Engineering Mechanics I, II
Mechanics of Materials I, II
Static Analysis:
Static Analysis:
(1) Equilibrium equations
(1) Equilibrium equations:
∑Fx = 0; ∑Fy = 0; ∑Fz = 0;
∑Fx = 0; ∑Fy = 0; ∑Fz = 0;
∑M = 0; ∑M = 0; ∑M = 0
∑M = 0; ∑M = 0; ∑M = 0
(2) Stress-strain relation: s = Ee
(3) Strain-displacement relation:
Rigid Body
e = DL/L
Elastic Body
Dynamics Analysis: ∑F = ma
1
Mechanics of Materials,
by Ferdinand. P. Beer, et. al.
Strength of Materials, 4th Edition,
by Andrew. Pytel & Ferdinand L.
Singer
2
Chapter 1: Simple Stress
(1) Analysis of Internal Forces
(2) Normal Stress
(3) Shearing Stress
(4) Bearing Stress
(5) Thin-Walled Pressure Vessels
3
Chapter 2: Simple Strain
(1) Stress-Strain Diagram
(2) Hooke’s Law: Axial & Shearing Deformations
P
P
(3) Poisson’s Ratio
 = −e y / e x = −e z / e x
(4) Statically Indeterminate Members
(5) Thermal Stresses
4
Chapter 3: Torsion
(1) Derivation of Torsion Formulas
(2) Flanged Bolt Couplings
(3) Longitudinal Shearing Stress
(4) Torsion of Thin-Walled Tubes; Shear Flow
(5) Helical Springs
5
Chapter 4: Shear & Moment in Beams
(1) Shear & Moment
(2) Moving Loads
6
Chapter 5: Stresses in Beam
(1) Derivation of Flexure Formula
Mc
s=
I
(2) Economic Sections
(3) Derivation of Formula for Horizontal Shearing Stress
(4) Design for Flexure and Shear
VQ
=
Ib
7
Chapter 6: Beam Deflections
d2 y M
=
2
dx
EI
Determinate Beams
(1) Double Integration Method
(2) Theorem of Area-Moment Method
(3) Moment Diagram by Parts
(4) Conjugate-Beam Method
(5) Superposition Method
F
y
= 0;
M
z
= 0,
8
Chapter 7: Restrained Beams
(1) Propped & Restrained Beams
Indeterminate Beams
not only  Fy = 0;
M
z
= 0,
(2) Double Integration & Superposition Methods
(3) Area-Moment Method
(4) Design of Restrained Beams
9
Chapter 8: Continuous Beams
Indeterminate Beams
not only
 F = 0;
 M = 0,
y
z
(1) Three-Moment Equation
(2) Moment Distribution
10
Chapter 9: Combined Stresses
(1) Axial + Flexural Loads
(2) Mohr’s Circle & Variation of Stresses at Point
(3) Transformation of Strain Components & Strain Rosette
11
Chapter 10: Reinforced Beams
(1) Composite Beams
(2) Reinforced Concrete Beams
(3) Design of RC Beam
(4) T Beam of Reinforced Concrete
12
Chapter 11: Columns
(1) Critical Load
(2) Long Columns & Euler’s Formula
(3) Intermediate Columns
(4) Eccentrically Loaded Columns
Short
Long
13
Chapter 13: Special Topics
(1) Repeated Loading; Fatigue
(2) Stress Concentration
(3) Theories of Failure
(4) Energy Methods
(5) Impact or Dynamic Loading
(6) Shearing Stresses in Thin-Walled Members
(7) Shear Center
(8) Unsymmetrical Bending
(9) Curved Beams
14
15
16
17
18
19
20
21
22
23
24
Review of statics
25
26
Stress
This is a measure of the internal resistance
in a material to an externally applied load.
The intensity of force at a point is called
stress.
Types of Stress:
- Normal stress,
- Shearing stress,
- Bearing stress.
27
Normal stress
Shear stress
28
29
stress
load P
s=
area A
30
Units
SI units
Average Normal stress
31
•Cross-sections of axial members have normals that are parallel to the
longitudinal axis.
•Purely axially loaded members undergo strictly axial loading.
