MECHANICS OF MATERIALS - Course Book

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MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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Course Book

MECHANICS OF MATERIALS
By
Prof. Dr. Jalal Ahmed Saeed
Department of Civil Engineering
College of Engineering
University of Sulaimani
2010 – 2011













MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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Course Book

MECHANICS OF MATERIALS
By
Prof. Dr. Jalal Ahmed Saeed
Department of Civil Engineering
College of Engineering
University of Sulaimani
2010 – 2011



BC 22 : MECHANICS OF MATERIALS:

[ October ; 2010 ]

General Information:

• Class Room : Studying Halls [ B1 to B5 ] in the department
• Class Hours : ( 2 Theory + 2 Applied + 1 Practical ) / Week
• Units : 7
• Instructor : Prof. Dr. Jalal Ahmed Saeed
• Assistants : Mr. Jaza Hassan Muhammed
Mr. Farhad Rahem Karem
• Office : Dean's Office, 1st floor , College Building
• Mobile : 0770 - - - - - - - ; e-mail:
jalal.saeed@univsul.net

• Course Coordinator: Mr. Hardy Kamal Karim

Goals:
This course has two specific goals:
( I ) To introduce students to concepts of stress and strain; shearing force and bending; as well as
torsion and deflection of some structural elements.

( II ) To develop theoretical and analytical skills relevant to the areas mentioned in ( I ) above.

Course Framework:


1.
BC 22
will consist of 30 studying weeks
2. Students will be assigned home works, exams, oral discussions, class and Lab activities.
3. The final grades will be based on the home works, exams, class and Lab activities as follows:
. Exam 1 = 12 %
. Exam 2 = 12 %
. Home works and Quizzes = 6 %
. Lab experiments and activities = 10 %
. Final Exam = ( 45 + 15 ) % = 60 %

Course Policies:
1. Students must attend all the lectures. They must adequately perform all the work assigned
by the instructor. { There is an allowance of 10 % disattending }
2. All the assigned work must be submitted by the due date and time.
( Exceptions can be made for students with emergencies or special circumstances ).
3. Students must take the exams on the assigned dates and times. Make-up exams may be
arranged for students with emergencies or special circumstances.
MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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4. Students are encouraged to submit their work on clean white engineering paper, A4.
They should explain and show all their calculations. Including final drawings.

5. The course will be taught through lectures. Lectures will also involve the solution of tutorial
questions. Tutorial questions are designed to complement and enhance both the lectures and
the students appreciation of the subject.
6 .Course work assignments will be reviewed with the students.

Course Syllabus and Outline:

• Introduction, Materials and Properties@
• Simple Stress
• Simple Strain
• Stress – Strain Relationship [ Hook’s Law ]
• Torsion of circular shafts
• Shear Force and Bending Moment in Beams
• Stresses in Beams
• Combined Stresses
• Beam Deflection
• Statically Indeterminate Beams
• Columns & Miscellaneous Topics

Course Objectives:
Upon successful completion of this course, students should be able to:
( i ) Understand and solve simple problems involving stresses and strains.

( ii ) Understand the difference between statically determinate and indeterminate problems.

( iii ) Understand and carry out simple experiments illustrating properties of materials in tension,
compression as well as hardness, impact and other tests.

( iv ) Analyze stresses in two dimensions and understand the concepts of principal stresses and the
use of Mohr circles to solve two – dimensional stress problems.

( v ) Draw shear force and bending moment diagrams of simple beams and understand the
relationships between loading intensity, shearing force and bending moment.

( vi ) Compute the bending stresses in beams with one or two materials.

( vii ) Calculate the deflection of beams using the direct integration, moment – area and other
methods.

Lecture Times:
• Details to be Announced by the department.

• Attendance at the Lectures and Labs is Compulsory.

