Deflection of Beam - mmmut

Deflection of Beam

Ankit Saxena ([email protected])

Beam Differential Equation

Elastic

Curve PQ

d

R

ds

dx

dyds

Where dx and dy represent the projected lengths of the segment ds along X

and Y axes.



22

2

22

2

3 2

2

32 22

2

.....................................(1)

tan .....................................(2)

. .

sec .

1sec . .

sec

1 tan

ArcAngle

Radius

dsd

R

dy

dx

differenting w r t x

d d y

dx dx

ds d y

R dx dx

d y

R dx

d y

dx R


2

2

2

2

1d y M

dx R EI

d yEI M

dx

Note :-

1) The above equation is the governing differential equation of the beam.

2) we only take the effect of bending moment. The effect of shear on the deflection

is extremely small and usually neglected.

3) EI is an index which is known as flexural strength of an element.


Boundary Condition for Beams

Case (1) Cantilever beam

- Transverse deflection

- Angular deflection of beam (or) slope of beam

AP

c

cyTransverse

Deflection

cy

c CC

c

Free

End

y y Maximum

Maximum

0

0

A

A

Fixed

End

y y


Cantilever beam concentrated load at free end

A

P

C

x

x

x

2

2

2

2

x x

x x

d yEI M

dx

M Px

d yEI Px

dx

Integrating w.r.t 'x' 2

12

,

0

dy PxEI C

dx

at x L

dy

dx

L


2

2

2 2

102

2

2 2

PLC

PLC

dy Px PLEI

dx

Again integrating w.r.t 'x' 3 2

3 3

3

3 2 3

2

2

2

.6 2

,

0

06 2

3

.6 2 3

Px PL xEI y C

at x L

y

PL PLC

PLC

Px PL x PLEI y


Note

3 2 3

.6 2 3

Px PL x PLEI y

1. The magnitude of the slope curve is slope of deflection curve.

2. The slope of slope curve is magnitude of bending moment.

3

max 0

2

max 0

3

2

C x

C xMax

PLy y

EI

dy PL

dx EI


A

A

A

C

C

CBMD

Slope

Curve

Deflection

Curve

PL

2

2

PL

EI

3

3

PL

EI


Case (2) Cantilever Beam concentrated Load not at free end

A

P

C

cy

c

cyB

B

CM

By

L

a


Case (3) Cantilever beam subjected to uniformly distributed load on whole span

length

A

L

/W N m

B

4

max 0

3

max 0

8

6

B x

B xMax

WLy y

EI

dy WL

dx EI


Case (4) Cantilever beam subjected to uniformly distributed load on a part of

span length

A

L

/W N m

B

a

4 34 .

8 8 6

W L a W L a aWLy

EI EI EI


Case (5) Cantilever beam subjected to a couple at the free end.

A

L

B

M

2

max 0

max 0

2B x

B xMax

MLy y

EI

dy ML

dx EI


Case (6) Cantilever beam subjected to a uniformly varying load having zero

intensity at the free end.

A

L

B

/W N m

4

max 0

3

max 0

30

24

B x

B xMax

WLy y

EI

dy WL

dx EI


Case (7) Cantilever beam subjected to a uniformly varying load having zero

intensity at the fixed end.

A

L

B

/W N m


Simply Supported Beam

W

A B

2

L

2

L

C

Symmetry in Bending moment diagram

1. Maximum deflection occurs at the centre or mid point of the beam axis.

2. At the mid point, point C [slope = 0]

3. At supports (A & B)


2

0

Max

Lx

y y

dy

dx

0

Max

y

dy

dx

No symmetry in Bending moment diagram

1. Maximum deflection occurs in a region between point of application of

load and mid point.

2. The maximum slope is occurred at that support which is nearer to the line

of action of force.



