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MFGT 290 MFGT Certification Class

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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

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