Theory of simple bending (assumptions)

  Material of beam is homogenous and isotropic => constant E in all direction

  Young’s modulus is constant in compression and tension => to simplify analysis

  Transverse section which are plane before bending before bending remain plain after bending. => Eliminate effects of strains in other direction (next slide)

  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 will generate additional internal stresses.

Key Points:

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

tension.

Bending in beams

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 to longitudinal axis.

Bending in beams

Key Points:

1.  Bending moment causes beam to deform.

2.  X = longitudinal axis 3.  Y = axis of symmetry 4.  Neutral surface – does

not undergo a change in length

Bending in beams

P A B

RB RA M M

Radius of Curvature, R

Deflected Shape

Consider the simply supported beam below:

M M What stresses are generated within, due to bending?

Bending Stress in beams

Neutral Surface

M=Bending Moment

M M

Beam

σx (Tension)

σx (Compression)

σx=0

(i) Bending Moment, M (ii) Geometry of Cross-section

σx is NOT UNIFORM through the section depth

σx DEPENDS ON:

Axial Stress Due to Bending:

stress generated due to bending:

Neutral Surface

Bending Stress in beams

Bending Stress in beams

Stresses due to bending

N’ N’

R

E F

A’ C’

B’ D’

Strain in layer EF

€

=yR

€

E = Stress_ in _ the_ layer _ EFStrain _ in _ the_ layer_ EF

E = σyR⎛

⎝ ⎜

⎞

⎠ ⎟

€

σy

=ER

€

σy

=ER

€

σ =ERy

Neutral axis

N A

y dy

dA force on the layer=stress on layer*area of layer

€

=σ × dA

=ER× y × dA

Total force on the beam section

€

=ER× y × dA∫

=ER

y × dA∫

For equilibrium forces should be 0

€

y × dA = 0∫Neutral axis coincides with the geometrical axis

Stress diagram

x

σx

M M

σx

Moment of resistance

N A

y dy

dA force on the layer=stress on layer*area of layer

€

=σ × dA

=ER× y × dA

Moment of this force about NA

€

=ER× y × dA × y

=ER× y 2 × dA

Total moment M=

€

ER× y 2 × dA = E

R∫ y2∫ × dA

Stress diagram

€

y 2 × dA∫ = I

€

M = ERI⇒ M

I=ER

x

σx

M M

σx

Flexure Formula

€

MI

=ER

=σy

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 in unit 1).

Resultant moments and stresses

Section Modulus

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

€

Z = Iymax

MI

=σy

MI

=σmaxymax

M =σmaxIymax

M =σmaxZ

Section Modulus of symmetrical sections

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

1

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.

2

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

SteelSteel

Aluminum

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 cross sectional 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.

3

Slide No. 4

Composite Beams

Foam Core with Metal Cover Plates

hf

tm

tmb

Foam Core

Metal FacePlates

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 zero because its modulus of elasticity Ef is small compared to the modulus of elasticity of the metal.

0≈= εσ ff E (53)

4

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

Foam Core with Metal Cover Plates

Neutral Axis

Compressive Strain

Tensile Strain

x

yM

Figure 27

5

Slide No. 8

Composite Beams

Foam Core with Metal Cover Plates– Using Hooke’s law, the stress in the metal

of the cover plates can be expressed as

xm

xm

mmm

IMy

IME

EyE

−=

=

−==

σ

ρρ

εσ

therefore,//but

(53)

(54)

Slide No. 9

Composite Beams

Foam Core with Metal Cover Plates– The relation for the stress is the same as

that established earlier; however, the foam does not contribute to the load carrying capacity of the beam because its modulus of elasticity is negligible.

– For this reason, the foam is not considered when determining the moment of inertia Ix.

