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CM 197
Mechanics 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
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Chap 9: Strength of Materials
Simple Stress
• Objectives
–
–
–
–
–
Introduction
Normal and Shear Stresses
Direct Normal Stresses
Direct Shear Stresses
Stresses on an Inclined Plane
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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
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Mechanical Test Considerations
• Normal and Shear Stresses
– Force per unit area
P
A
• 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
P P
P
 
A
P
P
• 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
shear
tension
compression
torsion
flexure
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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.
P
– Note: P is sometimes called force, F.
• Stress units
–
–
–
–
Eqn 9-1
 
A
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
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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?
0.1 in
1 in
10in
100 lbs
1 cm
10cm 5cm
• 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
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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
Pallow
– Eqn 9-3
A
 allow
– 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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