4 Pure Bending JKF

MECHANICS OFThird Edition

MECHANICS OF MATERIALS

CHAPTER

MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T. DeWolf

Pure Bending

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.


ThirdEdition

Beer • Johnston • DeWolf

Pure BendingPure BendingOther Loading TypesSymmetric Member in Pure BendingB di D f ti

Example 4.03Reinforced Concrete BeamsSample Problem 4.4St C t tiBending Deformations

Strain Due to BendingBeam Section PropertiesProperties of American Standard Shapes

Stress ConcentrationsPlastic DeformationsMembers Made of an Elastoplastic MaterialPlastic Deformations of Members With a SingleProperties of American Standard Shapes

Deformations in a Transverse Cross SectionSample Problem 4.2Bending of Members Made of Several

Plastic Deformations of Members With a Single Plane of S...

Residual StressesExample 4.05, 4.06

MaterialsExample 4.03Reinforced Concrete BeamsSample Problem 4 4

Eccentric Axial Loading in a Plane of SymmetryExample 4.07Sample Problem 4.8Unsymmetric BendingSample Problem 4.4

Stress ConcentrationsPlastic DeformationsMembers Made of an Elastoplastic Material

Unsymmetric BendingExample 4.08General Case of Eccentric Axial Loading

© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 2


ThirdEdition


Pure Bending

Pure Bending: Prismatic members subjected to equal and opposite couples


j q pp pacting in the same longitudinal plane


ThirdEdition


Other Loading Types

• Eccentric Loading: Axial loading which does not pass through section centroid produces internal forces equivalent to an axial force and a couple

• Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear

• Principle of Superposition: The normal

force and a couple

stress due to pure bending may be combined with the normal stress due to axial loading and shear stress due to h l di t fi d th l t t t


shear loading to find the complete state of stress.


ThirdEdition


Symmetric Member in Pure Bending• Internal forces in any cross section are equivalent

to a couple. The moment of the couple is the section bending moment.

• From statics, a couple M consists of two equal and opposite forces.

• The sum of the components of the forces in anyThe sum of the components of the forces in any direction is zero.

• The moment is the same about any axis perpendicular to the plane of the couple and

• These requirements may be applied to the sums of the components and moments of the statically

perpendicular to the plane of the couple and zero about any axis contained in the plane.

∫∫ ==

dAMdAF xx σ

00

of the components and moments of the statically indeterminate elementary internal forces.


∫ =−=

∫ ==

MdAyM

dAzM

xz

xy

σ

σ 0


ThirdEdition


Bending DeformationsBeam with a plane of symmetry in pure

bending:• member remains symmetricmember remains symmetric

• bends uniformly to form a circular arc

• cross-sectional plane passes through arc center• cross-sectional plane passes through arc centerand remains planar

• length of top decreases and length of bottom increases

• a neutral surface must exist that is parallel to the upper and lower surfaces and for which the lengthupper and lower surfaces and for which the length does not change

• stresses and strains are negative (compressive) b th t l l d iti (t i )


above the neutral plane and positive (tension) below it


ThirdEdition


Strain Due to Bending

Consider a beam segment of length L.

After deformation, the length of the neutral surface remains L. At other sections,

( )( ) yyLL

yL

θρθθρδ

θρ

′

−=′

( )

xyy

L

yyLL

ρρθθδε

θρθθρδ

−=−==

−=−−=−=

linearly) ries(strain va

mm

y

cρc

εε

ερε

−=

== or

mx cεε =



ThirdEdition


Stress Due to Bending• For a linearly elastic material,

mxx EcyE εεσ −==

linearly) varies(stressmcyσ−=

• For static equilibriumFor static equilibrium,

∫

∫∫ −=== dAcydAF mxx

σ

σσ0• For static equilibrium,

dAcyydAyM mx ⎟

⎠⎞

⎜⎝⎛−−=−= ∫∫ σσ

∫−= dAycmσ0

First moment with respect to neutral plane is zero Therefore the neutral

MMcc

IdAyc

M mm

==

== ∫

σ

σσ 2

plane is zero. Therefore, the neutral surface must pass through the section centroid.

cy

SI

mx

m

−=

==

σσ

σ

ngSubstituti


IMy

x −=σ


ThirdEdition


Beam Section Properties• The maximum normal stress due to bending,

inertiaofmomentsection

==

ISM

IMc

mσ

modulussection

inertiaofmoment section

==

=

cIS

I

A beam section with a larger section modulusA beam section with a larger section modulus will have a lower maximum stress

• Consider a rectangular beam cross section,3

Ahbhhbh

cIS 6

1361

3121

2====

Between two beams with the same crossBetween two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bending.