6

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 mfm

mfmNA th

btthbtAdI +=

=≅ (55)

( )( )2max

2

mfm

mf

thbtthM

+

+=σ (56)

Slide No. 11

Composite Beams

Foam Core with Metal Cover Plates– The maximum stress in the metal plates of

the beam is given by

( )( )2max

2

mfm

mf

thbtthM

+

+=σ

(56)

hf

tm

tmb

Foam Core

Metal FacePlates

7

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 (cont’d)The maximum moment for a simply supported beam is given by

When the composite beam yields, the stresses in the cover plates are

( ) qqqLM 18008

12108

22

max =×

==

psi 000,32max == yFσ

8

Slide No. 14

Composite Beams

Example 1 (cont’d)Substituting above values for Mmax and σmax into Eq. 56, we get

Or

( )( )

( )( )[ ]2

2max

063.06063.010063.0261800000,32

2

+×+

=

+

+=

qthbtthM

mfm

mfσ

ftlb806

inlb 2.67 ==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 Ef of the foam is so negligible,that is, it does not contribute to the load-carrying capacity of the composite beam.

9

Slide No. 16

Composite Beams

Bending of Members Made of Several Materials– 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 will deform as described earlier.

– Thus the normal strain εx still varies linearly with the distance y from the neutral axis of the section (see Fig 28b), and the following formula holds:

ρε yx −= (57)

10

Slide No. 18

Composite Beams

Transformed Section

M

σxεx

y y

ρε yx −= ρ

σ yE11 −=

ρσ 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.

11

Slide No. 20

Composite Beams

Transformed SectionWe can write:

From Eq. 58, it follows thatρ

εσ

ρεσ

yEE

yEE

x

x

222

111

−==

−== (58a)

(58b)

dAyEdAdF

dAyEdAdF

ρσ

ρσ

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.

( ) ( )ndAyEdAynEdFρρ11

2 −=−= (60)

12

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

13

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 of normal stresses on the section.

Slide No. 25

Composite Beams

Transformed Section

σx

y

IMy

x −=σ

N.A.

y

C

Figure 30. Distribution of Fictitious Normal Stress on Cross Section

14

Slide No. 26

Composite Beams

Stresses on Transformed Section1. To obtain the stress σ1 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 σ2 at a point located in the upper portion of the cross section of the original composite beam, stress σxcomputed from My/I is multiplied by n.

Slide No. 27

Composite Beams

Example 2A 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 200 GPa. 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.

15

Slide No. 28

Composite Beams

Example 2 (cont’d)

M 20 mm

40 mm

30 mm

Steel

Aluminum

Slide No. 29

Composite Beams

Example 2 (cont’d)First, because we have different materials, we need to transform the section into a section that represents a section that is made of homogeneous material, either steel or aluminum.We have

857.270200

===a

s

EEn

16

Slide No. 30

Composite Beams

Example 2 (cont’d)

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 (cont’d)Consider the transformed section of Fig. 31b, therefore

( ) ( )( ) ( )

( ) ( )( )

( ) 49433

33

m1042.852mm1042.8523

353.22204030

320353.223071.85

3353.2271.85

topfrom mm 353.2240302071.85

4030402071.8510

−×=×=−+

+

−−−=

=×+××+×

=

NA

C

I

y

17

Slide No. 32

Composite Beams

Example 2 (cont’d)

30 mm

85.71 mm

yC = 22.353 mm 20 mm

40 mmC

N.A.

Slide No. 33

Composite Beams

Example 2 (cont’d)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) MPa 253.66

Pa10253.661042.852

1065.371500 69

3

+=

×=××−

−=−= −−

IMy

alσ

18

Slide No. 34

Composite Beams

Example 2 (cont’d)b) Maximum normal stress in stelel occurs

at extreme upper fiber of the cross section, that is. at y =+ 22.353 mm.

( )

(C) MPa 8.112

Pa108.1121042.852

10353.221500)867.2( 693

=

×−=××

−=−= −−

IMynStσ

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.