Course References:

1. Singer and Pytel, “ Strength of Materials “.
2. E Popove, “ Mechanics of Materials “.
3. Beer & Johnson , “ Mechanics of Materials “.
4. Any textbook under the title of : “ Mechanics of Materials [ MOM ] “ or “ Strength of Materials
[ SOM ] “
MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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Course Details


Week No,
Details
Week One
02 / 10 – 07 / 10 / 2010
A Review on Engineering Mechanics: Equilibrium, Reactions …….
Week Two
09 / 10 – 14 / 10 / 2010
Introduction to Mechanics of Materials & Analysis of Internal Forces
Week Three
16 / 10 – 21 / 10 / 2010
Simple Stress: Axial Stress, Types, Units, Examples and Problems …..…
Week Four
23 / 10 – 28 / 10 / 2010
Simple Stress: Shear Stress, Bearing Stress, Examples and Problems …...
Week Five
30 / 10 – 04 / 11 / 2010
Thin – Walled Cylinders
Week Six
06 / 11 – 11 / 11 / 2010
Simple Strain: Axial Strain, Shear Strain, Examples and Problems ……..
Week Seven
13 / 11 – 18 / 11 / 2010
Stress – Strain Relationship, Hook's Law, Examples and Problems ……..
Week Eight
20 / 11 – 25 / 11 / 2010
Review Problems
Week Nine
27 / 11 – 02 / 12 / 2010
Poission's Ratio, Examples and Problems …….
Week Ten
04 / 12 – 09 / 12 / 2010
Thermal Stresses, Examples and Problems …………..
Week Eleven
11 / 12 – 16 / 12 / 2010
Review Problems
Week Twelve
18 /12 – 23 / 12 / 2010
Torsion of Circular Shafts, Examples and Problems ……………
25 / 12 – 30 / 12 / 2010
Vacation
Week Thirteen
03 / 01 – 06 / 01 / 2011
Reactions, Shear Force, Bending Moments,
Week Fourteen
08 / 01 – 13 / 01 / 2011
Shear Force and Bending Moment Diagrams,
Week Fifteen
15 / 01 – 20 / 01 / 2011
Shear Force and Bending Moment Diagrams, Continued
Week Sixteen
22 / 01 – 27 / 01 / 2011
Stresses in Beams, Flexural Stresses, Examples and Problems …..…….
29 / 01 – 03 / 02 / 2011
First Examination
05 / 02 – 10 / 02 / 2011
First Examination
Week Seventeen
12 / 02 – 17 / 02 / 2011
Shear Stresses in Beams, Examples and Problems …..…
Week Eighteen
19 / 02 – 24 / 02 / 2011
Combined Flexural – Shear Stresses, Examples and Problems …..…
Week Nineteen
26 / 02 – 03 / 03 / 2011
Economical Sections, Examples and Problems …..…
Week Twenty
05 / 03 – 10 / 03 / 2011
Combined Stresses, Solving by Equations
12 / 03 – 17 / 03 / 2011
Spring & Nawroz Vacations
19 / 03 – 24 / 03 / 2011
Spring & Nawroz Vacations
Week Twenty – One
26 / 03 – 31 / 03 / 2011
Combined Stresses, Solving by Mohr's Circle
Week Twenty – Two
02/ 04 – 07 / 04 / 2011
Beam Deflections, Double Integration Method
Week Twenty – Three
09 / 04 – 14 / 04 / 2011
Area – Moment Method, Moment Diagram by Parts
16 / 04 – 21 / 04 / 2011
Second Examination
Week Twenty – Four
23 / 04 – 28 / 04 / 2011
Deflection Applications
Week Twenty – Five
30 / 04 – 05 / 05 / 2011
Three – Moment Equations, Examples and Problems …..…
Week Twenty – Six
07 / 05 – 12 / 05 / 2011
Review Problems
Week Twenty – Seven
14 / 05 – 19 / 05 / 2011
Statically Indeterminate Beams
Week Twenty – Eight
21 / 05 – 26 / 05 / 2011
Applications
Week Twenty - Nine
28 / 05 – 02 / 06 / 2011
Columns
04 / 06 – 23 / 06 / 2011
Final Examinations
Note:
Examination times may change due to unexpected circumstances.
MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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Introduction to MOM
• When an external force acts on a body, the body tends to undergo some deformation.
• Due to cohesion between the molecules , the body resists deformation.
• This resistance by which material of the body oppose the deformation is known as

