lecture7b bending 2

8/12/2019 Lecture7b Bending 2

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Theory of simple

bending (assumptions)! Material of beam is homogenous and isotropic => constant E in all

direction

! Youngs modulus is constant in compression and tension => to simplifyanalysis

! Transverse section which are plane before bending before bending remainplain after bending. => Eliminate effects of strains in other direction (nextslide)

! Beam is initially straight and all longitudinal filaments bend in circular arcs=> simplify calculations

! Radius of curvature is large compared with dimension of cross sections =>simplify calculations

! Each layer of the beam is free to expand or contract => Otherwise they willgenerate additional internal stresses.


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Key Points:

1. Internal bending moment causes beam to deform.2. For this case, top fibers in compression, bottom in

tension.

Bending in beams


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Key Points:

1. Neutral surface no change in length.2. Neutral Axis Line of intersection of neutral surface

with the transverse section.

3. All cross-sections remain plane and perpendicular tolongitudinal axis.

Bending in beams


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Key Points:

1. Bending momentcauses beam to deform.

2. X = longitudinal axis3. Y = axis of symmetry4. Neutral surface does

not undergo a changein length

Bending in beams


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PA B

RBRAM M

Radius of Curvature, R

Deflected

Shape

Consider the simply supportedbeam below:

M M

What stresses are generated

within, due to bending?

Bending Stress in beams

Neutral Surface


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M=Bending Moment

M M

Beam

x (Tension)

x (Compression)

x=0

(i) Bending Moment, M

(ii) Geometry of Cross-section

xis NOT UNIFORMthrough

the section depth

xDEPENDS ON:

Axial Stress Due to Bending:

stress generated due to bending:

Neutral Surface


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Stresses due to bending

N N

R

E F

A C

B D

Strain in layer EF =y

R

E =Stress_ in _ the_ layer_EF

Strain_ in _ the_ layer_EF

E=

"

y

R

#

$% &

'(

"

y=

E

R

"

y=

E

R "=

E

Ry


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Neutral axis

N A

y

dy

dAforce on the layer=stress on layer*area of layer

="# dA

=

E

R# y # dA

Total force on the beam section

=

E

R

" y" dA#

=

E

Ry" dA#

For equilibrium forces should be 0

y" dA =0#

Neutral axis coincides with the geometrical

axis

Stress diagram

x

x

M M

x


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Moment of resistance

N A

y

dy

dAforce on the layer=stress on layer*area of layer

="# dA

=

E

R# y # dA

Moment of this force about NA

=

E

R" y " dA " y

=

E

R" y

2" dA

Total moment M=E

R" y 2 " dA =

E

R# y 2# " dA

Stress diagram

y 2 "dA# =I

M =E

RI"

M

I=

E

R

x

x

M M

x


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Flexure Formula

M

I=

E

R=

"

y


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Beam subjected to 2 BM

In this case beam is subjected to

moments in two directions y and z.The total moment will be a resultant

of these 2 moments.

You can apply principle of superposition

to calculate stresses. (topic covered inunit 1).

Resultant moments and stresses


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Section Modulus

Section modulus is defined as ratio of moment of inertia about the neutral axis tothe distance of the outermost layer from the neutral axis

Z =I

ymax

M

I=

"

y

M

I=

"max

ymax

M ="max

I

ymax

M ="maxZ


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Section Modulus of

symmetrical sections

Source:-http://en.wikipedia.org/wiki/Section_modulus


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BEAMS: COMPOSITE BEAMS;

STRESS CONCENTRATIONS

Slide No. 1

Composite Beams

Bending of Composite Beams

In the previous discussion, we have

considered only those beams that are

fabricated from a single material such as

steel. However, in engineering design there is an

increasing trend to use beams fabricated

from two or more materials.


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Slide No. 2

Composite Beams

Bending of Composite Beams

These are called composite beams.

They offer the opportunity of using each of

the materials employed in their

construction advantage.Concrete

Steel

SteelAluminum

Slide No. 3

Composite Beams

Foam Core with Metal Cover Plates

Consider a composite beam made of metal

cover plates on the top and bottom with a

plastic foam core as shown by the crosssectional area of Figure 26.

The design concept of this composite

beam is to use light-low strength foam to

support the load-bearing metal plates

located at the top and bottom.


