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# 14 static equilibrium and elasticity

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• Figure 12-14. Hooke’s law: The fractional change in length is proportional to the applied force.
• Figure 12-15. Applied force vs. elongation for a typical metal under tension.
• Solution: The elastic modulus for steel is in the table; F = 980 N.
• Figure 12-16. Stress exists within the material.
• Figure 12-16. Stress exists within the material.
• Figure 12-17. This Greek temple, in Agrigento, Sicily, built 2500 years ago, shows the post-and-beam construction. The columns are under compression.
• Figure 12-16. Stress exists within the material.
• Figure 12-19. The fatter book (a) shifts more than the thinner book (b) with the same applied shear force.
• Figure 12-20. Fracture as a result of the three types of stress.
• Solution: Using the tensile strength of steel from the table, we find F = 1600 N.
• ### 14 static equilibrium and elasticity

1. 1. Static Equilibrium and Elasticity Topic 12 (cont.)
2. 2. Lecture Outline 12.4 Hooke’s Law 12.5 Young Modulus 12.6 Stress and Strain 12.7 Fracture
3. 3. 12.4 Hooke’s Law <ul><li>An object exerted by force – pulling force – the length will change </li></ul><ul><li>If the amount of elongation is small compare to the length of the object initial length, Hooke’s Law state that: </li></ul><ul><li>The change in length is proportional to the applied force. </li></ul>
4. 4. F – pulling force Δ l – change in length k – proportionality constant First noted by Robert Hooke (1635-1703) Valid for almost any solid material But applicable for only to a certain point – force is not to strong Up to a point – Hooke’s Law is not valid
5. 5. Ultimate Strength – Maximum force the object can sustain This proportionality, holds until the force reaches the proportional limit . Beyond that, the object will still return to its original shape up to the elastic limit . Beyond the elastic limit, the material is permanently deformed More force is apply it breaks at the breaking point.
6. 6. 12.5 Young Modulus <ul><li>The change in length of a stretched object depends not only on the applied force, but also on its length , cross-sectional area and the material from which it is made. </li></ul><ul><li>The elongation of an object is proportional to the original length and inversely proportional to the cross-section area of the object </li></ul><ul><li>E – Elastic modulus/Young Modulus </li></ul>
7. 8. Example 12-7: Tension in piano wire. A 1.60-m-long steel piano wire has a diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.25 cm when tightened? (E = 2.0 × 10 11 N/m 2 )
8. 9. 12.6 Stress and Strain Stress is defined as the force per unit area . [unit: N/m 2 ] Strain is defined as the ratio of the change in length to the original length . (how much an object is deform) Therefore, the elastic modulus is equal to the stress divided by the strain:
9. 10. Tension/tensile stress In tensile stress, forces tend to stretch the object. Three type of stress:
10. 11. Compression stress Compression – force act inwardly on the object.
11. 12. Compressional stress is exactly the opposite of tensional stress. These columns are under compression.
12. 13. Sheer stress Equal and opposite force acted on the object – across its opposite faces G – Sheer Modulus
13. 14. The shear strain, where G is the shear modulus:
14. 15. If an object is subjected to inward forces on all sides, its volume changes depending on its bulk modulus . This is the only deformation that applies to fluids. or
15. 16. <ul><li>Example: Squeezing a Brass Sphere </li></ul><ul><li>A solid brass sphere is initially surrounded by air, and the air pressure exerted on it is 1.0 × 10 5 N/m 2 . The sphere is lowered into the ocean to depth where the pressure is 2.0 × 10 7 N/m 2 . The volume of the sphere in air is 0.50 m 3 . By how much does this volume change once the sphere is submerged? </li></ul><ul><li>(Bulk Modulus for brass = 80 × 10 9 N/m 2 ) </li></ul>
16. 17. 12.7 Fracture If the stress on an object is too great, the object will fracture. The ultimate strengths of materials under tensile stress, compressional stress, and shear stress have been measured. When designing a structure, it is a good idea to keep anticipated stresses less than 1/3 to 1/10 of the ultimate strength.
17. 19. Example 12-8: Breaking the piano wire. A steel piano wire is 1.60 m long with a diameter of 0.20 cm. Approximately what tension force would break it? (Tensile strength of steel = 500 × 10 6 N/m 2 )