Mechanics or Strength of Materials “

• “ Mechanics of Materials (
MOM
) “ is an extension of “ Engineering Mechanics “ ; i.e “
Statics and Dynamics “ which you studied last year.
• In Engineering Mechanics bodies are assumed to be rigid and in equilibrium.
• In MOM bodies are in equilibrium but no longer rigid ……… Deformations are of great
interest.

CHAPTER ONE
STRESS
DIRECT OR NORMAL STRESS

• When a force is transmitted through a body, the body tends to change its shape or deform.
The body is said to be under Stress or strained.

Applied Force ( F or P )
• Direct Stress =
Cross Sectional Area ( A )


Units:
Usually N/m
2
( Pa ), kN/mm
2
, MPa, GPa

σ = P / A
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Where:
- σ = axial or normal stress [ T or C ]
- P = applied force
- A = cross sectional area

Direct stress may be tensile, T or compressive, C and result from forces acting perpendicular to
the plane of the cross-section







SHEARING STRESS

Forces parallel to the area resisting the force cause shearing stress. It differs to tensile and
compressive stresses, which are caused by forces perpendicular to the area on which they act.
Shearing stress is also known as
Tangential stress
.

τ = V / A

If the area resisting the shear force is ONE area its called
Single Shear
, and if the resisting area is
Two its called
Double Shear
.
τ = V / 2A

MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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BEARING STRESS:
Bearing stress is the contact pressure between the separate bodies. It differs from compressive
stress, as it is an internal stress caused by compressive forces.

















Thin-Walled Pressure Vessels

A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which resist
bursting, developed across longitudinal and transverse sections.


TANGENTIAL STRESS Or ( Circumferential Stress ):

Consider the tank shown below being subjected to an internal pressure ( p ). The length of the tank
is ( L ) and the wall thickness is ( t ). Isolating the right half of the tank, we get:











MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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LONGITUDINAL STRESS,:
Consider the free body diagram in the transverse section of the tank













CHAPTER TWO
STRAIN
AXIAL STRAIN:
Also known as unit deformation, strain is the ratio of the change in length caused by the applied
force, to the original length.








Stress-Strain Diagram:















PROPORTIONAL LIMIT (HOOKE'S LAW)

From the origin O to the point called proportional limit, the stress-strain curve is a straight line. This
linear relation between elongation and the axial force causing was first noticed by Sir Robert Hooke
in 1678 and is called Hooke's Law that within the proportional limit, the stress is directly
proportional to strain. The constant of proportionality is called the Modulus of Elasticity E or
Young's Modulus and is equal to the slope of the stress-strain diagram from O to P. Then :
σ = E ε @
P / A = E ( δ / L ) or δ = PL / AE
MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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SHEAR STRAIN:
Shearing forces cause shearing deformation. An element subject to shear does not change in length
but undergoes a change in shape.




The change in angle at the corner of an original rectangular element is called the shear strain and is
expressed as:



The ratio of the shear stress and the shear strain is called the
Modulus of Elasticity in Shear
or
Modulus of Rigidity
, and is denoted as G, in GPa.


Poisson's Ratio:

When a bar is subjected to a tensile loading there is an increase in length of the bar in the direction of the applied load,
but there is also a decrease in a lateral dimension perpendicular to the load. The ratio of the sidewise deformation (or
strain) to the longitudinal deformation (or strain) is called the Poisson's ratio and is denoted by ν. For most steel, it lies
in the range of 0.25 to 0.3, and 0.20 for concrete.








