Transcript Mock Course – Plane Stress Transformation (ref MCHT 213

Mock Course – Plane Stress Transformation
(ref MCHT 213, Hibbeler text)
• 1 hr 15 min format:
– Present Theory/Overview with Power Point, 15
min.
– Problem solving on board (example problems),
45 min
– Industry Examples, 5 min
– Answer HW questions
Chapter 9 – Plane Stress Transformations
Stress Analysis - Review
1. Determine critical point
2. Solve for internal forces at that point (or reduce
to cantilever)
3. Solve for stresses at that point
4. Add “like stresses, i.e.:
Chapter 8
1. stotal = s1 + s2 + s3 + ….………..
2. ttotal = t1 + t2 + t3 + ……………
5. Summarize stresses at that point on a stress
element.
6. May be necessary to use Stress Transformation
Chapter 9
or Mohr’s circle to get max stresses!
Recall from Chapter 8, already did steps 1 – 5:
Recall from Chapter 8, already did steps 1 – 5:
Drill Bit
Drill Bit Isolator (
Thrust Load = 8,000 to
10,000 lb
Bending Load= 125 lbs)=
K*d
Drill Rod
Torsion Load= 300 lb-ft
Chuck
Isolator
HOT
SPOT!
Now What????
Solve for stress at a point
using standard SoM
Equations.
Summarize these
stresses on an initial
stress element or aligned
stress element.
step 6
Must find MAXIMUM
stresses at that point,
may be different then the
applied stresses can
occur at some other
orientation plane or
angle.
Compare
max stresses
to material
allowables to
determine:
Is it safe,
will it fail???
Step 6. Stresses on other planes? Really nothing new, recall
Chapter 1: Average normal stress and shear stress (see
example 1-10):
Now, instead of stresses on “planes” transform stress at a point:
9.1 Stress Elements:
a) In general, can
have 6 independent
stresses (3 normal
and 3 shear) acting
at a point.
a) Many practical
engineering problems
involve only three
independent stresses –
called plane stress.
a) Stress state for
plane stress can be
summarized on a 2D
element.
9.2 Plane Stress Transformation:
In this course –force equilibrium in 2D)
Graduate course –force equilubruim in 3D but math solved with
matrices and tensors!!/ solve eigenvalue problem)
Derivation of the plane-stress transformation equations:
A. Given Plane Stress State:
B. What are new stresses at
element rotation of q ??:
Note, positive stress
directions shown.
Note, positive angle (ccw)
shown.
Easy, cut element, sum forces in the x’ and
y’ directions:
  FX '  0
s x ' A  t xy A sin q cosq  s y A sin q sin q
 t xy A cosq sin q  s x A cosq  cosq  0
s x' 
s x s y
2

s x s y
2
cos 2q  t xy sin 2q
s x' 
s x s y
2

s x s y
2
cos 2q  t xy sin 2q
(9.1)
  FY '  0
t x' y'  
s x s y
2
(9.2)
sin 2q  t xy cos 2q
Substitute q q  90 for q in equation9.1
s y' 
s x s y
2

s x s y
2
cos 2q t xy sin 2q
(9.3)
9.3 Principal Stresses and Maximum
In-Plane Shear Stresses
• What are the maximum stresses at a point?
• Will they be different than what’s shown
on the initial stress element?
Recall equation for normal stress at new angle q:
s x' 
s x s y
2

s x s y
2
cos 2q  t xy sin 2q
(9.1)
Want max and min stresses so what do we do?
ds x '
0
dq
t xy
tan2q P 
s x  s y / 2
This is the principal plane
Solving for q:
(9.4)
Next, plug 9.4 into 9.1 and get:
s 1, 2 
s x s y
2
Max
principal
 s x s y 
  t xy2
 /  
 2 
2
s1  s 2
Min
principal
(9.5)
So, to summarize:
Principal
Stress/Princip
al plane –
Note-shear
stresses are
zero on
principal
plane!
s 1, 2 
s x s y
2
 s x s y 
 /  

2
t xy
tan2q P 
s x  s y / 2
2
  t xy2

(9.5)
(9.4)
Next, do the same thing for shear stress:
t x' y'  
s x s y
dt x ' y '
dq
2
0
sin 2q  t xy cos 2q
(9.2)
Solve for angle, then plug
into eqn 9.2 to get:
tan2q S 
 s x  s y / 2
(9.6)
t xy
 s x s y 
  t xy2
t max  
in plane
 2 
2
s avg 
s x s y
(9.7)
(9.8)
2
1. On planes of max shear, normal stress is not zero but savg as shown in 9.8.
2. The planes for max shear stress can be determined by orienting an element
45deg from the position of an element of max principal stress!
Summary:
t xy
tan2q P 
s x  s y / 2
Max principal stress
s 1, 2 
s x s y
2
tan2q S 
t max
 s x s y 
  t xy2
 /  
 2 
2
 s x  s y / 2
t xy
 s x s y 
 
in plane

s avg 
2
s x s y
2
2
  t xy2

Max shear stress
Important Points:
• The principal stresses represent the maximum and minimum
normal stress at the point. These stresses are shown on the
principal stress element.
• When the state of stress is represented by the principal stresses,
no shear stress will act on the element.
• The state of stress at the point can also be represented in terms of
the maximum in-plane shear stress. In this case an average
normal stress will also act on the element. This is called the
maximum in-plane shear stress element.
• The maximum in-plane shear stress element is oriented 45
degrees from the principal stress element.
210/418/470 – Capstone Design Projects!!
D
K 1 / 3 Fy1 / 2
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