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Russell C. Hibbeler
Chapter 1: Stress
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2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
1 - 2
Contents
Concept of Stress
Review of Statics
Structure Free-Body Diagram
Component Free-Body Diagram
Method of Joints
Stress Analysis
Design
Axial Loading: Normal Stress
Centric & Eccentric Loading
Shearing Stress
Shearing Stress Examples
Bearing Stress in Connections
Stress Analysis & Design Example
Rod & Boom Normal Stresses
Pin Shearing Stresses
Pin Bearing Stresses
Stress in Two Force Members
Stress on an Oblique Plane
Maximum Stresses
Stress Under General Loadings
State of Stress
Factor of Safety
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MECHANICS OF MATERIALSThird
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Concept of Stress
The main objective of the study of 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 thedetermination ofstressesand deformations. This chapter is
devoted to the concept of stress.
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2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7thEdition
Introduction
Mechanics of materials is a study of the relationshipbetween the external loads on a body and the
intensity of the internal loads within the body.
This subject also involves the deformationsand
stability of a body when subjected to external forces.
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Chapter 1: Stress
Mechanics of Material 7thEdition
Equilibrium of a Deformable Body
External Forces1.Surface Forces
- caused by direct contact
of other bodys surface
2.Body Forces
- other body exerts a force
without contact
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Chapter 1: Stress
Mechanics of Material 7thEdition
Equilibrium of a Deformable Body
Reactions Surface forces developed at the supports/points of
contact between bodies.
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Chapter 1: Stress
Mechanics of Material 7thEdition
Equilibrium of a Deformable Body
Equations of Equilibrium Equilibrium of a body requires a balance of forces
and a balance of moments
For a body withx, y, z coordinate system with origin
O,
Best way to account for these forces is to draw
the bodys free-bo dy diagram (FBD).
0M0F O
0,0,0
0,0,0
zyx
zyx
MMM
FFF
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MECHANICS OF MATERIALSThird
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Beer Johnston DeWolf
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Review of Statics
The structure is designed tosupport a 30 kN load
Perform a static analysis to
determine the internal force in
each structural member and the
reaction forces at the supports
The structure consists of a
boom and rod joined by pins
(zero moment connections) at
the junctions and supports
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Structure Free-Body Diagram
Structure is detached from supports and
the loads and reaction forces are indicated
Ayand Cycan not be determined from
these equations
kN30
0kN300
kN40
0
kN40
m8.0kN30m6.00
yy
yyy
xx
xxx
x
xC
CA
CAF
AC
CAF
A
AM
Conditions for static equilibrium:
MECHANICS OF MATERIALSTE
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MECHANICS OF MATERIALSThird
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Component Free-Body Diagram
In addition to the complete structure, each
component must satisfy the conditions forstatic equilibrium
Results:
kN30kN40kN40 yx CCA
Reaction forces are directed along boom
and rod
0
m8.00
y
yB
A
AM
Consider a free-body diagram for the boom:
kN30yC
substitute into the structure equilibrium
equation
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MECHANICS OF MATERIALSThird
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Method of Joints
The boom and rod are 2-force members, i.e.,
the members are subjected to only two forceswhich are applied at member ends
kN50kN40
3
kN30
54
0
BCAB
BCAB
B
FF
FF
F
Joints must satisfy the conditions for static
equilibrium which may be expressed in the
form of a force triangle:
For equilibrium, the forces must be parallel to
to an axis between the force application points,
equal in magnitude, and in opposite directions
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Stress Analysis
Conclusion: the strength of memberBCisadequate
MPa165all
From the material properties for steel, theallowable stress is
Can the structure safely support the 30 kN
load?
MPa159m10314
N105026-
3
A
PBC
At any section through member BC, theinternal force is 50 kN with a force intensity
or stress of
dBC= 20 mm
From a statics analysis
FAB= 40 kN (compression)
FBC= 50 kN (tension)
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MECHANICS OF MATERIALSThird
Edition
Beer Johnston DeWolf
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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?
mm2.25m1052.2m10500444
m10500Pa10100
N1050
226
2
26
6
3
Ad
dA
PA
A
P
allall
An aluminum rod 26 mm or more in diameter is
adequate
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Chapter 1: Stress
Mechanics of Material 7thEdition
Equilibrium of a Deformable Body
Internal Resultant Loadings Objective of FBD is to determine the resultant force
and moment acting within a body.