Case 1. Simply Supported beam subjected to a point load at mid span

A B

2

L

2

L

Cx

x2

A

WR

2B

WR

Bending Equation for the section BC On Integration w.r.t 'x' Again integrating w.r.t 'x'

2

2

2

2

x x

WxM

d y WxEI

dx

2

1.....................(1)4

dy WxEI C

dx

3

1 2. .............(2)12

WxEI y C x C

Wx


2

1

3

1 2

.....................(1)4

. .............(2)12

dy WxEI C

dx

WxEI y C x C

Apply boundary condition, At From equation (2) From equation (1) Substitute the value of C1 & C2 in equation (1) & (2)

, 02

&

0 / , 0

L dyx

dx

x L y

2 0C

2

116

WLC


2 2

3 2

.....................(3)4 16

. ...................(4)12 16

dy Wx WLEI

dx

Wx WL xEI y

For maximum deflection and slope

3

max2

2

max 0

48

16

LC x

B Ax x LMax

WLy y

EI

dy WL

dx EI


Home Work

4

max2

3

max 0

5

384

24

LC x

B Ax x LMax

WLy y

EI

dy WL

dx EI

2

max2

max 0

8

2

LC x

B Ax x LMax

MLy y

EI

dy ML

dx EI

Case (1) S.S.B subjected to point moment at its both ends. Case (2) S.S.B subjected to UDL (Uniformly distributed load ) over it length.

L

MM

L

W N/m


Case (2) S.S.B subjected to Eccentric concentrated load

L

W

a b

Macaulay's Method

1. In Macaulay's method, a single equation is written for the bending

moment for all the portions of the beam.

2. Same integration constants of integration are applicable for all portions.


a b c d1W2W

3W

AB C D

E

AR ER


a b c d1W2W

3W

AB C D

E

AR ERx

x

x

1 2 3. .x x E DE CD BC ABM R x W x d W x c d W x b c d

Note

In the bending moment equation by substituting any value of x, if the form in the

bracket become negative, delete that term completely.


2

1 2 32

2 2 22

1 1 2 3

3 3 33

1 2 1 2 3

. .

. .2 2 2 2

. . .6 6 6 6

E DE CD BC AB

E

DE CD BC AB

E

DE CD BC AB

d yEI R x W x d W x c d W x b c d

dx

x d x c d x b c ddy xEI R C W W W

dx

x d x c d x b c dxEI y R C x C W W W


Q. A simply Supported beam carry 2 point loads 64 KN & 48 KN at B and C points.

Find the deflection under each load.

Given E = 210 GPa and I = 180 * 106 mm4.

A

B CD

64KN 48KN

1m 3m 4m

Area Moment Method (Mohr's Theorem)

Statement [1]

The difference between the slope of any two point is equal to the area of

(M/EI) diagram.

Statement [2]

The difference between the deflection of any two point is equal to the moment

of area of (M/EI) diagram.



Note

1. For the two points, one point should be point of zero slope and other point

should be point of non-zero slope.

2. The point of zero slope is known as reference point and point of non zero slope

is known as origin point.

3. As per statement [2]

Difference b/w deflection of any two point = moment of area of [M/EI] diagram

= A * x of [M/EI] diagram

x = it is the distance b/w the centroid of area and the point of non zero slope point or

origin point.


L

W


Q. For the cantilever beam as known in the figure determine the maximum slope

and deflection. For section AC the flexural rigidity is 2EI and for CB section is

EI.

L L

2EI EI

WA B

C


Q. For the given propped cantilever beam as shown in the figure, determine the support

reaction at the simple support B. W load is acting at mid point of the given beam.

L

W

A B

2

WM

Castigliano's Theorem

If a structure is subjected to a number of external loads (or couples), the partial

derivative of the total strain energy with respect with respect to any load (or

couple) provides the deflection in the direction of that load (or couple).

U = strain energy [S.E] of beam due to bending moment


2

0

.2

L

x x

x x

M dxS E U

EI


L

W

AB

Statement [1]

Statement [2]

Note - W is concentrated point load and M is the concentrated point moment.

M

.B

dy S E

dW

.B

dS E

dM


Note

1. For calculating deflection at a point if there is no point load at that point,

introduce a dummy point load at that point and do the complete

calculation.

2. In the final step, put the dummy load value equal to the zero.


Gate [2014] / 2 Marks

Q. A frame is subjected to a load P as shown in the figure. The frame has a constant

flexural rigidity EI. The effect of axial load is neglected. The deflection at point A

due to the applied load P is -

L W

L

3

3

4

3

1( )

3

2( )

3

( )

4( )

3

PLa

EI

PLb

EI

PLc

EI

PLd

EI


L W

L

Fixed Beam



Fixed Beam

L

W

A B

Symmetrical Loading

0

0

A

Ay

0

0

B

By

Case 1. Symmetrical loading on fixed beam (A fixed beam having a concentrated point load at mid span)

Step (1) A fixed beam is considered as a simply beam with the given loading

condition.


