Bending stresses Presentation

7/27/2019 Bending stresses Presentation

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1

3a. Bending Stresses

1Prepared by: Prof. Nabil El-Tayeb

OUTLINE


1. Bending Deformation of aStraight Member

2. The Flexure Formula3. Unsymmetrical Bending

4. Curved Beams5. Stress Concentrations



OBJECTIVES

To establish shear and moment diagrams for a

beam or shaft,

To determine largest shear and moment in a

member, and specify where they occur

To determine stress in the x-sec of memberscaused by bending,

To study special cases of unsymmetrical bending

To study the effect of stress concentrations.

Done

in

Ch3


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+SFD,N

-

+BMD,

N.m

-

Pure Bending: Prismatic memberssubjected to equal and opposite

couples acting in the same

longitudinal plane. CD

subjected to ZERO SF



Pri nciple of Superposit ion: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.

Eccentri c L oading: 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.

Other Loading Types

M

P


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Bending Deformations

length of top layer

decreases and length of

bottom one increases

a neutral sur facemust

exist that is parallel to

the upper and lower

surfaces and for which

the length does not

change

Neutral layer



BENDING DEFORMATION OF A STRAIGHT MEMBER

A neutral surface is where longitudinal fibers of thematerial will not undergo a change in length.


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THE FLEXURE FORMULA

By mathematical expression,equilibrium equations of

moment and forces, we get

Ay dA = 0

max

cM =A y

2

dA

The integral represents the moment of inertia of x-sectional area, computed about the neutral axis.

We symbolize its value asI.



THE FLEXURE FORMULA

Assume that material behaves in a

linear-elastic manner so that

Hookes law applies.

A linear variation of normal strainmustthen be the consequence of

a linear variation in normal stress

Applying Hookes law to Eqn 6-8,

= (y/c)max


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THE FLEXURE FORMULA

The following Equations are often referred toas the f lexure formula.

Mc

Imax =

max = maximum normal stress in member,at a pt on x-sectional area farthest away fromneutral axis at the edge of the x-sec (c)

M= resultant internal moment, computed aboutneutral axis at distance y in the x-section

I= moment of inertia ofx-sectional area computedabout neutral axis

c = perpendicular distance from neutral axis to a ptfarthest away from neutral axis, where max acts

My

I=



THE FLEXURE FORMULA

Normal stress at intermediate distancey can bedetermined from:

MyI

a =

is -ve as it acts in the -ve direction (compression)

Equations 6-12 and 6-13 are often referred to asthe flexure formula.

A

B


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THE FLEXURE FORMULA

IMPORTANT

X-section of straight beam remains plane whenbeam deforms due to bending.

The neutral axis is subjected to zero stress

Due to deformation, longitudinal strain varies

linearlyfrom zero at neutral axis to maximum at

outer fibers of beam

Provided material is homogeneous and Hookeslaw applies, stress also varies linearlyover the x-

section



FLEXURE FORMULA

Stress Due to Bending (THE FLEXURE FORMULA)

y

xx

y

y=+ve is ve & y=-ve is +ve


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Example 3.8

A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E= 165 GPa and neglecting

the effects of fillets, (a) Draw the bending stress

distribution in the x-sec., (b) Determine the

maximum tensile and compressive stresses, and

(c) Determine the radius of curvature.

Based on the cross-section

geometry, calculate the location of

the centroid of the section and

moment of inertia.

Parallel-axis theorem applied

2dAII

A

AyY x

(a). Apply the elastic flexural formula to

draw and find the maximum tensile

and compressive stresses.

I

Mcm

(b). Calculate the curvature EI

M

1

SOLUTION:

Y



Example 3.8 (Cont.)

Based on the cross-section geometry, calculate

the location of the section centroid and

moment of inertia.

mm383000

10114 3

A

AyY

3

3

3

32

101143000

104220120030402

109050180090201

mm,mm,mmArea,

AyA

Ayy

49-43

23

12

123

12

1

23

12

12

m10868mm10868

18120040301218002090

I

dAbhdAIIx

I

Mcm

I=868x103mm4


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Example 3.8 (Cont.)

Apply the elastic flexural formulato find the maximum tensile and

compressive stresses.

49

49

m10868

m038.0mkN3

m10868

m022.0mkN3

I

cM

I

cM

I

Mc

BB

AA

m

MPa0.76A

MPa3.131B

Calculate the curvature

m7.47

m1095.201 1-3

- +76MPa

131.3MP

49- m10868GPa165mkN31

EI

M

Normal Stress



Example 3.9

Beam shown has x-sec. area in the shape of a channel.

(a) Determine the normal stresses (maximumstresses) that occurs in the beam at section a-a.

