Transcript Normal Stress (1.1-1.5) - Department of Mechanical and Aerospace

Stress and Strain
(3.8-3.12, 3.14)
MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical & Aerospace Engineering
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Stress and Strain
Introduction

MAE 316 is a continuation of MAE 314 (solid mechanics)

Review topics




Beam theory
Columns
Pressure vessels
Principle stresses
New topics

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
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Contact Stress
Press and shrink fits
Fracture mechanics
Fatigue
Stress and Strain
Normal Stress (3.9)
Normal Stress (axial loading)

F

A
Sign Convention



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σ > 0 Tensile (member is in tension)
σ < 0 Compressive (member is in compression)
Stress and Strain
Shear Stress (3.9)

Shear stress (transverse loading)

“Single” shear
average shear stress
 ave 
P F

A A
“Double” shear

 ave
4
F
P
F
2
 

A
A 2A
Stress and Strain
Stress and Strain Review (3.3)
Bearing stress

P P
b 

Ab td
Single shear case
Bearing stress is a normal stress

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Stress and Strain
Strain (3.8)
Normal strain (axial loading)



L
Hooke’s Law

  E
Where E = Modulus of Elasticity (Young’s modulus)
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Stress and Strain
Stress and Strain Review (3.12)
Torsion (circular shaft)


Shear strain


T
L

Shear stress
T

J

Angle of twist
TL

GJ
Where G = Shear modulus (Modulus of rigidity)
and J = Polar moment of inertia of shaft cross-section
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Stress and Strain
Thin-Walled Pressure Vessels (3.14)
Thin-walled pressure vessels


Cylindrical
 t  1 

Spherical
pri
t
Circumferential “hoop”
stress
1   2 
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Stress and Strain
l  2 
pri
2t
Longitudinal stress
pri
2t
Beams in Bending (3.10)
Beams in pure bending
Strain


x  
y

y
 y  z 

Where ν = Poisson’s Ratio and
ρ = radius of curvature
Stress

My
x  
I
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Where I = 2nd moment of inertia of the
cross-section
Stress and Strain
Beam Shear and Bending (3.10-3.11)


Beams (non-uniform bending)
Shear and bending moment
dM
V
dx
dV
 w
dx

Shear stress
 avg 

VQ
It
Where Q = 1st moment of the cross-section
Design of beams for bending
 max 
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M
max
I
c

M
max
Factor of Safety 
S
Stress and Strain
allowable
applied
Example
Draw the shear and bending-moment diagrams for the beam and loading
shown.
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Stress and Strain
Example Problem
An extruded aluminum beam has the cross section shown. Knowing that
the vertical shear in the beam is 150 kN, determine the shearing stress
a (a) point a, and (b) point b.
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Shear Stress in Beams
Combined Stress in Beams

In MAE 314, we calculated stress and strain for each type
of load separately (axial, centric, transverse, etc.).

When more than one type of load acts on a beam, the
combined stress can be found by the superposition of
several stress states.
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Stress and Strain
Combined Stress in Beams

Determine the normal and shearing stress at point K.
The radius of the bar is 20 mm.
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Stress and Strain
2D and 3D Stress
(3.6-3.7)
MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical & Aerospace Engineering
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Stress and Strain
Plane (2D) Stress (3.6)

Consider a state of plane stress: σz = τxz = τyz = 0.
φ
φ
φ
φ
φ
φ
φ
Slice cube at an angle φ to the x-axis
(new coordinates x’, y’)
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Stress and Strain
φ
φ
Plane (2D) Stress (3.6)

Sum forces in x’ direction and y’ direction and use trig
identities to formulate equations for transformed stress.
 x' 
x  y
2

 x  y
2
cos  2    xy sin  2 
φ
 y' 
x  y
2
 x' y'  
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
 x  y
2
 x  y
2
cos  2    xy sin  2 
φ
φ
φ
φ
sin  2    xy cos  2 
Stress and Strain
Plane (2D) Stress (3.6)

Plotting a Mohr’s Circle, we can also develop equations
for principle stress, maximum shearing stress, and the
orientations at which they occur.
 max,min 
 x  y
2
  x  y 
   xy2
 
