Chapter 3 – Stress and
Deformation Analysis (ref
MCHT 213!!)
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Strength of Materials can really
be divided into 2 categories:
1. Stress analysis:
– Structure exists, material and loading known
• IS IT SAFE?????
2. Design:
– Determine geometry OR material based on an
allowable stress (i.e. Sy/4).
– Generally certain aspects are fixed.
– Much more involved than analysis – why??
– Ref Chapter 1 – start with design
requirements/functions/evaluation criteria,etc.
Show overhead – trailer – analysis or design? how would you analyze it?
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Possible modes of failure
(mechanical)???
1.
2.
3.
4.
5.
6.
Fracture (s >> Su)
Yield (s >> Sy
Instability (buckling)
Fatigue and wear
Excessive deformation (i.e. too soft)
Creep or stress relaxation (polymers)
1,2,4 – most important parameter? STRESS!!!!!!
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Definition of
Definition of Stress:
•Internal Force per unit area
•Intensity of internal force on a specific plane (area) passing
through a point
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Stress States:
Preferred stress
element for 2D stress:
a) In general, can have 6
independent stresses (3
normal and 3 shear)
acting at a point.
b) Many practical
engineering problems
involve only three
independent stresses –
called plane stress.
b) Stress state for plane
stress can be
summarized on a 2D
element.
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Quick Review: Basic Types of Stress (ref: MCHT213)
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AXIAL MEMBERS: Average Normal Stress (aka Direct
Normal Stress):
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3.4 Average Normal Stress:
Requirements for Average Normal Stress, s = P/A:
1. Member starts out straight and remains straight after loading
2. Homogenous, isotropic
3. Invoke St. Venant’s Principal
EXAMPLES of AVERAGE NORMAL STRESS:
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Example Normal Stress:
If P = 20K lbs
and A = 2 in2
s=?
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Average Shear Stress (AKA Direct
Shear Stress):
t = P/As
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Average Shear Stress:
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Example Shear Stress:
If load = 2,000 lb and bolt
diameter = ½”, Find shear
stress in bolt. What if
double shear, what would
the new shear stress
be??
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Example: normal stress and shear stress:
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Example 2 – direct normal and
shear stress
DISCUSS ONLY!
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1.119
St. Venant’s Principle and Stress Concentration Factors, Kt:
=
St Venant’s:
1.
Stress profile, sufficiently removed from
the local effect of loads will be uniform
(i.e. = P/A)
2.
Stress and strain produced by statically
equivalent load system will be the same.
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St. Venant’s Principal:
Note, def’m of grid uniform
at middle, therefore strain
and stress will be uniform.
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Example:
Look at deformation in
vicinity of hole. Is it
uniform???
Stress profile is not
uniform. smax occurs at area
of discontinuity.
smax > savg
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smax = Kt* savg
smax = maximum stress
savg = average stress (P/Amin)
Kt = stress concentration factor
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smax = Kt* savg
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Find: Max Stress:
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Deformation of Axial Member with Constant Load and
Cross-Sectional Areas:
Deformatio n   
PL
AE
Or, for multiple sections:
 total 

PL
AE
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Example: Multiple sections. Find total deformation of end
A with respect to D. Area = 20 mm2. Material is steel w/
E = 200 GPa = 200 x 109 Pa:
= 100 mm
= 150 mm
PL
 A/D 

 A/D 
PAB L AB
AE
= 200 mm
AE

PBC L BC
AE

PCD L CD
AE
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First, solve for
internal loads:
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 A/D 
 A/D 
 A/D 
PAB L AB

PBC L BC
AE
AE
( 5 , 000 N )(. 1m )

AE

AE
(  3 , 000 N )(. 15 m )
AE
 1,350

PCD L CD

(  7 , 000 N )(. 2 m )
AE
 1,350
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AE
  3 . 38 x10
4
m   0 . 338 mm
(. 00002 )( 200 x10 )
How would this answer change if aluminum instead of steel????
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TORSION:
Key points:
Varies linearly with radius, .
Zero at center
Max at outer fiber ( = c)
Constant for given .
Solid vs. hollow
Torsion of non-circular
sections.
7. Now how to calculate torque
given power and rotational
speed.
1.
2.
3.
4.
5.
6.
 