Str ct ral steel beams are designed to ha e a


• Structural steel beams are designed to have a large section modulus.


ThirdEdition


Properties of American Standard Shapes



ThirdEdition


Deformations in a Transverse Cross Section• Deformation due to bending moment M is

quantified by the curvature of the neutral surfaceMcmm ===

11 σε

EIM

IEcEcc

=

ρ

• Although cross sectional planes remain planar when subjected to bending moments, in-plane deformations are nonzero,

ρννεε

ρννεε yy

xzxy =−==−=

• Expansion above the neutral surface andExpansion above the neutral surface and contraction below it cause an in-plane curvature,

curvature canticlasti 1==

′ ρν

ρ


ρρ


ThirdEdition


Sample Problem 4.2SOLUTION:

• Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.

( )∑ +=∑∑= ′

2dAIIAAyY x

• Apply the elastic flexural formula to find the maximum tensile and compressive stresses

A cast-iron machine part is acted upon

compressive stresses.

IMc

m =σ

p pby a 3 kN-m couple. Knowing E = 165 GPa and neglecting the effects of fillets, determine (a) the maximum

• Calculate the curvature

EIM

=ρ1


tensile and compressive stresses, (b) the radius of curvature.


ThirdEdition


Sample Problem 4.2SOLUTIONSOLUTION:

Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.

×=× 3

32

109050180090201mm ,mm ,mm Area, Ayy

∑ ×==∑

×=××=×

3

3

101143000104220120030402109050180090201

AyA

mm 383000

10114 3=

×=

∑∑=

AAyY

( ) ( )( ) ( )23

12123

121

231212

18120040301218002090 ×+×+×+×=

∑ +=∑ +=′ dAbhdAIIx


49-31212

m10868 mm10868 ×=×=I


ThirdEdition


Sample Problem 4.2

• Apply the elastic flexural formula to find the maximum tensile and compressive stresses.

Mc

49 mm10868m022.0mkN 3

−×

×⋅==

=

IcM

IA

A

m

σ

σ

MPa0.76+=Aσ

49 mm10868m038.0mkN 3

−×

×⋅−=−=

IcM B

Bσ MPa3.131−=Bσ

• Calculate the curvature1=

EIM

ρ

( )( )49- m10868GPa 165mkN 3×

⋅=

m 7.47

m1095.201 1-3

=

×= −

ρρ



ThirdEdition


Bending of Members Made of Several MaterialsC id i b f d f• Consider a composite beam formed from two materials with E1 and E2.

• Normal strain varies linearly.y

ρε y

x −=

• Piecewise linear normal stress variation.

ρεσ

ρεσ yEEyEE xx

222

111 −==−==

Neutral axis does not pass through section centroid of composite section.

• Elemental forces on the section areyEyE 21 dAyEdAdFdAyEdAdF

ρσ

ρσ 2

221

11 −==−==

( ) EyEynE• Define a transformed section such thatx I

Myσ −=


( ) ( )1

2112 E

EndAnyEdAynEdF =−=−=ρρ

xx nσσσσ == 21


ThirdEdition


Example 4.03SOLUTION:

• Transform the bar to an equivalent cross section made entirely of brasssection made entirely of brass

• Evaluate the cross sectional properties of the transformed sectione s o ed sec o

• Calculate the maximum stress in the transformed section. This is the correct

Bar is made from bonded pieces of

maximum stress for the brass pieces of the bar.

steel (Es = 29x106 psi) and brass (Eb = 15x106 psi). Determine the maximum stress in the steel and b h f 40 ki *i

• Determine the maximum stress in the steel portion of the bar by multiplying the maximum stress for the transformed

i b h i f h d li f


brass when a moment of 40 kip*in is applied.

section by the ratio of the moduli of elasticity.


