Mechanics of structures - module3

Mechanics of structuresMechanics of structures

Stresses in beams Stresses in beams, Inelastic bending,

Deflections

Dr. Rajesh K. N.Assistant Professor in Civil EngineeringAssistant Professor in Civil EngineeringGovt. College of Engineering, Kannur

1

Module III

Bending stresses in beams - shear flow - shearing stress formulae for

Module III

Bending stresses in beams shear flow shearing stress formulae for beams –

Inelastic bending of beams –Inelastic bending of beams –

Deflection of beams - direct integration method - singularity f ti iti t h i t th d functions - superposition techniques - moment area method -conjugate beam ideas –

Elementary treatment of statically indeterminate beams - fixed and continuous beams

Dept. of CE, GCE Kannur Dr.RajeshKN

2

Important assumptions in the theory of simple bending

• Material of the beam is homogeneous and isotropic

p p y p g

• The stress is proportional to strain and stress is within elastic limit

• Modulus of elasticity is same for tension and compression• Modulus of elasticity is same for tension and compression

• Plane vertical sections remain plane after bending

• Loads are applied in the plane of bending

• Cross section is symmetrical about vertical axis


3

Theory of simple bending

r syp q

O

θ p q Rθ′ ′ =

( )r s R y θ′ ′ = −R ( )r s R y θ= −

p q pq′ ′ =

r’ s’

p’ q’y

p q pq


4

p q

Strain for the fibre rs

rs r s p q r srs rs′ ′ ′ ′ ′ ′− −

=( )R R y yR R

θ θθ

− −= =

rs rs R Rθ

yσ=

Eyσ =If σ is the stress in fibre rs, and E is the Young’s modulus, strain

E R RE is the Young s modulus, strain

O l f

. .EydA dAR

σ =

On any cross section, normal force on an elemental area dA,

dAR

Moment about NA due to normal force on dA, yNA2

. . .EydA y dAR

σ =

Sum of all such elemental moments is the

NA


5

Sum of all such elemental moments is the moment of resistance of the cross section.

Moment of resistance balances the bending moment at the cross section.g

2

. . .RA A

EyM M dA y dAR

σ= = =∫ ∫2.

A

EM y dAR

= ∫

2.A

y dA∫ is the moment of inertia of the cross section about the neutral axis

EIMR

∴ = M EI R y

σ= = Bending equationR I R y


6

Myσ Along a cross section, variation of bending stress is I

σ =

For a beam with transverse downward loading, compressive force

g , glinear.

g, pacts above NA, tensile force acts below NA

C

T

MMTotal compressive force = Total tensile force


7

C dA

NA

y

T

0 . . 0C T dA dAσ σ+ = ⇒ + =∫ ∫

Total force on the cross section is zero

. 0dAσ =∫1 2A A∫ ∫

0 0MydA dA∫ ∫ ∫

A∫

. 0 . 0A A

ydA dAI

σ = ⇒ =∫ ∫ .A

y dA

dA

∫

∫

Distance from a point to the centroidal axis

. 0A

y dA⇒ =∫ A

dA∫ the centroidal axis


8i.e., NA coincides with the centroidal axis

Section Modulus

( )MI y

σ =

MmaxZ I y=

Section modulus for various shapes of cross section

maxMZ

σ =Section modulus

b

( )3 212bdI bdd

( )( )

2

max

122 6

bdI bdZy d

= = =

d ( )4 3642 32

d dZd

π π= =


9

B

( )( )

3 33 3

112

2 6

BD bdI BD bdZD D

− −= = =

b

( )max 2 6y D DD d

( ) ( )4 4 4 464

2 32

D d D dZ

D D

ππ− −

= =2 32D D


10

2h3

CG axis h

b

3bh⎛ ⎞⎜ ⎟ 2

max

362 243

bhZ I y h

⎜ ⎟⎝ ⎠= = =


11

Problem 1. A beam 500mm deep of symmetrical rectangulary gsection has I = 1x108 mm4 and is simply supported over a spanof 10 m. If the beam carries a central point load of 25 kN,calculate the bending stress at 100 mm above neutral axis andcalculate the bending stress at 100 mm above neutral axis andthe maximum bending stress on the beam.

