1
1
Introduction to Deformable
Mechanics
By the end of this lesson, you should be able to:
• Calculate Normal and Shear stress for simple
problems
2
Outline
• Stress
• Strain
• Material Properties
• Shear
3
Names of “Mechanical” Courses
• Biomechanics
– 1
st
half of course: called Statics
– 2
nd
half of course: called:
• Mechanics of Materials
• Strength of Materials
• Mechanics of Deformable bodies
– 2
nd
semester: called Dynamics
– Later? Material Properties
Fluid Dynamics
2
4
Mechanics of Materials
• Any force deforms a load
• Mechanics of Materials calculates:
– Deformation
– Internal forces
These calculations let you know if your object will
break, and where it will break
Finite Element Analysis (FEA) is often more accurate,
but you should still know the basic equations
5
Normal Stress
• Externally, we look at:
– Forces
– Displacements
• Internally, we look at
– Normalized Forces
– Normalized Displacements
A
F
=
σ
Stress
Force
Area
6
Normal Strain
• Normal strain is dimensionless
L
δ
ε
=
Strain
Deformation
Length
L
δ
F
3
7
Conventions
• Tension
– Object becomes longer
– Positive
• Compression
– Object becomes
shorter
– Negative
• Normal stress / strain
– stress acts
perpendicular to
surface
8
Example
• A short post constructed from a hollow circular tube
of aluminum supports a compressive load of 54 N.
The inner and outer diameters are:
• D1=3.6 cm
• D2 = 5 cm
• Length = 50 cm
• The post shortens .022 cm
• What is the compressive stress and strain
54 N
40 cm
( )
( )
2 2 2 2 2
2 1
5 3.6 9.456 cm
4 4
A d d
π π
= − = − =
2
2
54
5710
9.456
F N
N cm
A cm
σ
= = =
6
0.022
550 10
40
cm
L cm
δ
ε
−
= = = ×
9
Measuring Stress and Strain
4
10
Stress Strain Diagram
11
Stress Strain Diagram
• P: Proportional Limit
• OP: Stress is linear to Strain
• OE: Object is Elastic. If you remove
force, it will return to original shape• Y: Yield point. Object will deform
without any added load• U: Ultimate Stress (Ultimate strength)
•Slope of OP is called Modulus of
Elasticity, or Young’s Modulus
12
Stress Strain Diagram
5
13
Plastic Deformation
• Object does not return to original shape
14
Creep
Common at high temperatures
15
Linear Elasticity
E
σ ε
=
Modulus of Elasticity (Young’s modulus)
6
16
Poisson’s Ratio
• Pulling a bar in one direction:
– Makes it longer in that direction
– Shorter in transverse direction
lateral strain
axial strain
ν
= −
Poisson’s ratio
ν= 0.5 : Incompressible (rubber)
ν = 0.4 – 0.45: Many biological materials
ν = 0.25 – 0.35: Metals
ν = 0.1: Concrete
ν = 0: Cork
17
Example:
A steel pipe has:
• length L = 40 cm
• OD = 6 cm
• ID = 4.5 cm.
• It is compressed by force F=100 N.
• The material has modulus of
elasticity E = 70 kPa
• Poisson’s ratio ν=0.30
• Determine the following:
– Shortening: δ
– Lateral strain: ε’
– New OD
– New ID
18
Solution
( )
( )
2 2 2 2 2
2 1
6 4.5 12.37 cm
4 4
A d d
π π
= − = − =
2
2
100
8.08 80.8
12.37
F N
N cm kPa
A cm
σ
= = = =
80.8
.00115
70
kPa
E MPa
σ
ε
= = =
(
)
.00115 40 0.046
L cm cm
δ ε
= = =
(
)
4
'3.46 10 6 0.0021
6.0021
OD OD cm cm
OD cm
ε
−
Δ = = × =
=
(
)
4
'0.30.00115 3.46 10
ε νε
−
= − = − = ×
(
)
4
'3.46 10 4.5 0.0016
4.5016
ID OD cm cm
OD cm
ε
−
Δ = = × =
=
7
20
Double Shear
Average Bearing Stress
b
b
b
F
A
σ
=
Area = Cross sectional area
Average Shear Force
2
P
V
=
Because V + V  P=0
21
Single Shear
Average Shear Force
V P
=
(because V – P = 0)
22
Shear Stress
• Normal Stress acts perpendicular to surface
• Shear Stress acts tangential to surface
V
A
τ
=
Shear Stress
Shear Force
Area
Normal Force
Shear Force
8
23
Free Body Diagram Observations
1) ΣF: Shear stress on opposite faces of an element
are equal in magnitude and opposite in direction
2) ΣM:Shear stresses on adjacent faces are equal in
magnitude. Both forces point toward, or both faces
point away, from the line of intersection of the
faces.
24
Shear Strain
• Shear stresses do not change length of sides
• Shear stress change shape of object
• Shear Strain: Angular deformation γ (gamma)
25
+

