EE035-3-2-SOM Strength of Materials
Topic 1: Stress
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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TOPIC LEARNING OUTCOMES
At the end of this topic, you should be able to:
1. draw proper free body diagram and write equations based on the free body diagram
2. solve problems pertaining to normal stress in various type of scenarios.
3. distinguish shearing between one and multiple cross sections.
4. differentiate bearing stress with normal stress.
5. distinguish cross sections that would be taken account of in the case of pin is
introduced in the system.
6. solve problems on factor of safety.
7. able to compute stress in truss structures using both Method of Joints and Method of
Sections
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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Contents & Structure
1. Types of forces
2. Normal Stress
3. Shearing Stress
4. Bearing Stress
5. Factor of Safety
6. Analysis of Structure
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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Recap From Last Lesson
• Nil
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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PHYSICAL EXAMPLES
POINT LOADS
DISTRIBUTED LOADS
• Hydrostatic Loading
• Pressure Vessels
P
• Aerodynamic
Wing Loading
P
P(x)
Loads applied to bodies cause deformation in shape and size of the body which is actually due to the
stress that the body experiences. Therefore, it is essential to study on the various kinds of stresses
that a body experiences from various kinds of loadings.
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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TYPES OF FORCES
• Normal Forces:
– Tensile Force, when a force extends the object, and increases
the length in the direction of the forces.
Fa
Fa
– Compressive Force, when a pair of forces applied to an object,
it will reduce the length in the direction of the force.
Fa
Fa
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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TYPES OF FORCES
• Shear Force:
– Is when a pair of forces caused on the face of the material and to
slide relative to an adjacent face.
Fs
Fs
• Bending Force:
– The loads are applied transversely to the longitudinal axis of the
member and the member subjected to this bending load will
bend along its length.
Fb
EE035-3-2-SOM STRENGTH OF MATERIALS
Fb
Chapter 1 - Stress
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TYPES OF FORCES
• Torsional Force/ load:
– Is one that is subjected to a shaft or a twisting member.
Mn
Mn
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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STRESSES
Cut
section
F
A→0 A
 = lim
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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NORMAL STRESS
4. Stress varies across the cross section.
5. The points very near the application of the loads experience
a larger stress value compared to the points faraway from
the application of load on the same section.
6. The variation of stress across the cross section is negligible
when the section is faraway about equal to the width of the
bar, from the application of loads.
7. When a section is assumed to have a uniform distribution of
stresses (internal forces are uniformly distributed across the
section), resultant, P of the internal forces act at the centroid
of the section. This type of loading is called centric loading.
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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SHEARING STRESS
1. Shear Force is introduced when transverse forces P and P’ are applied to a member.
2. Internal forces are called shearing forces and magnitude, P of the resultant
of the shearing forces is called the shear in the section.
 ave =
P
A
3. Actual value of τ of shearing stress varies from zero at the surface of the member to a
maximum value τmax that may be larger than the average value τave .
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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SHEARING STRESS
Single Shear
 ave =
EE035-3-2-SOM STRENGTH OF MATERIALS
Double Shear
P F
=
A A
 ave =
Chapter 1 - Stress
P F
=
A 2A
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BEARING STRESS
Diameter of bolt X Thickness of plate
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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FACTOR OF SAFETY
Selection of an appropiate Factor of safety depends on:
1. Manufacturing errors – composition, material strength, dimensions and material
properties altered eg. through heating
2. Damage during storage and transportation.
3. Fatigue – over time, ultimate stress decreases as loads are increased and applied to
member repeatedly.
4. Possibility of sudden failures through buckling and stability failures.
5. Not designed for certain loadings
6. Deterioration that may occur because of poor maintenance (corrosion, decay) or
because of unpreventable natural causes such as storms.
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE
5.2 Trusses
A truss is a straight link connected together at joints to form a structure. Examples of trusses
structure are bridges, roof supports, derricks etc. The loads applied to the structure will be
distributed to each truss. In truss analysis, two assumptions are made
i) All loadings are applied at the joints.
ii) Bolted or welded connections are assumed to be pinned connections.
