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1 CM 197 Mechanics of Materials Chap 11: Mechanical Properties of Materials Professor Joe Greene CSU, CHICO Reference: Statics and Strength of Materials,

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Presentation on theme: "1 CM 197 Mechanics of Materials Chap 11: Mechanical Properties of Materials Professor Joe Greene CSU, CHICO Reference: Statics and Strength of Materials,"— Presentation transcript:

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

2 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 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 9-1 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 4 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) Where, –  = linear strain (  is Greek for epsilon) –  = total axial deformation (elongation of contraction) = L final –L initial = L f - 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 5 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 = (L f - L 0 )/L 0 = ( )/(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? 10in 1 in 0.1 in 100 lbs

6 6 Stress-Strain Diagrams 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 Test Sample Forces Fixed

7 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 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 –ModulusYield Strength Stainless SteelE= 28.5 million psi (196.5 GPa) Aluminum E= 10 million psi BrassE= 16 million psi CopperE= 16 million psi MolybdenumE= 50 million psi NickelE= 30 million psi TitaniumE= 15.5 million psi TungstenE= 59 million psi Carbon fiberE= 40 million psi Glass E= 10.4 million psi CompositesE= 1 to 3 million psi PlasticsE= 0.2 to 0.7 million psi

9 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 10-4 Eqn 10-5 Eqn 10-3

10 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 1.Make a Force balance with Free body diagram Identify normal forces Identify shear forces 2.Write stress as Force per unit area Calculate area from set-up, or Calculate force from set-up 3.Write Hooke’s law Rearrange for deflections Write deflections balance 4.Solve for problem unknowns Eqn 10-4 Eqn 10-5 Eqn 10-3

11 11 Allowable Stresses and Factor of Safety Factor of Safety

12 12 Stress Concentrations Stress is concentrated near holes

13 13 Elastic Design versus Plastic Design Ignore this section


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