1. Mechanics of Materials – MAE 243 (Section 002) Spring 2008
Aaron Kessman, substitute for Dr. K.A. Sierros

2. CHAPTER 7
ANALYSIS OF STRESS

3. 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?

4. 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.

5. 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… *

6. 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, sx, sy
One shear stress component txy
Which btw, txy = tyx

7. 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 q from original For now we’ll deal with plane stress, the 2-D biaxial stress projection of the 3-D cube

8. Plane Stress – How do we look at stresses in rotation?
If you were to rotate that little square stress element some angle q, 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

9. 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 (SF = 0 and SM = 0),

10. Stress Transformation - Ramifications
Given stresses at one angle we can calculate stresses at any arbitrary angle
Even a uniaxial loading (sx) will create both perpendicular (sy) and shear (txy) loadings upon rotation
Why this is important:
If any of the transformed stresses at angle q
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. *

11. Stress Transformations – Example 7.2-11
Approach:
Determine sx, sy, txy, q
Plug
Chug

12. 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 (q) at which these occur
These are called the principal stresses (s1, s2) and maximum shear stress (txy).
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.

13. s1, s2, txy, and q - equations qp = planes of principal stresses
qp = qp1, qp2, 90º apart
no shear stress acts on the principal planes
qs = planes of max shear stress
qs = qs1, qs2, 90º apart, 45º offset qp
tmaxIP = max in-plane shear stress

14. 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 *

15. Principal Stresses and Max. Shear Stress - Example 7.3-18
Approach:
Determine sx, txy
Find sy(sx, txy, t0)
Find numerical range
sy cannot be = 0 because at some angles the combined effect will raise txy above t0.