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Russell C. Hibbeler
Chapter 1: Stress
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Introduction•
Mechanics of materials is a study of the relationship between the external loads on a body and the intensity of the internal loads within the body.
•
This subject also involves the deformations
and stability of a body when subjected to external forces.
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Equilibrium of a Deformable BodyExternal ForcesExternal Forces1.Surface Forces
-
caused by direct contact of other body’s surface
2.Body Forces-
other body exerts a force without contact
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Equilibrium of a Deformable BodyReactionsReactions
Surface forces developed at the supports/points of contact between bodies.
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Equilibrium of a Deformable BodyEquations of EquilibriumEquations of Equilibrium
Equilibrium of a body requires a balance of forcesand a balance of moments
For a body with x, y, z coordinate system with origin O,
Best way to account for these forces is to draw the body’s free-body diagram (FBD).
0M 0F == ∑∑ O
0 , 0 , 0
0 , 0 , 0
===
===
∑∑∑∑∑∑
zyx
zyx
MMM
FFF
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Equilibrium of a Deformable BodyInternal Resultant LoadingsInternal 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, Nb) Shear force, Vc) Torsional moment or torque, Td) Bending moment, M
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 1.1Determine the resultant internal loadings acting on the cross section at C of the beam.
Solution:Free body Diagram
mN1809
2706
=⇒= wwDistributed loading at C is found by proportion,
Magnitude of the resultant of the distributed load,
( )( ) N540618021 ==F
which acts from C( ) m2631 =
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:Equations of Equilibrium
( )(Ans) mN 0108
02540 ;0 (Ans) 540
0540 ;0
(Ans) 0 0 ;0
⋅−=
=−−=+
=
=−=↑+
=
=−=→+
∑
∑
∑
C
CC
C
Cy
C
Cx
MMMV
VF
NNF
Applying the equations of equilibrium we have
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
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 of 50 N and a couple moment of 70 N·m
at its end A. It is fixed to the wall at C.
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
FBD
.
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
SolutionFree-Body Diagram ( )( )( )
( )( )( ) N 525.2481.925.12N 81.981.95.02
====
AD
BD
WW
Calculating the weight of each segment of pipe,
Applying the six scalar equations of equilibrium,( )( )( )( ) (Ans) N 3.84
050525.2481.9 ;0
(Ans) 0 ;0
(Ans) 0 ;0
=
=−−−=
==
==
∑∑∑
xB
zBz
yBy
xBx
FFF
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
MM
MMM
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
StressDistribution of internal loading is important in mechanics of materials.We will consider the material to be continuous.This intensity of internal force at a point is called stress.
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
StressNormal Stress Normal Stress σσ
Force per unit area acting normal to ΔA
Shear StressShear Stress
ττForce per unit area acting tangent to ΔA
AFz
Az ΔΔ
=→Δ 0
limσ
AFAF
y
Azy
x
Azx
Δ
Δ=
ΔΔ
=
→Δ
→Δ
0
0
lim
lim
τ
τ
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Average Normal Stress in an Axially Loaded Bar
When a cross-sectional area bar is subjected to axial force through the centroid, it is only subjected to normal stress.Stress is assumed to be averaged over the area.
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Average Normal Stress in an Axially Loaded Bar
Average Normal Stress DistributionAverage Normal Stress DistributionWhen a bar is subjected to a constant deformation,
EquilibriumEquilibrium2 normal stress components that are equal in magnitude but opposite in direction.
APAP
dAdFA
=
=
= ∫∫
σ
σ
σ
σ
= average normal stress
P = resultant normal force
A = cross sectional area of bar
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
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 loading shown.
Solution:By inspection, different sections have different internal forces.
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Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Graphically, the normal force diagram is as shown.Solution:
By inspection, the largest loading is in region BC,
kN 30=BCP
Since the cross-sectional area of the bar is constant, the largest average normal stress is
( )( )( ) (Ans) MPa 7.85
01.0035.01030 3
===A
PBCBCσ
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Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
3kN/m 80=stγ
Example 1.8The casting is made of steel that has a specific weight of
. Determine the average compressive stress acting at points A and B.
Solution:By drawing a free-body diagram of the top segment,
the internal axial force P at the section is
( )( ) ( )kN 042.8
02.08.080
0 ;02
==−
=−=↑+ ∑
PP
WPF stz
π
The average compressive stress becomes
( )(Ans) kN/m 0.64
2.0042.8 2
2 ===π
σAP
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Average Shear StressThe average shear stress distributed over each sectioned area that develops a shear force.
2 different types of shear:
AV
avg =τ
τ
= average shear stress
P = internal resultant shear force
A = area at that section
a) Single Shear b) Double Shear
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
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 by AB and BC, and the average shear stress along the horizontal plane defined by EDB.
Solution:The compressive forces acting on the areas of contact are
( )( ) N 240003000 ;0
N 180003000 ;0
54
53
=⇒=−=↑+
=⇒=−=→+
∑∑
BCBCy
ABABx
FFF
FFF
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
The shear force acting on the sectioned horizontal plane EDB isSolution:
N 1800 ;0 ==→+ ∑ VFx
Average compressive stresses along the AB and BC planes are
( )( )
( )( ) (Ans) N/mm 20.14050
2400
(Ans) N/mm 80.14025
1800
2
2
==
==
BC
AB
σ
σ
( )( ) (Ans) N/mm 60.04075
1800 2==avgτ
Average shear stress acting on the BD plane is
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Allowable StressMany unknown factors that influence the actual stress in a member.A factor of safety is needed to obtained allowable load.The factor of safety (F.S.) is a ratio of the failure load divided by the allowable load
allow
fail
allow
fail
allow
fail
SF
SF
FF
SF
ττσσ
=
=
=
.
.
.
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
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
MPa 55=allowableτ
( ) ( ) ( )( )( )( ) kN 3002515 ;0
kN 502515 ;0
kN 150125.025075.0152.0 ;0
53
54
53
=⇒=−−=+↑
=⇒=+−−=+→
=⇒=−==+
∑∑∑
yyy
xxx
ABABC
CCF
CCF
FFM
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:The pin at C resists the resultant force at C. Therefore,
( ) ( ) kN 41.30305 22 =−=CF
mm 8.18
mm 45.2462
m 1045.2761055
205.15
2
263
2
=
=⎟⎠⎞
⎜⎝⎛
×=×
== −
d
d
VAallowable
π
τ
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)
©© 2008 Pearson Education South Asia 2008 Pearson Education South Asia PtePte LtdLtd
Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 1.17The rigid bar AB supported by a steel rod AC having a diameter of 20 mm and an aluminum block having a cross sectional area of 1800 mm2. The 18-mm-diameter pins at A 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
( ) MPa 680=failstσ
( ) ( )
( ) ( )
MPa 4502
900..
MPa 352
70..
MPa 3402
680..
===
===
===
SF
SF
SF
failallow
failalallowal
failstallowst
ττ
σσ
σσ
( ) MPa 70=failalσMPa 900=failτ
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Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
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, ( ) ( ) ( ) ( )[ ] kN 8.10601.010340 26 === πσ ACallowstAC AF
Using Eq. 1, ( )( ) kN 17125.1
28.106==P
For block B, ( ) ( ) ( )[ ] kN 0.631018001035 66 === −BallowalB AF σ
Using Eq. 2, ( )( ) kN 16875.0
20.63==P
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Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:For pin A or C, ( ) ( )[ ] kN 5.114009.010450 26 ==== πτ AFV allowAC
Using Eq. 1,( )( ) kN 183
25.125.114==P
When P reaches its smallest value (168 kN), it develops the allowable normal stress in the aluminium block. Hence,
(Ans) kN 168=P
Russell C. Hibbeler
Chapter 2: Strain
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 2: StrainChapter 2: StrainMechanics of Material 7Mechanics of Material 7thth EditionEdition
DeformationWhen a force is applied to a body, it will change the body’s shape and size. These changes are deformation.
Note the before and after positions of 3 line segments where the material is subjected to tension.
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Chapter 2: StrainChapter 2: StrainMechanics of Material 7Mechanics of Material 7thth EditionEdition
StrainNormal StrainNormal Strain
The elongation / contraction of a line segment per unit of length is referred to as normal strain.Average normal strain is defined as
If the normal strain is known, then the approximate final length is
sss
avg ΔΔ−Δ
='ε
( ) ss Δ+≈Δ ε1'
+ε line elongate-ε line contracts
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Chapter 2: StrainChapter 2: StrainMechanics of Material 7Mechanics of Material 7thth EditionEdition
StrainUnitsUnits
Normal strain is a dimensionless quantity since it is a ratio of two lengths.