Myσ8 41 10 mmI = × 100 mmy =y

Iσ =

25 10 62 5 kNWLM ×

1 10 mmI = × 100 mmy

6262.5 10 100 62 5 N/× ×

∴62.5 kNm4 4

M = = = 28 62.5 N/mm

1 10σ∴ = =

×

maxmax

MyI

σ =max

500 250 mm2

y = =

62

max 8

62.5 10 250 156.25 N/mm1 10

σ × ×∴ = =


12

max 81 10×

Problem 2. For the above problem, calculate the UDL it mayycarry, if the maximum bending stress is not to exceed 150N/mm2.

IIM

yσ

= max

.allowR

IMy

σ=Moment of resistance

2150 N mmallowσ = max 250 mmy =

8 41 10 mmI = ×

86150 1 10 60 10 Nmm 60 kNm

250RM × ×∴ = = × =

2. 60 kNm

8alloww l

= . 2

60 8 4.8 kN/m10alloww ×

∴ = =


13

Problem 3. Find the maximum BM that should be imposed onhi i f il if h il i h fl ithis section of a cantilever, if the tensile stress in the top flange is

not to exceed 40 MPa. What is then the value of compressivestress in the bottom flange?stress in the bottom flange?

200 40 260 200 40 140 120 40 20y × × + × × + × ×=

200 mm

200 40 200 40 120 40y =

× + × + ×

158.5 mm=

i.e., 158.5 mmcy =

3 3200 40 40 200

280 158.5 121.5 mmty∴ = − =

( ) ( )

( )

3 32 2

32

200 40 40 200200 40 121.5 20 40 200 158.5 14012 12

120 40 120 40 158.5 20

I × ×= + × × − + + × × − +

×+ + × × −


14

( )120 40 158.5 2012

+ + × ×8 42.056 10 mmI = ×

.ll Iσ 2 8 440 N/mm 2.056 10 mm× ×

,max

.allow

t

IMyσ

=Max. BM 121.5 mm=

667.6 10 Nmm= × 67.6 kNm=

MyCompressive stress in the bottom flange ,maxcc

MyI

σ =

667.6 10 Nmm 158.5 mm× × 28 4

67.6 10 Nmm 158.5 mm2.056 10 mmcσ =

×252.19 N mm=

240 N mmtσ =t

40 158.5σ ×=

Alternatively,

121.

5 m

m

NA121.5cσ =

252.19 N mm=

158

.5 m

m


15252.19 N mmcσ =

1

Theory of simple bending is applicable to pure bending. y g g

But since the effect of shear on bending stress is negligible, the theory can be applied generally

The effect of shear on bending stress is not of practical

theory can be applied generally.

The effect of shear on bending stress is not of practical importance, but shearing stresses must be considered for their own importance.


16

Shear stresses in bending Myσ c Iσ =

( )d

M dM yσ

+=

p’ q’p’ q’

c d d Iσ

m ny1M M dM+

yc m ny1 y

c d

M dM+

p qdx

p q

. .c cy y

cMydA dAI

σ =∫ ∫( ). .

c cy y

d

M dM ydA dA

Iσ

+=∫ ∫


17

1 1y y I1 1y y I

c cy y

dA dAσ σ−∫ ∫ ( )c cy yM dM y MydA dA+

∫ ∫cy dM y dA= ∫

1 1

. .d cy y

dA dAσ σ∫ ∫ ( )1 1

. .y y

dA dAI I

= −∫ ∫1

.y

y dAI∫

.cy

dAσ∫

Acy

dAσ∫1

.cy

dAσ∫b

d

1

.dy

dAσ∫. .b dxτ

NAdx NA

cy dM 1 cydM d M

1

. . .y

dMb dx y dAI

τ = ∫1

1 .y

dM y dAIb dx

τ = ∫cyV

d M Vd x

=

( )V A


181

.y

V y dAIb

τ = ∫ ( )V AyIb

τ =

Shear stress distribution Rectangular sectionShear stress distribution – Rectangular section

( )V Ayτ =

b IbA

1d dAy b y y y⎡ ⎤⎛ ⎞ ⎛ ⎞= − × + −⎜ ⎟ ⎜ ⎟⎢ ⎥d y

y 2 2 2Ay b y y y= − × + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

1V d d⎛ ⎞ ⎛ ⎞2

2V d⎛ ⎞

maxτ

12 2 2

V d db y yIb

τ ⎛ ⎞ ⎛ ⎞= − × +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2 4V d yI⎛ ⎞