•The resultant normal force at any crosssection passes through the centroid of
that section
32
Summary
Normal
33
34
35
36
37
38
Summary of lecture 1
Normal
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Single Shear
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Double Shear
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Bearing stress:
It is the pressure resulting from the connection of separate
bodies.
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Example:
The connection shown in the figure consists of two steel
plates, each 2.5 mm thick, to be joined by a single bolt
7.14mm diameter. Determine the maximum shear stress
produced in bolt?
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Example:
Determine the max. shear stress in bolts for given
data shown in the figure?
max. = 3600/314 = 11.46 MPa
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Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Strength of Materials I
CH.1
Lecture 1 (Tutorial): stresses
Prepared
by
Asst. Lecturer: Shireen Taha Saadullah
Email: Shireen.taha@uod.ac
6-OCT. 2020
Supports and their reactions
Example 1
Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Knowing that d1 = 50 mm and d2 = 30 mm, find the average normal
stress at the midsection of (a) rod AB, (b) rod BC.
Solution
Example 2
Determine the average shear stress in the 20-mm-diameter pin at A and
the 30-mm-diameter pin at B that support the beam shown.
Solution
Example 3
Solution of Example 3
(continued)
(continued)
Example 4
The inclined member shown is subjected to a compressive force of 600 lb.
Determine the average compressive stress along the smooth areas of contact
at AB and BC, and the average shear stress along the horizontal plane DB.
Solution of Example 4
From the free-body diagram of the inclined member, the compressive
forces acting on the areas of contact are
(continued)
Also, from the free-body diagram of the segment ABD, the shear force
acting on the sectioned horizontal plane DB is
(continued)
Average Stress: The average compressive stresses
along the vertical and horizontal planes
of the inclined member are
The average shear stress acting on the
horizontal plane defined by DB is
Example 5
A steel pipe column shown (6.5 in. outside diameter; 0.25 in. wall
thickness) supports a load of 11 kips. The steel pipe rests on a square
steel base plate, which in turn rests on a concrete slab.
(a) Determine the bearing stress between the steel pipe and the steel plate.
(b) If the bearing stress of the steel plate on the concrete slab must be limited to
90 psi, what is the minimum allowable plate dimension a?
Solution of Example 5
(continued)
The bearing stress between the pipe and the base plate is
(b) The minimum area required for the steel plate in order to limit
the bearing stress to 90 psi is
Since the steel plate is square, its area of contact with the concrete
slab is
Example 6
Solution of Example 6
(continued)
Strain
Strain, represented by the Greek letter ε, is a term
used to measure the deformation or extension of a
body that is subjected to a force or set of forces.
The strain of a bar is generally defined as the change
in length divided by the initial length.
strain
change in length  L
=
original length L
Often, the change in length of the bar, ΔL, is simply
referred to as the total displacement, or δ. In that
case, strain is then
strain
=

L
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In engineering strain is not a measure of force but
is a measure of the deformation produced by the
influence of stress.
* Strain is dimensionless,
Similar to stress there are two types:
-Normal strain
- Shear strain
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Normal Strain
The elongation of the bar is assumed normal, or
perpendicular, to the cross section. Therefore, like
stress, the strain is called a normal strain. Similar to
stress, a tensile strain is generally considered positive
and a compressive strain is considered negative.
Bar in tension
P
L
Tensile strain ε t 
L
Bar in compression
L
Compressive strain ε c 
L
P
L
P
L
P
L
L
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Example
P
Given data
P
L
L
Tensile load P = 15,000 kg
Steel rod diameter, d = 2 cm
Length = 10 cm
Extension = 20 mm
Find
Normal Stress?
Normal Strain?
Tensile stress σ t 
P
A
15000 kg
π/4  22 cm2
 47747 kg/cm2
σt 
L
Tensile strain ε t 
L
 t  2  0.2
10
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Example
P
P
L
L
Given data
A wire , Tensile strain t = 0.0002
Elongation
L= 0.75 mm
To find
Length of wire L=?