In general, there are 4 different types of resultant
loadings:
a) Normal force, N
b) Shear force, V
c) Torsional moment or torque, Td) Bending moment, M
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Chapter 1: Stress
Mechanics of Material 7thEdition
Example 1.1Determine the resultant internal loadings acting on the cross section at C of the
beam.
Solution:Free body Diagram
mN1809
270
6 w
w
Distributed loading at C is found by proportion,
Magnitude of the resultant of the distributed load, N5406180
21 F
which acts from C m2631
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Chapter 1: Stress
Mechanics of Material 7thEdition
Solution:Equations of Equilibrium
(Ans)mN0108
02540;0
(Ans)540
0540;0
(Ans)0
0;0
C
CC
C
Cy
C
Cx
M
MM
V
VF
N
NF
Applying the equations of equilibrium we have
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Chapter 1: Stress
Mechanics of Material 7thEdition
Example 1.5Determine the resultant internal loadings acting on the cross section at B of the
pipe. The pipe has a mass of 2 kg/m and is subjected to both a vertical force of50 N and a couple moment of 70 Nm at its endA. It is fixed to the wall at C.
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Chapter 1: Stress
Mechanics of Material 7thEdition
SolutionFree-Body Diagram
N525.2481.925.12
N81.981.95.02
AD
BD
WW
Calculating the weight of each segment of pipe,
Applying the six scalar equations of equilibrium,
(Ans)N3.84
050525.2481.9;0
(Ans)0;0
(Ans)0;0
xB
zBz
yBy
xBx
F
FF
FF
FF
(Ans)0;0
(Ans)mN8.77
025.150625.0525.24;0(Ans)mN3.30
025.081.95.0525.245.05070;0
zBzB
yB
yByB
xB
xBxB
MM
M
MMM
MM
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Chapter 1: Stress
Mechanics of Material 7thEdition
Stress
Distr ibut ionof internal loading is important inmechanics of materials.
We will consider the material to be con t inuous.
Thisin tensi tyof internal force at a point is called
stress.
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Chapter 1: Stress
Mechanics of Material 7thEdition
Stress
Normal Stress Force per unit area acting normal toA
Shear Stress
Force per unit area acting tangent toA
A
FzA
z
0lim
A
F
A
F
y
Azy
x
Azx
0
0
lim
lim
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Chapter 1: Stress
Mechanics of Material 7thEdition
Average Normal Stress in an Axially Loaded Bar
When a cro ss -sect io nal area bar is subjected toaxial force through the centroid, it is only subjected
to normal stress.
Stress is assumed to be averaged over the area.
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Chapter 1: Stress
Mechanics of Material 7thEdition
Average Normal Stress in an Axially Loaded Bar
Average Normal Stress Distribution When a bar is subjected to a
constant deformation,
Equilibrium
2 normal stress componentsthat are equal in magnitude
but opposite in direction.
A
P
AP
dAdFA
= average normal stress
P = resultant normal force
A = cross sectional area of bar
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MECHANICS OF MATERIALShird
dition
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The normal stress at a particular point may not be
equal to the average stress but the resultant of the
stress distribution must satisfy
Aave dAdFAP
Axial Loading: Normal Stress
The resultant of the internal forces for an axially
loaded member is normalto a section cutperpendicular to the member axis.
A
P
A
FaveA
0lim
The force intensity on that section is defined as
the normal stress.
The detailed distribution of stress is statically
indeterminate, i.e., can not be found from statics
alone.
MECHANICS OF MATERIALSTh
Ed
B J h t D W lf
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MECHANICS OF MATERIALSirddition
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If a two-force member is eccentrically loaded,
then the resultant of the stress distribution in asection must yield an axial force and a
moment.
Centric & Eccentric Loading
The stress distributions in eccentrically loaded
members cannot be uniform or symmetric.
A uniform distribution of stress in a section
infers that the line of action for the resultant of
the internal forces passes through the centroidof the section.
A uniform distribution of stress is only
possible if the concentrated loads on the end
sections of two-force members are applied atthe section centroids. This is referred to as
centric loading.
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Chapter 1: Stress
Mechanics of Material 7thEdition
Example 1.6The 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 loadingshown.
Solution:By inspection, different sections have different internal forces.
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Chapter 1: Stress
Mechanics of Material 7thEdition
Graphically, the normal force diagram is as shown.
Solution:
By inspection, the largest loading is in region BC,
kN30BCP
Since the cross-sectional area of the bar is constant,
the largest average normal stress is
(Ans)MPa7.8501.0035.0
10303
A
PBCBC
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Chapter 1: Stress
Mechanics of Material 7thEdition
3kN/m80st
Example 1.8The casting is made of steel that has a specific weight of
. Determine the average compressive stressacting at pointsA and B.
