Chapter 9: Mechanical Properties of Matter

1. Chapter 9: Mechanical Properties of Matter Properties of matter and matter in motion Density

3. Example: A 50 g bracelet is suspected of being gold-plated lead instead of pure gold. When it is dropped into a full bucket of water, 4.0 cm3 overflows. What is the density of the bracelet? Example: A well pump 20.0 m deep is to supply .50 m3 of water per minute. How much mechanical power must be provided?

4. Elasticity • “stretchiness/springiness” • -how materials respond to stress compression tension shear

5. “Stretchability” = amount of stress (applied force) produces a strain (elongation/compression/shear) Hooke’s Law: the amount of stretching is proportional to the applied force. F = k x The details of such springiness depends upon the size and shape of the material as well as how the forces are applied x x 1 Ton 2 Tons

6. Elastic Limit: the maximum stress (force) which can be applied to an object without resulting in permanent deformation. Plastic Deformation: the permanent deformation which results when a materials elastic limit has been exceeded. Ultimate strength: greatest tension (or compression or shear) the material can withstand. *snap* A malleable or ductile material has a large range of plastic deformation. Fatigue: small defects reduce materials strength well below original strength.

7. L0 L0 A A A A • Young’s Modulus: how things stretch (elastically) • stress: force per area = F/A compression tension L L • strain: fractional change in length • = change in length per original length = DL/Lo • Elastic modulus = stress/strain • Young’s modulus (for stretching in one direction)

8. From Lab • Force: suspended weight • Area: cross section of wire • elongation causes cylinder to rotate d 2q DL q D

9. Example 9.4: Stress as force per area • A 70 kg swami lies on a bed of nails 1.0 cm apart whose points have an area of 1.0 mm2. If the area of the swami’s body in contact with the bed is 0.50 m2 and the threshold for pain is 1MPa (1,000,000 Pa, 1Pa = 1 Pascal = 1N/m2 ), is the swami actually subjected to a painful experience? (i.e. F/A =?) • - # nails • - total area • - force (=weight) • -F/A

10. Example 9.4’ (+): A copper wire1.0 mm in diameter and 2.0 m long is used to support a mass of 5.0 kg. By how much does this wire stretch under this load? What is the maximum mass which can be supported without exceeding copper’s elastic limit? Y = 1.1x1011 Pa elastic limit = 1.5x108 Pa Example 9.6: A sagging floor is jacked up and a steel girder 3 m long whose cross sectional area is 30 cm2 is placed underneath. When the jack is removed, a sensitive strain gage shows that the girder has been compressed by .0020 cm. Find the load supported by the girder. Y = 20x1011 Pa

11. Shear Modulus: • stress: force per area = F/A s F A d F • shear strain cannot be supported in fluids (gas, liquid)

12. Bulk Modulus: • stress: force per area = F/A s • compression of solids, liquids and gases

13. Structure of solids • crystalline solids (most solids are crystalline) • atoms arranged in regular patterns • bonds hold atoms together like microscopic springs • amorphous solid • atoms are not in a well defined ordering • defects in crystalline structure allow permanent deformation