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE
5.2.1 Method of Joints
D
B
A
C
P
Figure 5.1 Simple trusses structure
Figure 5.1 shows a structure consists of 4 truss members of ABC, AD, CD and BD. External load
P is applied at joint B. This load will be transferred to each member. Procedure of analyzing
trusses with method of joints are as follows
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE
i)
Draw the FBD of the whole structure
and identify all external and reaction
D
forces. Determine these forces using the
static equilibrium analysis.
ii) Explode the FBD to obtain individual
RAx
A
FBD for the joints and trusses. All
B
RAy
C
members may be assumed to be in
tension at this stage.
RC
P
iii) Now all joints have become particles.
Do force analysis at each joint, starting
with the joint with least unknown
D
forces. Since entire structure is in
RAD
RDC
equilibrium,
RBD
RAD
RAx
RAC
A
RAy
EE035-3-2-SOM STRENGTH OF MATERIALS
joint
also
in
equilibrium.
RDC
RBD
B
each
RAC
P
Chapter 1 - Stress
C
RC
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ANALYSIS OF STRUCTURE
5.2.2 Special Loadings
Special loadings condition is when members at a joint carry either zero load or similar load.
These conditions are identified by visual inspection and are used to simplify the analysis
significantly. The trusses with special loadings are shown as follows.
If two members connected at a joint and no
FB
external or reaction force at the joint, the
members are considered zero-load member.
FB
FA = FB = 0
FA
FA
If three members connected at a joint, and
FB
FB
FB
FB
two members are collinear, the third is zeroload member.
FA
FA
FA = 0
If two members connected at a joint, and
FA
FB
collinear, both member shared the same
load.
FA = FB
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE
Example
Note
Determine the force in each member of the structure as
•
shown.
First draw the FBD
structure to determine th
B
and C. At A, the con
350N
therefore two reaction f
1.8m
the connection is roller
30o
one normal force.
65o
C
A
•
joint.
2m
•
Solution
The reactions at joints A and C are
EE035-3-2-SOM STRENGTH OF MATERIALS
B
Chapter 1 - Stress
Once done, draw indi
You don’t have to solve
obtain all internal forces
350N
Once all reaction forces are
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ANALYSIS OF STRUCTURE
Example
Note
Determine the force in each member of the structure as
•
shown.
First draw the FBD for the whole
structure to determine the reactions at A
and C. At A, the connection is pin,
350N
B
therefore two reaction forces and at C,
1.8m
the connection is roller, therefore only
o
30
one normal force.
65o
C
A
•
Once done, draw individual FBD at
joint.
2m
•
Solution
The reactions at joints A and C are
B
You don’t have to solve for all joints to
obtain all internal forces.
Once all reaction forces are identified, draw
350N
FBD for each joint, assuming all trusses are
1.8m
in tension.
30o
RAx
B 350N
65o
C
RAy
A
RAB
RC
RBC
2m
350N
A
→ Rx = 0 :
RA + 350 N = 0
x
30o
65o
RAC
157.5N
RA = −350N
C
157.5N
x
Joint A has two unknowns so do joints B and
Taking moment about A
C, therefore any joint can be chosen to start
M A = 0;
the analysis.
(
)
350N 1.8m sin 30 − RC ( 2m ) = 0
RC = 157.5N
Joint C:
 Ry = 0 :
 Ry = 0 :
157.5N + RBC sin 65 = 0
RA + 157.5 N = 0
RBC = −173.782N
RA = −157.5 N
Therefore link BC is in compression with a
y
y
magnitude of 173.782N
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE
→ Rx = 0 :
Note
− RAC − RBC cos 65 = 0
•
Make sure you stick with the reference
− RAC − ( −173.782N ) cos 65 = 0
direction for the positive x and y
RAC = 73.4435N
respectively.