= −⎜ ⎟⎝ ⎠

Hence, variation of shear stress over the cross section is parabolic2 2

0V d dτ⎛ ⎞⎜ ⎟2 0

2 4 4y d Iτ =± = − =⎜ ⎟

⎝ ⎠

2 2 3V d Vd V⎛ ⎞ 3


19( )max 0 3

32 4 28 12yV d Vd VI bdbd

τ τ =

⎛ ⎞= = = =⎜ ⎟

⎝ ⎠

32 meanτ=

Shear stress distribution I sectionShear stress distribution – I section

( )V Ayτ =

B

Shear stress in flangesIb

12 2 2D DAy B y y y⎡ ⎤⎛ ⎞ ⎛ ⎞= − × + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

D d b

g

2 2 2⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

1V D DB y yτ ⎛ ⎞ ⎛ ⎞= − × +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

22

2 4V D yI⎛ ⎞

= −⎜ ⎟⎝ ⎠2 2 2

yB

yI ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ 2 4y

I ⎜ ⎟⎝ ⎠

Hence, variation of shear stress over the flange is parabolic

2 2

2 02 4 4y DV D DI

τ =±

⎛ ⎞= − =⎜ ⎟

⎝ ⎠

y

2 2 4 4y D I=± ⎜ ⎟⎝ ⎠

( )2 2

2 2V D d V D dτ⎛ ⎞

= − = −⎜ ⎟


20

( )/2 2 4 4 8y d D dI I

τ = = =⎜ ⎟⎝ ⎠

( )V Ayτ =Shear stress in web

Ib

( ) ( ) 12 4 2 2 2

D d D d d dAy B b y y− + ⎛ ⎞ ⎛ ⎞= + × − × +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠y

2 4 2 2 2⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) ( ) 12 4 2 2 2

D d D dV d db y yIb

Bτ− +⎡ ⎤⎛ ⎞ ⎛ ⎞= + × − × +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦2 4 2 2 2

y yIb ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

( ) ( )2 2 2 248V B D d

bb d y

Iτ ⎡ ⎤= − + −⎣ ⎦

Hence, variation of shear stress over the web is parabolic

( ) ( ) ( )2 2 2 2 2 2V VB⎡ ⎤

( ) ( )8 bI ⎣ ⎦

( ) ( ) ( )2 2 2 2 2 22 8 8y d

V VBB D d b db

d D dI bI

τ =±⎡ ⎤= − + − = −⎣ ⎦

( )2 2 2V ⎡ ⎤


21

( )2 2 2max 0 8y

V B dIb

D bdτ τ =⎡ ⎤= = − +⎣ ⎦

maxτ

Shear stress distribution


22

Shear stress distribution – Circular section( )V AyIb

τ =R

y

Rb

dy2 22b R y= −

( )2 2 24b R y= −. .

y R

Ay b dy y=

= ∫y ( )4b R y= −

2 . 8 .b db y dy= −1

y y=

1. .4

y dy b db= −

When ,y y b b= =When , 0y R b= =0 1 .

4

b

b b

Ay b b db=

=

⎛ ⎞∴ = −⎜ ⎟⎝ ⎠∫

0 321 .

4 12

b

b b

bb db=

=

= − =∫

3 2V b Vbτ⎛ ⎞

= =⎜ ⎟( ) ( )2 2 2 24V R y V R y× − −

= =

b b b b


23

12 12Ib Iτ = =⎜ ⎟

⎝ ⎠ 12 3I I

Hence variation of shear stress over the cross section is parabolicHence, variation of shear stress over the cross section is parabolic

( )2 2

0V R R

τ−

= =m a xτ

03y R I

τ = = =

2VR 2 4 4 4VD V Vmax 0 3y

VI

τ τ == = 4 23 3 31264 4

meanV V V

AreaD Dτ

π π= = = =

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


24

Find shear stress distribution for the sections shown below:Find shear stress distribution for the sections shown below:


25

Inelastic Bending of Beams(Plastic Analysis)

g

ress

str

strainOIdealised stress-strain curve of elastic-plastic materialp

AssumptionsAssumptions

• Plane sections remain plane in plastic conditionS i l i i id i l b h i i d i


• Stress-strain relation is identical both in compression and tension

• Let M at a cross-section increases gradually.

