Mechanics of Materials – MAE 243 (Section 002) Spring 2008

1. Mechanics of Materials – MAE 243 (Section 002) Spring 2008 Dr. Konstantinos A. Sierros

2. Problem 1.2-4 A circular aluminum tube of length L = 400 mm is loaded in compression by forces P (see figure). The outside and inside diameters are 60 mm and 50 mm, respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal direction. (a) If the measured strain is 550 x 10-6 , what is the shortening of the bar? (b) If the compressive stress in the bar is intended to be 40 MPa, what should be the load P?

3. Problem 1.2-7 Two steel wires, AB and BC, support a lamp weighing 18 lb (see figure). Wire AB is at an angle α = 34° to the horizontal and wire BC is at an angle β = 48°. Both wires have diameter 30 mils. (Wire diameters are often expressed in mils; one mil equals 0.001 in.) Determine the tensile stresses AB and BC in the two wires.

4. Problem 1.2-11 A reinforced concrete slab 8.0 ft square and 9.0 in. thick is lifted by four cables attached to the corners, as shown in the figure. The cables are attached to a hook at a point 5.0 ft above the top of the slab. Each cable has an effective cross-sectional area A = 0.12 in2 . Determine the tensile stress σt in the cables due to the weight of the concrete slab. (See Table H-1, Appendix H, for the weight density of reinforced concrete.)

5. 1.3 Mechanical properties of materials • In order to understand the mechanical behaviour of materials we need to perform experimental testing in the lab • A tensile test machine is a typical equipment of a mechanical testing lab • ASTM (American Society for Testing and Materials)

6. Stress (σ) – strain (ε) diagrams FIG. 1-10 Stress-strain diagram for a typical structural steel in tension (not to scale) • Nominal stress and strain (in the calculations we use the initial cross-sectional area A) • True stress (in the calculations we use the cross-sectional area A when failure occurs) • True strain if we use a strain gauge • Stress-strain diagrams contain important information about mechanical properties and behaviour

7. Stress (σ) – strain (ε) diagrams FIG. 1-10 Stress-strain diagram for a typical structural steel in tension (not to scale) OA: Initial region which is linear and proportional Slope of OA is called modulus of elasticity BC: Considerable elongation occurs with no noticeable increase in stress (yielding) CD: Strain hardening – changes in crystalline structure (increased resistance to further deformation) DE: Further stretching leads to reduction in the applied load and fracture OABCE’: True stress-strain curve

8. Stress (σ) – strain (ε) diagrams FIG. 1-12 Stress-strain diagram for a typical structural steel in tension (drawn to scale) • The strains from zero to point A are so small as compared to the strains from point A to E and can not be seen (it is a vertical line…) • Metals, such as structural steel, that undergo permanent large strains before failure are ductile • Ductile materialsabsorb large amounts of strain energy • Ductile materials: aluminium, copper, magnesium, lead, molybdenum, nickel, brass, nylon, teflon

9. Aluminium alloys FIG. 1-13 Typical stress-strain diagram for an aluminum alloy. • Although ductile…aluminium alloys typically do not have a clearly definable yield point… • However, they have an initial linear region with a recognizable proportional limit • Structural alloys have proportional limits in the range of 70-410 MPa and ultimate stresses in the range of 140-550 MPa

10. Offset method • When the yield point is not obvious, like in the previous case, and undergoes large strains, an arbitrary yield stress can be determined by the offset method • The intersection of the offset line and the stress-strain curve (point A) defines the yield stress

11. Rubber (elastomers) FIG. 1-15 Stress-strain curves for two kinds of rubber in tension • Rubber maintains a linear relationship between stress and strain up to relatively, as compared to metals, large strains (up to 20%) • Beyond the proportional limit, the behaviour depends on the type of rubber (soft rubber stretches enormously without failure!!!) • Rubber is not ductile but elastic material • Percent elongation = (L1-Lo)/ Lo % • Percent reduction in area = (Ao-A1)/ Ao % Measure of the amount of necking Parameters that characterize ductility

12. Brittle materials FIG. 1-16 Typical stress-strain diagram for a brittle material showing the proportional limit (point A) and fracture stress (point B) • Brittle materials fail at relatively low strains and little elongation after the proportional limit • Brittle materials: concrete, marble, glass, ceramics and metallic alloys • The reduction in the cross-sectional area until fracture (point B) is insignificant and the fracture stress (point B) is the same as the ultimate stress

13. Plastics • Viscoelasticity • Time and temperature dependence • Some plastics are brittle and some are ductile • COMPOSITES (glass fiber reinforced plastics) combine high strength with light weight Glass fiber Polymer matrix

14. Compression FIG. 1-17 Stress-strain diagram for copper in compression • Stress-strain curves in compression are different from those in tension • Linear regime and proportional limit are the same for tension and compression for materials such as steel, aluminium and copper (ductile materials) • However, after yielding begins the behaviour is different. The material bulges outward and eventually flattens out (curve becomes really steep) • Brittle materials have higher ultimate compressive stresses than when they are under tension. They do not flatten out but break at maximum load.

15. Tables of mechanical properties Appendix H contains tables that list materials properties. Please make sure that you use these tables when solving problems that require input of material properties data.

16. Have a good weekend… Wednesday (23 January 2008): Quiz on Statics, I will send you e-mail with further details…