Shear StrainShear StrainChange in angle between 2 line segments that were perpendicular to one another refers to shear strain.
'lim2
along along θπγ
tACnABnt
→→
−=
θ<90 +shear strainθ>90 -shear strain
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Chapter 2: StrainChapter 2: StrainMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 2.1The slender rod creates a normal strain in the rod of where z is in meters. Determine (a) displacement of end B due to the temperature increase, and (b) the average normal strain in the rod.
Solution:Part (a)Part (a)Since the normal strain is reported at each point along the rod,
it has a deformed length of
The sum along the axis yields the deformed length of the rod is
The displacement of the end of the rod is therefore
( ) 2/131040 zz−=ε
( )[ ]dzzdz 2/1310401' −+=
( )[ ] m 20239.010401'2.0
0
2/13 =+= ∫ − dzzz
(Ans) mm39.2m00239.02.020239.0 ↓==−=ΔB
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Chapter 2: StrainChapter 2: StrainMechanics of Material 7Mechanics of Material 7thth EditionEdition
Part (b)Part (b)Assumes the rod has an original length of 200 mm and a change in
length of 2.39 mm. Hence,
Solution:
(Ans) mm/mm 0119.0200
39.2'==
ΔΔ−Δ
=s
ssavgε
Example 2.3The plate is deformed into the dashed shape. If, in this deformed shape, horizontal lines on the plate remain horizontal and do not change their length, determine (a) the average normal strain along the side AB, and (b) the average shear strain in the plate relative to the x and y axes.
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Chapter 2: StrainChapter 2: StrainMechanics of Material 7Mechanics of Material 7thth EditionEdition
Part (a)Part (a)Line AB, coincident with the y axis, becomes line after deformation, thus the length of this line is
The average normal strain for AB is therefore
The negative sign indicates the strain causes a contraction of AB.
Solution:
( ) mm 018.24832250' 22 =+−=AB
( ) ( ) (Ans) mm/mm 1093.7240
250018.248' 3−−=−
=−
=AB
ABABavgABε
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Chapter 2: StrainChapter 2: StrainMechanics of Material 7Mechanics of Material 7thth EditionEdition
Part (b)Part (b)As noted, the once 90°
angle BAC between the sides of the plate, referenced from the x, y axes, changes to θ’
due to the displacement of B to B’.
Since then is the angle shown in the figure. Thus,
Solution:
'2 θγ π −=xy xyγ
(Ans) rad 121.02250
3tan 1 =⎟⎠⎞
⎜⎝⎛
−= −
xyγ
Russell C. Hibbeler
Chapter 3: Mechanical Properties
of Materials
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Tension and Compression TestThe strength of a material depends on its ability to sustain a load.This property is to perform under the tension or compression test.The following machine is designed to read the load required to maintain specimen stretching.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Stress–Strain DiagramConventional StressConventional Stress––Strain DiagramStrain Diagram
Nominal or engineering stress is obtained by dividing the applied load P by the specimen’s original cross-sectional area.
Nominal or engineering strain is obtained by dividing the change in the specimen’s gauge length by the specimen’s original gauge length.
0AP
=σ
0Lδε =
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Stress–Strain DiagramConventional StressConventional Stress––Strain DiagramStrain DiagramStress-Strain Diagram
Elastic BehaviourStress is proportional to the strain.Material is said to be linearly elastic.
YieldingIncrease in stress above elastic limit will cause material to deform permanently.
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Stress–Strain DiagramConventional StressConventional Stress––Strain DiagramStrain DiagramStress-Strain Diagram
Strain Hardening.After yielding a further load will reaches a ultimate stress.
NeckingAt ultimate stress, cross-sectional area begins to decrease in a localized region of the specimen.Specimen breaks at thefracture stress.
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Stress–Strain DiagramTrue StressTrue Stress––Strain DiagramStrain Diagram
The values of stress and strain computed from these measurements are called true stress and true strain.Use this diagram since most engineering design is done within the elastic range.
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Stress–Strain Behavior of Ductile and Brittle Materials
Ductile MaterialsDuctile MaterialsMaterial that can subjected to large strains before it ruptures is called a ductile material.
Brittle MaterialsBrittle MaterialsMaterials that exhibit little or no yielding before failure are referred to as brittle materials.
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Hooke’s LawHooke’s Law defines the linear relationship between stress and strain within the elastic region.
E can be used only if a material has linear–elastic behaviour.
εσ E=σ
= stress
E
= modulus of elasticity or Young’s modulus
ε
= strain
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Hooke’s LawStrain HardeningStrain Hardening
When ductile material is loaded into the plastic region and then unloaded, elastic strain is recovered.The plastic strain remains and material is subjected to a permanent set.
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Strain EnergyWhen material is deformed by external loading, it will store energy internally throughout its volume. Energy is related to the strains called strain energy.
Modulus of ResilienceModulus of ResilienceWhen stress reaches the proportional limit, the strain-energy density is the modulus of resilience, ur.
Eu pl
plplr
2
21
21 σ
εσ ==
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Strain EnergyModulus of ToughnessModulus of Toughness
Modulus of toughness, ut, represents the entire area under the stress–strain diagram.It indicates the strain-energy density of the material just before it fractures.
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 3.2The stress–strain diagram for an aluminum alloy that is used for making aircraft parts is shown. When material is stressed to 600 MPa, find the permanent strain that remains in the specimen when load is released. Also, compute the modulus of resilience both before and after the load application.
Solution:When the specimen is subjected to the load,the strain is approximately 0.023 mm/mm.
The slope of line OA is the modulus of elasticity,
From triangle CBD,
( ) ( ) mm/mm 008.0100.7510600 96
=⇒=== CDCDCD
BDE
GPa 0.75006.0
450==E
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:This strain represents the amount of recovered elastic strain.
The permanent strain is
( ) ( )( )
( ) ( )( ) (Ans) MJ/m 40.2008.060021
21
(Ans) MJ/m 35.1006.045021
21
3
3
===
===
plplfinalr
plplinitialr
u
u
εσ
εσ
(Ans) mm/mm 0150.0008.0023.0 =−=OCε
Computing the modulus of resilience,
Note that the SI system of units is measured in joules, where 1 J = 1 N •
m.
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Poisson’s RatioPoisson’s ratio, v (nu), states that in the elastic range, the ratio of these strains is a constant since the deformations are proportional.
Negative sign since longitudinal elongation (positive strain) causes lateral contraction (negative strain), and vice versa.
long
latvεε
−= Poisson’s ratio is dimensionless.Typical values are 1/3 or 1/4.
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 3.4A bar made of A-36 steel has the dimensions shown. If an axial force of is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. The material behaves elastically.
Solution:The normal stress in the bar is
( )( ) ( )mm/mm 108010200100.16 6
6
6−===
st
zz E
σε
( )( )( ) ( )Pa 100.16
05.01.01080 6
3
===AP
zσ
From the table for A-36 steel, Est
= 200 GPa
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
The axial elongation of the bar is therefore
Solution:
The contraction strains in both the x and y directions are
( )[ ] m/m 6.25108032.0 6 μεεε −=−=−== −zstyx v
( )( )[ ] (Ans) m1205.11080 6z μεδ === −
zz L
The changes in the dimensions of the cross section are
( )( )[ ]( )( )[ ] (Ans) m28.105.0106.25
(Ans) m56.21.0106.256
6
μεδ
μεδ
−=−−==
−=−==−
−
yyy
xxx
L
L
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Shear Stress–Strain DiagramFor pure shear, equilibrium requires equal shear stresses on each face of the element.
When material is homogeneous and isotropic, shear stress will distort the element uniformly.
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Shear Stress–Strain DiagramFor most engineering materials the elastic behaviour is linear, so Hooke’s Law for shear applies.
3 material constants, E, and G are actually related by the equation
γτ G=
G
= shear modulus of elasticity or the modulus of rigidity
( )vEG+
=12
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 3.5A specimen of titanium alloy is tested in torsion and the shear stress–strain diagram is shown. Find the shear modulus G, the proportional limit, and the ultimate shear stress. Also, find
the maximum distance d that the top of a block of this material could be displaced horizontally if the material behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement?
Solution:The coordinates of point A are (0.008 rad, 360 MPa).
