1. 1 CM 197 Mechanics of Materials Chap 9: Strength of Materials Simple Stress 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 9: Strength of Materials Simple Stress Objectives –Introduction –Normal and Shear Stresses –Direct Normal Stresses –Direct Shear Stresses –Stresses on an Inclined Plane

3. 3 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. 4 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 torsion flexure P P PPP P A

5. 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/m 2 –Pounds per square inch = Psi Note: 1MPa = 1 x10 6 Pa = 145 psi –1 kPa = 10 3 Pa, 1 MPa = 106Pa, 1GPa = 109Pa –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/  r 2 = 100/(3.1415927 * 0.05 2 )= 12,739 psi = 87.86 MPa

6. 6 Stress 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/m 2 = 100,000 Pa = 0.1 MPa= 0.1 MPa *145psi/MPa = 14.5 psi Load = 100 N 10in 1 in 0.1 in 10cm 5cm 1 cm 100 lbs

7. 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