Stress and Strain
MAE 314 – Solid Mechanics
Y. Zhu
Slide 1
Stress and Strain
Introduction to Normal Stress
Stress = Force per unit area
F
σ=
A
Slide 2
Stress and Strain
Introduction to Normal Stress – cont’d
• If the stress varies over the cross-section, we can write
the stress at a point as
σ = lim
ΔA→0
ΔF
ΔA
• We assume the force F is evenly distributed over the
cross-section of the bar. In reality F = resultant force
over the end of the bar.
∫ σdA = F
A
Slide 3
Stress and Strain
Introduction to Normal Stress – cont’d
Sign convention
σ >0
σ <0
Tensile (member is in tension)
Compressive (member is in compression)
Units (force/area)
English: lb/in2 = psi
kip/in2 = ksi
SI:
Slide 4
Tensile
Compressive
N/m2 = Pa (Pascal)
kN/ 2 = kPa
kN/m
kP
MPa, GPa, etc.
Stress and Strain
Introduction to Normal Stress – cont’d
•
•
Homogenous: material is the same throughout the bar
Cross-section: section p
perpendicular
p
to longitudinal
g
axis of bar
A
P’
•
P
Prismatic: cross-section does not change along axis of bar
Prismatic
Slide 5
Non-Prismatic
Stress and Strain
Definitions and Assumptions - cont’d
•
Uniaxial bar: a bar with only one axis
•
Normal Stress (σ): stress acting perpendicular to the cross-section.
•
Deformation of the bar is uniform throughout. (Uniaxial Stress State)
•
Stress is measured far from the point of application
application.
•
Loads must act through the centroid of the cross-section.
Let’s expand
p
these last two assumptions…
p
Slide 6
Stress and Strain
Definitions and Assumptions - cont’d
• The uniform stress state does not apply near the ends of
the bar.
• Assume the distribution of normal
stresses in an axiallyy loaded
member is uniform, except in the
immediate vicinity of the points of
application of the loads.
“Uniform” Stress
Saint-Venant’s Principle
Slide 7
Stress and Strain
Definitions and Assumptions - cont’d
• How do we know all loads must act through the centroid
of the cross-section?
cross section?
• Let us represent P, the resultant force, by a uniform
stress
t
over the
th cross-section
ti (so
( that
th t they
th are statically
t ti ll
equivalent).
Slide 8
Stress and Strain
Definitions and Assumptions - cont’d
• Moments due to σ:
M x = ∫ yσdA
A
M y = − ∫ xσdA
A
• Set M
Myy == M
Myy
Mx x== M
Mxx and M
1
1
y = ∫ yσdA = ∫ ydA
PA
AA
1
1
x = ∫ xσdA = ∫ xdA
PA
AA
Slide 9
Stress and Strain
Right hand rule
Equations
q
for the centroid
Example Problem 1
Boom AB with a 30x50-mm rectangular cross section, rod BC
with a 20-mm-diameter circular cross section. Calculate forces
AB and BC.
Slide 10
Stress and Strain
Procedure
0a. Given/Wanted + Sketch
0b. Plan
• Coordinate System
• Free Body Diagram
• Equilibrium Equations (Statics)
• Count number of unknowns and equations
• Solution
S l ti
• Check your solution !!
N t
Note:
• Use variables instead of numbers
• Units!
Slide 11
Stress and Strain
Example Problem 1 Solution
Slide 12
Stress and Strain
Example Problem 2
•
Two solid cylindrical rods AB and BC are welded together at B
and loaded as shown. Determine the average normal stress at
the midsection of (a) rod AB and (b) rod BC.
BC
40 kips
Slide 13
Stress and Strain
Example Problem 2 Solution
Slide 14
Stress and Strain
What is Shearing Stress?
• Last time we talked
about normal stress (σ),
which acts
perpendicular to the
cross-section.
• Shear stress (τ) acts
tangential to the surface
off a material
t i l element.
l
t
Slide 15
Normal stress results
in a volume change.
Shear stress results
in a shape change.
Stress and Strain
Where Do Shearing Stresses Occur?
• Shearing stresses are commonly found in bolts, pins, and
rivets.
Bolt is in “single” shear
Free Body Diagram of Bolt
Force P results in shearing stress
Force F results
F
lt in
i bearing
b i stress
t
(will discuss later)
Slide 16
Stress and Strain
Shear Stress Defined
• We do not assume τ is uniform over the cross-section,
because this is not the case.
• Therefore, τ is the average shear stress.
P F
τ ave = A = A
• Th
The maximum
i
value
l off τ may be
b considerably
id bl greater
t
than τave, which is important for design purposes.
Slide 17
Stress and Strain
Double Shear
Bolt is in “double”
double shear
Free Body Diagram of Bolt
Free Body Diagram of Center
of Bolt
τ ave =
Slide 18
Stress and Strain
F
P
F
= 2=
A
A 2A
Bearing Stress
•
•
•
Bearing stress is a normal stress, not a shearing stress.
B i stress
Bearing
t
i in
is
i th
the members
b
th t a bolt
that
b lt connects
t (not
( t
in the bolt itself), along a bearing surface.