Thus, shear modulus is
( ) (Ans) MPa 1045008.0
360 3==G
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:By inspection, the graph ceases to be linear at point A. Thus, the proportional limit is
(Ans) MPa 504=uτ
(Ans) MPa 360=plτ
This value represents the maximum shear stress,point B. Thus the ultimate stress is
Since the angle is small, the top of the will be displaced horizontally by
( ) mm 4.0mm 50
008.0rad 008.0tan =⇒=≈ dd
The shear force V needed to cause the displacement is
( )( ) (Ans) kN 270010075
MPa 360 ; =⇒== VVAV
avgτ
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
*Failure of Materials Due to Creep and FatigueCreepCreep
When material support a load for long period of time, it will deform until a sudden fracture occurs. This time-dependent permanent deformation is known as creep.Both stress and/or temperature play a significant role in the rate of creep.Creep strength will decrease for higher temperatures or higher applied stresses.
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Chapter 3: Mechanical Properties of MaterialsChapter 3: Mechanical Properties of MaterialsMechanics of Material 7Mechanics of Material 7thth EditionEdition
*Failure of Materials Due to Creep and FatigueFatigueFatigue
When metal subjected to repeated cycles of stress or strain, it will ultimately leads to fracture.This behaviour is called fatigue.Endurance or fatigue limit is a limit which no failure can be detected after applying a load for a specified number of cycles.This limit can be determined in S-N diagram.
Russell C. Hibbeler
Chapter 4: Axial Load
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Saint-Venant’s PrincipleSaint-Venant’s principle states that both localized deformation and stress tend to “even out” at a distance sufficiently removed from these regions.
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Elastic Deformation of an Axially Loaded Member
Using Hooke’s law and the definitions of stress and strain, we are able to develop the elastic deformation of a member subjected to axial loads.Suppose an element subjected to loads,
( )( ) dx
dδεxAxP
== and σ( )( )∫=
L
ExAdxxP
0
δ
= small displacementL = original length
P(x) = internal axial forceA(x) = cross-sectional area
E = modulus of elasticity
δ
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Elastic Deformation of an Axially Loaded Member
Constant Load and CrossConstant Load and Cross--Sectional AreaSectional AreaWhen a constant external force is applied at each end of the member,
Sign ConventionSign ConventionForce and displacement is positive when tension and elongation and negative will be compression and contraction.
AEPL
=δ
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 4.2The assembly consists of an aluminum tube AB having a cross-sectional area of 400 mm2. A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod. (Est = 200 GPa, Eal = 70 GPa )
Solution:Find the displacement of end C with respect to end B.
( )[ ]( )( )[ ] ( )[ ] →=−=
−== − m 001143.0001143.0
1070104004.01080
96
3
AEPL
Bδ
( )[ ]( )( ) ( )[ ] →+=+
== m 003056.010200005.06.01080
9
3
/ πδ
AEPL
BC
Displacement of end B with respect to the fixed end A,
Since both displacements are to the right, →==+= mm 20.4m 0042.0/BCCC δδδ(Ans)
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 4.4A member is made from a material that has a specific weight and modulus of and elasticity E. If it is formed into a cone, find how far its end is displaced due to gravity when it is suspended in the vertical position.
Solution:Radius x of the cone as a function of y is determined by proportion,
yLrx
Lr
yx oo == ;
γ
The volume of a cone having a base of radius x and height y is
32
22
33y
LryxV oππ
==
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Since , the internal force at the section becomes
Solution:
( ) 32
2
3 ;0 y
LryPF o
yγπ
==↑+ ∑
VW γ=
( ) 22
22 y
LrxyA oππ ==
The area of the cross section is also a function of position y,
Between the limits of y =0 and L yields
( )( )
( )[ ]( )[ ] (Ans)
63 2
022
22
0 EL
ELrdyLr
EyAdyyP L
o
oL γ
γπγπδ === ∫∫
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Principle of SuperpositionPrinciple of superposition is to simplify stress and displacement problems by subdividing the loading into components and adding the results.
A member is statically indeterminate when equations of equilibrium are not sufficient to determine the reactions on a member.
Statically Indeterminate Axially Loaded Member
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 4.5The steel rod has a diameter of 5 mm. It is attached to the fixed wall at A, and before it is loaded, there is a gap between the wall at and B’ and the rod of 1 mm. Find the reactions at A and B’ if the rod is subjected to an axial force of P = 20 kN. Neglect the size of the collar at C. (Est = 200 GPa)
Solution:Equilibrium of the rod requires
( ) (1) 01020 ;0 3 =+−−=→+ ∑ BAx FFF
( ) ( ) (2) mN 0.39278.04.0
001.0/
⋅=−
−==
BA
CBBACAAB
FFAELF
AELFδ
The compatibility condition for the rod is .m 001.0/ =ABδBy using the load–displacement relationship,
Solving Eqs. 1 and 2 yields FA = 16.6 kN and FB = 3.39 kN. (Ans)
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 4.8The bolt is made of 2014-T6 aluminum alloy and is tightened so it compresses a cylindrical tube made of Am 1004-T61 magnesium alloy. The tube has an outer radius of 10 mm, and both the inner radius of the tube and the radius of the bolt are 5 mm. The washers at the top and bottom of the tube are considered to be rigid and have a negligible thickness. Initially the nut is hand-tightened slightly; then, using a wrench, the nut is further tightened one-half turn. If the bolt has 20 threads per inch, determine the stress in the bolt.
Solution:
(1) 0 ;0 =−=↑+ ∑ tby FFF
( ) bt δδ −=↑+ 5.0
Equilibrium requires
When the nut is tightened on the bolt, the tube will shorten.
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Taking the 2 modulus of elasticity,Solution:
( )[ ] ( )[ ]
( )[ ] ( )[ ]( ) (2) 911251255
10755605.0
104551060
32322
bt
bt
FF
FF
−=
−=−
πππ
kN 56.3131556 === tb FF
The stresses in the bolt and tube are therefore
Solving Eqs. 1 and 2 simultaneously, we get
( )
( ) (Ans) MPa 9.133N/mm 9.133510
31556
(Ans) MPa 8.401N/mm 8.4015
31556
222
2
==−
==
====
πσ
πσ
t
ts
b
bb
AFAF
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 4.9The A-36 steel rod shown has a diameter of 5 mm. It is attached to the fixed wall at A, and before it is loaded there is a gap between the wall at and the rod of 1 mm. Determine the reactions at A and B’.
Solution:
(1) 0.001 Bp δδ −=→+
( )[ ]( )( ) ( )[ ]
( )( ) ( )[ ] ( ) B
BABBB
ACP
FFAELFAEPL
69
9
3
103056.0102000025.0
2.1
m 002037.0102000025.04.01020
−===
===
πδ
πδ
Consider the support at B’ as redundant and using principle of superposition,
Thus,
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
By substituting into Eq. 1,
Solution:
( )( ) (Ans) kN 39.31039.3
103056.0002037.0001.03
6
==
−= −
B
B
F
F
(Ans) kN 6.16 039.320 ;0
=
=−+−=→+ ∑A
Ax
FFF
From the free-body diagram,
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Thermal StressChange in temperature cause a material to change its dimensions.Since the material is homogeneous and isotropic,
TLT Δ−= αδ
= linear coefficient of thermal expansion, property of the material= algebraic change in temperature of the member= original length of the member= algebraic change in length of the member
αTΔTTδ
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 4.12The rigid bar is fixed to the top of the three posts made of A-36 steel and 2014-T6 aluminum. The posts each have a length of 250 mm when no load is applied to the bar, and the temperature is T1 = 20°C. Determine the force supported by each post if the bar is subjected to a uniform distributed load of 150 kN/m and the temperature is raised to T2 = 20°C.
Solution:
( ) (2) alst δδ =↓+
( ) (1) 010902 ;0 3 =−+=↑+ ∑ alsty FFF
From free-body diagram we have
The top of each post is displaced by an equal amount and hence,
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Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth EditionEdition
The final position of the top of each post is equal to its displacement caused by the temperature increase and internal axial compressive force.
Solution:
( ) ( ) ( ) ( )FalTstFstTst δδδδ +−=+−
( ) ( ) ( )( ) ( ) ( )FalTalal
FstTstst
δδδ
δδδ
+−=↓+
+−=↓+
Applying Eq. 2 gives
With reference from the material properties, we have
( )[ ]( )( ) ( )( ) ( )[ ] ( )[ ]( )( ) ( )
( ) ( )[ ]( ) (3) 109.165216.1
101.7303.025.025.020801023
1020002.025.025.020801012
3
926
926
−=
+−−=+−− −−
alst
alst
FF
FFππ
Solving Eqs. 1 and 3 simultaneously yields (Ans) kN 123 and kN 4.16 =−= alst FF
Russell C. Hibbeler
Chapter 5: Torsion
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
Torsional
Deformation of a Circular Shaft
Torque is a moment that twists a member about its longitudinal axis.If the angle of rotation is small, the length of the shaft and its radius will remain unchanged.