Solution:By drawing a free-body diagram of the top segment,
the internal axial force P at the section is
kN042.8
02.08.080
0;0
2
P
P
WPF stz
The average compressive stress becomes
(Ans)kN/m0.64
2.0
042.8 22
A
P
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Chapter 1: Stress
Mechanics of Material 7thEdition
Average Shear Stress
The average shear stress distributed over eachsectioned area that develops a shear force.
2 different types of shear:
A
Vavg
= average shear stressP = internal resultant shear force
A = area at that section
a) Single Shear b) Double Shear
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Shearing Stress Examples
A
F
A
Pave
Single Shear
A
F
A
P
2ave
Double Shear
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Bearing Stress in Connections
Bolts, rivets, and pins create
stresses on the points of contactor bearing surfacesof the
members they connect.
dt
P
A
Pb
Corresponding average force
intensity is called the bearing
stress,
The resultant of the force
distribution on the surface isequal and opposite to the force
exerted on the pin.
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Would like to determine the
stresses in the members and
connections of the structure
shown.
Stress Analysis & Design Example
Must consider maximum
normal stresses inABand
BC, and the shearing stress
and bearing stress at each
pinned connection
From a statics analysis:
FAB= 40 kN (compression)
FBC= 50 kN (tension)
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Rod & Boom Normal Stresses
The rod is in tension with an axial force of 50 kN.
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.
MPa167m10300
1050
m10300mm25mm40mm20
26
3
,
26
N
A
P
A
endBC
At the flattened rod ends, the smallest cross-sectional
area occurs at the pin centerline,
At the rod center, the average normal stress in thecircular cross-section (A= 314x10-6m2) is BC= +159
MPa.
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MECHANICS OF MATERIALSrdtion
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Pin Shearing Stresses
The cross-sectional area for pins atA,B,
and C,26
22
m104912
mm25
rA
MPa102m10491
N105026
3
,
A
PaveC
The force on the pin at Cis equal to the
force exerted by the rodBC,
The pin atAis in double shear with atotal force equal to the force exerted by
the boomAB,
MPa7.40m10491
kN2026,
A
PaveA
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Divide the pin atBinto sections to determine
the section with the largest shear force,
(largest)kN25
kN15
G
E
P
P
MPa9.50m10491
kN2526,
A
PGaveB
Evaluate the corresponding averageshearing stress,
Pin Shearing Stresses
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Pin Bearing Stresses
To determine the bearing stress atAin the boomAB,
we have t= 30 mm and d= 25 mm,
MPa3.53
mm25mm30
kN40
td
Pb
To determine the bearing stress atAin the bracket,
we have t= 2(25 mm) = 50 mm and d= 25 mm,
MPa0.32
mm25mm50
kN40
td
Pb
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Stress in Two Force Members
Will show that either axial or
transverse forces may produce bothnormal and shear stresses with respect
to a plane other than one cut
perpendicular to the member axis.
Axial forces on a two force
member result in only normal
stresses on a plane cut
perpendicular to the member axis.
Transverse forces on bolts and
pins result in only shear stresseson the plane perpendicular to bolt
or pin axis.
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Pass a section through the member forming
an angle q with the normal plane.
qq
q
q
q
q
q
q
q
cossin
cos
sin
cos
cos
cos
00
2
00
A
P
A
P
A
V
A
P
A
P
A
F
The average normal and shear stresses on
the oblique plane are
Stress on an Oblique Plane
qq sincos PVPF
ResolvePinto components normal andtangential to the oblique section,
From equilibrium conditions, the
distributed forces (stresses) on the plane
must be equivalent to the forceP.
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The maximum normal stress occurs when the
reference plane is perpendicular to the member
axis,
00
m A
P
The maximum shear stress occurs for a plane at
+ 45owith respect to the axis,
00 2
45cos45sinA
P
A
Pm
Maximum Stresses
qqq cossincos0
2
0 A
P
A
P
Normal and shearing stresses on an oblique
plane
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Stress Under General Loadings
A member subjected to a general
combination of loads is cut into
two segments by a plane passing
through Q
For equilibrium, an equal and
opposite internal force and stress
distribution must be exerted on
the other segment of the member.