TRIAXIAL DEFORMATION:
If an element is subjected simultaneously by three mutually perpendicular normal stresses σ
x
, σ
y
, and σ
z
, which are
accompanied by strains ε
x
, ε
y
, and ε
z
, respectively,






Thermal Stress:
Temperature changes cause the body to expand or contract. The amount δ
T
, is given by




deformation due to equivalent axial stress;









MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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CHAPTER THREE
TORSION
Consider a bar to be rigidly attached at one end and twisted at the other end by a torque or twisting moment T
equivalent to F × d, which is applied perpendicular to the axis of the bar, as shown in the figure. Such a bar is said to be
in torsion.








For a solid or hollow circular shaft subject to a twisting moment T, the torsional shearing stress
τ
at a distance ρ from
the center of the shaft is:





where J is the polar moment of inertia of the section and r is the outer radius.


For solid cylindrical shaft:




For hollow cylindrical shaft:





ANGLE OF TWIST:
The angle ¸ through which the bar length L will twist is
:


POWER TRANSMITTED BY A SHAFT:












MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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CHAPTER FOUR
SHEAR AND BENDING MOMENTS IN BEAMS


A beam is a bar subject to forces or couples that lie in a plane containing the longitudinal of the bar. According to
determinacy, a beam may be determinate or indeterminate.


STATICALLY DETERMINATE BEAMS:














STATICALLY INDETERMINATE BEAMS:


















TYPES OF LOADING
:











MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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Shear and Moment Diagrams:
Consider a simple beam shown of length L that carries a uniform load of w (N/m) throughout its length and is held in
equilibrium by reactions R
1
and R
2
.

Assume that the beam is cut at section [ C ] a distance of ( x ) from the left support and the portion of the beam to the
right of C be removed. The portion removed must then be replaced by vertical shearing force V together with a couple
M to hold the left portion of the bar in equilibrium under the action of [ R
1

] and ( wx ). The couple
M
is called the
resisting moment or
moment
and the force
V
is called the resisting shear or
shear
.

















Shear Force:
is the algebraic sum of the vertical forces acting to the left or right of the cut section
Bending Moment:
is the algebraic sum of the moment of the forces to the left or to the right of the section taken about
the section





The following are some important properties of shear and moment diagrams:

1. The area of the shear diagram to the left or to the right of the section is equal to the moment at that section.

2. The slope of the moment diagram at a given point is the shear at that point.

3. The slope of the shear diagram at a given point equals the load at that point.

The maximum moment occurs at the point of zero shears.
4.

When the shear diagram is increasing, the moment diagram
5. is concave upward.

6. When the shear diagram is decreasing, the moment diagram is concave downwar
d.























MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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CHAPTER FIVE

STRESSES IN BEAMS


orces and couples acting on the beam cause
bending
or
flexural stresses
and
shearing stresses
on any cross
SSUMPTIONS:
F
section of the beam and deflection perpendicular to the longitudinal axis of the beam. If couples are applied to the ends
of the beam and no forces act on it, the bending is said to be pure bending. If forces produce the bending
,
the bending is
called ordinary bending.


A

ulas for flexural and shearing stresses, it is assumed that:
ains plane after the forces and
2- d of uniform cross section and

LEXURAL STRESS:
In using the following form
1- a plane section of the beam normal to its longitudinal axis prior to loading rem
couples have been applied, and
that the beam is initially straight an
3- that the modulii of elasticity in tension and compression are equal.
F

ing moment are known as flexural or bending stresses.

HEAR STRESS:
Stresses caused by the bend


















S
Beam shear is defined as the
internal shear stress of a beam caused by the shear force applied to the beam.

where
V = total shear force at the location in question;
Q = statical moment of area
= A` y'
hear;
t = thickness in the material perpendicular to the s
I =
Moment of Inertia
of the entire c
ross sectiona
l area.
This fo
rmula is also known as the Jourawski formula



MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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CHAPTER SIX

BEAMS DEFLECTIONS


- The deformation of a beam is usually expressed in terms of its deflection from its original unloaded
position.
- The deflection is measured from the original neutral surface of the beam to the neutral surface of
the deform
ed beam.
- The configuration assumed by the deformed neutral surface is known as the elastic curve of the
beam.