Calculation
L
Tensile strain ε t 
L
0.75 mm
L
L  3750 mm
length of wire  3.75 m
 0.0002  
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Shear strain
=
l
h
 tan( )  
Shear strain is defined as change of angle of side
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faces that are originally perpendicular
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Dr. Abdulhameed A. Yaseen
Robert Hooke, FRS (18 July 1635 – 3 March 1703) was
an English physicist.
Hooke is known principally for his law of elasticity
(Hooke's Law) in 1676 .
Hooke's Law states that the restoring force of a spring is
directly proportional to a small displacement. [
0 grams
0 cm displacement
50 grams
2 cm displacement
100 grams
4 cm displacement
150 grams
6 cm displacement
∝ ]
200 grams
8 cm displacement
Universal tensile testing machine
Specimen and extensometer
A – Proportional limit
B – Elastic limit
C – Upper yield point
D – Lower yield point
E – Ultimate limit
F - Fracture point
Strength of Materials I
Lecture 2 (Tutorial): normal strain, shear strain and
Hooks law
Prepared
by
Asst. Lecturer: Shireen Taha Saadullah
Email: Shireen.taha@uod.ac
18-OCT. 2020
Example 1
The steel propeller shaft ABCD carries the axial loads shown. Determine the
change in the length of the shaft caused by these loads. Use E = 29× 106 psi
for steel.
Solution
Noting that tension causes elongation and compression results in shortening, we
obtain for the elongation of the shaft
Example 2
Determine the average normal strains in the two wires in the Figure shown if the
ring at A moves to A′.
Solution
Example 3
The rectangular plate is deformed into the shape of a parallelogram shown by
the dashed lines. Determine the average shear strain 𝛾xy at corners A and B.
Solution
Example 4
The corners of the square plate are given the displacements indicated.
a. Determine the shear strain along the edges of the plate at A and B.
b. Determine the average normal strains along side AB and diagonals
AC and DB.
Solution
a.
b.
Example 5
The piece of plastic is originally rectangular. Determine the average
normal strain that occurs along the diagonals AC and DB.
Solution
Example 6
The rigid bar BDE is supported by two links AB and CD. Link AB is
made of aluminum (E = 70 GPa) and has a cross-sectional area of 500
mm2; link CD is made of steel (E = 200 GPa) and has a cross-sectional
area of 600 mm2. For the 30-kN force shown, determine the deflection
(a) of B, (b) of D, (c) of E.
Solution of Example
Example 7
Rigid beam AB rests on the two short posts shown in Figure.. AC is
made of steel and has a diameter of 20 mm, and BD is made of
aluminum and has a diameter of 40 mm. Determine the displacement of
point F on AB if a vertical load of 90 kN is applied over this point. Take
Est = 200 GPa, Eal = 70 GPa.
Solution
Example 8
Each of the links AB and CD is made of aluminum (E = 10.9 x 106 psi)
and has a cross-sectional area of 0.2 in2. Knowing that they support the
rigid member BC, determine the deflection of point E.
Solution
Robert Hooke, FRS (18 July 1635 – 3 March 1703) was an
English physicist.
Hooke is known principally for his law of elasticity
(Hooke's Law) in 1676 .
Hooke's Law states that the restoring force of a spring is
directly proportional to a small displacement.
0 grams
0 cm displacement
50 grams
2 cm displacement
100 grams
4 cm displacement
150 grams
6 cm displacement
200 grams
8 cm displacement
W  X
W =KX
Weight
(dynes or g.cm/sec2)
Displacement
(cm)
49000
2
98000
4
147000
6
196000
8
Hook’s law
stress ( load)  strain (extension)
L
Stress
(load)
Stress – strain graph
x
P
Hook’s law
Born
Died
13 June 1773 England
10 May 1829 (aged 55)
Strain ( elongation)
Why is
Young’s modulus different for different
materials?
Why is
Young’s modulus important?
The Relation Between Shearing Stress and Shearing Strain
Hook’s law
stress ( load)  strain (extension)
Modulus of rigidity
The more elastic the member, the higher the modulus of rigidity or shear modulus
Poisson’s Ratio
• Element loaded in a given direction will also
produce lateral strains in mutually perpendicular
directions.
• Ratio of lateral to axial strain is Poisson’s Ratio ().