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Slide No. 4

Composite Beams


hf

tm

tmb

Foam Core

Metal Face

Plates

Figure 26

Slide No. 5

Composite Beams

Foam Core with Metal Cover Plates The strain is continuous across the

interface between the foam and the cover

plates. The stress in the foam is given by

The stress in the foam is considered zerobecause its modulus of elasticity Ef is smallcompared to the modulus of elasticity ofthe metal.

0= ff E (53)


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Slide No. 6

Composite Beams

Foam Core with Metal Cover Plates Assumptions:

Plane sections remain plane before and

after loading.

The strain is linearly distributed as shown

in Figure 27.

Slide No. 7

Composite Beams


Neutral Axis

Compressive Strain

Tensile Strain

x

y

M

Figure 27


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Slide No. 8

Composite Beams


Using Hookes law, the stress in the metal

of the cover plates can be expressed as

x

m

xm

mmm

I

My

IME

Ey

E

=

=

==

therefore,//but

(53)

(54)

Slide No. 9

Composite Beams


The relation for the stress is the same as

that established earlier; however, the foam

does not contribute to the load carryingcapacity of the beam because its modulus

of elasticity is negligible.

For this reason, the foam is not considered

when determining the moment of inertiaIx.


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Slide No. 10

Composite Beams

Foam Core with Metal Cover Plates Under these assumptions, the moment of

inertia about the neutral axis is given by

Combining Eqs 54 and 55, the maximum

stress in the metal is computed as

( )22

2

2222 mf

mmf

mNA thbtth

btAdI +=

= (55)

( )2max2

mfm

mf

thbt

thM

++= (56)

Slide No. 11

Composite Beams


The maximum stress in the metal plates of

the beam is given by

( )2max2

mfm

mf

thbt

thM

++=

(56)

hf

tm

tm

b

Foam Core

Metal Face

Plates


22/43

Slide No. 12

Composite Beams

Example 1A simply-supported, foam core, metal

cover plate composite beam is subjected to

a uniformly distributed load of magnitude q.

Aluminum cover plates 0.063 in. thick, 10

in. wide and 10 ft long are adhesively

bonded to a polystyrene foam core. The

foam is 10 in. wide, 6 in. high, and 10 ft

long. If the yield strength of the aluminum

cover plates is 32 ksi, determine q.

Slide No. 13

Composite Beams

Example 1 (contd)

The maximum moment for a simply

supported beam is given by

When the composite beam yields, the

stresses in the cover plates are

( )q

qqLM 1800

8

1210

8

22

max =

==

psi000,32max == yF


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Slide No. 14

Composite Beams

Example 1 (contd)

Substituting above values for Mmax and

max into Eq. 56, we get

Or

( )( )

( )( )[ ]2

2max

063.06063.010

063.0261800000,32

2

+

+=

+

+=

q

thbt

thM

mfm

mf

ft

lb806

in

lb2.67 ==q

Composite Beams

Bending of Members Made of Several

Materials

The derivation given for foam core with

metal plating was based on the assumption

that the modulus of elasticity Efof the foam

is so negligible,that is, it does not

contribute to the load-carrying capacity of

the composite beam.


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Slide No. 16

Composite Beams

Bending of Members Made of SeveralMaterials

When the moduli of elasticity of various

materials that make up the beam structure

are not negligible and they should be

accounted for, then procedure for

calculating the normal stresses and

shearing stresses on the section will follow

different approach, the transformed section

of the member.

Slide No. 17

Composite Beams

Transformed Section

Consider a bar consisting of two portions of

different materials bonded together as

shown in Fig. 28. This composite bar willdeform as described earlier.

Thus the normal strain xstill varies linearly

with the distance yfrom the neutral axis of

the section (see Fig 28b), and the following

formula holds:

yx = (57)


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Slide No. 18

Composite Beams

Transformed Section

M

xx

y y

yx =

yE11 =

yE22 =

N.A

Figure 28 (a) (b) (c)

1

2

Slide No. 19

Composite Beams

Transformed Section

Because we have different materials, we

cannot simply assume that the neutral axis

passes through the centroid of the

composite section.

In fact one of the goal of this discussion will

be to determine the location of this axis.


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Slide No. 20

Composite Beams

Transformed SectionWe can write:

From Eq. 58, it follows that

yEE

yEE

x

x

222

111

==

== (58a)

(58b)

dAyE

dAdF

dAyE

dAdF

222

111

==

== (59a)

(59b)

Slide No. 21

Composite Beams

Transformed Section

But, denoting by n the ratio E2/E1 of the two

moduli of elasticity, dF2 can expressed as

Comparing Eqs. 59a and 60, it is noted

that the same force dF2 would be exerted

on an element of area n dA of the first

material.