-
Methods of Determining Beam Deflections:

Numerous methods are available for the determination of b
eam deflections. These methods include:
1. Double-integration method

2. Area-moment method

3.
Strain-energy method (Castig
liano's Theorem)

4.
Conjugate-beam metho
d

5.
Method of superposition

Of t e the ones that are commonly used.
hes methods, the first two are
1 – Double Integration Method:

The double integration method is a powerful tool in solv
ing deflection and slope of a beam at any point
because we will be able to get the equation of the elastic curve.

If EI is constant, the equation may be written as:

- The first integration ( y' ) yields the slope of the elastic curve and the second integration ( y ) gives
the deflection of the beam at any distance x.

- The resulting solution must contain two constants of integration since EI y" = M is of second order.
- These two constants must be evaluated from known conditions concerning the slope deflection at
certain points of the beam.


2 – Area–Moment Method:

Another method of determining the slopes and d
eflections in beams is the area-moment method, which
involves the area of the moment diagram.
MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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Theorems of Area-Moment Method

Theorem I:

The change in slope between the tangents drawn to the elastic curve at any two points A and B is equal to
the product of 1/EI multiplied by the area of the moment diagram between these two points.



Theorem II:

The deviation of any point B relative to the tangent drawn to the elastic curve at any other point A, in a
direction perpendicular to the original position of the beam, is equal to the product of 1/EI multiplied by the
moment of an area about B of that part of the moment diagram between points A and B.


and


MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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Rules of Sign

1. The deviation at any point is positive if the point lies above the tangent, negative if the point is below
the tangent.
2. Measured from left tangent, if θ is counterclockwise, the change of slope is positive, negative if θ is
clockwise.
Deflections in Simply Supported Beams : Area-Moment Method:

The deflection ( δ ) at some point B of a simply supported beam can be obtained by the following steps:


1. Compute

2. Compute

3. Solve ( δ ) by ratio and proportion (see figure above).





The
moment-area method
of finding the deflection of a beam will demand the accurate computation
of the area of a moment diagram, as well as the moment of such area about any axis. To pave its
way, this section will deal on how to draw moment diagrams by parts and to calculate the moment
of such diagrams about a specified axis.

MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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CHAPTER SEVEN

COMBINED STRESSES


Mohr's Circle:

Introduced by Otto Mohr in 1882, Mohr's Circle illustrates principal stresses and stress transformations via a graphical
format,


The two principal stresses are shown in red, and the maximum shear stress is shown in orange. Recall that the normal
stresses equal the principal stresses when the stress element is aligned with the principal directions, and the shear stress
equals the maximum shear stress when the stress element is rotated 45° away from the principal directions.
As the stress element is rotated away from the
principal
(or maximum shear) directions, the normal and shear stress
components will always lie on Mohr's Circle.
Mohr's Circle was the leading tool used to visualize relationships between normal and shear stresses, and to estimate the
maximum stresses, before hand-held calculators became popular. Even today, Mohr's Circle is still widely used by
engineers all over the world.
To establish Mohr's Circle, we first recall the stress
transformation formulas
for plane stress at a given location,


Using a
basic trigonometric relation
(cos
2
2θ + sin
2
2θ = 1) to combine the two above equations we
have,

This is the equation of a circle, plotted on a graph where the abscissa is the normal stress and
the ordinate is the shear stress. This is easier to see if we interpret σ
x
and σ
y
as being the two
principal stresses
, and τ
xy
as being the maximum shear stress. Then we can define the average
stress, σ
avg
, and a radius" R (which is just equal to the maximum shear stress),
MECHANICS OF MATERIALS COURSE BOOK { 2010 – 2011 } – Second Year
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The circle equation above now takes on a more familiar form,

The circle is centered at the average stress value, and has a radius R equal to the maximum shear stress, as shown in the
figure below,

The formulas used in this article are,




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