CM 197 Mechanics of Materials Chap 9: Strength of Materials Simple Stress

1. CM 197Mechanics of Materials Chap 9: Strength of Materials Simple Stress Professor Joe Greene CSU, CHICO Reference: Statics and Strength of Materials, 2nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill, Westerville, OH (1997) CM 197

2. Chap 9: Strength of Materials Simple Stress • Objectives • Introduction • Normal and Shear Stresses • Direct Normal Stresses • Direct Shear Stresses • Stresses on an Inclined Plane

3. Introduction • Introduction • Statics: first 8 chapters • Strength of Materials: Rest of book • Relationships between external loads applied to an elastic body • Intensity of the internal forces within the body • Statics: all bodies are rigid. • Strength of materials: all bodies are deformable • Terms • Strain: deformation per unit length • Stress: Force per unit area from an external source • Strength: Amount of force per unit area that a material can support without breaking. • Stiffness: A material’s resistance to deformation under load

4. Mechanical Test Considerations P P A P P P P shear • Normal and Shear Stresses • Force per unit area • Normal force per unit area • Forces are perpendicular (right angle) to the surface • Shear force per unit area • Forces are parallel (in same direction) 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 compression tension torsion flexure

5. Stress • Stress: Intensity of the internally distributed forces or component of forces that resist a change in the form of a body. • Tension, Compression, Shear, Torsion, Flexure • Stress calculated by force, P, per unit area. Applied force divided by the cross sectional area of the specimen. • Note: P is sometimes called force, F. Eqn 9-1 • Stress units • Pascals = Pa = Newtons/m2; MegaPascal=MPa= Newton/mm2 • Pounds per square inch = Psi Note: 1MPa = 1 x106 Pa = 145 psi • 1 kPa = 1x103 Pa, 1 MPa = 1x106Pa, 1GPa = 1x109 Pa • 1 psi = 6.895kPa, 1ksi = 6.895MPa, 1 psf = 47.88 Pa • Example • Wire 12 in long is tied vertically. The wire has a diameter of 0.100 in and supports 100 lbs. What is the stress that is developed? • Stress = P/A = P/r2 = 100/(3.1415927 * 0.052 )= 12,739 psi = 87.86 MPa

6. Stress 0.1 in 1 in 10in 1 cm 5cm 10cm • Example • Tensile Bar is 10in x 1in x 0.1in is mounted vertically in test machine. The bar supports 100 lbs. What is the stress that is developed? What is the Load? • Stress = F/A = F/(width*thickness)= 100lbs/(1in*.1in )= 1,000 psi = 1000 psi/145psi = 6.897 MPa • Load = 100 lbs • Block is 10 cm x 1 cm x 5 cm is mounted on its side in a test machine. The block is pulled with 100 N on both sides. What is the stress that is developed? What is the Load? • Stress = F/A = F/(width*thickness)= 100N/(.01m * .10m )= 100,000 N/m2 = 100,000 Pa = 0.1 MPa= 0.1 MPa *145psi/MPa = 14.5 psi • Load = 100 N 100 lbs

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 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 • Internal Axial Force Diagram • Varaition 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 9-2