( )( )ndA

yEdA

ynEdF

112 == (60)


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Slide No. 22

Composite Beams

Transformed Section This mean that the resistance to bending of

the bar would remain the same if both

portions were made of the first material,

providing that the width of each element of

the lower portion were multiplied by the

factor n.

The widening (if n>1) and narrowing (n


28/43

Slide No. 24

Composite Beams

Transformed Section

Since the transformed section represents

the cross section of a member made of a

homogeneous material with a modulus of

elasticity E1,the previous method may be

used to find the neutral axis of the section

and the stresses at various points of the

section.

Figure 30 shows the fictitious distribution ofnormal stresses on the section.

Slide No. 25

Composite Beams

Transformed Section

x

y

IMyx =

N.A.

y

C

Figure 30. Distribution of Fictitious Normal Stress on Cross Section


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Slide No. 26

Composite Beams

Stresses on Transformed Section1. To obtain the stress 1 at a point located

in the upper portion of the cross section

of the original composite beam, the

stress is simply computed from My/I.

2. To obtain the stress 2 at a point located

in the upper portion of the cross section

of the original composite beam, stress xcomputed from My/Iis multiplied by n.

Slide No. 27

Composite Beams

Example 2

A steel bar and aluminum bar are bonded

together to form the composite beam

shown. The modulus of elasticity for

aluminum is 70 GPa and for streel is 200GPa. Knowing that the beam is bent about

a horizontal axis by a moment M= 1500 N-

m, determine the maximum stress in (a)

the aluminum and (b) the steel.


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Slide No. 28

Composite Beams

Example 2 (contd)

M20 mm

40 mm

30 mm

Steel

Aluminum

Slide No. 29

Composite Beams

Example 2 (contd)

First, because we have different materials,

we need to transform the section into a

section that represents a section that ismade of homogeneous material, either

steel or aluminum.

We have

857.270

200===

a

s

E

En


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Slide No. 30

Composite Beams

Example 2 (contd)

20 mm

40 mm

30 mm

Steel

Aluminum

30 mm

30 mm n = 85.71 mm

Figure 31a Figure 31b

Aluminum

Aluminum

Slide No. 31

Composite Beams

Example 2 (contd)

Consider the transformed section of Fig.

31b, therefore

( ) ( )( ) ( )

( ) ( )( )

( ) 49433

33

m1042.852mm1042.8523

353.22204030

3

20353.223071.85

3

353.2271.85

topfrommm353.2240302071.85

4030402071.8510

==+

+

=

=+

+=

NA

C

I

y


32/43

Slide No. 32

Composite Beams

Example 2 (contd)

30 mm

85.71 mm

yC= 22.353 mm20 mm

40 mmC

N.A.

Slide No. 33

Composite Beams

Example 2 (contd)

a) Maximum normal stress in aluminum

occurs at extreme lower fiber of section,

that is at y= -(20+40-22.353) = -37.65

mm.

( )

(T)MPa253.66

Pa10253.661042.852

1065.371500 69

3

+=

=

==

I

Myal


33/43

Slide No. 34

Composite Beams

Example 2 (contd)b) Maximum normal stress in stelel occurs

at extreme upper fiber of the cross

section, that is. at y=+ 22.353 mm.

( )

(C)MPa8.112

Pa108.1121042.852

10353.221500)867.2(

6

9

3

=

=

==

I

MynSt

Slide No. 35

Composite Beams

Reinforced Concrete Beam

An important example of structural

members made of different materials is

demonstrated by reinforced concrete

beams. These beams, when subjected to positive

bending moments, are reinforced by steel

rods placed a short distance above their

lower face as shown in Figure 33a.


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Slide No. 36

Composite Beams


Dead and Live Loads

M

Figure 32

Slide No. 37

Composite Beams


b

d

b

x x2

1

N.A.

Fxd - x

n As

(a) (b) (c)

Figure 33

C


35/43

Slide No. 38

Composite Beams


Concrete is very weak in tension, so it will

crack below the neutral surface and the

steel rods will carry the entire tensile load.

The upper part of the concrete beam will

carry the compressive load.

To obtain the transformed section, the total

cross-sectional areaAs of steel bar is

replaced by an equivalent area nAs.