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Torsion Formula
When material is linear-elastic, Hooke’s law applies. A linear variation in shear strain leads to a corresponding linear variation in shear stress along any radial line on the cross section.
JTp
JTc
== ττ or max
= maximum shear stress in the shaft= shear stress= resultant internal torque= polar moment of inertia of cross-sectional area= outer radius of the shaft= intermediate distance
maxττTJcp
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Torsion Formula
If the shaft has a solid circular cross section,
If a shaft has a tubular cross section,
4
2cJ π
=
( )44
2 io ccJ −=π
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 5.2The solid shaft of radius c is subjected to a torque T. Find the fraction of T that is resisted by the material contained within the outer region of the shaft, which has an inner radius of c/2 and outer radius c.
Solution:
( ) ( ) ( )ρπρτρρτρ dcdAdT 2' max==
( ) maxτρτ c=
For the entire lighter-shaded area the torque is
(1) 32
152' 3max
2/
3max cdc
Tc
c
τπρρπτ== ∫
Stress in the shaft varies linearly, thus
The torque on the ring (area) located within the lighter-shaded region is
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
Using the torsion formula to determine the maximum stress in the
shaft, we have
Solution:
(Ans) 1615' TT =
( )
3max
4max
22
cT
cTc
JTc
πτ
πτ
=
==
Substituting this into Eq. 1 yields
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 5.3The shaft is supported by two bearings and is subjected to three
torques. Determine the shear stress developed at points A and B, located at section a–a of the shaft.
Solution:From the free-body diagram of the left segment,
( ) mm 1097.4752
74 ×==πJ
kNmm 1250030004250 ;0 =⇒=−−=∑ TTM x
The polar moment of inertia for the shaft is
Since point A is at ρ
= c = 75 mm,( )( ) (Ans) MPa 89.1
1097.4751250
7 =×
==J
TcBτ
Likewise for point B, at ρ
=15 mm, we have( )( ) (Ans) MPa 377.0
1097.4151250
7 =×
==J
TcBτ
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
Power TransmissionPower is defined as the work performed per unit of time.For a rotating shaft with a torque, the power is
Since , the power equation is
For shaft design, the design or geometric parameter is
dtdTP / locity,angular veshaft where θωω ==
f2rad 2cycle 1 πωπ =⇒=
fTP π2=
allow
TcJ
τ=
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 5.5A solid steel shaft AB is to be used to transmit 3750 W from the motor M to which it is attached. If the shaft rotates at w =175 rpm and the steel has an allowable shear stress of allow τallow
=100 MPa, determine the required diameter of the shaft to the nearest mm.
Solution:The torque on the shaft is
Nm 6.20460
21753750 =⇒⎟⎠⎞
⎜⎝⎛ ×
=
=
TT
TPπ
ω
Since
( )( )( ) mm 92.10100
10006.20422
23/13/1
4
=⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
==
ππτ
τπ
allow
allow
Tc
Tcc
cJ
As 2c = 21.84 mm, select a shaft having a diameter of 22 mm.
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
Angle of TwistIntegrating over the entire length L of the shaft, we have
Assume material is homogeneous, G is constant, thus
Sign convention is determined by right hand rule,
( )( )∫=
L
GxJdxxT
0
φΦ
= angle of twistT(x) = internal torqueJ(x) = shaft’s polar moment of inertia
G = shear modulus of elasticity for the material
JGTL
=φ
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 5.8The two solid steel shafts are coupled together using the meshed
gears. Determine the angle of twist of end A of shaft AB when the torque 45 Nm is applied. Take G to be 80 GPa. Shaft AB is free to rotate within bearings E and F, whereas shaft DC is fixed at D. Each shaft has a diameter of 20 mm.
Solution:From free body diagram,
( ) ( ) Nm 5.22075.0300N 30015.0/45==
==
xDTF
Angle of twist at C is( )( )
( )( ) ( )[ ] rad 0269.01080001.02
5.15.2294 +=
+==π
φJG
TLDCC
Since the gears at the end of the shaft are in mesh,
( ) ( )( ) rad 0134.0075.00269.015.0 ⇒=Bφ
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:Since the angle of twist of end A with respect to end B of shaft AB caused by the torque 45 Nm,
( )( )( )( ) ( )[ ] rad 0716.0
1080010.02245
94/ +=+
==π
φJGLT ABAB
BA
The rotation of end A is therefore
(Ans) rad 0850.00716.00134.0/ +=+=+= BABA φφφ
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 5.11The solid steel shaft has a diameter of 20 mm. If it is subjected to the two torques, determine the reactions at the fixed supports A and B.
Solution:By inspection of the free-body diagram,
(1) 0500800 ;0 =−−+−=∑ Abx TTM
Since the ends of the shaft are fixed, 0/ =BAφ
Using the sign convention,
( ) ( )( ) ( )
(2) 7502.08.1
03.05.15002.0
−=−
=++
+−
BA
AAB
TTJG
TJG
TJG
T
Solving Eqs. 1 and 2 yields TA = -345 Nm and TB = 645 Nm.
Russell C. Hibbeler
Chapter 6: Bending
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Shear and Moment DiagramsMembers with support loadings applied perpendicular to their longitudinal axis are called beams.Beams classified according to the way they are supported.
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Shear and Moment DiagramsShear and moment functions can be plotted in graphs called shear and moment diagrams.Positive directions indicate the distributed load acting downward on the beam and clockwise rotation of the beam segment on which it acts.
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.1Draw the shear and moment diagrams for the beam shown.
Solution:From the free-body diagram of the left segment, we apply the equilibrium equations,
( ) (4) 222
;0
(3) 2
02
;0
L-xPMxPLxPMF
PVVPPF
y
y
=⇒−⎟⎠⎞
⎜⎝⎛ −+=
−=⇒=−−=↑
∑
∑
∑
∑
==↑+
==↑+
(2) 2
;0
(1) 2
;0
xPMM
PVFy
Left segment of the beam extending a distance x within region BC is as follow,
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
The shear diagram represents a plot of Eqs. 1 and 3
The moment diagram represents a plot of Eqs. 2 and 4
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.4Draw the shear and moment diagrams for the beam shown.
Solution:The distributed load is replaced by its resultant force and the reactions.
LwwL
wx
w 00 or ==
( )
( )∑
∑
=+⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛+−=+
−=⇒=−⎟⎠⎞
⎜⎝⎛−=↑+
(2) 031
21
23 ;0
(1)2
021
2 ;0
002
0
22000
MxxL
xwxLwLwM
xLL
wVVxL
xwLwFy
Intensity of the triangular load at the section is found by proportion,
Resultant of the distributed loading is determined from the area under the diagram,
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
The shear diagram represents a plot of Eqs. 1
The moment diagram represents a plot of Eqs. 2
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.6Draw the shear and moment diagrams for the beam shown.
Solution:2 regions of x must be considered in order to describe the shear and moment functions for the entire beam.
( ) (2) kNm 8075.5075.580 ;0
(1) kN 75.5075.5 ;0
m, 50
11
1
+=⇒=+−−=+
=⇒=−=↑+
<≤
∑∑
xMMxM
VVF
x
y
( ) ( )
( )
( ) (4) kNm 5.9275.155.2
02
5551575.580 ;0
(3) kN 575.150551575.5 ;0m, 10m 5
222
221
22
1
++−=
=+⎟⎠⎞
⎜⎝⎛ −
−+++−−=+
−=⇒=−−−−=↑+
<≤
∑
∑
xxM
MxxxM
xVVxFx
y
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
The shear diagram represents a plot of Eqs. 1 and 3
The moment diagram represents a plot of Eqs. 2 and 4
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Graphical Method for Constructing Shear and Moment Diagrams
Regions of Distributed LoadRegions of Distributed LoadThe following 2 equations provide a convenient means for quickly obtaining the shear and moment diagrams for a beam.
( )xwdxdV
−=
Slope of the shear
diagram at each point
-distributed load intensity at each point
Vdx
dM=
Slope of moment
diagram at each point
Shear at each point
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Graphical Method for Constructing Shear and Moment Diagrams
We can integrate these areas between any two points to get change in shear and moment.
( )dxxwV ∫−=ΔChange in
shear-area under distributed loading
( )dxxVM ∫=ΔChange in
momentArea under shear diagram
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Graphical Method for Constructing Shear and Moment Diagrams
Regions of Concentrated Force and MomentRegions of Concentrated Force and MomentSome of the common loading cases are shown below.
