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 1

Chap 9: Strength of Materials Simple Stress

• Objectives – Introduction – Normal and Shear Stresses – Direct Normal Stresses – Direct Shear Stresses – Stresses on an Inclined Plane 2

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

3

Mechanical Test Considerations

• Normal and Shear Stresses P – Force per unit area • Normal force per unit area P – Forces are perpendicular (right angle) to the surface • Shear force per unit area – Forces are parallel (in same direction) to the surface A  • Direct Normal Forces and Primary types of loading P  P

P A

P P – 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 shear tension compression torsion flexure 4

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  

P

• Stress units

A

– Pascals = Pa = Newtons/m 2 ; MegaPascal=MPa= Newton/mm 2 – Pounds per square inch = Psi Note: 1MPa = 1 x10 6 Pa = 145 psi – 1 kPa = 1x10 3 Pa, 1 MPa = 1x10 6 Pa, 1GPa = 1x10 9 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/  r 2 = 100/(3.1415927 * 0.05

2 )= 12,739 psi = 87.86 MPa 5

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 0.1 in 1 in 10in 100 lbs 1 cm 10cm 5cm 6

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  

P A

can be rewritten

P allow

 

allow A

• 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

A



P allow



allow

– 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 7