1. Mechanics of Materials Dr. Konstantinos A. Sierros
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 x , 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 in.) Determine the tensile stresses AB and BC in the two wires.

4. Problem
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
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
FIG Stress-strain diagram for a typical structural steel in tension (not to scale)

7. Stress (σ) – strain (ε) diagrams
FIG 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
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 materials absorb large amounts of strain energy
Ductile materials: aluminium, copper, magnesium, lead, molybdenum, nickel, brass, nylon, teflon
FIG Stress-strain diagram for a typical structural steel in tension (drawn to scale)

9. FIG. 1-13 Typical stress-strain diagram for an aluminum alloy.
Aluminium alloys
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 MPa and ultimate stresses in the range of MPa
FIG Typical stress-strain diagram for an aluminum alloy.

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. FIG. 1-15 Stress-strain curves for two kinds of rubber in tension
Rubber (elastomers)
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 %
FIG Stress-strain curves for two kinds of rubber in tension
Measure of the amount of necking
Parameters that characterize ductility

12. Brittle materials
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
FIG Typical stress-strain diagram for a brittle material showing the proportional limit (point A) and fracture stress (point B)

13. Time and temperature dependence
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. 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.
FIG Stress-strain diagram for copper in compression

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