*Note: Not to be memorized*Note: Not to be memorized
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.7Draw the shear and moment diagrams for the beam shown.
Solution:The reactions are shown on a free-body diagram.
PVLxPVx +==+== ,at and ,0at For shear diagram according to the sign convention,
Since w = 0, the slope of the shear diagram will be zero, thus
points allat 0=−= wdxdV
For moment diagram according to the sign convention,
0 ,at and ,0at ==−== MLxPLMx
points allat PVdxdM +==
The shear diagram indicates that the shear is constantPositive, thus
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.8Draw the shear and moment diagrams for the beam shown.
Solution:The reaction at the fixed support is shown on the free-body diagram.
Since no distributed load exists on the beam the sheardiagram will have zero slope, at all points.
From the shear diagram the slope of the moment diagram will be zero since V = 0.
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Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.10Draw the shear and moment diagrams for the beam shown.
Solution:The reaction at the support is calculated and shown on the free-body diagram.
The distributed loading on the beam is positive yetDecreasing, thus negative slope.
The curve of the moment diagram having this slope behaviour is a cubic function of x.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.11Draw the shear and moment diagrams for the beam shown.
Solution:The reaction at the support is calculated and shown on the free-body diagram.
The slope of the shear diagram will vary from zero at x = 0 to 2 and at x = 4.5. The point of zero shear can be found by
Slope of the moment diagram will begin at 1.5, then becomes decreasingly positive until it reaches zero at 2.6 m.
m 6.205.4
221.51 ;0 =⇒=⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=↑+ ∑ xxxFy
It then becomes increasingly negative reaching 3 at x = 4.5 m.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Bending Deformation of a Straight MemberCross section of a straight beam remains plane when the beam deforms due to bending.There will be tensile stress on one side and compressive stress on the other side.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Bending Deformation of a Straight MemberLongitudinal strain varies linearly from zero at the neutral axis.Hooke’s law applies when material is homogeneous.Neutral axis passes through the centroid of the cross-sectional area for linear-elastic material.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Flexure FormulaResultant moment on the cross section is equal to the moment produced by the linear normal stress distribution about the neutral axis.
By the right-hand rule, negative sign is compressive since it acts in the negative x direction.
IMy
−=σ
σ
= normal stress in the memberM = resultant internal momentI = moment of inertiay = perpendicular distance from the neutral axis
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.15The simply supported beam has the cross-sectional area as shown. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this location.
Solution:The maximum internal moment in the beam is kNm 5.22=M
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:By symmetry, the centroid C and thus the neutral axis pass through the mid-
height of the beam, and the moment of inertia is
( )( )( ) ( )( )( ) ( )( )
( ) 46
323
2
m 103.301
3.002.012116.0002.025.002.025.0
1212
−=
⎥⎦⎤
⎢⎣⎡+⎥⎦
⎤⎢⎣⎡ +=
+=∑ AdII
Applying the flexure formula where c = 170 mm,
( )( ) (Ans) MPa 7.12103.301
17.05.22 ; 6maxmax === −σσI
Mc
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.17The beam has a cross-sectional area in the shape of a channel. Determine the maximum bending stress that occurs in the beam at section a–a.
Solution:The resultant internal moment must be computed about the beam’s neutral axis at section a–a. Since the neutral axis passes through the centroid,
( )( )( ) ( )( )( )( )( ) ( )( )
mm09.59m05909.0 25.002.0015.02.02
25.002.001.0015.02.01.02
==
++
==∑∑
AAy
y
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:Applying the moment equation of equilibrium about the neutral axis, we have
( ) ( ) kNm 859.4005909.00.124.2 ;0 =⇒=−+=+∑ MMM NA
The moment of inertia about the neutral axis is
( )( ) ( )( )( )
( )( ) ( )( )( )
( ) 46
23
23
m 1026.42
05909.01.02.0015.02.0015.01212
01.005909.002.025.002.025.0121
−=
⎥⎦⎤
⎢⎣⎡ −++
⎥⎦⎤
⎢⎣⎡ −+=I
The maximum bending stress occurs at points farthest away from the neutral axis.
( )( ) (Ans) MPa 2.161026.42
05909.02.0859.46max =
−== −I
Mcσ
Russell C. Hibbeler
Chapter 6: Bending
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Shear and Moment DiagramsMembers with support loadings applied perpendicular to their longitudinal axis are called beams.Beams classified according to the way they are supported.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Shear and Moment DiagramsShear and moment functions can be plotted in graphs called shear and moment diagrams.Positive directions indicate the distributed load acting downward on the beam and clockwise rotation of the beam segment on which it acts.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.1Draw the shear and moment diagrams for the beam shown.
Solution:From the free-body diagram of the left segment, we apply the equilibrium equations,
( ) (4) 222
;0
(3) 2
02
;0
L-xPMxPLxPMF
PVVPPF
y
y
=⇒−⎟⎠⎞
⎜⎝⎛ −+=
−=⇒=−−=↑
∑
∑
∑
∑
==↑+
==↑+
(2) 2
;0
(1) 2
;0
xPMM
PVFy
Left segment of the beam extending a distance x within region BC is as follow,
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
The shear diagram represents a plot of Eqs. 1 and 3
The moment diagram represents a plot of Eqs. 2 and 4
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.4Draw the shear and moment diagrams for the beam shown.
Solution:The distributed load is replaced by its resultant force and the reactions.
LwwL
wx
w 00 or ==
( )
( )∑
∑
=+⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛+−=+
−=⇒=−⎟⎠⎞
⎜⎝⎛−=↑+
(2) 031
21
23 ;0
(1)2
021
2 ;0
002
0
22000
MxxL
xwxLwLwM
xLL
wVVxL
xwLwFy
Intensity of the triangular load at the section is found by proportion,
Resultant of the distributed loading is determined from the area under the diagram,
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
The shear diagram represents a plot of Eqs. 1
The moment diagram represents a plot of Eqs. 2
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.6Draw the shear and moment diagrams for the beam shown.
Solution:2 regions of x must be considered in order to describe the shear and moment functions for the entire beam.
( ) (2) kNm 8075.5075.580 ;0
(1) kN 75.5075.5 ;0
m, 50
11
1
+=⇒=+−−=+
=⇒=−=↑+
<≤
∑∑
xMMxM
VVF
x
y
( ) ( )
( )
( ) (4) kNm 5.9275.155.2
02
5551575.580 ;0
(3) kN 575.150551575.5 ;0m, 10m 5
222
221
22
1
++−=
=+⎟⎠⎞
⎜⎝⎛ −
−+++−−=+
−=⇒=−−−−=↑+
<≤
∑
∑
xxM
MxxxM
xVVxFx
y
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
The shear diagram represents a plot of Eqs. 1 and 3
The moment diagram represents a plot of Eqs. 2 and 4
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Graphical Method for Constructing Shear and Moment Diagrams
Regions of Distributed LoadRegions of Distributed LoadThe following 2 equations provide a convenient means for quickly obtaining the shear and moment diagrams for a beam.
( )xwdxdV
−=
Slope of the shear
diagram at each point
-distributed load intensity at each point
Vdx
dM=
Slope of moment
diagram at each point
Shear at each point
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Graphical Method for Constructing Shear and Moment Diagrams
We can integrate these areas between any two points to get change in shear and moment.
( )dxxwV ∫−=ΔChange in
shear-area under distributed loading
( )dxxVM ∫=ΔChange in
momentArea under shear diagram
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Graphical Method for Constructing Shear and Moment Diagrams
Regions of Concentrated Force and MomentRegions of Concentrated Force and MomentSome of the common loading cases are shown below.
*Note: Not to be memorized*Note: Not to be memorized
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.7Draw the shear and moment diagrams for the beam shown.
Solution:The reactions are shown on a free-body diagram.
PVLxPVx +==+== ,at and ,0at For shear diagram according to the sign convention,
Since w = 0, the slope of the shear diagram will be zero, thus
points allat 0=−= wdxdV
For moment diagram according to the sign convention,
0 ,at and ,0at ==−== MLxPLMx
points allat PVdxdM +==
The shear diagram indicates that the shear is constantPositive, thus
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.8Draw the shear and moment diagrams for the beam shown.
Solution:The reaction at the fixed support is shown on the free-body diagram.
Since no distributed load exists on the beam the sheardiagram will have zero slope, at all points.
From the shear diagram the slope of the moment diagram will be zero since V = 0.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.10Draw the shear and moment diagrams for the beam shown.
