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CM 197 Mechanics of Materials Chap 9: Strength of Materials Simple StressPowerPoint Presentation

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

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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

Chap 9: Strength of Materials Simple Stress

- Objectives
- Introduction
- Normal and Shear Stresses
- Direct Normal Stresses
- Direct Shear Stresses
- Stresses on an Inclined Plane

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

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

- Normal force per unit area

- Force per unit area
- 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

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

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

- 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?

100 lbs

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

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