A B

2

L

2

L

W

12

WR 2

2

WR


Step (2) Draw the free moment diagram.

Step (3) A fixed beam is considered as a simply supported beam having fixing

moments at both the ends. Since it is a case of symmetrical loading so equal fixing

moments are required at both ends.

A BL

4

WL

A B

3 0R 4 0R

MM


Step (4) Draw the fixing moment diagram.

area of (M/EI) Diagram = 0

Area of free bending moment diagram = area of fixing bending moment diagram

A B

L

-M

B A

. .2 4

8

L WLM L

WLM


Step (5) Maximum deflection will be at point C.

moment of area of (M/EI) diagram about any end support.

{ moment of area of [M/EI] diagram for free moment diagram about any

support} - { moment of area of [M/EI] diagram for fixing moment diagram about any

support.

area between A and C in free moment diagram.

Distance between centroid of area between A and C in free moment diagram to

fixed support A.

area between A and C in fixing moment diagram.

Distance between centroid of area between A and C in fixing moment diagram

to fixed support A.

C Ay y

1A

1x

2A

2x

1 1 2 2C Ay y A x A x

C Ay y


3

1 1 2. . . . .

2 4 2 3 2 8 2 4

192

WL L L WL L Ly

EI

WLy

EI


Q. Determine the maximum bending moment and the deflection of a beam of

length L and flexural rigidity EI. The beam is fixed horizontally at both ends and

carries a uniformly distributed load w over the entire length.

L

A B

0

0

A

Ay

0

0

B

By

w

4

384

wLy

EI

Unsymmetrical Loading

Step (1) A fixed beam is considered as a simply beam with the given loading

condition.


L

W

A B

0

0

A

Ay

0

0

B

By

a b

W

a b1

WbR

L 2

WaR

L

A B

Step (2) Draw free moment diagram.

Step (3) A fixed beam is considered as a simply supported beam having fixing

moments at both the ends. Since it is a case of unsymmetrical loading so

unequal fixing moments are required at both ends.

a b2

WaR

L1

WbR

L

A B

a b

A B

3R4R

AM BM

Wab

L


3 4

4

4

3

0

0

0

A

B A

B A

A B

R R

M

M R l M

M MR

L

M MR

L

Step (4) Draw the fixing moment diagram.

Let,

a bA B

AMBM


area of (M/EI) Diagram = 0

…………………..(1)

BA

Step (5) deflection at points A & B is zero. So net moment of area in between A

& B is also zero.

Moment of area in between A & B in free moment diagram = moment of area in

between A & B in fixing moment diagram

1.

2 2

A B

A B

M M WabL L

L

WabM M

L


2

2

2

2

A

B

WabM

L

Wa bM

L

………..(2)

From equation (1) & (2)

1 1 2 2

1 2 1 1. . . . . . . .

2 3 2 3 2 2 3B A B

A x A x

Wab Wab b L La a b a M L L M M

L L

Macaulay's Method

Due to symmetry,


W

A B

0

0

A

Ay

0

0

B

By

2

L

2

L

AM BM

AR BR

A B

A B

R R R

M M M

Q. Determine the maximum bending moment and deflection of a beam of

length L and flexural rigidity is EI. The beam is fixed horizontally at both ends

and carries a concentrated load w at the mid span.

Writing general equation -

On integrating,


W

A B

0

0

A

Ay

0

0

B

By

2

L

2

L

BRAR

C

AM BM

x

x

x

2

2

2

2

. 2

. 2

x x B B CB AC

x x

B B CB AC

M M R x W x L

d yEI M

dx

d yEI M R x W x L

dx

22 2

. .2 2

B B

CB AC

x Ldy xEI M x R A W

dx


, 0

0

0

At x

dy

dx

A

22 2

. .2 2

B B

CB AC

x Ldy xEI M x R W

dx

Again, integrating w.r.t 'x'

32 3

32 3

2. .

2 6 6

, 0

0

0

2. .

2 6 6

B B

CB AC

B B

CB AC

x Lx xEI y M R B W

At x

y

B

x Lx xEI y M R W


So the general equation of slope and deflection for the fixed beam for given loading

condition is given as -

At x = L, y = 0 & dy/dx = 0

From the above 2 equations we get -

22

32 3

2. .

2 2

2. .