• Within elastic limit, M = σ.ZZ i ti d l I/ • Z is section modulus, I/y

• Elastic limit – yield stresses reached Elastic limit yield stresses reached My = σy.Z

• When moment is increased, yield spreads into inner fibres. Remaining portion still elastic

• Finally, the entire cross-section yields, at a moment of MP

yσσ

σyσ

yσ yσyσσ

σ


yσ yσ yσσ

σy σy σy σyσy σy y y

σyσy


Plastic moment• M – Moment corresponding to working load

• My – Moment at which the section yieldsMy Moment at which the section yields

• MP – Moment at which entire section is under yield stress – plastic moment

yσ

CC

T

yσMP


• At plastic moment the entire section is under yield stressAt plastic moment, the entire section is under yield stress

C TA Aσ σ=

=AA A⇒ = =

•NA divides cross-section into 2 equal parts

c y t yA Aσ σ=2c tA A⇒ = =

AC T σ= =•NA divides cross-section into 2 equal parts2 yC T σ= =

yσ

y2 yAC = σ

yt

yc

AT σ=

yσ

2 yT σ

Dept. of CE, GCE Kannur Dr.RajeshKN31

ZσSimilar to

•Couple due to ( )A A Z⎛ ⎞⎜ ⎟

Plastic modulus

yZσ•Couple due to ( )2 2y y c t y py y Zσ σ σ⎛ ⎞ = + =⎜ ⎟

⎝ ⎠

Plastic modulus

Shape factor ( )1pZZ

= > b

Rectangular cross-section:

p ( )Z b

Section modulus ( )( )

3 2122 6

bdI bdZy d

= = =d

( )y

( )2A bd d d bdZ y y ⎛ ⎞+ +⎜ ⎟Plastic modulus ( )

2 2 4 4 4p c tZ y y= + = + =⎜ ⎟⎝ ⎠

Plastic modulus


Shape factor 1.5pZZ

= =

Shape factor for circular section

d( )2p c tAZ y y= +

2 32 28 3 3 6d d d d⎛ ⎞⎛ ⎞= + =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

ππ π

1 7pZS∴ = =( )4 364d dZ

π π

⎝ ⎠⎝ ⎠

1.7SZ

∴ = =( )2 32

Zd

= =

Sh f t f t i ul ti

2h

Shape factor for triangular section

( )2p c tAZ y y= +

hEqual area axis

cy

23h

3bh⎛ ⎞⎜ ⎟

( )2p c t

Equal area axistyCG axis

2362 243

bhZ h

⎜ ⎟⎝ ⎠= =


bS = 2.346

20mmShape factor for I section

10mm250mm

p

20mm200mm

S = 1.132


Z

Load factor

P y Pcollapse load MLoad factorworking load M Z

Zσσ

= = =

Factor of safety and load factor

Yield LoadFactor of Safety= =Working Load

Yield Stress

yWWσ

Elastic Analysis : Factor of Safety

Pl i A l i L d FYield Stress= =Working Stress

yσσ

Plastic Analysis : Load Factor


Plastic moment capacity of a rectangular cross-section:

2

P PbdM Zσ σ= =

2

4pbdZ =Plastic modulus Hence, plastic

t it 4P y P yM Zσ σmoment capacity

2 2

6 1.5 6ybd bdM Z

σσ σ= = =

Elastic moment capacitywith a factor of safety of 1.5

2

2

9 2.254P y P

M bdM Z Mbd

σ= = =2

9y

Mbd

σ∴ = Hence,4bdbd

2 2bd bd Elastic moment capacity

6 6ybd bdM Zσ σ σ= = =

p ywith a factor of safety of 1.0

26 bd6M


2

2

6 1.54P y P

M bdM Z Mbd

σ= = =2

6y

Mbd

σ∴ = Hence,

Plastic hinge

• When the section is completely yielded, the section is fully plastic• A fully plastic section behaves like a hinge – Plastic hinge

Plastic hinge is defined as an yielded zone due to bending in a structural member, at which large rotations can occur at a section

at constant plastic moment, MP

Mechanical hinge Plastic hinge

Mechanical hinge and Plastic hinge

Mechanical hinge Resists zero moment

Plastic hinge Resists a constant moment MP


Mechanism of failure in a simple beamp

D t i t b & Determinate beams & frames: Collapse after first plastic hinge

Simple beam

MPM

.uW lMEquilibrium4u

PM =

4M

Equilibrium:

8 PM


4 Pu

MWl

∴ = 2P

uwl

∴ =If UDL,

fl i f bDeflection of beams

• Differential equation of the elastic curve

• Slope and deflection of beams by method of successive integration

• Macaulay’s method

• Moment area method

• Conjugate beam method


39

Differential equation of the elastic curveDifferential equation of the elastic curve

yy

ds Rdα=

2 2

q

Rdα

ds

2 2ds dx dy= +

tan dyα =p

dy

dx

ds adx

α

22sec d d yαα

α α+dα

x

2secdx dx

α =

x

( )2

221 tan d yd dx

dxα α+ =


40

2 2dy d y⎧ ⎫⎪ ⎪⎛ ⎞

2

2d y dxdx

21 dy d yd dxdx dx

α⎧ ⎫⎪ ⎪⎛ ⎞+ =⎨ ⎬⎜ ⎟

⎝ ⎠⎪ ⎪⎩ ⎭2

1

dxddydx

α =⎛ ⎞+ ⎜ ⎟⎝ ⎠

2

22 22

d yR dxdxds Rd dx dy

dα= ⇒ + =

⎛ ⎞

2

22

21

d yR dxdy dxdxd

⎛ ⎞+ =⎜ ⎟⎝ ⎠ ⎛ ⎞

2

1 dydx

⎛ ⎞+ ⎜ ⎟⎝ ⎠

2

1dx dy

dx

⎜ ⎟⎝ ⎠ ⎛ ⎞+ ⎜ ⎟

⎝ ⎠2d 2

2

32 2

1

1

d ydx

R dy=⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟

2

2

1 d yR d=

1 dydx

⎛ ⎞+⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

2R dx

21M E M d 2d2

2

1M E M d yI R EI R dx= ⇒ = =

2

2

d yM EIdx

=

I th b d i ti ff t f h d fl ti i NOT


41

•In the above derivation, effect of shear on deflection is NOTtaken into account, since it is assumed to be very small.

Method of successive integrationsMethod of successive integrations2d 2

2

d yM EIdx

=2

2

d yEI Mdx

=

2d yEI M C= +∫ ∫ dyEI M C x C= + +∫ ∫ ∫12EI M Cdx

dyEI M C

= +

= +

∫ ∫

∫

1 2

1 2

EI M C x Cdx

EIy M C x C

= + +

= + +

∫ ∫ ∫

∫ ∫

{ }1

1

EI M Cdx

dy M C

= +

= +

∫

∫ { }1 2

1 21

y

y M C x CEI

= + +

∫ ∫

∫ ∫{ }1M Cdx EI

= +∫ { }EI ∫ ∫

42

Slope at B of the deflected beam = (dy/dx) at B

( )2

2

d yEI M P l x= = − −l B

Deflection at B = y at B

0 0 0dyAt x C= = ∴ =

( )2dx

2dy Px

lx

10, 0 0At x Cdx

= = ∴ =12

dy PxEI Plx Cdx

= − + +

2 3

22 6Plx PxEIy C−

= + +2

2dy PxEI Plxdx

= − + 2 62dx

2 3Plx Px−20, 0 0At x y C= = ∴ = 2 6

Plx PxEIy∴ = +


43

2 2

,2 2

dy Px dy PlAt x l EI Plxdx dx EI

= = − + ∴ = −2 2x ldx dx EI

=

2 3Plx Px−

3

,2 6 x l

Plx PxAt x l EIy=

= ∴ = +

3

3PlyEI

∴ = −

If when x increases, y also increases, then slope (dy/dx) is +ve.

If when x increases, y decreases, then slope (dy/dx) is -ve.


