BEAMS: BENDING STRESS Beams

1

• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering

Third EditionLECTURE

94.1 – 4.54.13

Chapter

BEAMS: BENDING STRESS

byDr. Ibrahim A. Assakkaf

SPRING 2003ENES 220 – Mechanics of Materials

Department of Civil and Environmental EngineeringUniversity of Maryland, College Park

LECTURE 9. BEAMS: BENDING STRESS (4.1 – 4.5, 4.13) Slide No. 1ENES 220 ©Assakkaf

Beams

Introduction– The most common type of structural

member is a beam.– In actual structures beams can be found in

an infinite variety of• Sizes• Shapes, and• Orientations

2


Beams

IntroductionDefinition

A beam may be defined as a member whoselength is relatively large in comparison withits thickness and depth, and which is loadedwith transverse loads that produce significantbending effects as oppose to twisting or axialeffects


Beams

Introduction

Beam

Figure 1

3


Beams

IntroductionFigure 2

(a)

(b)

(c)

(d)

a

b

c

d e

CableLoad


Beams

Introduction– Beams can be

• Straight as shown in Figure 1c– For example the straight member bde

• Curved as shown in Figure 1c– For example the curved member abc

4


Beams

Introduction– Beams are generally classified according

to their geometry and the manner in which they are supported.

– Geometrical classification includes such features as the shape of the cross section, whether the beam is

• straight or • curved


Beams

Introduction– Or whether the beam is

• Tapered, or• Has a constant cross section.

– And some other features that will be discussed later

5


Beams

Introduction– Beams can also be classified according to

the manner in which they are supported. Some types that occur in ordinary practice are shown in Figure 3, the names of some of these being fairly obvious from direct observation.

– Note that the beams in (d), (e), and (f) are statically indeterminate.


Beams

Introduction

(a) Cantilever

(c) Overhanging

(e) Fixed ended

(b) Simply supported

(d) continuous

(f) Cantilever, simply supported

Figure 3

6


Pure Bending

Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane


Other Loading Types

• Principle of Superposition: The normal stress due to pure bending may be combined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state of stress.

• 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 force and a couple

7

LECTURE 9. BEAMS: BENDING STRESS (4.1 – 4.5, 4.13) Slide No. 12ENES 220 ©AssakkafSymmetric Member in Pure Bending

∫ =−=

∫ ==∫ ==

MdAyM

dAzMdAF

xz

xy

xx

σ

σσ

00

• These requirements may be applied to the sums of the components and moments of the statically indeterminate elementary internal forces.

• 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 any direction is zero.

• The moment is the same about any axis perpendicular to the plane of the couple and zero about any axis contained in the plane.


Normal and Shearing Stress

Normal Stress σ in beamsThe normal stress on plane a-a is related to the resisting moment Mr as follows (see Figure 4):

∫−= area dAyM r σ (1)

8



Stresses in beams

x

y

b

b

a

aP

x

y Ph

w

∆x

R

Mr

VO

dA στ

+y

Figure 4


Shearing Stress τ in beamsThe shearing stress on plane a-a is related to the resisting shear V as follows (see Figure 4):

∫−= area dAVr τ (2)


9


Flexural Strains

Deformation of Beam due to Lateral Loading

wP

Figure 5



Figure 6

c

Flexural Strains

10

LECTURE 9. BEAMS: BENDING STRESS (4.1 – 4.5, 4.13) Slide No. 18ENES 220 ©AssakkafBending Deformations

Beam with a plane of symmetry in pure bending:

• member remains symmetric

• bends uniformly to form a circular arc

• 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 length does not change

• stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it

LECTURE 9. BEAMS: BENDING STRESS (4.1 – 4.5, 4.13) Slide No. 19ENES 220 ©AssakkafStrain Due to Bending

Consider a beam segment of length L.

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

( )( )

mx

mm

x

cy

cρc

yyL

yyLL

yL

εε

ερε

ρρθθδε

θρθθρδ

θρ

−=

==

−=−==

−=−−=′−=

−=′

or

linearly) ries(strain va

11


Flexural Strains

Deformation of Beam due to Lateral Loading– A segment of the beam of Fig. 4 between

planes a-a and b-b is shown in Figure 7 with the deformation (distortion) is greatly exaggerated

– Assumption is made that a plane section before bending remains plane after bending.


Flexural Strains


ρ - yρ

∆θ

∆x`

∆x

yc

xy

b

b

a

aP

x

y Ph

w

∆x

R

Mr

VO

dA στ

+y

Figure 4

Figure 7

12


Stresses in beams

x

y

b

b

a

aP

x

y Ph

w

∆x

R

Mr

VO

dA στ

+y

Figure 4

Flexural Strains


Flexural Strains

Experimental Results– Precise experimental measures suggest that at

some distance c (see Figure 7) above the bottom of the beam, longitudinal elements undergo no change in length.

