Chapter objectives
• Review some important principles of
statics
• Use the principles to determine internal
resultant loadings in a body
• Introduce concepts of normal and shear
stress
• Discuss applications of analysis and
design of members subjected to an
axial load or direct shear
Chapter 1: Stress
1
Chapter Outline
1.
2.
3.
4.
Introduction
Equilibrium of a deformable body
Stress
Average normal stress in an
axially loaded bar
5. Average shear stress
6. Allowable stress
7. Design of simple connections
Chapter 1: Stress
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1
Introduction
Read sections: 1.1, 1.2, 1.3, 1.4, 1.5, and 1.6
Mechanics of Materials: is a branch of mechanics that
studies the internal effects of stress and strain in a solid
body. It relates the effect of external loads to the
intensity of the internal loads within the body.
Stress: is associated with the strength of the material
from which the body is made.
Chapter 1: Stress
3
Equilibrium of a Body
External Forces
1. Surface Forces
- Caused by direct contact of
other body’s surface
2. Body Forces
- Other body exerts a force
without contact
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2
Equilibrium of a Body
Reactions: Forces developed at the contact surfaces
between two bodies.
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Equilibrium of a Body
Equations of Equilibrium
 Equilibrium of a body requires a balance of forces and a
balance of moments
F  0
M  0
O
 For a body with x, y, z coordinate system with origin O,
F  0, F  0, F  0
M  0,M  0,M  0
x
y
x
z
y
z
 Best way to account for these forces is to draw the body’s
free-body diagram (FBD).
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3
Free-body Diagram
• A free body diagram is a schematic drawing of a
particular object showing the external forces and
couples acting on it.
• The object is said isolated or freed from its
surroundings.
 Identify the object
 Sketch the isolated object
 Show the external forces and couples acting on
object
Chapter 1: Stress
7
Equilibrium of a Body
Equations of Equilibrium:
𝐹
0
𝐹
0
𝐹
0
𝑀
0
𝑀
0
𝑀
0
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4
Equilibrium of a Body
Example 1.1: Determine the resulting internal loadings
at point “C”
Chapter 1: Stress
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Equilibrium of a Body
Example 1.2: Determine the resulting internal loadings
at point “B”. Ignore the self-weight of the pipe.
The direction of each moment is determined using the right-hand rule:
positive moments (thumb) directed along positive coordinate axis
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5
Equilibrium of a Body
Example 1.3: Determine the resulting internal loadings
at point “C”
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11
Stress
 Distribution of internal loading is important in
mechanics of materials.
 We will consider the material to be continuous.
 The intensity of internal force at a point is called
stress.
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6
Stress
Normal stress
• Intensity of force, or force per unit area, acting normal
to ΔA
• Symbol used for normal stress, is σ (sigma)
σz =
lim ΔFz
ΔA →0
ΔA
• Tensile stress: normal force “pulls”
or “stretches” the area element ΔA
• Compressive stress: normal force “pushes” or
“compresses” area element ΔA
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Stress
Shear stress
• Intensity of force, or force per unit area, acting tangent
to ΔA
• Symbol used for shear stress is  (tau)
τzx =
τzy =
Chapter 1: Stress
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lim ΔFx
ΔA →0
ΔA
lim ΔFy
ΔA →0
ΔA
14
7
Stress
General state of stress
• Figure shows the state of stress
acting around a chosen point in
a body
Units (SI system)
• Newtons per square meter (N/m2) or
a Pascal (1 Pa = 1 N/m2)
x xy xz
• kPa = 103 N/m2 (kilo-pascal)
zx zy z
yx y yz
• MPa = 106 N/m2 (mega-pascal)
• GPa = 109 N/m2 (giga-pascal)
Chapter 1: Stress
15
Stress
Complementary nature of shear stresses
For equilibrium of forces and moments
So only 6 stress components needed to describe
general state of stress: 3 and 3
x xy xz
yx y yz
zx zy z
Chapter 1: Stress
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reduces to
x xy xz
y yz
z
16
8
Stress
Average Normal Stress in an Axially Loaded Bar
 When a cross-sectional area bar is subjected to
axial force through the centroid, it is only subjected
to normal stress.
 Stress is assumed to be averaged over the area.
Chapter 1: Stress
17
Average Normal Stress in an Axially Loaded Bar
Average Normal Stress Distribution
 When a bar is subjected to a
constant deformation,
P   dF    dA
A
P  A
P