44

Slope at B of the deflected beam = (dy/dx) at B

Deflection at B = y at B

( )22 w l xd yEI M−

= = −lB

d

2 2EI M

dx= = −l

x

10, 0 0dyAt x Cdx

= = ∴ =3

2 212 3

dy w xEI l x lx Cdx

⎛ ⎞= − − + +⎜ ⎟

⎝ ⎠

32 2

2 3dy w xEI l x lxdx

⎛ ⎞∴ = − − +⎜ ⎟

⎝ ⎠

2 2 3 4

2 2 3 12w l x lx xEIy⎛ ⎞

= − − +⎜ ⎟⎝ ⎠2 3dx ⎝ ⎠ ⎝ ⎠

2 2 3 4w l x lx x⎛ ⎞20, 0 0At x y C= = ∴ = 2 2 3 12

w l x lx xEIy⎛ ⎞

∴ = − − +⎜ ⎟⎝ ⎠


45

3 33 3,

2 3 6dy w l dy wlAt x l EI l ldx dx EI

⎛ ⎞= = − − + ∴ = −⎜ ⎟

⎝ ⎠2 3 6dx dx EI⎝ ⎠

4 4 4 4w l l l wl⎛ ⎞,

2 2 3 12 8w l l l wlAt x l EIy y

EI⎛ ⎞

= ∴ = − − + ∴ = −⎜ ⎟⎝ ⎠


46

Moment area methodMoment area methodCharles E. Greene

•For deflections of beams especially cantilever beams•For deflections of beams, especially cantilever beams

•Suitable when slopes and deflections at particular points are required not the complete equation of the deflection curverequired, not the complete equation of the deflection curve


47

Moment area theoremsMoment area theoremsO

RdαA

B

dα

dsmn

α

dα

x.dα y

xdx xdx

B’


48

B

O 1 Md ds dsα = =

RdαA

B R EIMds dx d dxEI

α≅ ⇒ =

dα

dsmn

EI

Md dxEI

α α= =∫ ∫α

x.dα yEI∫ ∫

M

xdx

Mxd xdsEI

Mds dx xd xdx

α

α

=

≅ ⇒

B’

ds dx xd xdxEI

α≅ ⇒ =

My xd xdxEI

α= =∫ ∫


49

M t A Th 1Moment Area Theorem 1

Angle between tangents at A & B = (Area of BMD between A & B) /EI

Moment Area Theorem 2Moment Area Theorem 2

Deviation of B from tangent at A = (Moment of BMD between A & B, b t B) /EI about B) /EI


50

Bl

Slope at B of the deflected beam = Area of M/EI

( )

p /diagram between A & B

2l Pl Pl= =(-)Pl

EI2 2EI EI

= − = −

Deflection at B of the beam = Moment of M/EI Deflection at B of the beam Moment of M/EI diagram between A & B about B 2 32

2 3 3Pl l PlEI EI

= − = −


l

Slope at B of the deflected beam = Area of M/EI

l

diagram between A & B 2 3

3 2 6l wl wl

EI EI= − = −

2l ( ) 3 2 6EI EI2

2wlEI

(-)

Deflection at B of the beam = Moment of M/EI 3 43l l l/

diagram between A & B about B3 43

6 4 8wl l wlEI EI

= − = −


52

C

CA B

Deflected shape

PL4PLEI

M/EI diagram C

A B

g C

Slope at A of the deflected beam = Area of M/EI di b t A & C

21 l Pl Pl= =diagram between A & C

2 2 4 16EI EI= =

Deflection at C of the beam = Moment of M/EI diagram 2 3


Deflection at C of the beam = Moment of M/EI diagram between A & C about A

2 323 2 16 48

l Pl PlEI EI

= =

Deflected shapeC

B

2

8wLEIA B8EI

M/EI diagram C

A B

Slope at A of the deflected beam = Area of M/EI diagram between A & C

2 323 2 8 24

l wl wlEI EI

= =

Deflection at C of the beam = Moment of M/EI3 45 5l wl wl

= =


54

/diagram between A & C about A 8 2 24 384EI EI

= =

Conjugate beam method

Actual beam

j g

l B

Conjugate beam(-)Pl

E I

BA

EI

Slope at B of the deflected beam Shear force at B of the conjugate Slope at B of the deflected beam = Shear force at B of the conjugate beam when conjugate beam is loaded with M/EI diagram 2