Solution:The reaction at the support is calculated and shown on the free-body diagram.
The distributed loading on the beam is positive yetDecreasing, thus negative slope.
The curve of the moment diagram having this slope behaviour is a cubic function of x.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.11Draw the shear and moment diagrams for the beam shown.
Solution:The reaction at the support is calculated and shown on the free-body diagram.
The slope of the shear diagram will vary from zero at x = 0 to 2 and at x = 4.5. The point of zero shear can be found by
Slope of the moment diagram will begin at 1.5, then becomes decreasingly positive until it reaches zero at 2.6 m.
m 6.205.4
221.51 ;0 =⇒=⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=↑+ ∑ xxxFy
It then becomes increasingly negative reaching 3 at x = 4.5 m.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Bending Deformation of a Straight MemberCross section of a straight beam remains plane when the beam deforms due to bending.There will be tensile stress on one side and compressive stress on the other side.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Bending Deformation of a Straight MemberLongitudinal strain varies linearly from zero at the neutral axis.Hooke’s law applies when material is homogeneous.Neutral axis passes through the centroid of the cross-sectional area for linear-elastic material.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Flexure FormulaResultant moment on the cross section is equal to the moment produced by the linear normal stress distribution about the neutral axis.
By the right-hand rule, negative sign is compressive since it acts in the negative x direction.
IMy
−=σ
σ
= normal stress in the memberM = resultant internal momentI = moment of inertiay = perpendicular distance from the neutral axis
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.15The simply supported beam has the cross-sectional area as shown. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this location.
Solution:The maximum internal moment in the beam is kNm 5.22=M
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:By symmetry, the centroid C and thus the neutral axis pass through the mid-
height of the beam, and the moment of inertia is
( )( )( ) ( )( )( ) ( )( )
( ) 46
323
2
m 103.301
3.002.012116.0002.025.002.025.0
1212
−=
⎥⎦⎤
⎢⎣⎡+⎥⎦
⎤⎢⎣⎡ +=
+=∑ AdII
Applying the flexure formula where c = 170 mm,
( )( ) (Ans) MPa 7.12103.301
17.05.22 ; 6maxmax === −σσI
Mc
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 6.17The beam has a cross-sectional area in the shape of a channel. Determine the maximum bending stress that occurs in the beam at section a–a.
Solution:The resultant internal moment must be computed about the beam’s neutral axis at section a–a. Since the neutral axis passes through the centroid,
( )( )( ) ( )( )( )( )( ) ( )( )
mm09.59m05909.0 25.002.0015.02.02
25.002.001.0015.02.01.02
==
++
==∑∑
AAy
y
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:Applying the moment equation of equilibrium about the neutral axis, we have
( ) ( ) kNm 859.4005909.00.124.2 ;0 =⇒=−+=+∑ MMM NA
The moment of inertia about the neutral axis is
( )( ) ( )( )( )
( )( ) ( )( )( )
( ) 46
23
23
m 1026.42
05909.01.02.0015.02.0015.01212
01.005909.002.025.002.025.0121
−=
⎥⎦⎤
⎢⎣⎡ −++
⎥⎦⎤
⎢⎣⎡ −+=I
The maximum bending stress occurs at points farthest away from the neutral axis.
( )( ) (Ans) MPa 2.161026.42
05909.02.0859.46max =
−== −I
Mcσ
Russell C. Hibbeler
Chapter 7: Transverse Shear
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 7: Transverse ShearChapter 7: Transverse ShearMechanics of Material 7Mechanics of Material 7thth EditionEdition
Shear in Straight MembersWhen a shear V is applied, non-uniform shear-strain distribution over the cross section will cause the cross section to warp.The relationship between moment and shear is dxdMV =
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 7: Transverse ShearChapter 7: Transverse ShearMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Shear FormulaThe shear formula is used to find the transverse shear stress on the beam’s cross-sectional area.
'' where
'
AyydAQIt
VQ
A
==
=
∫
τ
τ
= the shear stress in the memberV = internal resultant shear forceI = moment of inertia of the entire cross-sectional areat = width of the member’s cross-sectional area
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 7: Transverse ShearChapter 7: Transverse ShearMechanics of Material 7Mechanics of Material 7thth EditionEdition
Shear Stresses in BeamsFor rectangular cross section, shear stress varies parabolically with depth and maximum shear stress is along the neutral axis.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 7: Transverse ShearChapter 7: Transverse ShearMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 7.1The beam is made of wood and is subjected to a resultant internal vertical shear force of V = 3 kN. (a) Determine the shear stress in the beam at point P, and (b) compute the maximum shear stress in the beam.
Solution:
( ) ( )( ) 34 mm 1075.18100505021125' ×=⎥⎦
⎤⎢⎣⎡ +== AyQ
( )( ) 4633 mm 1028.16125100121
121
×=== bhI
(a) The moment of inertia of the cross sectional area computed about the neutral axis is
Applying the shear formula, we have
( )( )( )( ) (Ans) MPa 346.0
1001028.161075.183
6
4
=×
×==
ItVQ
pτ
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 7: Transverse ShearChapter 7: Transverse ShearMechanics of Material 7Mechanics of Material 7thth EditionEdition
(b) Maximum shear stress occurs at the neutral axis, since t is constant throughout the cross section,
Solution:
( )( )( )( ) (Ans) MPa 360.0
1001028.161053.193
6
4
max =×
×==
ItVQτ
( )( ) 34 mm 1053.195.621002
2.65'' ×=⎟⎠⎞
⎜⎝⎛== AyQ
Applying the shear formula yields
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 7: Transverse ShearChapter 7: Transverse ShearMechanics of Material 7Mechanics of Material 7thth EditionEdition
Shear Flow in Built-Up MembersFor fasteners it is necessary to know the shear force by the fastener along the member’s length.This loading is referred as the shear flow q, measured as a force per unit length.
IVQq =
q = shear flowV = internal resultant shearI = moment of inertia of the entire cross-sectional area
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 7: Transverse ShearChapter 7: Transverse ShearMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 7.4The beam is constructed from four boards glued together. If it is subjected to a shear of V = 850 kN, determine the shear flow at B and C that must be resisted by the glue.
Solution:
( ) 46 m 1052.87 −=I
m 1968.0~
==∑∑
AAy
y
The neutral axis (centroid) will be located from the bottom of the beam,
The moment of inertia computed about the neutral axis is thus
[ ]( )( ) ( ) 33 m 10271.001.0250.01968.0305.0'' −=−== BBB AyQ
Since the glue at B and holds the top board to the beam
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 7: Transverse ShearChapter 7: Transverse ShearMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
( )( )
( )( ) MN/m 0996.01052.87
1001026.0850'
MN/m 63.21052.87
10271.0850'
6
3
6
3
=×
×==
=××
==
−
−
−
−
IVQq
IVQq
CC
BB
(Ans) MN/m 0498.0 and MN/m 31.1 == CB qq
Likewise, the glue at C and C’
holds the inner board to the beam
Therefore the shear flow for BB’
and CC’,
[ ]( )( ) ( ) 33 m 1001026.001.0125.01968.0205.0'' −=−== CCC AyQ
Since two seams are used to secure each board, the glue per meter length of beam at each seam must be strong enough to resist one-half of each calculated value of q’.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 7: Transverse ShearChapter 7: Transverse ShearMechanics of Material 7Mechanics of Material 7thth EditionEdition
For point B, the area thus q’B = 0.
Example 7.7The thin-walled box beam is subjected to a shear of 10 kN. Determine the variation of the shear flow throughout the cross section.
Solution:
0'≈A
( )( ) ( )( ) 433 mm 1846412186
121
=−=IThe moment of inertia is
( ) N/mm 5.91 kN/cm 951.0184
2/5.1710====
IVQq C
CFor point C,
The shear flow at D is
( ) N/mm 163 kN/cm 63.1184
2/3010====
IVQq D
D
Also, ( )( )( )( )( )( ) 3
3
cm 304122'
cm 5.17155.3'
===
===
∑ AyQ
AyQ
D
C
Russell C. Hibbeler
Chapter 8: Combined Loading
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 8: Combined LoadingsChapter 8: Combined LoadingsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Thin-Walled Pressure VesselsThin wall refers to a vessel having an inner-radius to-wall-thickness ratio of 10 or more.For cylindrical vessels under normal loading, there are normal stresses in the circumferential or hoop direction and in the longitudinal or axial direction.