2 6 6

B B

CB AC

B B

CB AC

x Ldy xEI M x R W

dx

x Lx xEI y M R W

22 2 2

32 3 3 2 3

0 . . .2 8 2 8

0 .2 6 48 2 6 48

BB B B

CB AC

BB B B

CB AC

R LL L WLM L R W M L

R LL L L L WLM R W M


On solving these 2 equations the value of RB and MB is given as -

If we take the section from the end A at a distance of x, we will get the RA and MA

which value is given as -

8

2

B

B

WLM

WR

8

2

A

A

WLM

WR

Continuous Beam


Continuous Beam

1. In continuous beam we used 3 or more than 3 simple supports.

2. The moment reactions only at the end supports are zero.

Note - [ W1 & W2 are acting at mid point of AB & BC respectively.]


2W1W

1L 2L

AM BMcM

ARBR cR

A B C

Clapeyron's Three-Moment Equation

(Procedure)

1. Between 3 consecutive supports continuous beam is treated as simply

supported beam with given loading condition.

2. Calculate the support reactions.


1W

1L1

2A

WR

1

2B

WR

. . 1S S B 2W

2L

2

2B

WR 2

2c

WR

A B CB

. . 2S S B


3. Draw the Bending moment diagram for SSB-1 and SSB-2

4. Apply Clapeyron's equation

1L 2LA B CB

. . 1

BMD

S S B . . 2

BMD

S S B

1 1

4

W L 2 2

4

W L

1 1 2 21 1 2 2

1 2

6 62A B C

A x A xM L M L L M L

L L


5. In continuous beam moment reactions at the end support is zero. So moment

reaction at A & C supports are zero.

6. A1 = Area of BMD for SSB-1

A2 = Area of BMD for SSB-2

7. = Distance between centroid of area A1 to the left hand side support in

BMD-1

. = Distance between centroid of area A2 to the right hand side support in

BMD-2

1x

2x

0

0

A

C

M

M


Applying the given conditions to Clapeyron's three moment equation

2

1 1 1 11 1

2

2 2 2 22 2

11

22

2 2

1 1 1 2 2 2

1 2

1 2

2 2

1 1 2 21 2

0

0

1. .

2 4 8

1. .

2 4 8

2

2

6. . 6. .8 2 8 22

3 32 . .

8 8

A

C

B

B

M

M

W L W LA L

W L W LA L

Lx

Lx

W L L W L L

M L LL L

W L W LM L L


Case [1] If,

From the above equation

1 2

1 2

L L L

W W W

2 23 3

2 . .8 8

3.

16

B

B

WL WLM L L

M WL


Case [2] when continuous beam subjected to uniformly distributed load over its

entire span length.

On simplification the term has become,

1L 2L

AM BMcM

ARBR cR

A B C1w 2w

1 1 2 2

1 2

6 6A x A x

L L

3 3

1 1 2 2 1 1 2 2

1 2

6 6

4 4

A x A x w L w L

L L

Springs


Spring

• Spring is a elastic member which deflect under the action of external load or

couple.

• Due to deflection, spring store the energy and at the required time released

the energy.

Note

Stiffness or spring constant : It is defined as the force required per unit

deflection.

Solid length : It is the length of a spring in the fully compressed state when

the coils touch each other.


Helical Spring

"A helical spring is a piece of wire coil in the form of helix."

Helix- when a right angle triangle is wrapped around the circumference of a cylinder

through its base, a helix profile is generated.

Helix angle - The angle made by plane of coil with the horizontal plane which is

perpendicular to the axis of the spring is known as helix angle.

d = wire diameter (mm)

R = mean coil radius (mm)

D = mean coil diameter (mm)



Spring index:

It is the ratio of mean coil diameter [D] to the wire diameter [d]. It is denoted by

'C'. D

Cd

Types of Helical Spring

Open coil helical spring Closed coil helical spring

1. Coils do not touch each other 1. Coils touch each other.

2. helix angle is generally greater than

10 degree.

2. helix angle generally very small

generally less than 5 degree


Note

1. If there are,

Number of active coil = n

Length of spring = π.D.n

Closed coil helical spring

1. The helix angle is very small.

2. The coils may be assumed to be in a horizontal plane.

3. These spring may be acted upon by axial load or axial toque.

4. Due to axial load, there is axial extension may take place in a spring.

5. Due to axial torque, there is a change in the radius of curvature of the

spring coils.

Note :- Due to axial torque, there is an angular rotation of the free end and the

action is known as wind-up.