ThirdEdition


Example 4.03SOLUTIONSOLUTION:• Transform the bar to an equivalent cross section

made entirely of brass.6

in 25.2in 4.0in 75.0933.1in 4.0

933.1psi1015psi1029

6

6

=+×+=

=×

×==

T

b

s

b

EEn

• Evaluate the transformed cross sectional properties( )( )312

13121 in 3in. 25.2== hbI T

4in063.5=

• Calculate the maximum stresses( )( )in51inkip40Mc ( )( ) ksi 85.11

in5.063in5.1inkip40

4 =⋅

==I

Mcmσ

( )max = mb σσ ( ) ksi85.11max =bσ


( ) ksi 85.11933.1max ×== ms nσσ ( ) ksi 22.9max =sσ


ThirdEdition


Reinforced Concrete Beams• Concrete beams subjected to bending moments are

reinforced by steel rods.

• The steel rods carry the entire tensile load below

• In the transformed section the cross sectional area

The steel rods carry the entire tensile load below the neutral surface. The upper part of the concrete beam carries the compressive load.

• In the transformed section, the cross sectional area of the steel, As, is replaced by the equivalent areanAs where n = Es/Ec.

T d t i th l ti f th t l i• To determine the location of the neutral axis,( ) ( )

0

0221 =−+

=−−

dAnxAnxb

xdAnxbx s

02 + dAnxAnxb ss

• The normal stress in the concrete and steel

xMyσ −=


xsxc

x

nI

σσσσ

σ

==


ThirdEdition


Sample Problem 4.4

SOLUTION:

• Transform to a section made entirely fof concrete.

• Evaluate geometric properties of transformed sectiontransformed section.

• Calculate the maximum stresses in the concrete and steel.

A concrete floor slab is reinforced with 5/8-in-diameter steel rods. The modulus

f l ti it i 29 106 i f t l dof elasticity is 29x106psi for steel and 3.6x106psi for concrete. With an applied bending moment of 40 kip*in for 1-ft width of the slab determine the maximum


width of the slab, determine the maximum stress in the concrete and steel.


ThirdEdition


Sample Problem 4.4SOLUTION:• Transform to a section made entirely of concrete.

6068psi 1029

=×

== sEn

( ) 2285

4

6

in95.4in 206.8

06.8psi 106.3

=⎥⎦⎤

⎢⎣⎡×=

=×

==

πs

c

nA

En

• Evaluate the geometric properties of the transformed section.

( ) in45010495412 ==−−⎟⎞

⎜⎛ xxxx ( )

( )( ) ( )( ) 422331 in4.44in55.2in95.4in45.1in12

in450.10495.42

12

=+=

==⎟⎠

⎜⎝

I

xxx

• Calculate the maximum stresses• Calculate the maximum stresses.

41

i552iki40in44.4

in1.45inkip40 ×⋅==

M

IMc

cσ ksi306.1=cσ


42

in44.4in55.2inkip4006.8 ×⋅

==I

Mcnsσ ksi52.18=sσ


ThirdEdition


Stress Concentrations

Stress concentrations may occur:I

McKm =σ

• in the vicinity of points where the loads are applied

I

in the icinit of abr pt changes


• in the vicinity of abrupt changes in cross section


ThirdEdition


Plastic Deformations• For any member subjected to pure bending

mx cy εε −= strain varies linearly across the section

• If the member is made of a linearly elastic material, the neutral axis passes through the section centroid

Myx −=σand Ixand

• For a material with a nonlinear stress-strain curve, the neutral axis location is found by satisfying

∫ −=∫ == dAyMdAF xxx σσ 0

• For a member with vertical and horizontal planes of symmetry and a material with the same tensile and compressive stress-strain relationship, the neutral axis is located at the section centroid and the stress-

i l i hi b d h i


strain relationship may be used to map the strain distribution from the stress distribution.