Shear Conventions
• A positive face has its outward normal directed
in the positive direction of a coordinate axis
x
y
z
+

+

9
26
Shear Conventions
• Shear Stress:
– Positive Face: Positive Shear stress points in positive
direction
– Negative Face: Positive Shear stress points in
negative direction
• Shear Strain:
– Positive if angle between 2 positive faces is reduced
– Positive if angle between 2 negative faces is reduced
27
Hook’s Law in Shear
G
τ γ
=
Shear Modulus of Elasticity
( )
2 1
E
G
ν
=
+
28
Example
Strut transmits compressive force of P=54 kN to deck. Wall
thickness t=12 mm. Angle θ=40°. A pin through the strut
transmits the force to 2 gussets G welded to base plate B. Four
anchor bolts fasten the base plate to the deck.D
pin
= 18 mm. t
G
= 15 mm. t
B
= 8 mm. d
bolt
= 12 mm.
Determine the following stresses:
a) bearing stress between strut and pin
b) shear stress in pin
c) bearing stres between pin
and gussets
d) bearing stress between
anchor bolts and base plate
e) shear stress in anchor bolts
10
29
Solution: a & b)
a) Bearing stress between strut
and pin
( )( )
1
54
125
2 2 12 18
b
pin
P kN
MPa
td mm mm
σ
= = =
b) Shear stress in pin
( )
22
54
106
2 2 4
2 18 4
pin
pin
P P kN
MPa
A d
mm
τ
π
π
= = = =
Double shear
30
Solution: c & d)
a) Bearing stress between pin
and gussets
( )( )
2
54
100
2 2 15 18
b
G pin
P kN
MPa
t d mm mm
σ
= = =
b) Bearing stress between anchor bolts and base plate
(
)
(
)
( )( )
3
54 cos 40
cos 40
108
4 4 8 12
b
B bolt
kN
P
MPa
t d mm mm
σ
= = =
4 bolts
Horizontal force
31
Solution: e)
e) Shear stress in anchor bolts
(
)
(
)
( )
22
54 cos 40
cos 40 cos 40
91.4
4 4 4
4 12 4
bolt
bolt
kN
P P
MPa
A d
mm
τ
π
π
= = = =
4 bolts
Note: any friction would reduce load on anchor bolts
11
32
Example
Obtain formulas for the average shear stress
and horizontal displacement d o the plate.
33
Solution
For small angles
V
ab
τ
=
V
G abG
τ
γ
= =
tan tan
V hV
d h h
abG abG
γ
= = ≈
34
Typical Exam question
A punch for making holes in steel plates is shown below.
The punch has diameter d = 0.75 in. It is used to punch a
hole in a ¼inch plate. If a force P = 28,000 lb is required
to create the hole, what is the average shear stress in
the plate and the average compressive stress in the
punch?
Hint (shear area is tricky!)
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