t   t max
 c 
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The torsion formula (see derivation):
Torque (N-m, Nmm or lb-in, lbft, etc)
t max 
Tc
Outer radius of
shaft (m or in)
J
Polar moment of
inertia (m4 or in4)
Max shear stress
in shaft (MPa,
psi/ksi, etc.)
or
t 
T
J
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J = polar moment of inertia
• Solid shaft:
J 

c
•Hollow shaft:
4
J 
2

2
c
4
o
c
4
i

W
For Design:
T = P/n
r/s
N-m
rpm
T = 63,000 P/n
lb-in
hp
Zp 
J
c


c
3
2
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Stress Profiles:
Shear stress profile – YOU
MUST UNDERSTAND
THIS!!!!
Where is shear stress max?
zero? How does it vary along
the length and
circumference?
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Angle of twist - For straight sections:
Torque
 
Angle of twist
(rad)
Polar moment of inertia
Length
TL
JG
Modulus of Rigidity
(Shear Modulus) – see
back of book
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Angle of twist for Multiple
Sections:
 
TL
JG
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If the distance between gear E and the middle gear is 12 inches, find the angle of twist between the two
gears. The shaft is steel and G = 11.5 x 106 psi.
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5.8 Stress Concentrations (last
topic we’ll cover in Chapter 5)
Consider the torsion member only
(shaft) where do you think the
stress concentrations are??
Again, stress concentrations occur
where there’s an abrupt change in
geometry!
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How do we deal with stress
concentrations??
t max  K  t nom  K
Tc
J
Based on smaller of two connected shafts
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Torsion of non-circular cross-sections:
t 
T
Q
TL
 
GK
Where Q and K are determined based on
cross-section from F3-10
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Also, see equations for closed thin walled tubes! See HO: Stress Analysis 2 examples
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Example: Torsion find max shear
stress for the three cross-sections:
a, b and c
T = 4.1 N-m =
4,100 N-mm
a. Circular w/
dia = 10 m
b. Square w/
side = 8.86
mm
c. Hollow w/
od = 12.8 mm
and id = 8
mm
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Also see HO: Comparison of Torsion Elements, also overhead fabricated beam
Shear and Moment diagrams:
 V    w ( x ) dx
 M   V ( x ) dx
Do not get actual equations, good if
just after Vmax and Mmax
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Example: Draw Shear & Moment
diagrams for the following beam
12 kN
8 kN
A
C
D
B
1m
RA = 7 kN 
3m
1m
RC = 13 kN 
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12 kN
8 kN
A
C
D
B
1m
3m
8
7
V
1m
8
7
-15
(kN)
-5
7
M
(kN-m)
2.4 m
-8
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Example: Beam w/ Concentrated Moment:
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Also see HO: Simple Beam with Uniform Load, Load-Shear-Moment Relationships
Beam Bending Stress: The Flexure Formula:
Internal bending
moment, lb-in
Max bending stress,
psi
s max 
Mc
I
Or in general:
Distance from NA to
outer fiber, in
Moment of inertia, in4
s 
My
I
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Examples:
• Find maximum moment
• Find area properties, I and c
• Calculate stress
See HO: Bending Stress Concepts
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WHERE IS
BENDING STRESS
MAXIMUM???
Answer:
•Outer surface
(furthest away from
Neutral Axis)
•Value of x along
length where moment
is maximum!!
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Beam Shear Stress:
See HO: Shear Stress
Calculations in Beams
Internal Shear (lb)
t 
VQ
First Moment of area
(in3) at point of interest
It
Moment of inertia of
entire cross section (in4)
Thickness of crosssection at point of
interest (in)
Q  y ' A '
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Find: Max Shear and Bending
Stress:
1.93”
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See HO: Stress Analysis 1 Examples, Steel Beam Selection, Steel W-Shape Selection Data
Combined Loading:
• Look at each load individually and solve
for stress at a given point due to that load.
• Repeat for all loads.
• Add like stresses
• Summarize stresses on an initial stress
element.
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Example: Combined normal stress – find stress in horizontal portion:
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Example: Combined normal stress – find stress in horizontal portion:
Reduce to simple
cantilever!
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Ultimate Combined Loading Problem!!
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Final Concept: Beam Deflection – Superposition:
No solution for this case
Known case from App C
Known case from App C
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