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School of Engineering ES95D

Lecture 4: Bending and shear of beams

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• Elastic Bending of beams-Neutral plane and the 1 st moment of area- nd -Combined bending and axial load

• as c en ng o eams-Neutral plane; plastic hinge

• Shear stress in beams

•• Combination of stresses

2



• Based on initial geometry

• Ignores any geometrical non-linearities

• Deformations proportional to applied loads(Princi le of su er osition a lies)

DRAWBACKS• When instability effects are present, this analysis will

underestimate forces and moments in a structure anddeformations in a structure

as c s a yEstimation of the instability effect using e.g. Euler Buckling

3

,




Bending of beams

4



Symmetrical bending (elastic)

Assumptions:Beam is initiall strai ht and unstressed.Material of the beam is perfectlyhomogeneous (same throughout) andisotropic ( E and are the same in alldirections).

SteelThe elastic limit is nowhere exceeded.

Young’s modulus ( E) is same in tensionand compression.Doubly

s mmetricSinglys mmetric

Plane cross-sections remain planebefore and after loading ( fundamental ).This is true only if there is no shear forcein the cross section (pure bending). Purebending is rare, so S force usually exist,but does not influence the stresses much

5

eam eng cross sec ona

dimensions.



Exam le 1Before deformation

Centre line o n c e p t s /

Equally spaced vertical lines / c i v i l / s t r u c t u r a l c

After deformation r o j e c t / t e a c h i n g

• er ca nes rema n s ra g anperpendicular to the neutral line

•Centre line reserves its ori inal len th c h e s t e r . a c . u k /

•Upper surface extends, bottom surfaceshortens (top side in tension, bottom side

w w w

. m a c e . m a

6

h t t p : / /



Exam le 2 Thin beam – sameas previous slide

n an c eam

Thick beam:•Horizontal lines

perpendicular to the

centre line anymore cannot be neglected

7

http://www.mace.manchester.ac.uk/project/teaching/civil/structuralconcepts/





If there is no axial load (pure bending) the

A dA 0 ResultantCompr. force

ydA E

0ResultantTensile force

dA the 1 st moment of the area A

,of the area is 0. So the neutral plane passesthrough the centroid of the cross-section.

9



Second moment of the areaIn pure bending there is no axial load, there isonly bending moment M acting in the cross

dA

secton. e ntegraton o moments over t earea should be equal to M:

E 2

A y

A y

R

dA y I 2 the 2 nd moment of the area

y2

EI M

E M y y1

My I

My11

(tension)

I My2

2

(compression)

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Section may not be symmetrical to neutral plane; material may not have thesame strength in tension and compression



So how can we determine I ?Pg 48. GeometricalProperties of Sections ,

book.

Often12

3db

For the rectangularsection, increasing the

- will increase I eight-fold!

How do you break a chocolate bar?

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Sections efficient in bending

x xFor the same basic shape, the

x x

stressed fibres are from the

neutral axis (N.A.), thes er e sec on s an ehigher the value of I is.

x x x x

.of these sections?

1) Reference tables (payattention to axis and units)

2) Calculation

x x ? Not one beam!

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Exam le

Megson Example 9.7t f y

Calculate Ixx, I x x

t wd d w

Method 1: Method 2:bt f y

x x = - x x = +

)( 33ww d t bbd How to calculate this?

13

1212 xx

Parallel axes theorem



Parallel axes theorem Y Y’ y’

'' ' Ax I I XX X X

2'' ' Ay I I YY Y Y

’

x’

YX’X’

Megson Example 9.7 Method 2:

X X

= +X’ X’ x

122122

33

ww f w f

f xx

d t t d bt

bt I

14

Ixx Ax’



How to determine the neutral plane ?