Slide No. 39

Composite Beams


The ratio n is given by

The position of the neutral axis is obtained

by determining the distancexfrom the

upper face of the beam (upper fiber) to the

centroid Cof the transformed section.

c

s

E

En ==

ConcreteforElasticityofModulus

SteelforElasticityofModulus


36/43

Slide No. 40

Composite Beams


Note that the first moment of transformed

section with respect to neutral axis must be

zero.

Since the the first moment of each of the

two portions of the transformed section is

obtained by multiplying its area by the

distance of its own centroid from the

neutral axis, we get

Slide No. 41

Composite Beams


Solving the quadratic equation for x, both

the position of the neutral axis in the beam

and the portion of the cross section of the

concrete beam can be obtained.

( ) ( )( )

021

or

02

2 =+

=

dnAxnAbx

nAxdbxx

ss

s

(61)


37/43

Slide No. 42

Composite Beams


The neutral axis for a concrete beam is

found by solving the quadratic equation:

02

1 2 =+ dnAxnAbx ss (62)

b

d

b

x x2

1

d - x

n As

C

Slide No. 43

Composite Beams

Example 3

A concrete floor slab is reinforced by

diameter steel rods placed 1 in. above the

lower face of the slab and spaced 6 in. on

centers. The modulus of elasticity is 3106psi for concrete used and 30 106 psi forsteel. Knowing that a bending moment of

35 kipin is applied to each 1-ft width of theslab, determine (a) the maximum stress in

concrete and (b) the stress in the steel.

in-8

5


38/43

Slide No. 44

Composite Beams

Example 3 (contd)

6 in.

6 in.

6 in.6 in.

5 in.

4 in.

M= 35 kip in

5 in.

12 in.

4 in.

Slide No. 45

Composite Beams

Example 3 (contd)

Transformed Section

Consider a portion of the slab 12 in. wide, in

which there are two diameter rods having a

total cross-sectional area

in-8

5

2

2

in614.04

8

5

2 =

=

sA

4 in.

4 -x

x N.A.

C

( ) 2

6

6

in14.6614.010

10103

1030

==

=

==

s

c

s

nA

E

En

12 in.


39/43

Slide No. 46

Composite Beams

Example 3 (contd)

Neutral Axis The neutral axis of the slab passes through the

centroid of the transformed section. Using Eq.

62:

( ) ( )056.2414.66

0414.614.6122

1

02

1

2

2

2

=+

=+

=+

xx

xx

dnAxnAbx ss

in575.1=xaacbb

x2

42

=

Quadratic

Formula

599.2

take575.1

2

1

=

=

x

x

Slide No. 47

Composite Beams

Example 3 (contd)

Moment of Inertia

The centroidal moment of inertia of the

transformed section is

4 in.

2.425

1.575

CN.A.

6.14 in2

( )( ) 42

3

in7.51425.214.63

575.112=+=I

12 in.


40/43

Composite Beams

Example 3 (contd)Maximum stress in concrete:

Stress in steel:

( )(C)ksi066.1

7.51

575.135===

I

Myc

( )(T)ksi42.16

7.51

425.235)10( +=

==

I

Myns

Slide No. 49

Stress Concentrations

Stress concentrations may occur:

in the vicinity of points where the

loads are applied

in the vicinity of abrupt changes in

cross section

I

McKm =

Figure 33

a b


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Slide No. 50


Example 4Grooves 10 mm deep are to be cut in a

steel bar which is 60 mm wide and 9 mm

thick as shown. Determine the smallest

allowable width of the grooves if the stress

in the bar is not to exceed 150 MPa when

the bending moment is equal to 180 Nm.

Slide No. 51


Example 4 (contd) Figure 34


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Slide No. 52


Example 4 (contd) From Fig. 34a:

The moment of inertia of the critical cross

section about its neutral axis is given by

( )

( ) mm20402

1

2

1

mm4010260

===

==

dc

d

( )( ) 493

333 m1048104010912

1

12

1 === bdI

Slide No. 53


Example 4 (contd) Therefore, the stress is

Using

Also

( )MPa75

1048

10201809

3

=

==

I

Mc

( ) 275150 ==

=

KK

I

McKm

5.140

60==

d

D


43/43

Slide No. 54


Example 4 (contd)

From Fig. 33b, and for values of D/d= 1.5 and K= 2,therefore

Thus, the smallest allowable width of the grooves is

( ) ( ) mm2.54013.013.0

13.0

===

=

dr

d

r

( ) mm4.102.522 ==r

Slide No. 55


Stress concentrations may occur:

in the vicinity of points where the

loads are applied

I

McKm =

a b

lecture7b bending 2

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