Applying a load in the x direction causes a normal
stress in that direction, and the same is true for
normal stresses in the y and z directions. And, as
we now known, stress in one direction causes
strain in all three directions.
So now we incorporate this idea into Hooke's law,
and write down equations for the strain in each
direction as:
E  2G (1   )
Homogeneous means properties are same on every point. (A material of uniform
composition throughout that cannot be mechanically separated into
different materials)
Isotropic means the property is same in all directions.
(Isotropic materials have identical material properties in all directions at every
given point. This means that when a specific load is applied at any point in the x, y
or z-axis, isotropic materials will exhibit the same strength, stress, strain, young's
modulus and hardness.)
If something shows same properties at all points, then it is homogeneous and if
the properties are same in all directions , then it is isotropic.
•
For most metals  is approximately 0.3
Ultimate Stress
Working Stress
Example
Given data two steel rods
A
Length l = 5 m
100
Working stress  = 560 kg/cm2
E = 2.1 x 106 kg/cm2
To find
y
PCA
PCB
50 50
Diameter of rod d = ?
C
5m
x
x
Elongation x =?
Calculation
Fx = 0
gives PCA = PCB
Fy = 0 gives 10,000 kg = 2 x PCA cos 50
PCA
A
7778.62 kg
7778.62


A
( / 4 ) d 2
working stress σ 
560 kg/cm 2
d  4.21 cm
10 tonnes
P = 10 x1000 kg
(Ans)
PCA = 77778.62 kgs
P
l σl
 
A E E
560 x 500 cm

2.1  10 6 kg/cm 2
 0.133 cm (Ans)
elongation x 
B
Strength of Materials I
Lecture 3 (Tutorial): - Hooks law- E, G, v ,
deformation, factor of safety
By: Shireen Taha Saadullah
Example 1
Compressive centric forces of 40 kips are applied at both ends of the assembly
shown by means of rigid end plates. Knowing that Es =29 x106 psi and Ea
10.1x106 psi, determine (a) the normal stresses in the steel core and the
aluminum shell, (b) the deformation of the assembly.
Solution:
Example 2
Solution:
Example 3
Solution:
Example 4
Solution
Example 5
Solution:
Example 6
Two 1.75-in.-thick rubber pads are bonded to three steel plates to form
the shear mount shown. Find the displacement of the middle plate when the
1200-lb load is applied. Consider the deformation of rubber only. Use E =500 psi
and 𝝊 = 0.48 for rubber.
Solution:
Each rubber pad has a shear area of A = 5 × 9 = 45 in.2 that carries half the 1200-lb
load. Hence, the average shear stress in the rubber is
The wires AB and BC have original lengths of 2 ft and 3 ft, and diameters of 1/8 in. and 3/16
in., respectively. Determine the elongations of the wires and change in their diameters
after the 1500-lb load is placed on the platform. Assume that these wires are made of a steel
material which its mechanical properties can be found from performing a tension test on a
steel specimen having an original diameter of 0.503 in. and gauge length of 2.00 in. The data
is listed in the table. (Use a scale of 1 in. 20 ksi and 1 in. = 0.05 in. in. Redraw the elastic
region, using the same stress scale but a strain scale of 1 in. 0.001 in. in). Assume the shear
modulus of the steel wires = 11500 ksi.
For BC Stress < yield stress :Hook’s law applied here so can find from PL/AE too
If the friction pad C which is used to support the wire BC
and is made from the same material which is used for
wires . With no slipping occur, determine the normal and
shear strains in the pad. The width is 2 in. Assume that
the material is linearly elastic. Also, neglect the effect of
the moment acting on the pad.
Compound or Composite Bars
In certain application it is necessary to use a combination of elements or bars
made from different materials, each material performing a different function.
In over head electric cables or Transmission Lines for example it is often
convenient to carry the current in a set of copper wires or aluminum wires
surrounding steel wires. The later being designed to support the weight of the
cable over large spans.
Such a combination of materials is generally termed compound bars.
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Compound or Composite Bars
When a system comprises two or more members of different materials,
the forces in various members cannot be determined by the principle of
statics alone. Such system are know as (Statically indeterminate).
Additional equations are required to determine the unknown forces.
