Transcript Lecture 22

Mechanics of Materials – MAE 243 (Section 002)
Spring 2008
Aaron Kessman, substitute for Dr. K.A. Sierros
CHAPTER 7
ANALYSIS OF
STRESS
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 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
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
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 (SF = 0 and SM = 0),
sx 
s x s y
2
1
t x y1  
sx sy
2
1
sy 
1

s x s y
2

sx sy
2
cos(2q )  t xy sin(2q )
sin(2q )  t xy cos(2q )
sx sy
2
cos(2q )  t xy sin(2q )  s x1 (q  90º )
Stress Transformation - Ramifications
sx 
1
s x s y
2

sx sy
2
cos(2q )  t xy sin(2q )
t x y1  
1
sx sy
2
sin(2q )  t xy cos(2q )
• 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.
*https://ecourses.ou.edu
Stress Transformations – Example 7.2-11
Approach:
1. Determine sx, sy, txy, q
2. Plug
3. Chug
sx 
s x s y
2
1
t x y1  
sx sy
2
1
sy 
1

s x s y
2

sx sy
2
cos(2q )  t xy sin(2q )
sin(2q )  t xy cos(2q )
sx sy
2
cos(2q )  t xy sin(2q )
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 ( ds  0 ) and
dq
some algebraic manipulation.
• This is really just a more general look at the material in the
previous section.
s1, s2, txy,
s avg 
and
s x s y
qp = planes of principal stresses
2
t xy
tan(2q p ) 
(s x  s y ) / 2
s 1,2 
s x s y
2
tan(2q s )  
t max 
IP
t max 

  t
2
t xy
  t
s x s y 2
s1  s 2
2
qp = qp1, qp2, 90º apart
s x s y 2
(s x  s y ) / 2
2
q - equations
2
xy
2
xy
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
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 sx, txy
• Find sy(sx, txy, t0)
• Find numerical range
t max 
IP
  t
s x s y 2
2
2
xy
sy cannot be = 0 because at
some angles the combined
effect will raise txy above t0.