ThirdEdition


Plastic Deformations• When the maximum stress is equal to the ultimate

strength of the material, failure occurs and the corresponding moment MU is referred to as the

l i b diultimate bending moment.

• The modulus of rupture in bending, RB, is found from an experimentally determined value of MUfrom an experimentally determined value of MU and a fictitious linear stress distribution.

IcMR U

B =I

• RB may be used to determine MU of any member made of the same material and with the same cross sectional shape but different dimensions.



ThirdEdition


Members Made of an Elastoplastic MaterialR l b d f l l i i l• Rectangular beam made of an elastoplastic material

=≤ mYx

II

Mcσσσ

moment elastic maximum === YYYm cIM σσσ

• If the moment is increased beyond the maximum l ti t l ti d l delastic moment, plastic zones develop around an

elastic core.

thickness-halfcoreelastic1 2

2

31

23 =⎟

⎟⎞

⎜⎜⎛−= Y

YY yyMM 232 ⎟

⎠⎜⎝

YY yc

• In the limit as the moment is increased further, the elastic core thickness goes to zero corresponding to aelastic core thickness goes to zero, corresponding to a fully plastic deformation.

moment plastic 23 == Yp

M

MM


shape)section crosson only (dependsfactor shape ==Y

pMM

k


ThirdEdition


Plastic Deformations of Members With a Single Plane of S mmetrSingle Plane of Symmetry

• Fully plastic deformation of a beam with only a vertical plane of symmetryvertical plane of symmetry.

• The neutral axis cannot be assumed to pass through the section centroid.

• Resultants R1 and R2 of the elementary compressive and tensile forces form a couple.

YY AARRσσ 21

21==

The neutral axis divides the section into equal areas.

• The plastic moment for the member,

( )


( )dAM Yp σ21=


ThirdEdition


Residual Stresses

• Plastic zones develop in a member made of an elastoplastic material if the bending moment is large enough.g g

• Since the linear relation between normal stress and strain applies at all points during the unloading

h i b h dl d b i h bphase, it may be handled by assuming the member to be fully elastic.

• Residual stresses are obtained by applying the• Residual stresses are obtained by applying the principle of superposition to combine the stresses due to loading with a moment M (elastoplastic deformation) and unloading with a moment -Mdeformation) and unloading with a moment M(elastic deformation).

• The final value of stress at a point will not, in


general, be zero.


ThirdEdition


Example 4.05, 4.06

A member of uniform rectangular cross section is subjected to a bending moment M = 36.8 kN-m. Th b i d f l t l ti t i lThe member is made of an elastoplastic material with a yield strength of 240 MPa and a modulus of elasticity of 200 GPa.

Determine (a) the thickness of the elastic core, (b) the radius of curvature of the neutral surface.

After the loading has been reduced back to zero,After the loading has been reduced back to zero, determine (c) the distribution of residual stresses, (d) radius of curvature.



ThirdEdition


Example 4.05, 4.06• Thickness of elastic core:

1 2

2

31

23

⎟⎟⎠

⎞⎜⎜⎝

⎛−= Y

YcyMM

( ) 1mkN28.8mkN8.36 2

2

31

23

⎟⎟⎠

⎞⎜⎜⎝

⎛−⋅=⋅

⎠⎝

Y

yycy

666.0mm60

== YY ycy

mm802 =Yy

• Radius of curvature:

( )( )10601050233

322

32 ××== −− mmbc

cI

• Maximum elastic moment:d us o cu v u e:

3

9

6

1021

Pa10200Pa10240

−×=

×

×== Y

Y Eσε

( )( )MPa240m10120

m10120

36

36

×==

×=

−

−

YY cIM

c

σ31040

102.1

−

=

×=

YY

yρ

ε


mkN 8.28 ⋅=c

3

3

102.1m1040

−×

×==

Y

Yyε

ρ m3.33=ρ


ThirdEdition


Example 4.05, 4.06

• M = 36.8 kN-mmm40=Yy

• M = -36.8 kN-mmkN8.36 ⋅′ Mc

• M = 0core,elastictheofedgeAt the

MPa240mm40

Y =σYy

Y

36

2MPa7.306m10120mkN8.36

σ

σ

<=×

==′I

Mcm

6

9

6

101

Pa10200Pa105.35

core,elastictheofedgeAt the

−

×

×−== x

x Eσε

6

3

6

105.177m1040

105.177

−

−

−

×

×=−=

×−=

x

Yyε

ρ


m225=ρ


ThirdEdition


Eccentric Axial Loading in a Plane of Symmetry• Stress due to eccentric loading found by

superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment

( ) ( )MyP

xxx += bendingcentric σσσ

Iy

A−=

• Eccentric loading

PF =• Validity requires stresses below proportional

limit deformations have negligible effect onPdMPF

== limit, deformations have negligible effect on

geometry, and stresses not evaluated near points of load application.



ThirdEdition



• Find the equivalent centric load and bending momentbending moment

• Superpose the uniform stress due to the centric load and the linear stress due to the bending moment.

• Evaluate the maximum tensile and

An open-link chain is obtained by

compressive stresses at the inner and outer edges, respectively, of the superposed stress distribution.

bending low-carbon steel rods into the shape shown. For 160 lb load, determine (a) maximum tensile and compressive

(b) di b i

• Find the neutral axis by determining the location where the normal stress is zero


stresses, (b) distance between section centroid and neutral axis

is zero.


ThirdEdition


Example 4.07

( )in25.0 22 == cA ππ

• Normal stress due to a centric load

in1963.0lb160

in1963.0

20

2

==

=

APσ

• Equivalent centric load

psi815=

• Normal stress due toqand bending moment

( )( )in60lb160lb160==

=PdM

P ( )25.043

4414

41 == cI ππ

Normal stress due to bending moment

( )( )inlb104

in6.0lb160⋅=

PdM

( )( )in10068.

in25.0inlb104in10068.3

43

43

×

⋅==

×=

−

−

IMc

mσ


psi8475=


ThirdEdition


Example 4.07

• Maximum tensile and compressive stresses

• Neutral axis locationstresses

84758150

=+=

+= mt

σσσ

σσσpsi9260=tσ

( )ilb105

in10068.3psi815

0

430

0

×==

−=

−

MI

APy

IMy

AP

84758150−=−= mc σσσ

psi7660−=cσ

( )inlb105

p0 ⋅MAy

in0240.00 =y



ThirdEdition


Sample Problem 4.8The largest allowable stresses for the cast iron link are 30 MPa in tension and 120 MPa in compression. Determine the largest force P which can be applied to the link.

SOLUTION:

• Determine an equivalent centric load and bending moment.

S h d i

• Evaluate the critical loads for the allowable

• Superpose the stress due to a centric load and the stress due to bending.

tensile and compressive stresses.

• The largest allowable load is the smallest f th t iti l l d

From Sample Problem 2.4,23

m0380m103 −

=

×=

YA


of the two critical loads.49 m10868

m038.0−×=

=

I

Y


ThirdEdition


Sample Problem 4.8• Determine an equivalent centric and bending loads.

load centricm028.0010.0038.0

==−=

Pd

momentbending 028.0 === PPdM

• Superpose stresses due to centric and bending loads( )( )( )( )

( )( ) PPPI

McAP

PPPI

McAP

AB

AA

1559022.0028.0

37710868

022.0028.0103

93

93

−=−−=−−=

+=×

+×

−=+−=

−−

−−

σ

σ

• Evaluate critical loads for allowable stresses.kN679MP30377+ PP

IA 10868103 93 ×× −−

kN6.79MPa1201559

kN6.79MPa30377

=−=−=

==+=

PP

PP

B

A

σ

σ

kN077PTh l ll bl l d


kN0.77=P• The largest allowable load


ThirdEdition


Unsymmetric Bending• Analysis of pure bending has been limited

to members subjected to bending couples acting in a plane of symmetry.

• Members remain symmetric and bend in the plane of symmetry.

Will id it ti i hi h th

• The neutral axis of the cross section coincides with the axis of the couple

• Will now consider situations in which the bending couples do not act in a plane of symmetry.

I l th t l i f th ti ill

• Cannot assume that the member will bend in the plane of the couples.