Identify the centroid of the cross-section, using 0d A Ay

Centroid“ ” A 1

.By taking moments of area (for

simple shapes) about this baseline

x x. .

determined, from n

ii y ANA

Yn

i AY

2T-section is sum of two “ simple”rectangular sections numbered 1

21

2211

A Ay Ay A

Yase ne

15

Millais, M., Chapter 4. Section 4.4, Effect of beam cross-section , 2005.



H w t t rmin I ?

Centroid

xx -

of the two simple shapes numbered 1

and 2. A l in arallel axis theorem

A 1I1

x xI

1 y

I2

2 y 2

222

2

111 y A I y A I I xx

2

Now we know to calculate I for a cross section, exposedto known pure bending moment M. Therefore we can

stress

My x

My

N.A.compression

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Resistance of steel beams (elastic)

There are two situations to consider.

1. Elastic failure will occur when x.max =

y or f y.

yymax

xxxx f Wf

yMMM elelmax.max G

x

. .

the elastic limit) .

For a rectangular section of depth h and width b show the elastic section

modulus Wel

is bh 2/6. On Pg. 48 of Data Book Wel

is zxx

(section modulus)

For moment > Mel the general expression for elastic bending is no longer

valid.

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Resistance of steel beams (plastic)

2. Mmax from 1 is NOT the highest M that thesteel beam can resist. This is f y

M

Mpl is the moment when the full section has

yielded (i .e. i t is ful ly plastic).

yxx plplmax.max .

For a rectangular section of depth h and width

b , the plastic section modulus Wpl is bh 2/4 (i.e.f y

Wpl = 1.5 Wel )f y

y y

Y

f

f y f y f y Idealized stress-strain curvefor mild steel

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as M is increased.





W W

MPI

MPI

Increase W untill the moment atmid- MPI span reaches the plastic moment.Unrestricted plastic flow at this sectionoccurs. Large deformation takes place

.

In this case the beam was originally

statically indeterminate, collapse will

Plastichinge

happen after 2 plastic hinges areformed.

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determinate, the presence of a plastichinge leads to collapse.

o e: a rea nge carr es no momen ,while the plastic hinge still carries M PI.




Shear of beams

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h r in m

23



Shear in beams

n ec ure we saw a ex erna s ear oa ng pro uces n erna s ear forces ( S (x)) and bending moment ( M(x)) in a cross-section of the beam.

,

expression in the previous lesson. We now need an expression that will

give the corresponding shear stress ( ) distr ibut ion caused by S .

Flexure of unconnected members ( freeto slide over each other ) so horizontalshear stress is zero here.

Flexure when members are ful lyconnected (NO relative slip);

This is a complementary shear stress to the vertical shear stress.

24

Megson, T. H.G., Chapter 10, Shear in Beams , 2005.



Shear in beams: example

Sliding occurs

Shear connector(prevents sliding of one surface over another)

No sliding => stiffer beam

25

http://www.mace.manchester.ac.uk/project/teaching/civil/structuralconcepts/



Shear in beams: example for frames

t s /

t r u c t u r a l c o n c e p

/ t e a c h i n g / c i v i l / s

Frame moves in the loading directioncolumn-beam angles are no longer 90 0

Shear wall (characterised with large in-plane stiffness)inserted into the lower half: deformation reduced

e r . a c . u k / p r o j e c t

a c e . m a n c h e s t

h t t p : / / w w w

.

Force applied to the 2nd frame (it is next to the Inserting the floor reduces deformation

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one with the shear wall: the 2 nd frame deforms)

The examples demonstrate why we need both walls and floors in the building frames.



Shear stress distributionMegson’s book, 10.1 illustrates the general case of anuns mmetrical section. We will consider a sim ler case. In civilengineering ‘thick’ sections usually possess at least one axisof symmetry and are subjected to shear force in that

.