– The curved surface formed by these elements (at radius ρ) is referred to as the neutral surface of the beam, and the intersection of this surface with any cross section is called the neutral axis of the section.

13


Flexural Strains

Experimental Results– All elements (fibers) on one side of the

neutral surface are compressed and those on the opposite side are elongated.

– In reference to Fig. 8, the fibers above the neutral surface of the beam are compressed, while those below the neutral axis are elongated.


Flexural Strains

Experimental Results

ρ - yρ

∆θ

∆x`

∆x

yc

dA στ

+y

Neutral Axis

Figure 8

14


Flexural Strains

Longitudinal Strain– The longitudinal strain εx experienced by a

longitudinal element that is located a distance y from the neutral axis (surface) of the beam is determined as follows:

f

ifx L

LLLL −=

∆=ε

Lf = final length after loadingLi = initial length before loading

(3)


Flexural Strains

Longitudinal Strain– From the geometry of the beam as shown

in Figure 7 and 8:

ρ - yρ ∆θ

∆x`∆x

yc

( )θρ

θρθρ

ε

∆∆−∆−

=

∆∆−′∆

=

yx

xxx

(4)

15


Flexural Strains

Longitudinal Strain

( )

y

yy

yx

xx

x

x

ρε

θρθ

θρθρθθρ

θρθρθρε

1

−=

∆∆−

=∆

∆−∆−∆=

∆∆−∆−

=∆∆−′∆

=

(5)


Flexural Strains

Longitudinal StrainThe normal longitudinal strain εx varies linearly, through the member, with the distance y from the neutral surface, and it is given by

ρε y

x −= (6)

16


Flexural Stress

For special case of linearly elastic deformation, the relationship between the normal stress σx and the normal strain εx is given by Hooke’s law as

slope==x

xEεσ

σx

εx

slope ==x

xEεσ(7)


Flexural Stress

Flexural Normal Stress– Eq. 7 can be rewritten as

– Recall Eq. (6) , therefore

Exx εσ = (8)

ρε y

x −=

EyExx ρεσ −== (9)

17


Flexural Stress

Flexural Normal StressDistribution of Normal Stress in a Beam Cross Section

x

y P

w

R

FC

FT

Vr

c

ydy

yC

dA

Neutral axis

Figure 9

Centroidal axis

cc


Flexural Stress

Flexural Normal Stress– The resisting moment Mr that can be

develop by the normal stress in a typical beam with loading in a plane of symmetry but arbitrary cross section (Fig. 9) is given by Eq. 1 as

∫−= area dAyM xr σ

18


Flexural Stress

Flexural Normal Stress– Since y is measured from the neutral axis

(surface), it is necessary to locate this axis by means of the equilibrium equation as follows:

0

0

=

=

∫∑

dA

F

A x

x

σ (10)


Flexural Stress

Flexural Normal Stress– Substituting for σx given by Eq. 9 into Eq.

10, yields

– But

0 =−=

−=

∫

∫∫

A

AA x

ydAE

EdAydA

ρ

ρσ

)( axis centroidal to

axis neutral from distance

c-c

ydAyAC == ∫

(11)

(12)

19


Flexural Stress

Flexural Normal Stress– Therefore, Eq. 11 becomes

– Since neither (E/ρ) nor A are zero, yC must equal zero.

0=−=∫ AyEdA CA x ρσ (12)


Flexural Stress

Flexural Normal Stress

For flexural loading and linearlyelastic action, the neutral axis passes through the centroid of the cross sectionof the beam

20


Flexural Stress

Flexural Normal Stress– The maximum normal stress on the cross

section is given by

– Combining Eqs. 9 and 13, hence

cEρ

σ −=max (13)

cx cy

cy σσσ == max (14)


Flexural Stress

Flexural Normal Stress– Substituting Eq. 14 into Eq. 1, gives

– The integral is called the second moment of area, and it is given the symbol I.

∫∫

∫

−=

−=

−=

Ac

A c

xr

dAyc

dAcyy

dAyM

2

area

σσ

σ

∫ dAy 2

(15)

21


Flexural Stress

Flexural Normal Stress– Substituting for the second moment of area

I of the cross section of the beam into Eq. 15, yields

maxc

2

or

σσ

σσ

==

−=−= ∫

cI

M

Ic

dAyc

M

r

cA

cr

(16)

(17)


Elastic Flexural Formula

The elastic flexural formula for normal stress is given by

IyM

IcM

rx

r

=

=

σ

σ

and

max (18)

(19)

22


Elastic Flexural Formula

An alternative form of the flexural formula for maximum normal stress is given by

Where S

M r=max σ (20)

cIS =

LECTURE 9. BEAMS: BENDING STRESS (4.1 – 4.5, 4.13) Slide No. 43ENES 220 ©AssakkafStress Due to Bending

• For a linearly elastic material,

linearly) varies(stressm

mxx

cy

EcyE

σ

εεσ

−=

−==

• For static equilibrium,

∫

∫∫

−=

−===

dAyc

dAcydAF

m

mxx

σ

σσ

0

0

First moment with respect to neutral plane is zero. Therefore, the neutral surface must pass through the section centroid.