(b) Draw the normal stress distribution over the x-sec area


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Example 3.9 (Cont.)

Internal loads: N, V, MCalculate the internal loads at sec a-a by using the sectionmethod. Note that the resultant internal forces N passesthrough centroid of x-section.

The resultant internal moment must be computed about thebeams neutral axis a section a-a.

IMcm



Example 3.9 (Cont.)

Internal moment

Apply moment equation of equilibrium about neutral axis,

+ MNA = 0; 2.4 kN(2 m) + 1.0 kN(0.05909 m) =M

M= 4.859 kNm

I

Mcm


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Example 3.9 (Cont.)

To find location of neutral axis, x-sectional area divided into 3 compositeparts as shown. Then by using Eqns:

y = = ... = 59.09 mmy A

A

2dAIIA

AyY

x

I

Mcm

I = 42.26x10-6m4Given



Example 3.9 (Cont.)

Section property

Moment of inertia about neutral axis is determinedusing parallel-axis theorem applied to each of thethree composite parts of the x-sectional area.

I= [1/12(0.250 m)(0.020 m)3

+ (0.250 m)(0.020 m)(0.05909 m 0.010 m)2]

+ 2[1/12(0.015 m)(0.200 m)3

+ (0.015 m)(0.200 m)(0.100 m 0.05909 m)2]

I= 42.26(10-6) m4

I

Mcm

I = 42.26x10-6m4Given


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Example 3.9 (Cont.)

Maximum bending stress occurs at points farthest awayfrom neutral axis. At bottom of beam,

c= 0.200 m 0.05909 m = 0.1409 m. Thus,

At top of beam, = 6.79MPa. In addition, normal

force of N = 1 kN andshear force V = 2.4 kN willalso contribute additional

stress on x-section.

Mc

Imax = = = -16.2 MPa

4.859 kNm(0.1409 m)

42.26(10-6) m4



Q? Determine and Plot the Stress distribution acting over entire x-sectional

Example 3.9 (Cont.)

A

P

I

yMx

- + - +


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Stress due to eccentric loading found bysuperposing the uniform stress due to a centric

load and linear stress distribution due to a pure

bending moment

Eccentric Axial Loading in a Plane of Symmetry

Eccentric loading

PdM

PF

I

yM

A

P

I

yM

A

P

t

C

xxx

)(

)(

bendingcentric

xx

y

M

F

Supplementary Examples

5.10-5.13



Example 3.10An open-link chain is obtained by bending

low-carbon steel rods into the shape

shown. For a 700-N load, determine (a)

maximum tensile and compressive

stresses, (b) distancebetween section

centroid and neutral axis

SOLUTION:

Find the equivalent centric load and

bending moment.

Superpose the uniform stress due to

the centric load and the linear stress

due to the bending moment.

Evaluate the maximum tensile and

compressive stresses at the inner

and outer edges, respectively, of thesuperposed stress distribution.

Find the neutral axis by

determining the location where thenormal stress is zero.


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Example 3.10 (Cont.)

Equivalent centric load

and bending moment

Normal stress due to acentric load

Normal stress due to

bending moment




Maximum

tensile and

compressive

stresses

Neutral axis

location


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The largest allowable stresses for the castiron 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 equivalent centric load and

bending moment.

Evaluate the critical loads for the

allowable tensile and compressive stresses.

The largest allowable load is the

smallest of the two critical loads.

From Sample Problem 4.2,

49

23

m10868

m038.0

m103

I

Y

A

Superpose the stress due to a centric

load and the stress due to bending.




Determine equivalent centric and bending loads.

momentbending028.0

loadcentric

m028.0010.0038.0

PPdM

P

d

Evaluate critical loads for allowable stresses.

kN0.77MPa1201559

kN6.79MPa30377

PP

PP

B

A

The largest allowable load

Superpose stresses due to centric and bending loads

P

PP

I

Mc

A

P

PPP

I

Mc

A

P

A

B

A

A

155910868

022.0028.0

103

37710868

022.0028.0

103

93

93

kN0.77P


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UNSYMMETRIC BENDING

Moment arbitrarily applied If a member is loaded such that

resultant internal moment does not act

about one of the principal axes of x-

section, resolve the moment into

components directed along the

principal axes

Use flexure formula todetermine normal stress

caused by each momentcomponent

Use principle ofsuperposition to

determine resultantnormal stress at the pt

Mz

My



UNSYMMETRIC BENDINGMoment arbitrarily appliedResultant general normal stress at any pt on x-section is:

Mzy

Iz=- +

Myz

Iy

= normal stress at the pt

y, z= coordinates of ptmeasured fromx, y, zaxes

having origin at centroid of x-

sectional area and forming a

right-handed coordinatesystem


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UNSYMMETRIC BENDING

Orientation of neutral axis

My, Mz = resultant internal moment components alongprincipal yand zaxes. Positive if directed along +yand +z

axes. Can also be stated asMy = M sin and Mz = M cos ,

where is measured positive from +zaxis toward +y axis

Iy, Iz = principal moments of inertia computed about the y and

zaxes, respectively

0

y

y

z

z

I

zM

I

yM

From which Iz

Iytan = tan



UNSYMMETRIC BENDINGOrientation of neutral axisAngle of neutral axis can be

determined by applying Eqn 6-17

with = 0, since no normal stress

acts on neutral axis.