2 2l Pl Pl

EI EI= =

Deflection at B of the beam = Bending moment at B of the conjugate beam when conjugate beam is loaded with M/EI diagram 2 32Pl l Pl


j g / g 22 3 3Pl l PlEI EI

= =

Conjugate beam method - proof2

2

d yEI Mdx

=

3 4

In the actual beam,

3

3 , Shear forced y dMEI Vdx dx

= =4

4 , Loadd y dVEI wdx dx

= =

load, MwEI

=0 0

Shear force,x x MV wdx dx

EI= =∫ ∫In the conjugate beam,

Bending moment,x x x MM Vdx dx

EI= =∫ ∫ ∫

0 0 0 EI

2

i h j b i h l b x xM d y dyd d∫ ∫

Hence,

20 0

in the conjugate beam in the actual beam

Shear force in the conjugate beam Slope in the actual b am

e

M d y dydx dxEI dx dx

= =

=

∫ ∫

2x x x xM d 2

20 0 0 0

in the conjugate beam in the actual beam

BM i th j t b D fl ti i th t l

b

x x x xM d ydx yEI dx

= =∫ ∫ ∫ ∫BM in the conjugate beam Deflection in the actual e m b a=


Conjugate supportsj g pp

Actual support. Conjugate support.


58

Example 1:

l

Actual beam

B

( )Conjugate beam

BA

(-)PlEI

Slope at B of the deflected beam = Shear force at B of the conjugate beam when conjugate beam is loaded with M/EI diagram

2

2 2l Pl Pl

EI EI= =

Deflection at B of the beam = Bending moment at B of the conjugate beam when conjugate beam is loaded with M/EI diagram

2 32Pl l Pl


59

2 322 3 3Pl l PlEI EI

= =

Example 2:

ll

( )Conjugate beam

BA

2

2wlEI

(-)

Slope at B of the deflected beam = Shear force at B of the conjugate beam when conjugate beam is loaded with M/EI diagram 2 3l wl wl

3 2 6EI EI= =

Deflection at B of the beam = = Bending moment at B of the conjugate beam when conjugate beam is loaded with M/EI diagram 3 43

6 4 8wl l wlEI EI

= =


60

6 4 8EI EI

Example 3:

4PlEIA B

C21 l Pl Pl⎛ ⎞

⎜ ⎟

2Pl2l

2l

C

Conjugate beam

2 2 4 16EI EI⎛ ⎞ =⎜ ⎟⎝ ⎠ 16EI22

j g

Slope at A of the deflected beam = Shear force at A of the conjugate beam when conjugate beam is loaded with M/EI diagram 21 l Pl Pl⎛ ⎞

⎜ ⎟12 2 4 16

l Pl PlEI EI

⎛ ⎞= =⎜ ⎟⎝ ⎠

Deflection at C of the beam = = Bending moment at C of the conjugate beam when conjugate beam is loaded with M/EI diagram

2 2 31l Pl l Pl Pl⎛ ⎞


12 16 3 2 16 48l Pl l Pl Pl

EI EI EI⎛ ⎞

= − =⎜ ⎟⎝ ⎠

Example 4:

2

8wLEI

A B

CC2 31 2l wl wl⎛ ⎞

=⎜ ⎟3wl

Conjugate beam

2 3 8 24EI EI=⎜ ⎟

⎝ ⎠ 24EI

Slope at A of the deflected beam = Shear force at A of the conjugate beam when conjugate beam is loaded with M/EI diagram 3lconjugate beam is loaded with M/EI diagram

fl f h b d f h b

3

24wl

EI=

Deflection at C of the beam = = Bending moment at C of the conjugate beam when conjugate beam is loaded with M/EI diagram

3 2 43 2 5l wl l l wl wl⎧ ⎫⎛ ⎞⎛ ⎞


3 2 52 24 8 2 3 2 8 384l wl l l wl wl

EI EI EI⎧ ⎫⎛ ⎞⎛ ⎞= − =⎨ ⎬⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎩ ⎭

Principle of SuperpositionPrinciple of Superposition

Statement: Deflection at a given point in a structure produced byseveral loads acting simultaneously on the structure can beseveral loads acting simultaneously on the structure can befound by superposing deflections at the same point producedby loads acting individually.

Applicable when there exists a linear relationship betweenexternal forces and corresponding structural displacements.p g p


63

SummarySummary

Bending stresses in beams - shear flow - shearing stress formulae for Bending stresses in beams shear flow shearing stress formulae for beams –

Inelastic bending of beams –Inelastic bending of beams –

Deflection of beams - direct integration method - singularity functions iti t h i t th d j t b - superposition techniques - moment area method - conjugate beam

ideas –

l f ll d b f d dElementary treatment of statically indeterminate beams - fixed and continuous beams


64