Closed coil helical spring under axial load

W = axial load (N)

D = mean coil diameter (mm)

R = mean coil radius (mm)

d = wire diameter (mm)

θ = total angle of twist along wire (radian)

δ = deflection of W along the spring axis (mm)

n = number of coils

L = length of wire



Torsional torque = W. D/2

Assumptions:

1. Neglect the effect of shear force.

2. Neglect the radius of curvature effect.

Torsional shear stress

3

3

3

16

16. .2

8

T

d

DW

d

WD

d


Angle of twist (θ)

4

4

4

2

4

. .2

.32

16

16

16 .

T G

J L

TL

GJ

DW L

G d

WDL

G d

WD Dn

G d

WD L n

Gd


Strain Energy [S.E]

Deflection of W along the shaft axis (mm) [δ]

2

22

4

2 3

4

1. .

2

. .1 4.2

. .32

4 ..

T LS E

GJ

DW Dn

G d

W D nS E

Gd

2 3

4

3

4

.

4 .

8

dS E

dW

d W D n

dW Gd

WD n

Gd


Note

Consider shear stress and radius of curvature effect.

Torsional shear stress is given as-

Where Kw = Wahl's factor

Where C = Spring index, it is given as -

3

8. w

wDK

d

4 1 0.615

4 4w

CK

C C

DC

d

Closed oil helical spring subjected to axial torque


T = Axial torque

This axial torque become act as bending moment for spring

wire.

2

2

2

2

4

1. .

2

1.

2

. .

1 . ..

2

32.

M LS E

EI

T L

EI

L D n

T D n

EI

T DnS E

Ed

Angle of twist (θ)


2

4

4

32.

64

d d T DnS E

dM dM Ed

TDn

Ed


A close-coiled helical spring is required to absorb 2.25*103 joules of energy.

Determine the diameter of the wire, the mean diameter of the spring and the number

of coils necessary if

1. The maximum stress is not to exceed 400 MN/m2

2. The maximum compression of the spring is limited to 250 mm

3. the mean diameter of the spring can be assumed to be eight times that of the wire

How would the answers change if appropriate Wahl factors are introduced?

For spring material G = 70 GN/m2.


Q. Show that the ratio of extension per unit axial load to angular rotation per unit

axial torque of a close-coiled helical spring is directly proportional to the square of

the mean diameter, and hence that the constant of proportionality is 1/4 *(1 + ν).

If Poisson's ratio ν = 0.3, determine the angular rotation of a close-coiled helical

spring of mean diameter 80 mm when subjected to a torque of 3 N m, given that

the spring extends 150 mm under an axial load of 250 N.

Springs in Series

1. When two springs of different stiffness are joined end to end, they are said

to be connected in series.

2. For spring in series, the load is the same for both the springs whereas the

deflection is the sum of deflection of each.


1 2

1 2

1 2

1 2

1 2

1 1 1

W W W

s s s

s s s

s ss

s s

Springs in Parallel

1. When two springs of different stiffness are joined in parallel, they have the

common deflection and the load is the sum of load taken by each,

i.e., common deflection


1 2

1 2

W WW

s s s

1 2

1 2

1 2

W W W

s s s

s s s

Open coiled helical spring subjected to axial load

(W) and axial Torque (T)

Due to this axial load (W) and axial torque (T) both, twisting couple and

bending couple will act in spring wire.



The combined twisting couple is given as,

The combined bending couple is given as,

Length of wire is given as,

' cos sinT WR T

cos sinM T WR

seccos

DnL Dn


Total strain energy (U) is given as -

Axial Deflection

2 2

2 2

2 2

cos sin cos sin

2 2

M L T LU

EI GJ

T WR L WR T LU

EI GJ

2 22

2 cos sin sin 2 cos sin cos

2 2

cos sin 1 1sin cos

dU

dW

L T WR R L WR T R

EI GJ

WR L TRLGJ EI GJ EI


Axial Rotation

2 2

2 cos sin cos 2 cos sin sin

2 2

1 1 sin cossin cos

dU

dT

L T WR L WR T

EI GJ

WRL TLGJ EI GJ EI


Stresses

1. The combined twisting couple is given as,

Torsional shear stress is given as,

2. The combined bending couple is given as,

Bending stress is given as,

'cossinTWRT3

16 'T

d

cossinMTWR3

32b

M

d


THE END

Deflection of Beam - mmmut

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