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MECHANICS OFFourth Edition
MECHANICS OF MATERIALS
CHAPTER
MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T DeWolf
Introduction –John T. DeWolf
Lecture Notes:Concept of Stress
J. Walt OlerTexas Tech University
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MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
Contents
Concept of Stress Bearing Stress in ConnectionsReview of StaticsStructure Free-Body DiagramC F B d Di
Stress Analysis & Design ExampleRod & Boom Normal StressesPi Sh i SComponent Free-Body Diagram
Method of JointsStress Analysis
Pin Shearing StressesPin Bearing StressesStress in Two Force MembersStress Analysis
DesignAxial Loading: Normal Stress
Stress in Two Force MembersStress on an Oblique PlaneMaximum StressesAxial Loading: Normal Stress
Centric & Eccentric LoadingShearing Stress
Maximum StressesStress Under General LoadingsState of Stressg
Shearing Stress Examples 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.g
• Both the analysis and design of a given structure involve the determination of stresses andinvolve the determination of stresses and deformations. This chapter is devoted to the concept of stress.
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MECHANICS OF MATERIALS
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Review of Statics
• The structure is designed to gsupport a 30 kN load
• The structure consists of a boom and rod joined by pins (zero moment connections) at the junctions and supports
• Perform a static analysis to determine the internal force in
j pp
each structural member and the reaction forces at the supports
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Structure Free-Body Diagram
• Structure is detached from supports and the loads and reaction forces are indicated
( ) ( )( )m8.0kN30m6.00 −==∑ xC AM
• Conditions for static equilibrium:( ) ( )( )
0
kN40
+==
=
∑
∑
xxx
x
xC
CAF
A
0kN300
kN40
=−+==
−=−=
∑ yyy
xx
CAF
AC
• Ay and Cy can not be determined from
kN30=+ yy CA
y y
these equations
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Component Free-Body Diagram
• In addition to the complete structure, each component must satisfy the conditions for p ystatic equilibrium
( )800∑ AM• Consider a free-body diagram for the boom:
( )
0
m8.00
=
−==∑
y
yB
A
AM
substitute into the structure equilibrium
kN30=yC
substitute into the structure equilibrium equation
• Results:↑=←=→= kN30kN40kN40 yx CCA
Reaction forces are directed along boom and rod
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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
• For equilibrium, the forces must be parallel to to an axis between the force application points, equal in magnitude, and in opposite directions
• Joints must satisfy the conditions for static ilib i hi h b d i th
0=∑ BFr
equilibrium which may be expressed in the form of a force triangle:
3kN30
54==
∑
BCAB
B
FF
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kN50kN40 == BCAB FF
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Stress AnalysisCan the structure safely support the 30 kN
load?• From a statics analysis• 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
BC ( )
MPa159N105026
3=
×==
AP
BCσ
internal force is 50 kN with a force intensity or stress of
dBC = 20 mm
• From the material properties for steel, the allowable stress is
m10314 26-×ABC
• Conclusion: the strength of member BC is d
MPa 165all =σallowable stress is
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adequate
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Design• Design of new structures requires selection of
appropriate materials and component dimensions to meet performance requirementsto meet performance requirements
• For reasons based on cost, weight, availability, etc., the choice is made to construct the rod from ,aluminum (σall= 100 MPa). What is an appropriate choice for the rod diameter?
3m10500
Pa10100N1050
2
266
3×=
×
×=== −
σσ
d
PAAP
allall
( ) mm225m10522m1050044
4
226
××
=
−−
π
Ad
dA
( ) mm2.25m1052.2 =×===ππ
d
• An aluminum rod 26 mm or more in diameter is
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An aluminum rod 26 mm or more in diameter is adequate
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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
AP
AF
aveA
=∆∆
=→∆
σσ0
lim
the normal stress.
• The normal stress at a particular point may not be equal to the average stress but the resultant of the
AAA ∆→∆ 0
q gstress distribution must satisfy
∫∫ ===A
ave dAdFAP σσA
• The detailed distribution of stress is statically indeterminate, i.e., can not be found from statics
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alone.
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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 centroidthe internal forces passes through the centroid of the section.
A if di t ib ti f t i l• 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 ymoment.
• The stress distributions in eccentrically loaded
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The stress distributions in eccentrically loaded members cannot be uniform or symmetric.
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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.
P=τ
• The corresponding average shear stress is,
and is equal to the load P.
A=aveτ
• Shear stress distribution varies from zero at the member surfaces to maximum values that may bemember surfaces to maximum values that may be much larger than the average value.
• The shear stress distribution cannot be assumed to
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The shear stress distribution cannot be assumed to be uniform.
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Shearing Stress Examples
Single Shear Double Shear
FP==aveτ FP
ave ==τ
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AAave AA 2aveτ
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Bearing Stress in Connections
• Bolts, rivets, and pins create stresses on the points of contactstresses on the points of contact or bearing surfaces of the members they connect.
