strengths of materials - 7e - Hibbeler

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.



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



Equilibrium of a Deformable BodyReactionsReactions

Surface forces developed at the supports/points of contact between bodies.



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



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



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 =



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



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.



FBD

.



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



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.



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



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.



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



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.



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



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



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



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



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



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

=

=

=

.

.

.



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



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)



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



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



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.



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



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



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



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.



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



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


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.



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 =



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.



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.



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.



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.



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



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.



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 ==



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.



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



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.



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.



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



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



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.



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



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



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



*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.



*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


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.



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



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



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)



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

==



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 === ∫∫



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



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)



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.



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



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,



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,



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



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,



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


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.



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



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



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



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



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



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=



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.



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



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



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



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


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.



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.



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,



Solution:

The shear diagram represents a plot of Eqs. 1 and 3

The moment diagram represents a plot of Eqs. 2 and 4




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,



Solution:

The shear diagram represents a plot of Eqs. 1

The moment diagram represents a plot of Eqs. 2




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



Solution:





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


Shear at each point




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




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




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




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.




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.





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.



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.



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.



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



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



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



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



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



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.



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.




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,



Solution:






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,



Solution:

The shear diagram represents a plot of Eqs. 1

The moment diagram represents a plot of Eqs. 2




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



Solution:






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


-distributed load intensity at each point

Vdx

dM=

Slope of moment


Shear at each point




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




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




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




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.





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.





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.



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.



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.



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



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



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



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



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

0002221s

Line

0002221s

Line

0002221s

Line

0002221s

Line

0002221s

Line

0002221s

Line

0002221s

Pencil

0002221s

Pencil

0002221s

Pencil

0002221s

Pencil

0002221s

Pencil

0002221s

Pencil

0002221s

Pencil

0002221s

Pencil

0002221s

Pencil

0002221s

Pencil

0002221s

Line

Russell C. Hibbeler

Chapter 7: Transverse Shear


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 =



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



Shear Stresses in BeamsFor rectangular cross section, shear stress varies parabolically with depth and maximum shear stress is along the neutral axis.



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



(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



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



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



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’.



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


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



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



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



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



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



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



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


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.



Plane-Stress TransformationStress components from one orientation of an element can transform to an element having a different orientation.

=



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.



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



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.



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

+−

−=

+−

++

=



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



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.



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

−=



.The solution has two roots, thus we obtain the following principle stress.

212

2

2,1 where22

σστσσσσ

σ >+⎟⎟⎠

⎞⎜⎜⎝

⎛ −±

+= xy

yxyx



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

−=



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



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



Mohr’s Circle—Plane StressPlane stress transformation is able to have a graphical solution that is easy to remember.



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 ,



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



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



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,



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


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.



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



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



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.



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

+−=

=−⎟⎠⎞

⎜⎝⎛+=+∑



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,



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,



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



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



Discontinuity FunctionsThe Macaulay functions below describe both the uniform load and triangular load.



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



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



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

strengths of materials - 7e - Hibbeler

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