A
σ = average normal stress
P = resultant normal force
A = cross sectional area of bar
Equilibrium
 2 normal stress components
that are equal in magnitude
but opposite in direction.
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9
Stress
Example 1.4: The bar has a constant width of 35 mm and a
thickness of 10 mm. Determine the maximum average normal
stress in the bar when it is subjected to the loading shown.
Chapter 1: Stress
19
Stress
Example 1.5: Determine the position “x” of the 3 kN
force so that the average compressive stress at support
“C” equals to the average tensile stress in “AB”. Th
area of AB is 400 mm2 and at C is 650 mm2.
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10
Average Shear Stress
 The average shear stress distributed over each
sectioned area that develops a shear force.
 avg 
 = average shear stress
V
A
V = internal resultant shear force
A = area at that section
 2 different types of shear:
a) Single Shear
b) Double Shear
Chapter 1: Stress
21
Single shear
• Steel and wood joints shown below are examples
of single-shear connections, also known as lap
joints.
• Since we assume members are thin, there are no
moments caused by F
F V
  FA
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11
Double shear
Double-shear connections,
often called double lap joints.
Pin is subjected to
shear on two planes:
𝑽
𝑭
2
𝜏
𝐹
2𝐴
F
F
Chapter 1: Stress
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Shear stress
Example 1.6: Determine the average shear stress along
shear plane a-a and b-b. The thickness of the wood
joint is 150 mm.
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12
Shear stress
Problem 1.7: If P = 15 kN, determine the average
shear stress in the Pins at points A, B, and C. All pins
are in double shear and has a diameter of 18 mm.
Chapter 1: Stress
25
Allowable Stress
•
When designing a structural member or mechanical
element, the stress in it must be restricted to a safe
level
•
Choose an allowable load that is less than the load
the member can fully support
•
One method used is the factor of safety (F.S.)
F.S. =
Chapter 1: Stress
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Ffail
Fallow
26
13
Allowable Stress
 Many unknown factors that influence the actual
stress in a member.
 A factor of safety is needed to obtain allowable load.
 The factor of safety (F.S.) is a ratio of the failure
load divided by the allowable load
F .S 
F fail
Fallow
 fail
 allow
 fail
F .S 
 allow
F .S 
Chapter 1: Stress
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Factor of safety
Factor of safety depends on:
1) Load type: static, sustained, impact, repeated
2) Material properties: e.g. steel less variable than
concrete
3) Likely failure: brittle or ductile, excessive
deformation, buckling (before y), etc.
4) Importance of member to structural integrity
(Primary, Secondary)
5) Risk from failure or level of assumed risk
6) Environment Harsh or mild, hot or cold, exposure to
corrosion, etc.
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14
Design of Simple Connections
•
To determine the area of a section subjected to a
normal force, use
P
A= σ
allow
•
To determine area of section subjected to a shear
force, use
A=
V
τallow
Chapter 1: Stress
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Design of Simple Connections
Cross-sectional area of a tension member
Condition:
The force has a line of action that passes through
the centroid of the cross-section.
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15
Design of Simple Connections
Cross-sectional area of a connecter subjected to shear
Assumption:
If bolt is loose or clamping force of bolt is unknown,
assume frictional force between plates to be negligible.
Chapter 1: Stress
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Design of Simple Connections
Required area to resist bearing
• Bearing stress is normal stress produced by the
compression of one surface against another.
Assumptions:
1. (σb)allow of concrete <
(σb)allow of base plate
2. Bearing stress is
uniformly distributed
between plate and
concrete
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16
Design of Simple Connections
Required area to resist shear caused by axial load
•
Although actual shear-stress distribution along rod
difficult to determine, we assume it is uniform.
•
Thus use A = V / τallow to calculate l, provided d and
τallow are known.
Chapter 1: Stress
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Design of Simple Connections
Example 1.8: Determine the required diameters of the
steel pins at A and C if the factor of safety is 1.5 and
the failure shear stress is 12 ksi.
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17
Design of Simple Connections
Example 1.9: The rigid bar AB supported by a steel rod AC having a diameter of 20
mm and an aluminum block having a cross sectional area of 1800 mm2. The 18mm-diameter pins at A and C are subjected to single shear. If the failure stress for
the steel and aluminum is  st  fail  680 MPa and  al  fail  70 MPa respectively, and
the failure shear stress for each pin is  fail  900 MPa , determine the largest load P
that can be applied to the bar. Apply a factor of safety of F.S. = 2.
Solution:
Chapter 1: Stress
35
Application Problem
The structure shown was
designed to support 30 kN
load.
50 mm
30 mm
600 mm
a) Draw the FBD.
b) Find internal forces
c) Determine the normal
stresses, shearing
stresses and bearing
stresses.
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18
Application Problem
FBD
Detach structure at A and
C and represent forces
50 mm
30 mm
600 mm
Ay
By
Use equilibrium equations to find internal forces
Chapter 1: Stress
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Application Problem
FBD
Detach structure at A and
C and represent forces
50 mm
30 mm
600 mm
Ay
By
Use equilibrium equations to find internal forces
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19
Application Problem
50 mm
30 mm
600 mm
FBD of joint B (force triangle)
Chapter 1: Stress
39
Application Problem
• FBD of force
members AB and BC
Chapter 1: Stress
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• FBD of sections of
rod BC
40
20
Application Problem
20-mm diameter rod BC
had flat ends 20x40 mm
rectangular cross section
Boom AB has a 30x50 mm
rectangular cross section
Note:
Connection at C is single
bracket and at B &A is
double bracket.
Chapter 1: Stress
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Application Problem
Shear stresses in single pin at C
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21
Application Problem
Shear stresses in
double pin at A
Chapter 1: Stress
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Application Problem
Shear stresses in
double pin at B
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22
Chapter Review
•
Internal loadings consist of
1.
2.
3.
4.
•
Normal force, N
Shear force, V
Bending moments, M
Torsional moments, T
Get the resultants using
1. Method of sections
2. Equations of equilibrium
Chapter 1: Stress
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Chapter Review
Assumptions for a uniform normal stress distribution
over x-section of member (σ = P/A)
•
1.
Member made from homogeneous isotropic material
2.
Subjected to a series of external axial loads,
3.
The loads must pass through centroid of cross‐section
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23
Chapter Review
Determine average shear stress by using equation
τ = V/A
•

V is the resultant shear force on cross‐sectional area A

Formula is used mostly to find average shear stress in
fasteners or in parts for connections
Chapter 1: Stress
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Chapter Review
Design of any simple connection requires that
•

Average stress along any cross‐section not exceed a
factor of safety (F.S.) or allowable value of σallow or
τallow

These values are reported in codes or standards and
are deemed safe on basis of experiments or through
experience
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