1. 1 CM 197 Mechanics of Materials Chap 10: Strength of Materials Strains 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 10: Strains Objectives –Introduction –Linear Strain –Hooke’s Law –Axial Deformation –Statically Indeterminate Problems –Thermal Stresses –Poisson’s ratio –Shear Strain

3. 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 tension compression shear P P PPP P A Normal ForcesShear Forces

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 = (10.2 -10)/(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 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)

8. 8 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

9. 9 Statically Indeterminate Problems Sometimes you might have forces in structural members that can’t be found with stress-strain equations –Statically indeterminate Those involving axially loaded members can be analyzed by –Introducing conditions of axial deformations –Example 10-5 –Example 10-6

10. 10 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 (T hot – T cold ) »L = length of member Example 10-7 and 10-8 Eqn 10-6

11. 11 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 0.002 and an axial strain of 0.006 Poisson’s ratio = 0.002/0.006 = 0.333 –Example 10-10 Note: For most materials, Poisson’s ratio is between 0.25 and 0.5 Axial Strain Transverse Strain P, Load

12. 12 Shear Strain Shear strain - occurs when a shear stress is applied –Shear stress is sliding force or tangential –Shear strain is deformation in tangential or side direction Shear strain = lateral deformation length of side Hooks Law for shear strain –Remember: Normal stress = normal modulus * normal strain –Shear stress = shear modulus * shear strain  = G  –Shear Modulus is found from the Normal Modulus, E, by dividing it by Poisson’s ratio, . G = __E_ 2(1+  ) For example, if the Normal modulus (Modulus of Elasticity) is 30 million psi (steel) and the shear modulus is 11.6 million psi, then the Poisson’s ratio is found from Equation 10-11 after you rearrange it. Poisson’s ratio, Example 10-11 Eqn 10-11 Eqn 10-10