ADU
Lecture
MEC 302
1
1
Mechanics Of Materials
MEC 302
Dr. Mohammad Alkhedher
MEC 302
2
ADU
Tentative Lectures Schedule
Topic
Reading
Introduction Concept of Stress,
1.1-1.7
Introduction Concept of Stress
Stress and Strain—Axial Loading,
Stress and Strain—Axial Loading,
Torsion,
Torsion
Pure Bending,
Analysis and Design of Beams for Bending,
Shearing Stresses in Beams and Thin-Walled Members
Deflections of Beams,
Deflections of Beams
Columns (Buckling) ,
Transformations of Stress and Strain
Transformations of Stress and Strain,
Final Exam
1.8-1.13
2.1-2.10
2.11-2.19
3.1-3.5
3.6-3.8
4.1-4.7,
5.1-5.4
6.1-6.7,
9.1-9.5
9.7
10.1-10.6
7.1-7.3
7.4-7.9
2
Lecture 1
ADU
3
MEC 302
MecMovies
Mechanics of Materials: An
Integrated Learning System
By Temothy A. Philpot
HTTP://WEB.MST.EDU/~
MECMOVIE/INDEX.HTML
3
Lecture 1
ADU
Lecture
MEC 302
4
1
Chapter 1
Introduction –
Concept of Stress
MEC 302
ADU
Contents
Concept of Stress
Bearing Stress in Connections
Review of Statics
Stress Analysis & Design Example
Structure Free-Body Diagram
Rod & Boom Normal Stresses
Component Free-Body Diagram
Pin Shearing Stresses
Method of Joints
Pin Bearing Stresses
Stress Analysis
Stress in Two Force Members
Design
Stress on an Oblique Plane
Axial Loading: Normal Stress
Maximum Stresses
Centric & Eccentric Loading
Stress Under General Loadings
Shearing Stress
State of Stress
Shearing Stress Examples
Factor of Safety
3- 5
Lecture 1
MEC 302
ADU
Concept of Stress
The main objective of the study of the
mechanics of materials is to provide the
future engineer with the means of analyzing
and designing various machines and load
bearing structures.
Both the analysis and design of a given
structure involve the determination of
stresses and deformations. This chapter is
devoted to the concept of stress.
3- 6
Lecture 1
ADU
Review of Statics
The structure is designed
to support a 30 kN load
The structure consists of a
boom and rod joined by
pins (zero moment
connections) at the
junctions and supports
MEC 302
Perform a static analysis to
determine the internal
force in each structural
member and the reaction
forces at the supports
3- 7
Lecture 1
ADU
Structure Free-Body Diagram
Structure is detached from supports
and the loads and reaction forces
are indicated
Conditions for static equilibrium:
 M C  0  Ax 0.6 m   30 kN 0.8 m 
Ax  40 kN
 Fx  0 Ax  C x
C x   Ax  40 kN
MEC 302
 Fy  0  Ay  C y  30 kN  0
Ay  C y  30 kN
Ay and Cy can not be determined
from these equations
3- 8
Lecture 1
ADU
Component Free-Body Diagram
In addition to the complete structure, each
component must satisfy the conditions for
static equilibrium
Consider a free-body diagram for the boom:
 M B  0   Ay 0.8 m 
Ay  0
substitute into the structure
equilibrium equation
C y  30 kN
MEC 302
Results:
A  40 kN  Cx  40 kN  C y  30 kN 
Reaction forces are directed along
boom and rod
3- 9
Lecture 1
ADU
Method of Joints
The boom and rod are 2-force members, i.e., the
members are subjected to only two forces
which are applied at member ends
For equilibrium, the forces must be parallel to to
an axis between the force application points,
equal in magnitude, and in opposite directions
MEC 302
Joints must satisfy the conditions for static
equilibrium which may be expressed in the
form of a force triangle:

F
 B 0
FAB FBC 30 kN


4
5
3
FAB  40 kN
3- 10
FBC  50 kN
Lecture 1
ADU
Stress Analysis
Can the structure safely support the
30 kN load?
From a statics analysis
FAB = 40 kN (compression)
FBC = 50 kN (tension)
At any section through member BC, the
internal force is 50 kN with a force intensity
or stress of
MEC 302
dBC = 20 mm
P
50 103 N
 BC  
 159 MPa
A 314 10-6 m 2
From the material properties for steel, the
allowable stress is
 all  165 MPa
Conclusion: the strength of member BC is
adequate
3- 11
Lecture 1
ADU
Design
Design of new structures requires selection of
appropriate materials and component dimensions
to meet performance requirements
For reasons based on cost, weight, availability, etc.,
the choice is made to construct the rod from
aluminum all= 100 MPa). What is an
appropriate choice for the rod diameter?
P
 all 
A
A
MEC 302
d2
A
4
d
4A