These equations are obtained from deformation condition of the system
and are known as compatibility equations (Kinematic Condition).
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Rigid two plates
Tube B
Equilibrium equation
P = PA
Rod A
+ PB
……
P
(1)
P
P = A AA + B AB
LA = L B
Compatibility equation
(strain equation)
A
=  B ….. (2)
A LA =  B LB
A = B
A
EA

B
EB
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Example
Concrete column
Given data
P = 100 tons
Concrete column 18 x 18 in. square
4 steel rods
d = 1 in.
EC = 2 x 106 lb/in2 ,ES = 30 x 106 lb/in2
Load P = 100 tons
A
A
= 100 x 2240 lbs
To find
4 steel rods
S = ? , C = ?
d = 1 in
Solution
18 in
Area of steel rod , AS = 4 x (/4) x d2
= 3.14 in2
Net area of concrete AC = 18 x 18 – 3.14
18 in
2
Section A -A
= 320.86 in
For compound bar P = S AS + C AC
224000 = S x 3.14 + C x 320.86 eq.(1)
S
ES

C
EC
S
30 x 10 6

C
2 x 10 6
eq.(2)
From equations (1) and (2), C = 608 lb/in2 , S = 9120 lb/in2 (Ans)
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x
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Combined E:
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Example:
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Strength of Materials I
Lecture 4 (Tutorial): - Compound Bars
(Statically Indeterminate Members)
By: Shireen Taha Saadullah
Example 1
Solution :
Example 2
Solution :
Example 3
Figure (a) shows a rigid bar that is supported by a pin at A and
two rods, one made of steel and the other of bronze. Neglecting
the weight of the bar, compute the stress in each rod caused
by the 50-kN load, the following data：
Solution :
Equilibrium in Fig. (b), contains four unknown forces.
Since there are only three independent equilibrium equations,
these forces are statically indeterminate. The equilibrium
equation that does not involve the pin reactions at A is
Compatibility The displacement of the bar, consisting of a rigid body
rotation about A, is shown greatly exaggerated in Fig. (c).
From similar triangles, we see that the elongations of the supporting
rods must satisfy the compatibility condition
Example 4
Solution :
Example 5
Solution :
A change in temperature can cause a body to change its dimensions.
Generally, if the temperature increases, the body will expand, whereas if
the temperature decreases, it will contract.
Ordinarily this expansion or contraction is linearly related to the temperature
increase or decrease that occurs. If this is the case, and the material is
homogeneous and isotropic, it has been found from experiment that the
displacement of a member having a length L can be calculated using the formula
EQ -1
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The change in length of a statically determinate member can easily be
calculated using Eq. 1 , since the member is free to expand or contract
when it undergoes a temperature change.
In a statically indeterminate member, these thermal displacements will be
constrained by the supports, thereby producing thermal stresses that must
be considered in design. Determining these thermal stresses is possible
using the methods outlined in the previous sections.
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The A-36 steel bar shown in Fig. is constrained to just fit
between two fixed supports when T 1 = 60oF. If the temperature
is raised to T 2 = 120oF, determine the average normal thermal
stress developed in the bar. 
Since there is no external load, the force at A is equal
but opposite to the force at B; that is,
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Or
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The assembly has the diameters and material
makeup indicated. If it fits securely between its
fixed supports when the temperature is
T1 = 70°F, determine the average normal stress
in each material when the temperature reaches
T2 = 110°F.
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Dr. Abdulhameed A. yaseen
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Strength of Materials I
Lecture 5 (Tutorial): - Thermal Stresses
By: Shireen Taha Saadullah
Example 1
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel (Es = 200 GPa, αs = 11.7× 10 -6 / °C) and
portion BC is made of brass (Eb = 105 GPa, αb = 20.9 ×10-6 / °C). Knowing
that the rod is initially unstressed, determine the compressive force induced in
ABC when there is a temperature rise of 50 C.
Solution
Example 2
At room temperature (20 C) ° a 0.5-mm gap exists between the ends of the
rods shown. At a later time when the temperature has reached 140°C,
determine (a) the normal stress in the aluminum rod, (b) the change in length of
the aluminum rod.
