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
1
Chap 11: Mechanical Properties
• Objectives
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Introduction
Tension Test
Stress-Strain
Mechanical Properties of Materials
Compression Test
Allowable Stresses and Factor of Safety
Stress Concentration
Elastic Design Versus Plastic Design
2
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 9-1
• Required Area
P
 
A
can be rewritten
Pallow   allow A
– The required minimum cross-sectional area A that a structural member needs to
support the allowable stress is from Equation 9-1
Pallow
– Eqn 9-3
A
– Example 9-1
 allow
– 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
3
Strain
• 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)

 ( Lf  L )

L

L
L
• Where,
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 = 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%
4
Strain
• 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 = (10.2 -10)/(10) = 0.02 in/in
0.1 in
1 in
10in
100 lbs
• Percent Elongation = 0.02 * 100 = 2%
• What is the strain if the bar grows to 10.5 inches?
• What is the percent elongation?
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Stress-Strain Diagrams
Forces
• Equipment
– Tensile Testing machine
• UTM- Universal testing machine
• Measures
Test Sample
– 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
Fixed
6
Modulus and Strength
• Modulus: Slope of the stress-strain curve
– Can be Initial Modulus, Tangent Modulus or Secant Modulus
• Secant Modulus is most common
Ultimate Strength
Modulus
• Strength
– Yield Strength
Stress

• Stress that the material starts to yield Yield
Strength
• Maximum allowable stress
– Proportional Limit
Proportional
Limit
Strain

• 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
7
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
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Stainless Steel
Aluminum
Brass
Copper
Molybdenum
Nickel
Titanium
Tungsten
Carbon fiber
Glass
Composites
Plastics
E= 28.5 million psi (196.5 GPa)
E= 10 million psi
E= 16 million psi
E= 16 million psi
E= 50 million psi
E= 30 million psi
E= 15.5 million psi
E= 59 million psi
E= 40 million psi
E= 10.4 million psi
E= 1 to 3 million psi
E= 0.2 to 0.7 million psi
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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.
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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
  E
P

( Lf  L )
E E
A
L
L
• With these equations you can find
PL
– How much a rod can stretch without breaking.

AE
Eqn 10-3
Eqn 10-4
Eqn 10-5
– 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
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Problem solving techniques
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Steps to solve most Statics problems
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Set-up problem
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Draw picture and label items (D, L, P, Stress, etc..)
List known values in terms of units.
– Solve problem
1. Make a Force balance with Free body diagram
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Identify normal forces
Identify shear forces
2. Write stress as Force per unit area
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Calculate area from set-up, or
Calculate force from set-up
3. Write Hooke’s law
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Rearrange for deflections
Write deflections balance
4. Solve for problem unknowns
  E
P
 
A
A
Pallow   allow A
P

( Lf  L )
E E
A
L
L
PL

AE
Pallow
 allow
Eqn 10-3
Eqn 10-4
Eqn 10-5
10
Allowable Stresses and Factor of Safety
• Factor of Safety
11
Stress Concentrations
• Stress is concentrated near holes
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Elastic Design versus Plastic Design
• Ignore this section
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