19

Slide No. 36

Composite Beams

Reinforced Concrete Beam

Dead and Live Loads

M

Figure 32

Slide No. 37

Composite Beams

Reinforced Concrete Beam

b

d

b

x x21

N.A.

Fxd - x

n As

σ

(a) (b) (c)

Figure 33

··C

20

Slide No. 38

Composite Beams

Reinforced Concrete Beam– 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 area As of steel bar is replaced by an equivalent area nAs.

Slide No. 39

Composite Beams

Reinforced Concrete Beam– The ratio n is given by

– The position of the neutral axis is obtained by determining the distance x from the upper face of the beam (upper fiber) to the centroid C of the transformed section.

c

s

EEn ==

Concretefor Elasticity of ModulusSteelfor Elasticity of Modulus

21

Slide No. 40

Composite Beams

Reinforced Concrete Beam– 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

Reinforced Concrete Beam

– 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)

22

Slide No. 42

Composite Beams

Reinforced Concrete BeamThe neutral axis for a concrete beam is found by solving the quadratic equation:

021 2 =−+ dnAxnAbx ss (62)

b

d

b

x x21

d - x

n As

··C

Slide No. 43

Composite Beams

Example 3A 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 3×106psi for concrete used and 30 ×106 psi for steel. Knowing that a bending moment of 35 kip⋅in is applied to each 1-ft width of the slab, determine (a) the maximum stress in concrete and (b) the stress in the steel.

in-85

23

Slide No. 44

Composite Beams

Example 3 (cont’d)

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 (cont’d)– 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-85

2

2

in 614.0485

2 =

=

π

sA4 in.

4 - x

x N.A.·C( ) 2

6

6

in 14.6614.010

101031030

==

=××

==

s

c

s

nAEEn

12 in.

24

Slide No. 46

Composite BeamsExample 3 (cont’d)– Neutral Axis

• The neutral axis of the slab passes through the centroid of the transformed section. Using Eq. 62:

( ) ( )

056.2414.66

0414.614.61221

021

2

2

2

=−+

=−+

=−+

xx

xx

dnAxnAbx ssin 575.1=xa

acbbx2

42 −±−=

QuadraticFormula

599.2 take575.1

2

1

−==

xx

Slide No. 47

Composite Beams

Example 3 (cont’d)– Moment of Inertia

• The centroidal moment of inertia of the transformed section is

4 in.2.425

1.575·C

N.A.

6.14 in2

( ) ( ) 423

in 7.51425.214.63575.112

=+=I

12 in.

25

Composite Beams

Example 3 (cont’d)Maximum stress in concrete:

Stress in steel:

( ) (C) ksi 066.17.51

575.135−=−=−=

IMy

cσ

( ) (T) ksi 42.167.51425.235)10( +=−−=−=

IMynsσ

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

IMcKm =σ

Figure 33

a b

26

Slide No. 50

Stress Concentrations

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 N⋅m.

Slide No. 51

Stress Concentrations

Example 4 (cont’d) Figure 34

27

Slide No. 52

Stress Concentrations

Example 4 (cont’d)– From Fig. 34a:

– The moment of inertia of the critical cross section about its neutral axis is given by

( )

( ) mm 204021

21

mm 4010260

===

=−=

dc

d

( )( ) 493333 m 10481040109121

121 −−− ×=××== bdI

Slide No. 53

Stress Concentrations

Example 4 (cont’d)– Therefore, the stress is

– Using

– Also

( ) MPa 751048

10201809

3

=××

== −−

IMcσ

( ) 275150 =⇒=

=

KKI

McKmσ

5.14060

==dD

28

Slide No. 54

Stress Concentrations

Example 4 (cont’d)– From Fig. 33b, and for values of D/d = 1.5 and K = 2,

therefore

– Thus, the smallest allowable width of the grooves is

( ) ( ) mm 2.54013.013.0

13.0

===

=

drdr

( ) mm 4.102.522 ==r

Slide No. 55

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

IMcKm =σ

Figure 33

a b