P
 all

50 103 N
100 106 Pa
4 500 10 6 m 2

 500 10 6 m 2
  2.52 102 m  25.2 mm
An aluminum rod 26 mm or more in diameter is
adequate
3- 12
Lecture 1
ADU
Axial Loading: Normal Stress
The resultant of the internal forces for an axially
loaded member is normal to a section cut
perpendicular to the member axis.
The force intensity on that section is defined as
the normal stress.
F
A0 A
  lim
 ave 
P
A
MEC 302
The normal stress at a particular point may not be
equal to the average stress but the resultant of the
stress distribution must satisfy
P   ave A   dF    dA
A
The detailed distribution of stress is statically
indeterminate, i.e., can not be found from statics
alone.
3- 13
Lecture 1
ADU
Centric & Eccentric Loading
A uniform distribution of stress in a section
infers that the line of action for the resultant of
the internal forces passes through the centroid
of the section.
MEC 302
A uniform distribution of stress is only possible
if the concentrated loads on the end sections
of two-force members are applied at the
section centroids. This is referred to as
centric loading.
If a two-force member is eccentrically loaded,
then the resultant of the stress distribution in a
section must yield an axial force and a
moment.
The stress distributions in eccentrically loaded
members cannot be uniform or symmetric.
3- 14
Lecture 1
ADU
Shearing Stress
Forces P and P’ are applied transversely to the
member AB.
Corresponding internal forces act in the plane of
section C and are called shearing forces.
The resultant of the internal shear force distribution
is defined as the shear of the section and is equal
to the load P.
The corresponding average shear stress is,
MEC 302
 ave 
P
A
Shear stress distribution varies from zero at the
member surfaces to maximum values that may be
much larger than the average value.
The shear stress distribution cannot be assumed to be
uniform.
3- 15
Lecture 1
MEC 302
ADU
Shearing Stress Examples
Single Shear
 ave 
Double Shear
P F

A A
 ave 
3- 16
P F

A 2A
Lecture 1
ADU
Bearing Stress in Connections
Bolts, rivets, and pins create
stresses on the points of contact
or bearing surfaces of the
members they connect.
The resultant of the force
distribution on the surface is
equal and opposite to the force
exerted on the pin.
MEC 302
Corresponding average force
intensity is called the bearing
stress,
b 
3- 17
P P

A td
Lecture 1
ADU
Stress Analysis & Design Example
MechMovies website
http://web.mst.edu/~
mecmovie/index.html
Would like to determine the
stresses in the members and
connections of the structure
shown.
From a statics analysis:
FAB = 40 kN (compression)
FBC = 50 kN (tension)
MEC 302
Must consider maximum
normal stresses in AB and
BC, and the shearing stress
and bearing stress at each
pinned connection
3- 18
Lecture 1
ADU
Rod & Boom Normal Stresses
The rod is in tension with an axial force of 50 kN.
At the rod center, the average normal stress in the circular
cross-section (A = 314x10-6m2) is BC = +159 MPa.
At the flattened rod ends, the smallest cross-sectional
area occurs at the pin centerline,
A  20 mm 40 mm  25 mm   300 106 m 2
MEC 302
P
50 103 N
 BC ,end  
 167 MPa
A 300 106 m 2
The boom is in compression with an axial force of 40 kN
and average normal stress of –26.7 MPa.
The minimum area sections at the boom ends are
unstressed since the boom is in compression.
3- 19
Lecture 1
ADU
Pin Shearing Stresses
The cross-sectional area for pins at A, B,
and C,
2
 25 mm 
6 2
A r 
  491 10 m
 2 
2
The force on the pin at C is equal to the
force exerted by the rod BC,
MEC 302
P
50 103 N
 C , ave  
 102 MPa

6
2
A 49110 m
The pin at A is in double shear with a total
force equal to the force exerted by the
boom AB,
 A, ave 
3- 20
P
20 kN

 40.7 MPa
A 491 10  6 m 2
Lecture 1
ADU
Pin Shearing Stresses
Divide the pin at B into sections to determine
the section with the largest shear force,
PE  15 kN
PG  25 kN (largest)
Evaluate the corresponding average shearing
stress,
PG
25 kN

 50.9 MPa
A 491 10  6 m 2
MEC 302
 B, ave 
3- 21
Lecture 1
ADU
Pin Bearing Stresses
To determine the bearing stress at A in the boom AB, we
have t = 30 mm and d = 25 mm,
b 
P
40 kN

 53.3 MPa
td 30 mm 25 mm 
To determine the bearing stress at A in the bracket, we
have t = 2(25 mm) = 50 mm and d = 25 mm,
P
40 kN