Solution
Example 3
The horizontal steel rod, 2.5 m long and 1200 mm2 in cross-sectional
area, is secured between two walls as shown in Fig.If the rod is
stress-free at 20 ℃ , compute the stress when the temperature has
dropped to -20℃ . Assume that (1) the walls do not move and (2) the
walls move together a distance △ = 0.5 mm. Use α = 11.7×10-6 /℃
and E =200 GPa.
Solution
Example 4
Figure shows a homogeneous, rigid block weighing 12 kips that is
supported by three symmetrically placed rods. The lower ends of the
rods were at the same level before the block was attached. Determine
the stress in each rod after the block is attached and the temperature of
all bars increases by 100 ℉. Use the following data:
Solution
Torsion
Torque is a moment that tends to twist a member about its longitudinal axis.
Its effect is of primary concern in the design of drive shafts used in
vehicles and machinery.
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It depends on the position x and will vary along the shaft as shown.
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Maximum shear strain on surface
Due to the proportionality of triangles
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Torsion formula
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Torsion formula
The shear stress at the intermediate distance  can be determined from
Recall that these equations are used only if the shaft is circular and the material is
homogeneous and behaves in a linear elastic manner, since the derivation is based on
Dr. Abdulhameed
Hooke’s
law A. Yaseen
The polar moment of inertia, also known as second polar
moment of area, is a quantity used to describe resistance
to torsional deformation
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Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
A torsionally loaded shaft may be classified as statically indeterminate
if the moment equation of equilibrium, applied about the axis of the shaft,
is not adequate to determine the unknown torques acting on the shaft.
Condition of compatibility, or the kinematic condition, requires the angle of twist of one
end of the shaft with respect to the other end to be equal to zero, since the end supports
are fixed. Therefore,
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Dr. Abdulhameed A. Yaseen
Strength of Materials I
Lecture 6 (Tutorial): -Torsion (stress and strain)
By: Shireen Taha Saadullah
Example 1
(a) Determine the maximum shearing stress caused by a 4.6-kN .m
torque T in the 76-mm-diameter solid aluminum shaft shown.
(b) Solve part a, assuming that the solid shaft has been replaced by
a hollow shaft of the same outer diameter and of 24-mm inner
diameter.
Solution:
Example 2
Knowing that each of the shafts AB, BC, and CD consists of a solid
circular rod, determine (a) the shaft in which the maximum shearing
stress occurs, (b) the magnitude of that stress.
Solution :
Example 3
The horizontal shaft AD is attached to a fixed base at D and is subjected to
the torques shown. A 44-mm-diameter hole has been drilled into portion
CD of the shaft. Knowing that the entire shaft is made of steel for which
G = 77 GPa, determine the angle of twist at end A.
Solution:
Example 4
The 1.5-in.-diameter shaft shown is supported by two bearings and is
subjected to three torques. Determine the shear stress developed at points
A and B, located at section a–a of the shaft.
Solution:
Example 5
Shaft BC is hollow with inner and outer diameters of 90 mm and 120
mm, respectively. Shafts AB and CD are solid and of diameter d. For the
loading shown, determine (a) the maximum and minimum shearing stress in
shaft BC, (b) the required diameter d of shafts AB and CD if the allowable
shearing stress in these shafts is 65 MPa.
Solution:
It was demonstrated in previous lectures that when a torque is applied to a shaft having a
circular cross section(one that is axisymmetric) the shear strains vary linearly from zero at
its center to a maximum at its outer surface.
Furthermore, due to the uniformity of the shear strain at all points on the same radius, the
cross sections do not deform, but rather remain plane after the shaft has twisted.
Shafts that have a noncircular cross section, however, are not axisymmetric, and so their
cross sections will bulge or warp when the shaft is twisted.
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In all other cases except circular shafts the maximum shear stress occurs at a point on the edge
of the cross section that is closest to the center axis of the shaft as shown in Table by points
that are indicated as “dots” on the cross sections
The most efficient shaft has a circular cross section,
since it is subjected to both a smaller maximum shear
stress and a smaller angle of twist than one having the
same cross sectional area, but not circular, and
subjected to the same torque.