Finally, we get

For unsymmetrical bending, angle defining direction of

moment M is not equal to angle , angle defining

inclination of neutral axis unless

Iz= Iy. Thus, 90o

Equation 6-19Iz

Iytan = tan 0

y

y

z

z

I

zM

I

yM


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Unsymmetrical Bending

Superposition is applied to determine stresses inthe most general case of unsymmetric bending.

Resolve the couple vector into components along

the principle centroidal axes.

sincos MMMM yz

Superpose the component stress distributions

y

y

z

zx

I

yM

I

yM

Along the neutral axis,

tantan

sincos

0

y

z

yzy

y

z

zx

I

I

z

y

I

yM

I

yM

I

yM

I

yM



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.

PbMPaMP

zy forcecentric

By the principle of superposition, the

combined stress distribution is

y

y

z

zx

I

zM

I

yM

A

P

If the neutral axis lies on the section, it may

be found from

A

PzI

M

yI

M

y

y

zz


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Example 3.11

A 180 N . m 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.

sincos MMMM yz

Combine the stresses from the

component stress distributions.

y

y

z

zx

I

zM

I

yM

Determine the angle of the neutral

axis.

tantany

z

I

I

z

y



Example 3.11 (Cont.) Resolve the couple vector into components and calculate

the corresponding maximum stresses.

The largest tensile stress due to the combined loading

occurs atA.


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Determine the angle of the neutral axis.



Stress Concentrations

Stress concentrations may occur:

in the vicinity of points where the

loads are applied I

McKm

in the vicinity of abrupt changes

in cross section


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Example 3.12

NmmNmmNmm

Nmm

(c):I=13.92x10-6m

4

z=0.089mM=5.11Nm

c=15 kN/m2

t=32.7kN/m2

IMc

a a

bbc c

d d

e e

cbottom

ctop

M=5.1Nm, I=13.92x10-6m4,

cbottom=.089m, ctop=.041m,

bQ

IV

IbVQ

bQ

V=5.15N, I=13.92x10-6m4,

to be calculated at aa, bb,

cc, dd, and ee.b

Q

c, I

Mc



(c):I=13.92x10-m

z=0.089mM=5.11Nm

c=25.7kN/m2

t=32.7kN/m2

I

Mc

Ib

VQ

a a

bbc c

d d

e e

a

b cd

e

, N/m2 x10

6

b

Q

I

V

Ib

VQ

b

Q7x103 6

0,0b

Q

b

Q

eeaa

eeaa

2666 /10017.016.0

)2/03.0)(03.016.0(1037

b

Q1037 mNx

xxx

bb

bb

2666/1007.0

04.0

)2/03.0)(03.016.0(1037

b

Q1037 mNx

xxx

cc

cc

2666 /10015.004.0

)2/089.0)(04.0089.0(1037bQ1037 mNxxxx

bb

dd

V= 515N, I=13.92x10-

6m4

Example 3.12 Cont.


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EXAMPLE 3.13 Combined stresses

A force of 15,000 N isapplied to the edge of themember shown. Neglect theweight of the member anddetermine the state of

stress at pts Band C.



EXAMPLE 3.13 (SOLN)

Internal loadingsSectioning the member through B and C. For equilibrium,axial force of 15 kN acting through the centroid and abending moment of 750. kNmm about the centroidal or

principal axis.


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EXAMPLE 3.13 (SOLN)

Stress components

1. Normal force

Uniform normal-stress distributiondue to normal force is shown.

=P/A = = 3.75 MPa

2. Bending moment

Normal stress distribution due tobending moment is shown.

max =Mc/I= = 11.25 MPa



EXAMPLE 3.13 (SOLN)

Superposition

If above normal-stress distributions

are added algebraically, resultantstress distribution is shown. Although

not needed here, the location of the

line of zero stress can be determined

by proportional triangles, i.e.,

7.5 MPa

x=

15 MPa

(100 mmx)

x = 33.3 mm


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EXAMPLE 3.13 (SOLN)

Superposition

Elements of material atB and Care subjected only

to normal or uniaxial stress as shown. Hence

B = 7.5 MPa C= 15 MPa

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