Thus,
P
σb =
Ab
Force F results in bearing stress
where
Ab = projected area where bearing pressure is applied
P = bearing force
Slide 19
Stress and Strain
Bearing Stress - cont’d
• For “single
g shear” case
P P
σb =
=
Ab td
Slide 20
Stress and Strain
Example Problem 1
•
Determine (a) the average shearing stress in the pin at B, (b) the
average bearing stress at B in member BD, and (c) the average
b i stress
bearing
t
att B iin member
b ABC
ABC.
Member ABC has
rectangular crosssection 10x50 mm
Each pin has
16 mm diameter
Each vertical member
has rectangular crosssection 8x36 mm
Slide 21
Stress and Strain
Example Problem 1 Solution
Slide 22
Stress and Strain
Example Problem 2
• A load P=10 kips is applied to a rod supported as shown
by a plate with a 0.6 in. diameter hole. Determine the
shear stress in the disk and the plate.
1.6 in
Disk
0.4 in
0.25 in
Plate
0.6 in
Rod
Slide 23
Stress and Strain
Example Problem 2 Solution
Slide 24
Stress and Strain
Stress on an Oblique Plane
• What have we learned so far?
– Axial forces in a two-force member cause normal stresses.
– Transverse forces exerted on bolts and pins cause shearing
stresses.
t
Slide 25
Stress and Strain
Stress on an Oblique Plane - cont’d
• However,, axial forces cause both normal and shearing
g
stresses on planes which are not perpendicular to the
axis.
• This is also the case for transverse forces exerted on a
bolt or pin.
Slide 26
Stress and Strain
Stress on an Oblique Plane - cont’d
•
Consider an inclined section of a uniaxial bar.
⇒
•
The resultant force in the axial direction must equal P to satisfy
equilibrium
equilibrium.
•
The force can be resolved into components perpendicular to the
section F
section,
F, and parallel to the section
section, V
V.
F = P cosθ
•
V = P sin θ
The area of the section is
A0 = Aθ cos θ ⇒ Aθ = A0 / cos θ
Slide 27
Stress and Strain
Stress on an Oblique Plane - cont’d
•
We can formulate the average normal stress on the section
as
σ=
•
The average shear stress on the section is
τ=
•
Slide 28
F
P cosθ
P
=
= cos 2 θ
Aθ A0 / cosθ A0
V
P sin θ
P
=
=
sin θ cos θ
Aθ A0 / cos θ A0
Thus, a normall force
Th
f
applied
li d tto a b
bar on an iinclined
li d section
ti
produces a combination of shear and normal stresses.
Stress and Strain
Stress on an Oblique Plane - cont’d
• Since σ and τ are functions of sine and cosine, we know the
maximum and minimum values will occur at θ = 00, 450, and
900.
τ=
σ=
Slide 29
P
sin θ cos θ
A0
P
cos 2 θ
A0
At θ=±900
σ=0
At θ=±900
τ=0
At θ=±450
σ=P/2A0
At θ=±450
τ=P/2A0 (max)
At θ=00
σ=P/A0 (max)
At θ=00
τ=0
Stress and Strain
Stress on an Oblique Plane - cont’d
• Let’s interpret this visually.
Slide 30
Stress and Strain
Stress on an Oblique Plane - cont’d
• Sign
g convention for p
positive normal and shear stresses:
• This fits with our previous convention for θ = 0º.
Slide 31
Stress and Strain
Design Considerations
• From a design perspective, it is important to know the
g
load which a material can hold before failing.
g
largest
• This load is called the ultimate load, Pu.
• The stress equations are the same as before with P=Pu.
Ultimate normal stress is denoted as σu and ultimate
shear stress is denoted as τu.
Slide 32
Stress and Strain
Design Considerations - cont’d
• Often the allowable load is considerably smaller than the
ultimate load.
• It is a common design practice to use factor of safety.
F .S . =
•
Slide 33
ultimate load
allowable load
F .S . =
ultimate stress
allowable stress
The two equations above are identical when a linear relationship
exists between the load and the stress. In real-world engineering
application, however, this relationship ceases to be linear as the load
approaches its ultimate value
value.
Stress and Strain
Example Problem 1
• Two wooden members are joined by the simple glued
scarf splice shown below. Knowing that the maximum
allowable shearing stress in the glued splice is 75 psi
psi,
determine the largest axial load P that can be safely
applied.
5.5 in
3.5 in
Slide 34
Stress and Strain
Example Problem 1 Solution
Slide 35
Stress and Strain
Example Problem 2
• Knowing the ultimate load for cable BD is 25 kips and a
factor of safety of 3.2 with respect to cable failure is
required determine the magnitude of the largest force P
required,
which can be safely applied to member ABC.
15 in
18 in
Slide 36
12 in
Stress and Strain
Example Problem 2 Solution
Slide 37
Stress and Strain
Introduction to Normal Strain
• Normal strain (ε) is defined as the deformation per unit length
of a member under axial loading.