S+ S

A’

M+ M

boz z x

S

y

xy

27



A’

Consider horizontal equilibrium:

''')('

Ao

A

xbdAdA

x

bo y '

'1

AodA

xb

Neutral plane zz z x I

y M Beam bending

y z S

dx

dM (Lecture 2, internal forces)

''

y ydA

S

28

zz



Shear stress distribution

A’centroid

''

A ydA 1st moment of area

oz z

The equation can be written as:

NOT in data book (need toremember)

zz

y I b

y AS 0

' y

bo

Shear stress in the vertical plane iscomplimentary to the shear stress inhorizontal lane

y

y

xx x

xy

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y

xy



Example: shear stress distribution

S y zz

y xy I b

y AS

0

'

xy (y)ybeam as shown in Figure. Value of

for the slice A’ is found easily by y

h

‘inspection’b 0 = b ; Izz = bh 3/12 ; A ’ = b (0.5 h – y) ;

z

y..y

yh yhbhb

S y xy 5.05.05.0

1232

Which simplifies to

2

2

3 4

6 y

h

hb

S y xy hb

y xy 2

max,

at y = h/2 to xy,max = (3 S y)/(2bh) at y = 0.Note xy,max = 1.5 av ( av is theaverage vertical shear stress)

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Example: Shear Stress Distribution

We shall now consider the distribution of vertical shear stress in an I-section. It is clear that geometry of areas and , formed by taking a' A f

' Aw

f w ,

different – so giving different distr ibutions of shear stress.

228 w

d ht

b

S y

Distribution in web

2

2

421

w

w

yd I

zzh z

222 d

d hbS

maxw, b

S y

zz

Design process: Web resists nearly all the shear force, it is common with I-

31Megson, T. H.G., Chapter 10, Example 10.2, 2005.

sec ons o assume a s ear orce s res s e y e we .



What about Shear Stresses in Flanges?

Awareness Only: Because an I-section is ‘made up’of flanges and a web the horizontal shear force hasto be transmitted from one art of the beam toanother. Figure to the right shows how these shear forces ‘join’ the section together. Force is used here S

.

This figure shows that thehorizontal shear stresses inthe web is balanced by the

2 f,max t f = w,y =d /2 tw

ver ca s ear s resses, uthe horizontal s in theflange are acting in the planeo e ange an ave o ebalanced by s actingacross the flange.

Slide 9for w

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Millais, M., Section 4.3. The role of shear stresses in beams , 2005, pg. 95-99.



Shear Flow

For a “ thick” I-section, with tw t f , equil ibr ium equation2 f,max t f = w,y =d /2 tw informs us that the horizontal flange

. ,section is “ thin” -wal led this can produce shear distortions

of sufficient magnitude to redistribute due to bending (this

V

is shear lag ).

per unit length ; this is known as the SHEAR FLOW . Thefigure shows the shear flow (which is a force) for a I-sectionand a channel section (which can be obtained by splittingthe I-section into two parts).

,i t make this member ’s response become complicated?

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Sh C t



Shear Centre

in each flange of the channel are in overallequilibrium, they are numerical ly equal and act in

A momentcausing a

e same sense, w c means, w e r separation, there is a moment. This moment is

trying to twist the channel rather than bend it.

twistingaction

For twisting not to be there, the shear forces ( S y (and S z)) must pass through.

must pass to avoid any twisting caused by the distribution.

e

Shear centre coincides with centroid .Shear centre on axis of symmetry; but not ingeneral coincident with the centroid .

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We have to appreciate that the cross section shape can cause complex structuralbehaviour.



Axial forces:

Shear forces:

Bending moments:

e way o com n ng s ressesis relatively straightforward.Bending and axial stresses can

,they act in the same direction.

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(M. Millais, chapter 4)



Review: stresses in a beam

Right : Stresses in beam of rectangular cross-section: (a) points A to E, (b)direct and shear stresses actin onhorizontal and vertical planes, (c)principal stresses, and (d) maximumshear stresses.

ou a rea y now ow o reacomplex 2D stress problems(Mohr’s circle of stress)

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(b) (c) (d)



• Read Chapters 9 and 10 of Megson’s

• Example class 3: questions 1 and 2

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