A
V
A
V
A
F
xz
Axz
xy
Axy
x
Ax
limlim
lim
00
0
The distribution of internal stress
components may be defined as,
MECHANICS OF MATERIALSThird
Editi Beer Johnston DeWolf
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2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSdon
1 - 40
Stress components are defined for the planes
cut parallel to thex,yandzaxes. For
equilibrium, equal and opposite stresses areexerted on the hidden planes.
It follows that only 6 components of stress are
required to define the complete state of stress
The combination of forces generated by the
stresses must satisfy the conditions for
equilibrium:
0
0
zyx
zyx
MMM
FFF
yxxy
yxxyz aAaAM
0
zyyzzyyz andsimilarly,
Consider the moments about thezaxis:
State of Stress
MECHANICS OF MATERIALSThird
Editi Beer Johnston DeWolf
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MECHANICS OF MATERIALSdon
1 - 41
Factor of Safety
stressallowablestressultimate
safetyofFactor
all
u
FS
FS
Structural members or machines
must be designed such that theworking stresses are less than the
ultimate strength of the material.
Factor of safety considerations:
uncertainty in material properties uncertainty of loadings
uncertainty of analyses
number of loading cycles
types of failure
maintenance requirements and
deterioration effects
importance of member to structures
integrity
risk to life and property
influence on machine function
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2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7thEdition
Example 1.12The inclined member is subjected to a compressive force of 3000 N. Determine
the average compressive stress along the smooth areas of contact defined byABand BC, and the average shear stress along the horizontal plane defined by
EDB.
Solution:The compressive forces acting on the areas of contact are
N240003000;0
N180003000;0
54
53
BCBCy
ABABx
FFF
FFF
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2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7thEdition
The shear force acting on the sectioned horizontal plane EDB is
Solution:
N1800;0 VFx
Average compressive stresses along the AB and BC planes are
(Ans)N/mm20.1
4050
2400
(Ans)N/mm80.14025
1800
2
2
BC
AB
(Ans)N/mm60.0
4075
1800 2avg
Average shear stress acting on the BD plane is
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2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7thEdition
Allowable Stress
Many unknown factors that influence the actualstress in a member.
A factor of safety is needed to obtained allowable
load.
The facto r o f safety (F.S.) is a ratio of the failureload divided by the allowable load
allow
fail
allow
fail
allow
fail
SF
SF
F
FSF
.
.
.
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2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7thEdition
Example 1.14The control arm is subjected to the loading. Determine to the nearest 5 mm the
required diameter of the steel pin at C if the allowable shear stress for the steel is. Note in the figure that the pin is subjected to double shear.
Solution:For equilibrium we have
MPa55allowable
kN3002515;0
kN502515;0kN150125.025075.0152.0;0
53
54
5
3
yyy
xxx
ABABC
CCF
CCFFFM
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2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7thEdition
Solution:
The pin at C resists the resultant force at C. Therefore,
kN41.30305 22 CF
mm8.18
mm45.2462
m1045.2761055
205.15
2
26
3
2
d
d
VA
allowable
The pin is subjected to double shear, a shear force of 15.205 kN acts over its cross-
sectional area between the arm and each supporting leaf for the pin.
The required area is
Use a pin with a diameter of d = 20 mm. (Ans)
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2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7thEdition
Example 1.17The rigid barAB supported by a steel rodAC having a diameter of 20 mm and an
aluminum block having a cross sectional area of 1800 mm2. The 18-mm-diameterpins atA and C are subjected to single shear. If the failure stress for the steel and
aluminum is and respectively, and the failure
shear stress for each pin is , determine the largest load P that can be
applied to the bar. Apply a factor of safety of F.S. = 2.
Solution:The allowable stresses are
MPa680failst
MPa4502
900
..
MPa352
70
..
MPa3402
680
..
SF
SF
SF
fail
allow
failal
allowal
failst
allowst
MPa70failal
MPa900fail
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2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7thEdition
There are three unknowns and we apply the equations of equilibrium,
Solution:
(2)075.02;0
(1)0225.1;0
PFM
FPM
BA
ACB
We will now determine each value of P that creates the allowable stress in the rod,
block, and pins, respectively.
For rod AC, kN8.10601.010340 26 ACallowstAC AF
Using Eq. 1,
kN17125.1
28.106P
For block B, kN0.63101800103566
BallowalB AF
Using Eq. 2,
kN16875.0
20.63P
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Chapter 1: Stress
Solution:
For pin A or C,
kN5.114009.010450 26 AFV
allowAC
Using Eq. 1,
kN18325.1
25.114P
When P reaches its smallest value (168 kN), it develops the allowable normal
stress in the aluminium block. Hence,
(Ans)kN168P