( )10/ ≥tr
direction allongitudinin stress normal 2
direction hoopin stress normal
2
1
tprtpr
=
=
σ
σ
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 8: Combined LoadingsChapter 8: Combined LoadingsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Thin-Walled Pressure VesselsThe following is a summary of loadings that can be applied onto a member:a)
Normal Forceb)
Shear Forcec)
Bending Momentd)
Torsional Momente)
Thin-Walled Pressure Vesselsf)
Superposition
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 8: Combined LoadingsChapter 8: Combined LoadingsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 8.2A force of 15 000 N is applied to the edge of the member. Neglect the weight of the member and determine the state of stress at points B and C.
Solution:
( )( ) MPa 75.340100
15000===
APσ
For equilibrium at the section there must be an axial force of 15 000 N acting through the centroid and a bending moment of 750 000 N•mm
about the centroidal
or principal axis.
The maximum stress is
( )( )( )
MPa 25.1110040
121
50750003
max ===I
Mcσ
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 8: Combined LoadingsChapter 8: Combined LoadingsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
( )mm3.33
1001575
=−
=
xxx
The location of the line of zero stress can be determined by proportional triangles
Elements of material at B and C are subjected only to normal or uniaxial stress.
(Ans) on)(compressi MPa 15(Ans) (tension) MPa 5.7
==
C
B
σσ
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 8: Combined LoadingsChapter 8: Combined LoadingsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 8.3The tank has an inner radius of 600 mm and a thickness of 12 mm.
It is filled to the top with water having a specific weight of γw
= 10 kN/m3. If it is made of steel having a specific weight of γst
= 78 kN/m3, determine the state of stress at point A. The tank is open at the top.
Solution:
( ) kN 56.311000600
100061278
22
=⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛== ππγ ststst VW
The pressure on the tank at level A is ( )( ) kPa 10110 === zp wγ
The weight of the tank is
For circumferential and longitudinal stress, we have( )( )
( ) ( )[ ] (Ans) kPa 9.7756.3
(Ans) kPa 50010
210006002
10006122
1000121000600
1
=−
==
===
πσ
σ
st
st
AWtpr
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 8: Combined LoadingsChapter 8: Combined LoadingsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 8.6The rectangular block of negligible weight is subjected to a vertical force of 40 kN, which is applied to its corner. Determine the normal-stress distribution acting on a section through ABCD.
Solution:
( )( ) kPa 1254.08.0
40===
APσ
For 8 kNm, the maximum stress is( )
( )( )[ ] kPa 3754.08.0
2.083
121max ===
x
yx
IcM
σ
For uniform normal-stress distribution the stress is
For 16 kNm, the maximum stress is( )
( )( )[ ] kPa 3758.04.0
4.0163
121max ===
y
xy
IcM
σ
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 8: Combined LoadingsChapter 8: Combined LoadingsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
kPa 125375375125kPa 875375375125kPa 125375375125
kPa 625375375125
−=−+−=−=−−−=−=+−−=
=++−=
D
C
B
A
σσσσ
The line of zero stress can be located along each side by proportional triangles
( ) ( ) m 133.0125625
8.0 and m 0667.0125625
4.0=⇒=
−=⇒=
− hheeee
Assuming that tensile stress is positive, we have
Russell C. Hibbeler
Chapter 9: Stress Transformation
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Plane-Stress TransformationGeneral state of stress at a point is characterized by 6 independent normal and shear stress components.It can be analyzed in a single plane of a body, the material can said to be subjected to plane stress.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Plane-Stress TransformationStress components from one orientation of an element can transform to an element having a different orientation.
=
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Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 9.1The state of plane stress at a point on the surface of the airplane fuselage is represented on the element oriented as shown. Represent the state of stress at the point on an element that is oriented 30°clockwise from the position shown.
Solution:The element is sectioned by the line a–a.
The free-body diagram of the segment is as shown.
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Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
( ) ( )( ) ( )
(Ans) MPa 8.68 030sin30cos2530cos30sin80
30cos30cos2530sin30cos50 ;0'
==°°Δ+°°Δ−
°°Δ−°°Δ−Δ=+∑
x'y'
x'y'y
AA
AAAF
τ
τ
Applying the equations of force equilibrium in the x’ and y’ direction,
( ) ( )( ) ( )
(Ans) MPa 15.4 030cos30sin2530sin30sin80 30sin30cos2530cos30cos50
;0
'
'
'
−==°°Δ+°°Δ+
°°Δ+°°Δ−Δ
=+∑
x
x
x
AAAAA
F
σ
σ
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
( ) ( )( ) ( )
(Ans) MPa 8.68 030cos30sin5030sin30sin25
30sin30cos8030cos30cos25- ;0'
==°°Δ+°°Δ−
°°Δ+°°Δ+Δ=+∑
x'y'
x'y'y
AA
AAAF
τ
τ
Repeat the procedure to obtain the stress on the perpendicular plane b–b.
( ) ( )( ) ( )
(Ans) MPa 8.25 030sin30sin5030cos30cos25 30cos30cos8030sin30cos25 ;0
'
''
−==°°Δ−°°Δ−
°°Δ+°°Δ−Δ=+∑
x
xx
AAAAAF
σ
σ
The state of stress at the point can be represented by choosing an element oriented.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
General Equations of Plane-Stress Transformation
Positive normal stress acts outward from all faces and positive shear stress acts upward on the right-hand face of the element.
θτθσσ
τ
θτθσσσσ
σ
2cos2sin2
2sin2cos22
''
'
xyyx
yx
xyyxyx
x
+−
−=
+−
++
=
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Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 9.2The state of plane stress at a point is represented by the element. Determine the state of stress at the point on another element oriented 30° clockwise from the position shown.
Solution:From the sign convention we have,
To obtain the stress components on plane CD,
°−=−==−= 30 MPa 25 MPa 50 MPa 80 θτσσ xyyx
(Ans) MPa 8.682cos2sin2
(Ans) MPa 8.252sin2cos22
''
'
−=+−
−=
−=+−
++
=
θτθσσ
τ
θτθσσσσ
σ
xyyx
yx
xyyxyx
x
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Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:To obtain the stress components on plane BC,
°−=−==−= 60 MPa 25 MPa 50 MPa 80 θτσσ xyyx
(Ans) MPa 8.682cos2sin2
(Ans) MPa 15.42sin2cos22
''
'
=+−
−=
−=+−
++
=
θτθσσ
τ
θτθσσσσ
σ
xyyx
yx
xyyxyx
x
The results are shown on the element as shown.
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Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Principal Stresses and Maximum In-Plane Shear Stress
InIn--Plane Principal StressesPlane Principal StressesOrientation of the planes will determine the maximum and minimum normal stress.
( ) 2/2tan
yx
xyp σσ
τθ
−=
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Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
.The solution has two roots, thus we obtain the following principle stress.
212
2
2,1 where22
σστσσσσ
σ >+⎟⎟⎠
⎞⎜⎜⎝
⎛ −±
+= xy
yxyx
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Principal Stresses and Maximum In-Plane Shear Stress
Maximum InMaximum In--Plane Shear StressPlane Shear StressOrientation of an element will determine the maximum and minimum shear stress.
The solution has two roots, thus we obtain the maximum in-plane shear stress and averaged normal stress.
( )xy
yxs τ
σσθ
2/2tan
−−=
22
plane-inmax 2 xyyx τ
σστ +⎟⎟
⎠
⎞⎜⎜⎝
⎛ −= 2
yxavg
σσσ
−=
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 9.3When the torsional loading T is applied to the bar it produces a state of pure shear stress in the material. Determine (a) the maximum in-plane shear stress and the associated average normal stress, and (b) the principal stress.
Solution:From the sign convention we have,
a) Maximum in-plane shear stress is
ττσσ −=== xyyx 0 0
(Ans) 02
2
22
plane-inmax =+
=±=+⎟⎟⎠
⎞⎜⎜⎝
⎛ −= yx
avgxyyx σσ
σττσσ
τ
b) For principal stress,
( )
(Ans) 22
135,452/
2tan
22
2,1
12
ττσσσσ
σ
σσσστ
θ
±=+⎟⎟⎠
⎞⎜⎜⎝
⎛ −±
+=
°=°=⇒−
=
xyyxyx
ppyx
xyp
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 9.6The state of plane stress at a point on a body is represented on the element. Represent this stress state in terms of the maximum in-plane shear stress and associated average normal stress.
Solution:Since we have60,90 ,20 ==−= xyyx τσσ
(Ans) MPa 352
(Ans) MPa 4.812
22
plane-inmax
=+
=
=+⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
yxavg
xyyx
σσσ
τσσ
τ
The maximum in-plane shear stress and average normal stress is
( )°=°=⇒
−−= 3.111,3.21
2/2tan 12 ss
xy
yxs σσ
τσσ
θ
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Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Mohr’s Circle—Plane StressPlane stress transformation is able to have a graphical solution that is easy to remember.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 9.8The torsional loading T produces the state of stress in the shaft as shown. Draw Mohr’s circle for this case.
