlMohimeed
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2. Video
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3. Deformation
When a object crash a car, the car may not move but it will noticeably change shape.
A change in the shape due to the application of a force is a deformation.
Even very small forces are known to cause some deformation.
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4. where F is force, ∆L is the change in length
Cont. Deformation
For deformations, two important characteristics are observed:
The object may returns to its original shape when the force is removed.
The size of the deformation is proportional to the force.
𝐹∝ ∆𝐿
where F is force, ∆L is the change in length
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5. Hooke’s law
where ΔL is the amount of deformation (the change in length) produced by the force F, and k is a proportionality constant that depends on the shape and composition of the object and the direction of the force.
𝐹=𝑘 ∆𝐿
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6. Cont. Hooke’s law
The proportionality constant k depends upon a number of factors for the material.
For example, a guitar string made of nylon stretches when it is tightened, and the elongation ΔL is proportional to the force applied. Thicker nylon strings and ones made of different material (steel) stretch less for the same applied force, implying they have a larger k.
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7. Hooke’s law Examples
Example 4.1: A force of 600 N will compress a spring 0.5 meters. What is the spring constant of the spring?
Example 4.2: A spring has spring constant 0.1 N/m. What force is necessary to stretch the spring by 2 meters?
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8. Elastic Modulus ΔL depends on the material of the subject.
ΔL is proportional to the force F.
ΔL is proportional to the original length L0.
ΔL is inversely proportional to the cross-sectional area of the subject
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9. Cont. Elastic Modulus
All these factors can be combined into one equation for ΔL :
∆𝐿= 1 𝑌 𝐹 𝐴 𝐿 0
ΔL the change in length.
F the applied force.
Y is Young’s modulus, or elastic modulus (that depends on the subject material).
A is the cross-sectional area,
L0 is the original length
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10. Stress and Strain
The ratio of force to area, 𝐹 𝐴 , is defined as stress. So the stress is the force per unit area (pressure!!!)
𝜎= 𝐹 𝐴 𝑁/ 𝑚 2 or Pa
The ratio of the change in length to length, ∆𝐿 𝐿 0 , is defined as strain (a unitless quantity)
𝜀= ∆𝐿 𝐿 0
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11. 𝝈 𝒔𝒕𝒓𝒆𝒔𝒔 =𝒀 𝒀𝒐𝒖𝒏 𝒈 ′ 𝒔 𝒎𝒐𝒅𝒖𝒍𝒖𝒔 𝜺 (𝒔𝒕𝒓𝒂𝒊𝒏)
Cont. Stress and Strain
So, the equation of the elastic modulus can be rearranged in term of stress and strain:
𝝈 𝒔𝒕𝒓𝒆𝒔𝒔 =𝒀 𝒀𝒐𝒖𝒏 𝒈 ′ 𝒔 𝒎𝒐𝒅𝒖𝒍𝒖𝒔 𝜺 (𝒔𝒕𝒓𝒂𝒊𝒏)
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12. Stress and Strain Example
Example 4.3: A steel wire 10 m long and 2 mm in diameter is attached to the ceiling and a 200-N weight is attached to the end. What is the applied stress?
If the wire stretches 3.08 mm. What is the longitudinal strain? L DL
A = π × r2
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13. Stress and Strain Curve
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14. Cont. Stress and Strain Curve
Flash
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15. Cont. Stress and Strain Curve
Elastic behavior: the object will returns to its original length (or shape) when the stress acting on it is removed.
Plastic behavior: the object will NOT returns to its original length (or shape) when the stress acting on it is removed.
Proportionality limit: the stress above is not longer proportional to strain.
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16. Cont. Stress and Strain Curve
Elastic behavior: the object will returns to its original length (or shape) when the stress acting on it is removed.
Elastic Limit: The maximum stress that can be applied without resulting permanent deformation.
Yield Stress: at which there are large increases in strain with little or no increase in stress. steel exhibits this type of response.
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17. Cont. Stress and Strain Curve
Strain Hardening: when yielding has ended, a further load can be applied to the object.
Ultimate Strength: the maximum stress the object can withstand.
Necking: after the ultimate stress, the cross-sectional area begins to decrease in a localized region of the object.
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18. Stress & Strain Curve Example
Example 4.4: The elastic limit for steel of 3.14 x 10-6 m2 is 2.48 x 108 Pa and The ultimate strength is 4.89 x 108 Pa.
a)What is the maximum weight that can be supported without exceeding the elastic limit?
b) What is the maximum weight that can be supported without breaking the wire?
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19. Stress & Strain Curve Example
a stress applied to an object was 6.37 x 107 Pa and its strain was 3.08 x Find the modulus of elasticity for object?
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20. Stress & Strain Curve Example
Young’s modulus for brass is 8.96 x 1011Pa. A 120-N weight is attached to an 8-m length of brass wire; find the increase in length. The diameter is 1.5 mm.
8 m DL
120 N
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21. Strain Hardening
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22. Cont. Strain Hardening Strain Hardening
- If the material is loaded again from the beginning stage of strain hardening, the curve will be the same Elastic Modulus (slope).
- The material will has a higher yield strength.
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23. Types of Material Isotropic materials:
have elastic properties that are independent of direction.
Anisotropic materials:
whose properties depend upon direction.
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24. Material Behavior
Behavior of materials can be broadly classified into two categories:
Brittle
(Example: glass, ceramics)
Ductile
(Example: Metals; Gold, silver, copper, iron )
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25. Brittle Behavior Brittle \fracture at much lower strains.
yielding region is nearly nonexistent.
often have relatively large Young's moduli and ultimate stresses.
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26. Ductile Behavior Ductile \withstand large strains before failure.
yielding region often takes up the majority of the stress-strain curve
capable of absorbing much larger quantities of energy before failure.
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27. Resilience & Toughness
\Amount of energy stored in material up to elastic limit up to elastic limit per unit volume.
Toughness:
Amount of energy stored in material up to fracture per unit volume
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28. Cont. Resilience & Toughness
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29. Shear modulus
The shear modulus (S) is the elastic modulus we use for the deformation which takes place when a force is applied parallel to one face of the object while the opposite face is held fixed by another equal force.
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30. Cont. Shear modulus
Shear deformation behaves similarly to tension and compression and can be described with similar equations.
𝜏 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 =𝑆 𝑠ℎ𝑒𝑎𝑟 𝑚𝑜𝑑𝑢𝑙𝑒 ×𝜙(𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑖𝑛)
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31. Bulk Modulus
An object will be compressed in all directions if inward forces are applied evenly on all its surfaces.
It is relatively easy to compress gases and extremely difficult to compress liquids and solids.
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32. Bulk Modulus 𝐵= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 =− 𝐹/𝐴 Δ𝑉/𝑉 19/11/2013
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33. Bulk Modulus Example Example 4.6:
A hydrostatic press contains 5 liters of oil. Find the decrease in volume of the oil if it is subjected to a pressure of 3000 kPa. (Assume that B = 1700 MPa.)
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34. End of the Chapter
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