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Transcript of Mechanic of Materials Slayt
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8/8/2019 Mechanic of Materials Slayt
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MECHANICS OFMATERIALS
Fourth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
2006 The McGraw-Hill Com anies, Inc. All ri hts reserved.
Introduction
Concept of Stress
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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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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 ofstresses and
deformations. This chapter is devoted to the
concept of stress.
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Review of Statics
The structure is designed to
support 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 aboom 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
Ay and Cy can 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:
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Component Free-Body Diagram
In addition to the complete structure, eachcomponent must satisfy the conditions for
static 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:
kN30=yC
substitute into the structure equilibrium
equation
o
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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
kN50kN40
3
kN30
54
0
==
==
=
BCAB
BCAB
B
FF
FF
Fr
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 toto an axis between the force application points,
equal in magnitude, and in opposite directions
o
B J h t D W lf
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Stress Analysis
Conclusion: the strength of memberBCis
adequate
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, the
internal 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)
o
B J h t D W lf
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Design
Design of new structures requires selection ofappropriate materials and component dimensions
to meet performance requirements
For reasons based on cost, weight, availability,
etc., the choice is made to construct the rod fromaluminum (all= 100 MPa). What is anappropriate choice for the rod diameter?
( )mm2.25m1052.2
m1050044
4
m10500Pa10100N1050
226
2
266
3
==
==
=
====
Ad
dA
PAAP
allall
An aluminum rod 26 mm or more in diameter isadequate
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Axial Loading: Normal Stress
The normal stress at a particular point may not be
equal to the average stress but the resultant of the
stress distribution must satisfy
===A
ave dAdFAP
The resultant of the internal forces for an axiallyloaded member is normal to a section cut
perpendicular to the member axis.
A
P
A
Fave
A=
=
0
lim
The force intensity on that section is defined asthe normal stress.
The detailed distribution of stress is statically
indeterminate, i.e., can not be found from statics
alone.
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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 loadedmembers cannot be uniform or symmetric.
A uniform distribution of stress in a sectioninfers that the line of action for the resultant of
the internal forces passes through the centroid
of the section.
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 ascentric loading.
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Shearing Stress Examples
A
F
A
P==
ave
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 contact
or bearing surfaces of the
members they connect.
dt
P
A
P==b
Corresponding average force
intensity is called the bearing
stress,
The resultant of the force
distribution on the surface is
equal and opposite to the force
exerted on the pin.
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Stress Analysis & Design Example
Would like to determine the
stresses in the members and
connections of the structure
shown.
Must consider maximum
normal stresses inAB and
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.
( )( )
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 the
circular cross-section (A = 314x10-6m2) is BC= +159MPa.
The boom is in compression with an axial force of 40kN and average normal stress of 26.7 MPa.
The minimum area sections at the boom ends are
unstressed since the boom is in compression.
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Pin Shearing Stresses
The cross-sectional area for pins atA,B,and C,
262
2m10491
2
mm25 =
== rA
MPa102m10491
N105026
3
, ===
A
PaveC
The force on the pin at Cis equal to the
force exerted by the rodBC,
The pin atA is in double shear with atotal force equal to the force exerted by
the boomAB,
MPa7.40m10491
kN20
26,=
==
A
P
aveA
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Pin Shearing Stresses
kN50=BCF
Divide the pin atB into sections to determinethe section with the largest shear force,
(largest)kN25
kN15
=
=
G
E
P
P
MPa9.50m10491
kN2526,=
==
A
PGaveB
Evaluate the corresponding average
shearing stress,
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Pin Bearing Stresses
To determine the bearing stress atA in the boomAB,
we have t= 30 mm and d= 25 mm,
( )( )MPa3.53
mm25mm30
kN40 ===td
Pb
To determine the bearing stress atA in the bracket,
we have t= 2(25 mm) = 50 mm and d= 25 mm,
( )( )MPa0.32
mm25mm50
kN40===
td
Pb
our
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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 stresses
on the plane perpendicular to boltor pin axis.
our
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Stress on an Oblique Plane
Pass a section through the member forming
an angle with the normal plane.
cossincos
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
sincos PVPF ==
Resolve P into components normal and
tangential to the oblique section,
From equilibrium conditions, the
distributed forces (stresses) on the plane
must be equivalent to the force P.
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Maximum Stresses
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
+ 45o with respect to the axis,
===00 2
45cos45sinA
P
A
Pm
cossincos0
2
0 A
P
A
P==
Normal and shearing stresses on an obliqueplane
ourt
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Stress Under General Loadings
A member subjected to a generalcombination 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,
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Factor of Safety
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 integrity of
whole structure
risk to life and property
influence on machine function
Structural members or machinesmust be designed such that the
working stresses are less than the
ultimate strength of the material.
stressallowable
stressultimate
safetyofFactor
all
u ==
=
FS
FS