• For static equilibrium,

IMy

cy

SM

IMc

cIdAy

cM

dAcyydAyM

x

mx

m

mm

mx

−=

−=

==

==

−−=−=

∫

∫∫

σ

σσ

σ

σσ

σσ

ngSubstituti

2

23


Deformation in a Symmetric Member in Pure Bending

• Deformation due to bending moment M is quantified by the curvature of the neutral surface

EIM

IMc

EcEccmm

=

===11 σε

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 and contraction below it cause an in-plane curvature,

curvature canticlasti 1==

′ ρν

ρ


Example 1

A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E = 165 GPa and neglecting the effects of fillets, determine (a) the maximum tensile and compressive stresses, (b) the radius of curvature.

SOLUTION:

• 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.

IMc

m =σ

• Calculate the curvature

EIM

=ρ1

24


Example 1 (cont’d)SOLUTION:

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

mm 383000

10114 3=

×=

∑∑=

AAyY

∑ ×==∑

×=××=×

3

3

3

32

101143000104220120030402109050180090201

mm ,mm ,mm Area,

AyA

Ayy

( ) ( )( ) ( )

49-3

2312123

121

231212

m10868mm10868

18120040301218002090

×=×=

×+×+×+×=

∑ +=∑ +=′

I

dAbhdAIIx


Example 1 (cont’d)• Apply the elastic flexural formula to find the

maximum tensile and compressive stresses.

49

49

mm10868m038.0mkN 3

mm10868m022.0mkN 3

−

−

×

×⋅−=−=

×

×⋅==

=

IcM

IcM

IMc

BB

AA

m

σ

σ

σ

MPa 0.76+=Aσ

MPa 3.131−=Bσ

• Calculate the curvature

( )( )49- m10868GPa 165mkN 3

1

×

⋅=

=EIM

ρ

m 7.47

m1095.201 1-3

=

×= −

ρρ

25


Unsymmetrical Bending• Analysis of pure bending has been limited

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

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

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

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

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

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


Unsymmetrical Bending

Wish 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.

•

couple vector must be directed along a principal centroidal axis

inertiaofproductIdAyz

dAcyzdAzM

yz

mxy

==∫=

∫

−=∫==

0or

0 σσ

• The resultant force and moment from the distribution of elementary forces in the section must satisfy

coupleappliedMMMF zyx ==== 0

•

neutral axis passes through centroid∫=

∫

−=∫==

dAy

dAcydAF mxx

0or

0 σσ

•

defines stress distribution

inertiaofmomentIIc

Iσ

dAcyyMM

zm

mz

===

∫

−−==

Mor

σ

26


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

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

θθ sincos MMMM yz ==

• Superpose the component stress distributions

y

y

z

zx I

yMI

yM+−=σ

• Along the neutral axis,( ) ( )

θφ

θθσ

tantan

sincos0

y

z

yzy

y

z

zx

II

zy

IyM

IyM

IyM

IyM

==

+−=+−==


Unsymmetrical Bending

The form of the flexural formula for normal stress in unsymmetrical bending is given by

Where

(21)y

y

z

zx I

yMI

yM+−=σ

Mz, Iz = moment and moment of inertia about z-axisMy, Iy = moment and moment of inertia about z-axis

27


Example 2

A 1600 lb-in couple is applied to a rectangular wooden beam in a plane forming an angle of 30 deg. with the vertical. Determine (a) the maximum stress in the beam, (b) the angle that the neutral axis forms with the horizontal plane.

SOLUTION:

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

θθ sincos MMMM yz ==

• Combine the stresses from the component stress distributions.

y

y

z

zx I

yMI

yM+−=σ

• Determine the angle of the neutral axis.

θφ tantany

zII

zy==


Example 2 (cont’d)• Resolve the couple vector into components and calculate

the corresponding maximum stresses.

( )( )

( )( )

( )( )

( )( )

( )( ) psi5.609in9844.0

in75.0inlb800

along occurs todue stress nsilelargest te The

psi6.452in359.5

in75.1inlb1386 along occurs todue stress nsilelargest te The

in9844.0in5.1in5.3

in359.5in5.3in5.1

inlb80030sininlb1600inlb138630cosinlb1600

42

41

43121

43121

=⋅

==

=⋅

==

==

==

⋅=⋅=⋅=⋅=

y

y

z

z

z

z

y

z

y

z

IzM

ADM

IyM

ABM

I

I

MM

σ

σ

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

5.6096.45221max +=+= σσσ psi1062max =σ

y

y

z

zx I

yMI

yM+−=σ

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BEAMS: BENDING STRESS Beams