 32.0 MPa
td 50 mm 25 mm 
MEC 302
b 
3- 22
Lecture 1
MEC 302
ADU
Sample Problem 1.1
26
Lecture 1
MEC 302
ADU
Sample Problem 1.1
27
Lecture 1
MEC 302
ADU
Sample Problem 1.1
28
Lecture 1
MEC 302
ADU
Sample Problem 1.2
29
Lecture 1
MEC 302
ADU
Sample Problem 1.2
30
Lecture 1
MEC 302
ADU
Sample Problem 1.2
31
Lecture 1
ADU
Stress in Two Force Members
Axial forces on a two force member
result in only normal stresses on a
plane cut perpendicular to the
member axis.
MEC 302
Transverse forces on bolts and pins
result in only shear stresses on the
plane perpendicular to bolt or pin
axis.
Will show that either axial or transverse
forces may produce both normal and
shear stresses with respect to a plane
other than one cut perpendicular to the
member axis.
3- 32
Lecture 1
ADU
Stress on an Oblique Plane
Pass a section through the member forming an
angle q with the normal plane.
From equilibrium conditions, the distributed
forces (stresses) on the plane must be
equivalent to the force P.
Resolve P into components normal and
tangential to the oblique section,
F  P cosq
V  P sin q
MEC 302
The average normal and shear stresses on
the oblique plane are


3- 33
F
P cosq
P


cos2 q
Aq A0
A0
cosq
V
P sin q
P


sin q cosq
Aq A0
A0
cosq
Lecture 1
ADU
Maximum Stresses
Normal and shearing stresses on an oblique
plane

P
P
cos2 q  
sin q cosq
A0
A0
The maximum normal stress occurs when the
reference plane is perpendicular to the member
axis,
MEC 302
m 
P
A0
  0
The maximum shear stress occurs for a plane at +
45o with respect to the axis,
m 
3- 34
P
P
sin 45 cos 45 
 
A0
2 A0
Lecture 1
ADU
Stress Under General Loadings
A member subjected to a general
combination of loads is cut into
two segments by a plane passing
through Q
The distribution of internal stress
components may be defined as,
F x
 x  lim
A0 A
 xy  lim
MEC 302
A0
V yx
A
Vzx
 xz  lim
A0 A
For equilibrium, an equal and
opposite internal force and stress
distribution must be exerted on
the other segment of the member.
3- 35
Lecture 1
ADU
State of Stress
Stress components are defined for the planes
cut parallel to the x, y and z axes. For
equilibrium, equal and opposite stresses are
exerted on the hidden planes.
The combination of forces generated by the
stresses must satisfy the conditions for
equilibrium:
 Fx   Fy   Fz  0
Mx  My  Mz  0
MEC 302
Consider the moments about the z axis:
 M z  0   xy Aa   yx Aa
 xy   yx
similarly,  yz   zy
and  yz   zy
It follows that only 6 components of stress are
required to define the complete state of stress
3- 36
Lecture 1
ADU
MEC 302
37
State of Stress
Lecture 1
 In the preceding sections you learned to determine
the stresses in rods, bolts, and pins under simple
loading conditions.
 In later chapters you will learn to determine stresses
in more complex situations.
 In engineering applications, however, the
determination of stresses is seldom an end in itself.
 Rather, the knowledge of stresses is used by
engineers to assist in their most important task,
namely, the design of structures and machines that
will safely and economically perform a specified
function.
MEC 302
38
ADU
DESIGN CONSIDERATIONS
Lecture 1
MEC 302
39
ADU
DESIGN CONSIDERATIONS
a. Determination of the Ultimate Strength of a Material
 An important element to be considered by a designer
is how the material that has been selected will
behave under a load.
 For a given material, this is determined by
performing specific tests on prepared samples of the
material.
 For example, a test specimen of steel may be
prepared and placed in a laboratory testing machine
to be subjected to a known centric axial tensile force,
as described in Sec. 2.3.
 As the magnitude of the force is increased, various
changes in the specimen are measured, for example,
changes in its length and its diameter.
Lecture 1

MEC 302
40
ADU
DESIGN CONSIDERATIONS
Lecture 1
a. Determination of the Ultimate Strength of a Material
 Several test procedures are available to determine
the ultimate shearing stress, or ultimate strength in
shear, of a material.
 The one most commonly used involves the twisting of
a circular tube (Sec.3.5).
 A more direct, if less accurate, procedure consists in
clamping a rectangular or round bar in a shear tool
(Fig. 1.39) and applying an increasing load P until the
ultimate load PU for single shear is obtained.
MEC 302
41
ADU
DESIGN CONSIDERATIONS
Lecture 1
a. Determination of the Ultimate Strength of a Material
 If the free end of the specimen rests on both of the
hardened dies (Fig. 1.40), the ultimate load for
double shear is obtained.
MEC 302
42
ADU
DESIGN CONSIDERATIONS
Lecture 1
MEC 302
ADU
DESIGN CONSIDERATIONS

c. Factor of safety
considerations:
1. uncertainty in material
properties .
2. uncertainty of loadings.
3. uncertainty of analyses.
4. number of loading cycles.
5. types of failure.
6. maintenance requirements
and deterioration effects.
7. importance of member to
integrity of whole structure.
8. risk to life and property.
9. influence on machine
function.
Lecture 1
MEC 302
ADU
Sample Problem 1.3
44
Lecture 1
MEC 302
ADU
Sample Problem 1.3
45
Lecture 1
MEC 302
ADU
Sample Problem 1.4
46
Lecture 1
MEC 302
ADU
Sample Problem 1.4
47
Lecture 1