 2 
2
 max   ave  R
 min   ave  R
 max  R
 min   R
  x  y 
   xy2
  
 2 
2
 max,min
tan  2P  
2 xy
 x  y
 x  y
tan  2S   
2 xy
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Stress and Strain
Plane (2D) Strain

Mathematically, the transformation of strain is the same as
stress transformation with the following substitutions.
   and    2
 x' 
 y' 
x  y
2
x  y
2


x y
2
x y
2
cos  2  
cos  2  
 xy
2
 xy
2
sin  2 
 max,min 
sin  2 
 x ' y '     x   y  sin  2    xy cos  2 
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x y
tan  2P  
2
  x   y    xy 
  

 
 2   2 
 xy
x  y
Stress and Strain
2
2
3D Stress (3.7)

Now, there are three possible principal stresses.

Also, recall the stress tensor can be expressed in matrix
form.
 x  xy  xz 


 yx  y  yz 
 zx  zy  z 


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Stress and Strain
3D Stress (3.7)

We can solve for the principle stresses (σ1, σ2, σ3) using a
stress cubic equation.
 i  I1 i  I 2 i  I3  0
3

2
Where i = 1,2,3 and the three constant I1, I2, and I3 are
expressed as follows.
I1   x   y   z
I 2   x y   x z   y z   xy   yz   xz
2
2
2
I 3   x y z  2 xy yz xz   x yz   y xz   z xy
2
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Stress and Strain
2
2
3D Stress
Let nx, ny and nz be the direction cosines of the normal vector to surface ABC
with respect to x, y, and z directions respectively.
F
F
F
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x
0

( nx ) A   x (nx A )   xy (n y A)   xz (nz A)  0
y
0

( n y ) A   xy (nx A)   y (n y A )   yz (nz A)  0
z
0

( nz ) A   xz (nx A)   yz (ny A)   z (nz A )  0
Stress and Strain
3D Stress (3.7)



How do we find the maximum shearing stress?
The most visual method is to observe a 3D Mohr's
Circle.
Rank principle stresses largest to smallest: σ1 > σ2 > σ3
 max 
σ2
σ3
σ1
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Stress and Strain
1   3
2
3D Stress

(5.5)
A plane that makes equal angles with the principal planes
is called an octahedral plane.
1
3
1

( 1   2 ) 2  ( 2   3 ) 2  ( 3   1 ) 2
3
 oct  ( 1   2   3 )
 oct
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Stress and Strain
3D Stress (3.7)

For the stress state shown below, find the principle
stresses and maximum shear stress.
Stress tensor
9 4 0 
4  6 0


0 0 0 
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Stress and Strain
3D Stress (3.7)

Draw Mohr’s Circle for the stress state shown below.
Stress tensor
0 
60 30
30 35 0 


 0 0  25
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Stress and Strain
Curved Beams
(3.18)
MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical & Aerospace Engineering
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Stress and Strain
Curved Beams (3.18)


Thus far, we have only analyzed stress in straight beams.
However, there many situations where curved beams are
used.
Hooks
Chain links
Curved structural beams
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Stress and Strain
Curved Beams (3.18)

Assumptions




Pure bending (no shear and axial forces present – will add
these later)
Bending occurs in a single plane
The cross-section has at least one axis of symmetry
What does this mean?



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σ = -My/I no longer applies
Neutral axis and axis of symmetry (centroid) are no longer the
same
Stress distribution is not linear
Stress and Strain
Curved Beams (3.18)
Flexure formula for tangential stress:

My
Ae(rn  y)
Where
M = bending moment about centroidal axis (positive M puts inner surface in
tension)
y = distance from neutral axis to point of interest
A = cross-section area
e = distance from centroidal axis to neutral axis
rn = radius of neutral axis
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Stress and Strain
Curved Beams (3.18)

If there is also an axial force
present, the flexure formula
can be written as follows.


P
My

A Ae(rn  y )
Table 3-4 in the textbook
shows rn formulas for several
common cross-section shapes.
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Stress and Strain
Example
Plot the distribution of stresses across section A-A of the crane hook shown
below. The cross section is rectangular, with b=0.75 in and h =4 in, and the
load is F = 5000 lbf.
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Stress and Strain
Curved Beams (3.18)

Calculate the tangential stress at A and B on the curved
hook shown below if the load P = 90 kN.
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Stress and Strain