• The resultant of the force distribution on the surface is
l d it t th f
C di f
equal and opposite to the force exerted on the pin.
• Corresponding average force intensity is called the bearing stress,
dtP
AP
==bσ
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Stress Analysis & Design Example
• Would like to determine the t i th b dstresses in the members and
connections of the structure shown.
• From a statics analysis:FAB = 40 kN (compression)
M t id i
FAB 40 kN (compression) FBC = 50 kN (tension)
• Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection
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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 theAt the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2) is σBC = +159 MPa.
( )( ) 10300254020 26−A
• At the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline,
( )( )
MPa167m10300
1050
m10300mm25mm40mm20
26
3,
26
=×
×==
×=−=
−N
AP
A
endBCσ
• The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa.
m10300×A
g
• The minimum area sections at the boom ends are unstressed since the boom is in compression.
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p
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Pin Shearing Stresses
• The cross-sectional area for pins at A, B, and C,
262
2 m104912mm25 −×=⎟
⎠⎞
⎜⎝⎛== ππ rA
• The force on the pin at C is equal to the force exerted by the rod BC,
MPa102m10491N1050
26
3, =
×
×== −A
PaveCτ
o ce e e ted by t e od C,
• The pin at A is in double shear with a total force equal to the force exerted bytotal force equal to the force exerted by the boom AB,
MPa7.40kN2026 ===
PaveAτ
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m10491 26,× −AaveA
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Pin Shearing Stresses
• Divide the pin at B into sections to determine the section with the largest shear force,
(largest) kN25
kN15
=
=
G
E
P
P
kN50=BCF
• Evaluate the corresponding average
MPa9.50m10491
kN2526, =
×== −A
PGaveBτ
shearing stress,
m10491×A
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Pin Bearing Stresses
• To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,
kN40P( )( ) MPa3.53
mm25mm30kN40
===tdP
bσ
T d t i th b i t t A i th b k t• To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm,
MPa032kN40===
Pbσ ( )( ) MPa0.32
mm25mm50tdbσ
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Stress in Two Force Members
• Axial forces on a two force b lt i l lmember 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 p yon the plane perpendicular to bolt or pin axis.
• Will show that either axial or transverse forces may produce both
l d h i hnormal and shear stresses with respect to a plane other than one cut perpendicular to the member axis.
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Stress on an Oblique Plane• Pass a section through the member forming
an angle θ with the normal plane.
• From equilibrium conditions, the distributed forces (stresses) on the plane
t b i l t t th f P
• Resolve P into components normal and i l h bli i
must be equivalent to the force P.
Th l d h
θθ sincos PVPF ==tangential to the oblique section,
θθσ coscos 2PPF
• The average normal and shear stresses on the oblique plane are
θθθτ
θ
θ
σθ
cossinsin
cos
cos 00
PPV
AAA
===
===
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θθ
θ
τθ
cossin
cos 00 AAA===
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Maximum Stresses
• Normal and shearing stresses on an oblique plane
θθτθσ cossincos0
2
0 AP
AP
==
plane
• The maximum normal stress occurs when the reference plane is perpendicular to the member axisaxis,
00
m =′= τσAP
• The maximum shear stress occurs for a plane at + 45o with respect to the axis,
PP στ ′===00 2
45cos45sinAP
AP
m
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Stress Under General Loadings• A member subjected to a general
combination of loads is cut into t t b l itwo segments by a plane passing through Q
F x∆
• The distribution of internal stress components may be defined as,
VV
AF
xxy
x
Ax
∆∆
∆∆
=→∆
lim0
σ
AV
AV z
Axz
y
Axy ∆
∆=
∆
∆=
→∆→∆limlim
00ττ
• For equilibrium, an equal and opposite internal force and stress distribution must be exerted on
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the other segment of the member.
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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 areequilibrium, equal and opposite stresses are exerted on the hidden planes.
• The combination of forces generated by the g ystresses must satisfy the conditions for equilibrium:
0=== ∑∑∑ zyx FFF
0
0
=== ∑∑∑
∑∑∑
zyx
zyx
MMM
• Consider the moments about the z axis:( ) ( )
yxxy
yxxyz aAaAM
ττ
ττ
=
∆−∆==∑ 0• Consider the moments about the z axis:
• It follows that only 6 components of stress are
yy
zyyzzyyz ττττ == andsimilarly,
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o ows a o y 6 co po e s o s ess a erequired to define the complete state of stress
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Factor of Safety
Structural members or machines must be designed such that the
Factor of safety considerations:• uncertainty in material propertiesmust be designed such that the
working stresses are less than the ultimate strength of the material.
• uncertainty in material properties • uncertainty of loadings• uncertainty of analyses
stress ultimate
safety ofFactor
u ==
=
σFS
FS • number of loading cycles• types of failure• maintenance requirements and
stress allowableall==
σFS maintenance requirements and
deterioration effects• importance of member to integrity of
h l t twhole structure• risk to life and property• influence on machine function
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