1. MFGT 290 MFGT Certification Class
8: Strength of Materials
Chapter 11: Material Properties
Professor Joe Greene
CSU, CHICO
MFGT 290

2. Chap 8: Strength of Materials
Stress and Strain
Axial Loading
Torsional Loading
Beam Loading
Column Loading
Practice Problems

3. Mechanical Test Considerations
Normal and Shear Stresses
Force per unit area
Normal force per unit area
Forces are normal (in same direction) to the surface
Shear force per unit area
Forces are perpendicular (right angle) to the surface
Direct Normal Forces and Primary types of loading
Prismatic Bar: bar of uniform cross section subject to equal and opposite pulling forces P acting along the axis of the rod.
Axial loads: Forces pulling on the bar
Tension= pulling the bar; Compression= pushing; torsion=twisting; flexure= bending; shear= sliding forces P A
Normal Forces
Shear Forces
tension
compression
shear

4. 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

5. Stress-Strain Diagrams
Stress-strain diagrams is a plot of stress with the corresponding strain produced.
Stress is the y-axis
Strain is the x-axis
Stress
Strain
Linear
(Hookean)
Non-Linear
(non-Hookean)

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
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

7. 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
Example Hinged Beam
Statics review
Sum of forces = 0
Sum of Moments = 0. Moment is Force time a distance to solid wall

8. 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 

9. 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

10. Tensile Modulus and Yield Strength
Modulus of Elasticity (E) (Note: Multiply psi by 7,000 to get kPa)
Also called Young’s Modulus is the ratio of stress to corresponding strain
A measure of stiffness
Yield Strength: (Note: Multiply psi by 7,000 to get kPa)
Measure of how much stress a material can withstand without breaking
Modulus (Table 8-1) Yield Strength
Stainless Steel E= 28.5 million psi (196.5 GPa) 36,000 psi
Aluminum E= 10 million psi 14,000 psi
Brass E= 16 million psi 15,000 psi
Copper E= 16 million psi
Molybdenum E= 50 million psi
Nickel E= 30 million psi
Titanium E= 15.5 million psi 120,000 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 15,000 psi
Plastics E= 0.2 to 0.7 million psi 5,000 to 12,000 psi

11. 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
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 8-3

12. 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 8-3

13. Safety Factor Allowable Stresses and Factor of Safety
Provide a margin of safety in design for bridges, cars, buildings, rockets, space shuttles, air planes, etc…
Structural members and machines are designed so that columns, plates, trusses, bolts, see much less than the stress that will cause failure.
Ductile materials: If the stress is greater than the yield strength or proportional limit of the material.
Brittle materials: If the stress is greater than the ultimate strength of the material.Since they do not show any yielding, just fracturing.

14. Stress Concentrations
Stresses can be higher near holes, notches, sharp corners in a part or structural member.
Stress concentration factor, K = stresses near hole
stresses far away from hole
K is looked up in a table or on a graph
Stress at hole can be calculated to see if part will fail.
Where b is the net width at hole section and t is the thickness.

15. Thermal Stresses
Most materials expand when heated as the temperature increases.
As the temperature goes up, the material expands and results in forces that cause stress in the part. As temperature increases the stresses increase in part.
Examples,
Cast iron engine block heat up to 500F and expands the cast iron block which causes stresses at the bolts. The bolts must be large enough to withstand the stress.
Aluminum heats up and expands and then cools off and contracts.
Sometimes the stresses causes cracks in the aluminum block.
Space shuttle blasts off and heats up, goes into space and cools down (-200F), and reenters Earths atmosphere and heats up (3000F)
Aluminum melts at 1300F so need ceramic heat shields
Aluminum structure expands and cools.
The amount the material expands is as follows:
Change in length that is causes by temperature change (hot or cold)
Where,
 = change in length
 = the CLTE (coefficient of linear thermal expansion
T = change in temperature (Thot – Tcold)
L = length of member
Examples

16. Strain and Poisson’s Ratio
Axial strain is the strain that occurs in the same direction as the applied stress.
Transverse strain is the strain that occurs perpendicular to the direction of the applied stress.
Poisson’s ratio is ratio of lateral strain to axial strain.
Poisson’s ratio = lateral strain
axial strain
Example
Calculate the Poisson’s ratio of a material with lateral strain of and an axial strain of 0.006
Poisson’s ratio = 0.002/0.006 = 0.333
Axial
Strain
Transverse
P, Load
Note: For most materials, Poisson’s ratio is between 0.25 and 0.5
Plastics: Poisson’s ratio 0.3
Table Metals: Poisson’s ratio = 0.3 steel, 0.33 Al, 0.35 Mg, 0.34 Ti

17. Chapter 11: Material Properties
Structure of Matter
Material Testing Agencies
Physical Properties
Mechanical Properties and Test Methods
Stress and Strain
Fatigue Properties
Hardness
Practice Problems