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Dr. Abdulhameed A. yaseen
Dr. Abdulhameed A. yaseen
Dr. Abdulhameed A. yaseen
Dr. Abdulhameed A. yaseen
Or “Tangential stress”
Or
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Or
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Dr. Abdulhameed A. yaseen
LON GITUDIN AL STESS (σ
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Dr. Abdulhameed A. yaseen
0.6*/180=(7.1*T*750)/(30^4*26)
0.0105=5325T/21060000
T=41.53 N.m
max(4.81*41.53)/(30^3)
= 7.4 MPa
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Dr. Abdulhameed A. yaseen
Dr. Abdulhameed A. yaseen
Dr. Abdulhameed A. yaseen
Dr. Abdulhameed A. yaseen
Dr. Abdulhameed A. yaseen
Dr. Abdulhameed A. yaseen
Dr. Abdulhameed A. yaseen
Dr. Abdulhameed A. yaseen
Bending
Beams and shafts are important structural and mechanical elements in engineering.
In this chapter we will determine the stress in these members caused by bending.
The chapter begins with a discussion of how to establish the shear and moment
diagrams for a beam or shaft.
Like the normal-force and torque diagrams, the shear and moment diagrams provide
a useful means for determining the largest shear and moment in a member, and they
specify where these maximums occur. Once the internal moment at a section is
determined, the bending stress can then be calculated.
First we will consider members that are straight, have a symmetric cross section,
and are made of homogeneous linear elastic material.
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Beams
Members that are slender and support loadings that are applied perpendicular to their
longitudinal axis are called beams.
In general, beams are long, straight bars having a constant cross-sectional area.
Classification:
Often they are classified as to how they are supported.
For example,
a simply supported beam is pinned at one end and roller
supported at the other,
A cantilevered beam is fixed at one end and free at
the other,
And an overhanging beam has one or both of its ends
freely extended over the supports.
Beams are considered among the most important of
all structural elements. They are used to support the floor of a building, the deck of a bridge,
or the wing of an aircraft. Also, the axle of an automobile, the boom of a crane, even many
of the bones of the body act as beams.
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Shear and moment diagrams
Because of the applied loadings, beams develop
an internal shear force and bending moment
that, in general, vary from point to point along
the axis of the beam.
In order to properly design a beam it therefore becomes
necessary to determine the maximum shear and moment
in the beam. One way to do this is to express V and M as
functions of their arbitrary position x along the beam’s
axis. These shear and moment functions can then be
plotted and represented by graphs called shear and
moment diagrams.
The maximum values of V and M can then be obtained from these graphs.
Dr. abdulhameed a. Yaseen
Also, since the shear and moment diagrams provide detailed information about the variation of
the shear and moment along the beam’s axis, they are often used by engineers to decide where to
place reinforcement materials within the beam or how to proportion the size of the beam at
various points along its length.
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The positive directions are as follows:
the distributed load acts upward on the beam;
the internal shear force causes a clockwise
rotation of the beam segment on which it acts; and
the internal moment causes compression in the top
fibers of the segment such that it bends the
segment so that it “holds water”.
Loadings that are opposite to these are
considered negative
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Dr. abdulhameed a. Yaseen
Example:
Draw the shear and moment diagrams for the beam shown
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The point of zero shear can be found from Eq.
From the moment diagram, this value of
x represents the point on the beam
where the maximum moment occurs
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Dr. abdulhameed a. Yaseen
Example:
Express the internal shear and moment in terms of x and then draw the shear
and moment diagrams for the overhanging beam.
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Dr. abdulhameed a. Yaseen
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Dr. abdulhameed a. Yaseen
Dr. abdulhameed a. Yaseen
Strength of Materials II
(Tutorial): Shear Force and Bending
moment Diagram
By: Shireen Taha Saadullah
Example 1
Draw the shear and moment diagrams for the beam
shown.
Solution:
Support Reactions
Shear and Moment Functions
Shear and Moment Diagrams
Example 2
Draw the shear and moment diagrams for the beam shown.
Solution :
Support Reactions
Shear and Moment Functions
Shear and Moment Diagrams
The point of zero shear can be found from
the shear function