• In general, the neutral axis of the section will not coincide with the axis of the couple.


ThirdEdition


Unsymmetric Bending⎞⎛•

∫=

∫ ⎟⎠⎞

⎜⎝⎛−=∫==

dAy

dAcydAF mxx

0or

0 σσ

i h d i h di i d

neutral axis passes through centroid

dAyyMM ∫ ⎟⎞

⎜⎛ σWish to determine the conditions under

which the neutral axis of a cross section of arbitrary shape coincides with the axis of the couple as shown

•

d fi di ib i

inertiaofmomentIIc

Iσ

dAc

yMM

zm

mz

===

∫ ⎟⎠

⎜⎝−−==

Mor

σ

axis of the couple as shown.

• dAyzdAzM ∫ ⎟⎞

⎜⎛=∫==0 σσ

• The resultant force and moment from the distribution of

defines stress distribution

•

couple vector must be directed along

inertiaofproductIdAyz

dAc

zdAzM

yz

mxy

==∫=

∫ ⎟⎠

⎜⎝−=∫==

0or

0 σσfrom the distribution of elementary forces in the section must satisfy

coupleappliedMMMF yx ==== 0


couple vector must be directed along a principal centroidal axis

coupleappliedMMMF zyx 0


ThirdEdition


Unsymmetric BendingSuperposition is applied to determine stresses in the most general case of unsymmetric bending.

• Resolve the couple vector into components along p p gthe principle centroidal axes.

θθ sincos MMMM yz ==

• Superpose the component stress distributionsSuperpose the component stress distributions

y

y

z

zx I

yMI

yM+−=σ

• Along the neutral axis,( ) ( )θθσ sincos0

yzy

y

z

zx I

yMI

yMI

yMI

yM+−=+−==

θφ tantany

zII

zy==



ThirdEdition



• Resolve the couple vector into components along the principlecomponents along the principle centroidal axes and calculate the corresponding maximum stresses.

θθ sincos MMMM == θθ sincos MMMM yz ==

• Combine the stresses from the component stress distributions.

A 1600 lb-in couple is applied to a rectangular wooden beam in a plane

y

y

z

zx I

yMI

yM+−=σ

• Determine the angle of the neutralforming an angle of 30 deg. with the vertical. Determine (a) the maximum stress in the beam, (b) the angle that the

t l i f ith th h i t l

Determine the angle of the neutral axis.

θφ tantany

zII

zy==


neutral axis forms with the horizontal plane.

y


ThirdEdition


Example 4.08• Resolve the couple vector into components and calculate

the corresponding maximum stresses.

( ) inlb138630cosinlb1600 ⋅=⋅=zM ( )( )

( )( )

( )( )

in359.5in5.3in5.1

inlb80030sininlb1600

431

43121 ==

⋅=⋅=

z

y

z

I

M

( )( )

( )( ) psi6452in75.1inlb1386 along occurs todue stress nsilelargest te The

in9844.0in5.1in5.3

1

43121

=⋅

==

==

z

z

y

yMABM

I

σ

( )( ) psi5609in75.0inlb800

along occurs todue stress nsilelargest te The

psi6.452in359.5 41

⋅

===

y

z

z

zMADM

I

σ

σ

( )( ) psi5.609in9844.0 42 ===

y

yI

σ

• The largest tensile stress due to the combined loading occurs at A


occurs at A.5.6096.45221max +=+= σσσ psi1062max =σ


ThirdEdition


Example 4.08

• Determine the angle of the neutral axis.

in3595 4I

143.3

30tanin9844.0

in359.5tantan 4

=

== θφy

zII

o4.72=φ



ThirdEdition


General Case of Eccentric Axial Loading• Consider a straight member subject to equal

and opposite eccentric forces.

• The eccentric force is equivalent to the system of a centric force and two couples.

bP = forcecentric

PbMPaM zy ==

• By the principle of superposition, the combined stress distribution iscombined stress distribution is

y

y

z

zx I

zMI

yMAP

+−=σ

• If the neutral axis lies on the section, it may be found from

PMM yz


Az

Iy

I y

y

z

z =−

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4 Pure Bending JKF