18. Fatigue Properties Fatigue Properties
All materials that are subjected to a cyclic loading can experience fatigue
Failure occurs through a maximum stress at any cycle.
Test methods
Subject the material to stress cycles and counting the number of cycles to failure, then
Fatigue properties are developed.
Table of properties for each material
How many cycles a material can experience at a certain stress level before failing.
S-N diagrams are developed (Stress and Number of cycles)
Specify fatigue as a stress value
Design for less than fatigue stress
N, number of cycles
S, stress

19. Fundamentals of Hardness
Hardness is thought of as the resistance to penetration by an object or the solidity or firmness of an object
Resistance to permanent indentation under static or dynamic loads
Energy absorption under impact loads (rebound hardness)
Resistance toe scratching (scratch hardness)
Resistance to abrasion (abrasion hardness)
Resistance to cutting or drilling (machinability)
Principles of hardness (resistance to indentation)
indenter: ball or plain or truncated cone or pyramid made of hard steel or diamond
Load measured that yields a given depth
Indentation measured that comes from a specified load
Rebound height measured in rebound test after a dynamic load is dropped onto a surface

20. Hardness Mechanical Tests
Brinell Test Method
One of the oldest tests
Static test that involves pressing a hardened steel ball (10mm) into a test specimen while under a load of
3000 kg load for hard metals,
1500 kg load for intermediate hardness metals
500 kg load for soft materials
Various types of Brinell
Method of load application:oil pressure, gear-driven screw, or weights with a lever
Method of operation: hand or electric power
Method of measuring load: piston with weights, bourdon gage, dynamoeter, or weights with a lever
Size of machine: stationary (large) or portable (hand-held)

21. Brinell Test Conditions
Brinell Test Method (continued)
Method
Specimen is placed on the anvil and raised to contact the ball
Load is applied by forcing the main piston down and presses the ball into the specimen
A Bourbon gage is used to indicate the applied load
When the desired load is applied, the balance weight on top of the machine is lifted to prevent an overload on the ball
The diameter of the ball indentation is measured with a micrometer microscope, which has a transparent engraved scale in the field of view

22. Brinell Test Example Brinell Test Method (continued)
Units: pressure per unit area
Brinell Hardness Number (BHN) = applied load divided by area of the surface indenter
Where: BHN = Brinell Hardness Number
L = applied load (kg)
D = diameter of the ball (10 mm)
d = diameter of indentation (in mm)
Example: What is the Brinell hardness for a specimen with an indentation of 5 mm is produced with a 3000 kg applied load.
Ans:

23. Brinell Test Method (continued)
Range of Brinell Numbers
90 to 360 values with higher number indicating higher hardness
The deeper the penetration the higher the number
Brinell numbers greater than 650 should not be trusted because the diameter of the indentation is too small to be measured accurately and the ball penetrator may flatten out.
Rules of thumb
3000 kg load should be used for a BHN of 150 and above
1500 kg load should be used for a BHN between 75 and 300
500 kg load should be used for a BHN less than 100
The material’s thickness should not be less than 10 times the depth of the indentation

24. Advantages & Disadvantages of the Brinell Hardness Test
Well known throughout industry with well accepted results
Tests are run quickly (within 2 minutes)
Test inexpensive to run once the machine is purchased
Insensitive to imperfections (hard spot or crater) in the material
Limitations
Not well adapted for very hard materials, wherein the ball deforms excessively
Not well adapted for thin pieces
Not well adapted for case-hardened materials
Heavy and more expensive than other tests ($5,000)

25. Rockwell Test
Hardness is a function of the degree of indentation of the test piece by action of an indenter under a given static load (similar to the Brinell test)
Rockwell test has a choice of 3 different loads and three different indenters
The loads are smaller and the indentation is shallower than the Brinell test
Rockwell test is applicable to testing materials beyond the scope of the Brinell test
Rockwell test is faster because it gives readings that do not require calculations and whose values can be compared to tables of results (ASTM E 18)

26. Rockwell Test Description
Specially designed machine that applies load through a system of weights and levers
Indenter can be 1/16 in hardened steel ball, 1/8 in steel ball, or 120° diamond cone with a somewhat rounded point (brale)
Hardness number is an arbitrary value that is inversely related to the depth of indentation
Scale used is a function of load applied and the indenter
Rockwell B- 1/16in ball with a 100 kg load
Rockwell C- Brale is used with the 150 kg load
Operation
Minor load is applied (10 kg) to set the indenter in material
Dial is set and the major load applied (60 to 100 kg)
Hardness reading is measured
Rockwell hardness includes the value and the scale letter

27. Rockwell Values
B Scale: Materials of medium hardness (0 to 100HRB) Most Common
C Scale: Materials of harder materials (> 100HRB) Most Common
Rockwell scales divided into 100 divisions with each division (point of hardness) equal to 0.002mm in indentation. Thus difference between a HRB51 and HRB54 is 3 x mm mm indentation
The higher the number the harder the number

28. Rockwell and Brinell Conversion
For a Rockwell C values between -20 and 40, the Brinell hardness is calculated by
For HRC values greater than 40, use
For HRB values between 35 and 100 use

29. Rockwell and Brinell Conversion
For a Rockwell C values, HRC, values greater than 40,
Example,
Convert the Rockwell hardness number HRc 60 to BHN
Review Questions