ε=
δ
L
• Normal strain is dimensionless but can be expressed in
several ways. Let’s say L = 100 mm and δ = 0.01 mm.
–
–
–
–
ε = 0.01 mm / 100 mm = 1 x 10-4 or 100 x 10-6
ε = 100 μ (read as 100 microstrain)
ε = 1 x 10-4 in/in (if using English units)
ε = 1 x 10-4 * 100 = 0.01%
• Normal strain is also called extensional strain
Slide 38
--- Uniaxial Loading ---
Thermal Strain
•
Changes in temperature produce expansion or compression,
which cause strain.
ε T = αΔT
–
–
–
•
α = coefficient of thermal expansion
ΔT = change in temperature
Sign convention: expansion is positive (+), contraction is
negative (-)
For a bar that is completely free to deform (one or both ends
free):
δ T = ε T L = α (ΔT ) L
•
Slide 39
In this case, there is thermal strain but no thermal stress!
--- Uniaxial Loading ---
Mechanical Properties of Materials
•
We want to
develop a method
of analysis that is
characteristic of
the properties of
materials (σ and ε)
rather than the
dimensions or load (δ and P) of a particular
specimen.
specimen
•
Why?
σ & ε are truly material properties
– P & δ are specimen properties
Slide 40
--- Uniaxial Loading ---
Mechanical Properties of Materials - cont’d
• Stress and strain can be measured, so we want to
develop a relationship between the two for a given
material.
material
• How do we calculate the elongation of a bar due to
l di ?
loading?
– Apply force P
– Calculate σ = P/A
– Use material relation ε = f(σ) to calculate ε
– Calculate δ = εL
Will expand on
this in a moment
Slide 41
--- Uniaxial Loading ---
Stress-Strain Diagram
•
Material behavior is generally represented by a stress-strain diagram,
which is obtained by conducting a tensile test on a specimen of material.
Slide 42
--- Uniaxial Loading ---
Stress-Strain Diagram - cont’d
•
Stress-strain diagrams of various materials vary widely.
•
Different tensile tests conducted on the same material may yield
different results depending on test conditions (temperature, loading
speed, etc.).
•
Divide materials into two broad categories:
•
Slide 43
–
Ductile material - Material that undergoes large permanent
strains before failure (e.g. steel, aluminum)
–
Brittle material - Material that fails with little elongation after
yield stress (e.g. glass, ceramics, concrete)
Let’s examine the stress-strain diagram for a typical ductile material
(low-carbon steel) region by region.
--- Uniaxial Loading ---
Stress-Strain Diagram - cont’d
• Linear region
– Stress-strain response
is linear
– Slope = Modulus of
Elasticity (Young’s
modulus) = E
– E has units of force
per unit area (same as
stress))
– We get a relation
between stress and
strain known as
H k ’ L
Hooke’s
Law.
σ = Eε
Slide 44
--- Uniaxial Loading ---
Stress-Strain Diagram - cont’d
• Yielding region
– Begins at yield stress σY
– Sl
Slope rapidly
idl d
decreases
until it is horizontal or
near horizontal
– Large strain increase
increase,
small stress increase
– Strain is permanent
Slide 45
--- Uniaxial Loading ---
Stress-Strain Diagram cont’d
• Strain Hardening
– After undergoing large
deformations the metal
deformations,
has changed its
crystalline structure.
– The material has
increased resistance
to applied stress
(it appears to be
“harder”).
Slide 46
--- Uniaxial Loading ---
Stress-Strain Diagram - cont’d
• Necking
– The maximum supported
stress value is called the
ultimate stress, σu.
– Loading beyond σu
results in decreased
load supported and
eventually rupture.
Slide 47
--- Uniaxial Loading ---
Stress-Strain Diagram - cont’d
• Why does the stress appear to drop during necking?
• If we measure the true area
area, the graph looks like:
true stress
•
The difference
is in the area:
t
true
stress
t
takes
t k
into account the
decreased crosssection area.
•
Thus, at the same
stress level, the
load drops.
Slide 48
--- Uniaxial Loading ---
x
Offset Method
•
For some materials (e.g. aluminum) there is not a clear yield stress.
•
We can use the offset method to determine σY.
•
Choose the offset (0.002 is shown here).
– Draw a line with slope E, through
the point (0.002, 0).
– σY is given by the intersection of
this line with the stress-strain curve.
Slide 49
--- Uniaxial Loading ---
Elastic vs. Plastic
•
A material is said to behave elastically if the strain caused by the
application of load disappear when the load is removed – it returns to
g
state.
its original
•
The largest value of stress for which the material behaves elastically
is called the elastic limit (essentially the same as σY in materials with
a well-defined yield point).
•
Once the yield stress has been obtained, when the load is removed,
the stress and strain decrease linearly but do not return to their
original state.
•
This indicates plastic deformation.
•
When a material does not have a well-defined yield point, the elastic
limit can be closely approximated using the offset method.
Slide 50
--- Uniaxial Loading ---
Elastic vs. Plastic - cont’d
σY
σY
Reload
Plastic deformation (Permanent strain)
Slide 51
--- Uniaxial Loading ---