Therefore link AC is in tension with a magnitude of
•
Since initial assumption is all internal
forces are in tension, thus, negative
73.4435N
values mean that the internal forces are
in compression.
Joint A:
EE035-3-2-SOM STRENGTH OF MATERIALS
 Ry = 0 :
Therefore
RAB sin 30 − 157.5N = 0
Link AB = 315N (Tension)
RAB = 315N
Link BC = 173.782N (Compression)
Link AB is in tension with a magnitude of 315N
Link AC = 73.4435N (Tension)
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE –
METHOD OF JOINTS
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE
5.2.3 Method of Sections
The method of joints is normally used when the load in all members are required to be
determined. If only load in specific members are required, the method of section may be utilized.
The procedure is to cut through the members of which the loads are to be determined, and apply
static equilibrium equations to the unknown members.
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE –
METHOD OF SECTIONS
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE
Example
Note
Determine the load in the members BE and FE for the
•
structure shown.
there is no zero-load
5kN
A
From inspection, it
B
2m
2m
•
8kN
C
2m
After cutting the FB
take either side to
D
FBD.
2m
2m
•
Since the system
equilibrium, thus, t
F
2m
assumed to be in equ
Substitute into first equat
Solution
EE035-3-2-SOM STRENGTH OF MATERIALS
E
The FBD for the whole structure is
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE
Example
Note
Determine the load in the members BE and FE for the
•
structure shown.
there is no zero-load link exists.
5kN
A
From inspection, it can be seen that
B
2m
C
2m
•
8kN
2m
After cutting the FBD into 2, we may
take either side to become our new
D
FBD.
•
2m
2m
Since the system is originally in
equilibrium, thus, the cut FBD also
E
2m
F
assumed to be in equilibrium
Substitute into first equation, RAy = -4kN
Solution
The FBD for the whole structure is
5kN
RAx
A
B
2m
2m
Since we are interested to determine the
8kN
C
2m
D
load in the links BE and FE, we cut the FBD
through these two links.
RAy
2m
2m
A
F
B
2m
FBC
E
2m
2m
RE
4kN
Applying static equilibrium to obtain all reaction forces
 Ry = 0 :
RA + RE − 5kN − 8kN = 0
y
M A = 0 :
5kN ( 4m ) + 8kN ( 6m ) − RE ( 4m ) = 0
F
FBE
FFE
Take moment about point B
M B = 0 :
− FFE ( 2m ) − 4kN ( 2m ) = 0
FFE = −4kN
RE = 17kN
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE
Note
Take moment about point B
•
Since initial assumption is all members
M B = 0 :
are in tension, all internal force in
− FFE ( 2m ) − 4kN ( 2m ) = 0
negatives are in compression.
FFE = −4kN
→ Rx = 0 :
FBC + FFE + FBE sin 45 = 0
 Fy = 0 :
FBC − 4kN − 5.6569kN sin 45 = 0
4kN + FBE cos 45 = 0
FBC = 8kN
FBE = −5.6569kN
Therefore
Link BE = 5.6569kN (compression)
Link FE = 4kN (compression)
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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ANALYSIS OF STRUCTURE –
METHOD OF SECTIONS
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
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Review Questions
• What is another name for normal stress?
• What is the difference between shear stress and normal stress?
• Which would be the area that would be considered for maximum tensile
stress when there is a pin in the link?
• What are zero load members?
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
SLIDE 30
Summary / Recap of Main Points
• Normal Stress
• Shear Stress
• Bearing Stress
• Factor of Safety
• Analysis of Structure
EE035-3-2-SOM STRENGTH OF MATERIALS
Chapter 1 - Stress
SLIDE 31
What To Expect Next Week
In Class
• Strain
EE035-3-2-SOM STRENGTH OF MATERIALS
Preparation for Class
• Free body diagram
• Equilibrium of Rigid Bodies
Chapter 1 - Stress
SLIDE 32