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…
*https://ecourses.ou.edu

Plane Stress – components and conventions
• And that’s what we mean by
: the 2-D representation of combined stresses on the four faces of a stress element
• Two normal stress components, s x
, s y
• One shear stress component t xy
– Which btw, t xy
= t 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 x
1 y
1 rotated by some angle q from original axes -
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 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

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 ( S
F = 0 and
S
M = 0), s x
1 t x
1 y
1
 s x
 s y
  s
2 x
 s y
2
 s x
 s y sin(
2 cos(
2 q
)
 t xy
2 q
) cos(
 t xy
2 q
) sin( 2 q
) s y
1
 s x
 s y
2
 s x
 s y
2 cos( 2 q
)
 t xy sin( 2 q
)
 s x
1
( q 
90 º )

Stress Transformation - Ramifications s x
1
 s x
 s y
2
 s x
 s y
2 cos( 2 q
)
 t xy sin( 2 q
) t x
1 y
1
  s x
 s y
2 sin( 2 q
)
 t xy cos( 2 q
)
• Given stresses at one angle we can calculate stresses at any arbitrary angle
• Even a uniaxial loading ( s x loadings upon rotation
) will create both perpendicular ( s y
) and shear ( t xy
)
• 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.
*https://ecourses.ou.edu

Stress Transformations – Example 7.2-11
Approach:
1. Determine s x
, s y
, t xy
, q
2. Plug
3. Chug s x
1 t x
1 y
1
 s x

  s
2 x s

2 y s y
 s x sin(
 s y
2
2 q cos(
)
 t xy
2 q
)
 t xy cos( 2 q
) sin( 2 q
) s y
1
 s x
 s y
2
 s x
 s y
2 cos( 2 q
)
 t xy sin( 2 q
)

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 ( s
1 maximum shear stress ( t xy
).
, s
2
) and
• The equations for these can be found from the stress transformation equations by differentiation ( ) and some algebraic manipulation.
d s d q
• This is really just a more general look at the material in the previous section.

s
1
, s
2
, t xy
,
q
s x
 s y s avg s
1,2
 tan( 2 q p
)
 s x

2
2
 s
( s y x
 t xy
 s
) / 2 y s
 
2 y
2
 t xy
2 q p
= planes of principal stresses q p
= q p 1
, q p 2
, 90º apart no shear stress acts on the principal planes tan( 2 q s
) t max
IP

 
( s x s
 
2 y
2
 s t xy y
) / 2
 t xy
2 s
1
 s
2 t max

2 q s
= planes of max shear stress q s
= q s 1
, q s 2
, 90º apart, 45º offset q p t 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
*https://ecourses.ou.edu

Principal Stresses and Max. Shear Stress - Example 7.3-18
Approach:
• Determine s x
, t xy
• Find s y
( s x
, t xy
, t
0
)
• Find numerical range t max
IP
 s
 
2 y
2
 t xy
2 s y cannot be = 0 because at some angles the combined effect will raise t xy above t
0
.