Solution:We first construct of the circle, ττσσ −=== xyyx and 0,0
0 , plane-inmax =−= avgσττ
The center of the circle C is on the axis at 02
=+
= yxavg
σσσ
Point A is the average normal stress and maximum in-plane shear stress,
Principal stresses are identified as points B and D on the circle.
τστσ −== 21 ,
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 9.10The state of plane stress at a point is shown on the element. Determine the maximum in-plane shear stresses and the orientation of the element upon which they act.
Solution:We first construct of the circle, 60 and 90,20 ==−= xyyx τσσ
MPa 4.815560 22 =+=R
The center of the circle C is on the axis at MPa 352
9020=
+−=avgσ
From point C and the A(-20, 60) are plotted, we have
Max in-plane shear stress and average normal stress are
(Ans) MPa 35 , MPa 81.4plane-inmax == avgστ
The counter-clockwise angle is
(Ans) 3.2160
3520tan2 11
°⇒⎟⎠⎞
⎜⎝⎛ +
= −sθ
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 9.12An axial force of 900 N and a torque 2.5 Nm of are applied to the shaft. The shaft diameter is 40 mm, find the principal stresses at a point P on its surface.
Solution:The stresses produced at point P is
( )( ) ( )
kPa 2.71602.0
900 kPa, 9.19802.002.05.2
242
======π
στπ A
PJ
Tc
kPa 1.3582
2.7160 =+
=avgσ
The principal stresses can be found using Mohr’s circle,
Principal stresses are represented by points B and D,
(Ans) kPa 5.517.4091.358(Ans) kPa 7.7677.4091.358
2
1
−=−==+=
σσ
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 9.13The beam is subjected to the distributed loading of w = 120 kN/m. Determine the principal stresses in the beam at point P, which lies at the top of the web. Neglect the size of the fillets and stress concentrations at this point. I = 67.4(10-6) m4.
Solution:kNm 6.30 kN 84 == MVThe equilibrium of the sectioned beam is as shown where
( )( )
( )( )( )[ ]( )( ) (Ans) MPa 2.35
01.0104.67015.0175.01075.084
(Ans) 4.45104.67
1.0106.30
6
6
3
===
−==−
=
−
−
ItVQ
IMy
τ
σ
At point P,
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Chapter 9: Stress TransformationChapter 9: Stress TransformationMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:
( )( ) MPa 6.649.417.22
MPa 2.197.229.41
2
1
−=+−==−=
σσ
7.222
04.45−=
+−
Thus the results are as follows,
Thus the radius is calculated as 41.9, thus the principle stresses are
The centre of the circle is and point A is (-45.4, -32.5).
The counter-clockwise angle is
°=⇒°= 6.282.572 22 pp θθ
Russell C. Hibbeler
Chapter 12: Deflection of Beams and Shafts
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Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Elastic CurveFor elastic curve, positive internal moment tends to bend the beam concave upward, and vice versa.There must be an inflection point at point C, where the curve changes from concave up to concave down, since this is a point of zero moment.
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Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
The Elastic CurveMomentMoment--Curvature RelationshipCurvature Relationship
Due to loading, deformation of the beam is caused by both the internal shear force and bending moment.If material is homogeneous and behaves in a linear-elastic manner, Hooke’s law applies thus,
EIM
=ρ1
ρ
= radius of curvature at a specific pointM = internal moment in the beam at the pointE = material’s modulus of elasticityI = beam’s moment of inertia computed about the neutral axisEI = flexural rigidity
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Slope and Displacement by IntegrationFor most problems the flexural rigidity will be constant along the length of the beam.The slope and displacement relationship of the beam is
Each integration is used to solve for all the constants to obtain a unique solution for a particular problem.
( ) ( ) ( )xMdx
vdEIxVdx
vdEIxwdx
vdEI ==−= 2
2
3
3
4
4
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Slope and Displacement by IntegrationBoundary and Continuity ConditionsBoundary and Continuity Conditions
The constants of integration are determined by evaluating the functions for shear, moment, slope, or displacement.These values are called boundary conditions.
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Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 12.2The simply supported beam supports the triangular distributed loading. Determine its maximum deflection. EI is constant.
Solution:Due to symmetry only one x coordinate is needed for the solution,
The equation for the distributed loading is .Hence,xLww 02
=
2/0 Lx ≤≤
( )
xLwLxwM
xLwxLxwMM NA
43
043
;0
02
0
02
0
+−=
=−⎟⎠⎞
⎜⎝⎛+=+∑
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:Integrating twice, we have
For boundary condition, 0,192
5 have we2,0 and 0,0 2
30
1 =−===== CLwCLxdxdvxv
213050
12040
0302
2
2460
812
43
CxCxLwxL
wEIv
CxLwxL
wdxdvEI
xLwxL
wMdx
vdEI
+++−=
++−=
+−==
(Ans) 120
40
max EILwv −=
Hence,xLwxLwx
LwEIv
1925
2460
303050 −+−=
For maximum deflection at x = L/2,
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 12.4The beam is subjected to a load P at its end. Determine the displacement at C. EI is constant.
Solution:Due to the loading 2 x coordinates will be considered,
Using the free-body diagrams, 2211 2
PxMxPM −=−=
axax ≤≤≤≤ 21 0 and 20
211311
121
1
1
121
12
12
4
2 20for
CxCxPEIv
CxPdxdvEI
xPdx
vdEIax
++−=
+−=
−=<≤Thus,
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Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
And
Solution:
34
232
2
1 67 0
3PaCPaCCPaC −====
axvaxvxv ====== 221111 ,0 ;2,0 ;0,0
423322
322
2
2
222
22
2
6
2
0for
CxCxPEIv
CxPdxdvEI
Pxdx
vdEIax
++−=
+−=
−=<≤
The four constants of integration are determined using three boundary conditions,
Solving, we obtain
Thus solving the equations,EIPax
EIPax
EIPv
32
2322 6
76
−+−=
(Ans) 3
EIPavc −=When x2 = 0, we get
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Discontinuity FunctionsWhen expressing load or internal moment of the beam, we need to use discontinuity functions.
1) Macaulay Functions1) Macaulay FunctionsX is the point along the beam and a is the location on the beam where a “discontinuity” occurs.General equation can used for distributed loadings:
( )an
axax
axax n
n
≥⎩⎨⎧
≥−
<=−
for
for 0
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Discontinuity FunctionsThe Macaulay functions below describe both the uniform load and triangular load.
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Discontinuity FunctionsSingularity FunctionsSingularity Functions
The functions are used to describe the point location forces or couple moments acting on a beam. i) To describe a force,
=
⎩⎨⎧
=≠
=−= −
axPax
axPwfor for 01
ii) To describe a couple moment,
=⎩⎨⎧
=≠
=−= −
axMax
axMwfor for 0
0
20
iii) Integration of both equations will give 2,1,1 −−=−=− +
∫ naxdxax nn
©© 2008 Pearson Education South Asia Pte Ltd2008 Pearson Education South Asia Pte Ltd
Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 12.5Determine the equation of the elastic curve for the cantilevered
beam, EI is constant.
Solution:The boundary conditions require zero slope and displacement at A.The support reactions at A have been calculated by statics and are shown on the free-body diagram,
02021 5855000258052 −−−−−+−+−−= −−−− xxxxxw
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Chapter 12: Deflection of Beams and ShaftsChapter 12: Deflection of Beams and ShaftsMechanics of Material 7Mechanics of Material 7thth EditionEdition
Solution:Since
Integrating twice, we have
2142432
13132
2022
2
531525
31
326129
534550
3426258
54550452258
CxCxxxxxEIv
CxxxxxdxdvEI
xxxxdx
vdEI
++−+−+−+−=
+−+−+−+−=
−+−+−+−=
( ) VdxdMxwdxdV =−= and 11110 58550080258052 −+−+−−−−−= −− xxxxxV
( ) ( )
( ) mkN 54550452258
582155008
210520258
202
20210
⋅−+−+−+−=
−+−+−−−+−−=
xxxx
xxxxxM
Since dv/dx
= 0, x = 0, C1
= 0; and v = 0, C2
= 0. Thus
(Ans) m 531525
31
3261291 42432 ⎟
⎠⎞
⎜⎝⎛ −+−+−+−= xxxxx
EIv
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