1. CM 197 Mechanics of Materials Chap 11: Mechanical Properties of Materials
Professor Joe Greene
CSU, CHICO
Reference: Statics and Strength of Materials, 2nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill, Westerville, OH (1997)
CM 197

2. Chap 11: Mechanical Properties
Objectives
Introduction
Tension Test
Stress-Strain
Mechanical Properties of Materials
Compression Test
Allowable Stresses and Factor of Safety
Stress Concentration
Elastic Design Versus Plastic Design

3. Allowable Axial Load
Structural members are usually designed for a limited stress level called allowable stress, which is the max stress that the material can handle.
Equation can be rewritten
Required Area
The required minimum cross-sectional area A that a structural member needs to support the allowable stress is from Equation 9-1
Eqn 9-3
Example 9-1
Example 9-2
Internal Axial Force Diagram
Variation of internal axial force along the length of a member can be detected by this
The ordinate at any section of a member is equal to the value of the internal axial force of that section
Example 9-3 and 9-4 – Statics review

4. Strain Units % Elongation = strain x 100%
Strain: Physical change in the dimensions of a specimen that results from applying a load to the test specimen.
Strain calculated by the ratio of the change in length, , and the original length, L. (Deformation)
Where,
 = linear strain ( is Greek for epsilon)
 = total axial deformation (elongation of contraction) = Lfinal –Linitial = Lf - L
L = Original length
Strain units (Dimensionless)
Units
When units are given they usually are in/in or mm/mm. (Change in dimension divided by original length)
% Elongation = strain x 100% L 

5. Strain
10in
1 in
0.1 in
Example
Tensile Bar is 10in x 1in x 0.1in is mounted vertically in test machine. The bar supports 100 lbs. What is the strain that is developed if the bar grows to 10.2in? What is % Elongation?
 =Strain = (Lf - L0)/L0 = ( )/(10) = 0.02 in/in
Percent Elongation = 0.02 * 100 = 2%
What is the strain if the bar grows to 10.5 inches?
What is the percent elongation?
100 lbs

6. Stress-Strain Diagrams
Test Sample
Forces
Fixed
Equipment
Tensile Testing machine
UTM- Universal testing machine
Measures
Load, pounds force or N
Deflection, inches or mm
Data is recorded at several readings
Results are averaged
e.g., 10 samples per second during the test.
Calculates
Stress, Normal stress or shear stress
Strain, Linear strain
Modulus, ratio of stress/strain

7. Modulus and Strength Modulus: Slope of the stress-strain curve
Can be Initial Modulus, Tangent Modulus or Secant Modulus
Secant Modulus is most common
Strength
Yield Strength
Stress that the material starts to yield
Maximum allowable stress
Proportional Limit
Similar to yield strength and is the point where Hooke’s Law is valid
If stress is higher than Hooke’s Law is not valid and can’t be used.
Ultimate strength
Maximum stress that a material can withstand
Important for brittle materials
Stress
Strain
Modulus
Yield
Strength
Proportional
Limit
Ultimate Strength

8. Tensile Modulus and Yield Strength
Modulus of Elasticity (E) or
Young’s Modulus is the ratio of stress to corresponding strain
A measure of stiffness
Yield Strength
Measure of how much stress a material can withstand without breaking
Modulus Yield Strength
Stainless Steel E= 28.5 million psi (196.5 GPa)
Aluminum E= 10 million psi
Brass E= 16 million psi
Copper E= 16 million psi
Molybdenum E= 50 million psi
Nickel E= 30 million psi
Titanium E= 15.5 million psi
Tungsten E= 59 million psi
Carbon fiber E= 40 million psi
Glass E= 10.4 million psi
Composites E= 1 to 3 million psi
Plastics E= 0.2 to 0.7 million psi

9. Hooke’s Law Hooke’s Law relates stress to strain by way of modulus
Hooke’s law says that strain can be calculated as long as the stress is lower than the maximum allowable stress or lower than the proportional limit.
If the stress is higher than the proportional limit or max allowable stress than the part will fail and you can’t use Hooke’s law to calculate strain.
Stress = modulus of elasticity, E, times strain
Stress=  = load per area, P/A
Strain=  = deformation per length,  /L
Rearrange Hooke’s law
Solving for deformation is Equation 10-5
With these equations you can find
How much a rod can stretch without breaking.
What the area is needed to handle load without breaking
What diameter is needed to handle load without breaking
Example 10-1
Example 10-3
Eqn 10-3
Eqn 10-4
Eqn 10-5

10. Problem solving techniques
Steps to solve most Statics problems
Set-up problem
Draw picture and label items (D, L, P, Stress, etc..)
List known values in terms of units.
Solve problem
Make a Force balance with Free body diagram
Identify normal forces
Identify shear forces
Write stress as Force per unit area
Calculate area from set-up, or
Calculate force from set-up
Write Hooke’s law
Rearrange for deflections
Write deflections balance
Solve for problem unknowns
Eqn 10-3
Eqn 10-4
Eqn 10-5

11. Allowable Stresses and Factor of Safety

12. Stress Concentrations
Stress is concentrated near holes

13. Elastic Design versus Plastic Design
Ignore this section