Mechanics of Materials – MAE 243 (Section 002) Spring 2008 Aaron Kessman, substitute for Dr. K.A. Sierros
Mechanics of Materials – MAE 243 (Section 002)
Spring 2008
Aaron Kessman, substitute for Dr. K.A. Sierros
Overview
• Introduction – haven’t we been analyzing stress the whole time???
• 7.1-2 Plane stress – how uniaxial normal stress creates a shear component
• Problem solving example
• 7.3 Principal stresses and max shear stress – will the material break under loading?
• Problem solving example
Introduction
• Up till now, you’ve been dealing mostly with the big picture: uniaxial loading, torsion, and some combined loading in 2-D and 3-D.
• The end result has been to solve for the stress/moment at a given location on some loaded object – either explicitly or by a shear/moment diagram.
• Now we’ll take a microscopic look at the combined stresses and the effects of those loadings on the fabric of the material that is being loaded.
Introduction – stresses at a point
• When a body is loaded by normal and shear stresses, we can consider any point in that body as a stress element.
• The stress element can be depicted by a little square (in 2-D – or more correctly a cube in 3-D) with the stresses acting upon it. We’ll just ignore 3-D for the meantime…
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Plane Stress – components and conventions
• And that’s what we mean by plane stress: the 2-D representation of combined stresses on the four faces of a stress element
• Two normal stress components, x, y
• One shear stress component xy
– Which btw, xy = yx
Elements in plane stress, note sign conventions:
(a) three-dimensional view of an element oriented to the xyz axes,
(b) two-dimensional view of the same element, and
(c) two-dimensional view of an element oriented to the x1y1 axes -
rotated by some angle from original
For now we’ll deal with plane stress, the 2-D biaxial stress
projection of the 3-D cube
Plane Stress – How do we look at stresses in rotation?
• If you were to rotate that little square stress element some angle , what would happen?
• Well, stresses aren’t vectors, so they can’t be resolved the same (easy) way.
• We have to account for:– Magnitude
– Direction
– AND the orientation of the area upon which the force component acts
Stress Transformation - equations
• The stress transformation is a way to describe the effect of combined loading on a stress element at any orientation.
• From geometry and equilibrium conditions (F = 0 and M = 0),
)2cos()2sin(2
)2sin()2cos(22
1
x
1
1
xyyx
yx
xyyxyx
)º90()2sin()2cos(22 11 xy
xyyxyx
Stress Transformation - Ramifications
• Given stresses at one angle we can calculate stresses at any arbitrary angle• Even a uniaxial loading (x) will create both perpendicular (y) and shear (xy)
loadings upon rotation
• Why this is important:If any of the transformed stresses at angle
exceed the material’s yield stress, the material will fail in this direction, even if it was loaded by lower stresses.
• Sometimes the way this works out is failure by shear, which is not obvious.Materials are often weaker in shear.
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)2sin()2cos(221x
xyyxyx
)2cos()2sin(
211
xy
yxyx
Stress Transformations – Example 7.2-11
)2cos()2sin(2
)2sin()2cos(22
1
x
1
1
xyyx
yx
xyyxyx
Approach:
1. Determine x, y, xy,
2. Plug
3. Chug)2sin()2cos(
221y
xyyxyx
Principal Stresses and Maximum Shear Stress
• If material failure is what we ultimately care about, then we really want to know what are the – maximum and minimum normal stresses– maximum shear stress– orientation () at which these occur
• These are called the principal stresses (1, 2) and maximum shear stress (xy).
• The equations for these can be found from the stress transformation equations by differentiation ( ) and some algebraic manipulation.
• This is really just a more general look at the material in the previous section.
0d
d
1, 2, xy, and - equations
22
21,2 2
2/)(
)2tan(
2
xyyx
yx
xyp
yxavg
yx
p = planes of principal stresses
p = p1, p2, 90º apart
no shear stress acts on the principal planes
2
2/)(
)2tan(
21max
22
2max
xy
xy
yxs
yx
IP
s = planes of max shear stress
s = s1, s2, 90º apart, 45º offset p
maxIP = max in-plane shear stress
Summary
• Principal stresses represent the max and min normal stresses at the point.
• At the orientation at which principal stresses act, there is no acting shear stress.
• At the orientation at which maximum in-plane shear stress acts, the average normal stress acts in both normal directions (x, y)
• The element acted upon by the maximum in-plane shear stress is oriented 45º from the element acted upon by the principal stresses
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