Strength of Materials and Failure Theories

Strength of Materials and Failure Theories

• Submitted By: dwyang
• Date Submitted: 04/15/2013 1:21 AM
• Category: Technology
• Words: 4846
• Page: 20
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Strength of Materials and Failure Theories

State of Stress

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This is a 2D state of stress – only the independent stress components are named. A third stress component σ3 can exist on the z-axis and the state of stress is still called 2D and the following equations apply. To relate failure to this state of status, three important stress indicators are derived: Principal stress, maximum shear stress, and VonMises stress.

Principal stresses:
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Maximum shear stress – Only the absolute values count.
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The Vom Mises stress:
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When σ3=0, the von Mises stress is:
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When only σx, and τxy are present (as in combined torsion and bending/axial stress), there is no need to calculate the principal stresses, the Von Mises stress is:
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Problem #S1

A member under load has a point with the following state of stress:
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Determine σ1, σ2, τmax
Answers: 11444 tensile, 6444 Compressive, 8944 psi

Strain (one dimensional)

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Total strain definition:

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Total strain is a combination of mechanical and thermal strains:

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Both the mechanical and the thermal strains are algebraic. DT is positive for an increase in temperature. F is positive when it is a tensile force.

Problem #S2

The end of the steel bar has a gap of 0.05” with a rigid wall. The length of the bar is 100” and its cross-sectional area is 1 in2. The temperature is raised by 100 degrees F. Find the stress in the bar. ANS: 4500 PSi Comp.

Bending of straight beams

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Bending stress for bending about the Z-axis:

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Bending stress for bending about the Y-axis:

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where Iz and Iy are area moments of inertias about the z and y axes. Use tables to look up moments of inertia for various cross-sections. The parallel axis theorem can be used to find moment of inertia w/r a parallel axis:

Problem #S3

The solid circular steel bar with R=2” (diameter 4”) is under two loads as shown....