Deflection of Beam- IsM

8/10/2019 Deflection of Beam- IsM

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DEFLECTION OF BEAM

x



BE M DEFLECTIONS

As a beam is loaded,Different regions are subjected to V and M

The beam will deflect

Recall the curvature equation

x x

The slope (q) & deflection (y) at any spanwise location

can be derived

R



e

Bending

e

e

e

CT

Shear



DESIGN OF BEAMS

The cross section of a beam has to be designed in such away that it is strong enough to limit the bending moment and

shear force that are developed in the beam. This criterion is

known as the STRENGTH CRITERION of design .

Another criterion for beam design is that the maximum

deflection of the beam must not exceed a given permissible

limit and the beam must be stiff enough to resist the

deflection caused due to loading. This criterion is known as„STIFFNESS CRITERION of design”



Definitions:-

(i) DEFLECTION :-The vertical distance in transverse direction

between positions of axis before and after loading at the

section of the beam, is defined as the deflection of beam at

that section.

(ii) ELASTIC CURVE OR DEFLECTION CURVE:-The neutral

axis in its deflected position after loading of the beam is knownas its elastic curve or deflection curve.

(iii) SLOPE:-The slope of the beam at any section is defined as

the angle (in radians) of inclination of the tangent drawn at

that section to the axis in its deflected position afterloading, measured w. r. t. the un-deformed axis.

(iv) FLEXURAL RIGIDITY(EI):-The product of modulus of elasticity

and Moment of Inertia is known as Flexural rigidity.



ASSUMPTIONS MADE IN THE DEFLECTION:-

(i) Axis of the beam is horizontal before loading.

(ii) Deflection due to S.F. is negligible.

(iii) (a) Simple Bending equation M/I=σ/y=E/R is

applicable and all the assumptions made in simple bending

theory are valid.

(b) Material of the beam is homogenous, isotropic and obey

Hook‟s law.

(c) The modulus of elasticity is same in compression as well as

in tension.

(d) Plane section remain plane before and after bending



DIFFERENTIAL EQUATION FOR DEFLECTION AND

SLOPE OF BEAMS

Below is shown the arc of the neutral axis of a beam

subject to bending.

For small angle dy/dx = tan θ = θ

The curvature of a beam is identified as dθ /ds =

1/R

In the figure δθ is small ,and δx=~ δs;

i.e ds /dx =1 6

x



Then dy /ds =sinθ

dx /ds= cosθ

dy/dx =tanθ

Differentiating equation (a) w.r.t. x,

we get

Sec2 θ.(dθ/dx)=d2y/dx2 Therefore,

dθ/dx =(d2y/dx

2)/sec

2θ –-(2)

Other Method

R = ds/dθ = ( ds /dθ) ×(dx/dx) =(ds /dx)/(dθ/dx)

= secφ/(d θ /dx) -----(1)



From equation(1),

R= secθ /(dθ /dx) = sec3θ /(d

2y/dx2)

1/R=(d2y/dx

2)/sec

3θ =(d2y/dx2)/ (sec2θ)3/2

= (d2y/dx2)/ (1+tan2θ)3/2

=(d2y/dx

2)/[1+(dy/dx)

2]3/2

In any practical case of bending of beams, the slope (dy/dx) is

very small (because curve is almost flat); hence (dy/dx)2 can be

ignored

so, 1/R=d2y/dx

2



From the pure bending equation, we know that.

Integrating between selected limits. Thedeflection between limits is obtained by further

integration.

9

R

E

I

M



SIGN CONVENTIONS:

Linear horizontal distance x: positive when measured from left

to right

Vertical distance or deflection y is positive when measured

above the axis and is negative when measured below theaxis of beam.

NOTE :



Deflection at support A and B are zero and maximum at the

middle ofSpan. slope θ is maximum at A and B and zero at middle of Span

(at point of symmetry); At point of maximum deflection slope is

zero

θB

C A B D

θ A A

NOTE : SUPPORT or BOUNDARY CONDITIONS:

(i) Simply supported beams:

Deflection at support A and B are zero but more at free end and

also at the centre of span . Slope is maximum at supports B and

A



(ii) Cantilever Beam:

Deflection and slope both are zero at fixed support.

Slope and deflections are maximum at free end

θ increases from

point A towards B.

B

θmax

ymax

A



Methods for finding slope and deflection of

beams:

(i) Double or Direct integration method

(ii) Macaulay’s method

(iii) Area moment method

(iv) Conjugate beam method

(v) Unit load method



We have from differential equation of flexure,

EI d2y/dx2=M

Integrating w. r. t.. x both sides, we get

EI (dy /dx) =∫M dx + C1

Integrating again w .r .t. x both sides ,we get

EI (y) = ∫ ∫ ( M dx) dx + C1(x) + C2

where C1 and C2 are constant of integration

Thus, integrating the differential equation w .r .t. x, we get the

equation for slope (dy /dx) ,and integrating it twice w. r .t. x, we

get the equation for deflection ( y).

DOUBLE INTEGRATION METHODS



The two constants of integration can be determined

using the known boundary conditions of deflectionand slope at the supports

The method of double or direct integration is

convenient only in few cases of loadings. Ingeneral case of loading, where several loads are

acting the B.M. expression for each of the different

regions of loading is different .Then the method of

double integration will become extremely lengthyand laborious. Therefore ,it is not generally used.



Case--1(i): Determine the slope and deflection equation for

the beam loaded as shown in fig.

(ii) Find the maximum deflection and maximum slope.

Solution:

B.M. at section 1-1

M= - P( x)

EI d

2

y/dx

2

= M = - P (x)EI (dy /dx ) = -P (x

2/2) + C1

EI y = -P (x3/6) + C1x + C2

P1

1

x

A B

θ A l

y A

At fixed end, when x = L ,(dy /dx) =0

(iii) What will be the deflection and slope at free end when

P=6kN, L=3m, E=210GPa, I=16x104cm4.



Therefore,

- (PL2/2) + C1= 0

C1= PL

2

/2 At x = L, y = 0,

-PL3/6+(PL

2/2) L+C2=0

or,C2= PL3/6 - PL

3/2

= PL3/6[1-3]= - PL3/3Therefore ,C2= - PL

3/3



Equation of slope; EI (dy/ dx) =-Px2 /2 + PL

2 /2-----(1)

Equation of deflection ,EI (y)=-Px3 /6 + PL

2x/2 - PL

3 /3-----(2)

Maximum deflection :

When x=0 (at free end) ,then from equation (2),

EI (y)=-0+0-PL3 /3

ymax= -PL3 /3EI

Maximum Slope:

Slope is maximum at free end (at x=0).hence fromequation(1),

EI (dy/ dx) = -0 + PL2 /2

(dy /dx) max = PL2 /2EI



Deflection at free end (i.e; at A):= y A = PL3/3EI

(iii)

θ A=PL2/2EI

slope θ A=(dy/dx) at A

radian5

83

233

100357.8

)1016()10210(2

)103(106

)1016()10210(3

)103(10683

333

= 0.161mm

P



Case (2) : When cantilever is subjected to an u .d.

L. of intensity w unit/m run over entire span

Here A is the origin. Take a

section X-X at a distance x

from A.

B.M at distance x from A= Mx = EI d2y/dx

2

=-w.x.x/2=-wx2/2

Integrating once,

EI (dy/dx) =-wx3/6 + C1 ------------------------(1)

where C1 is constant of integration

B A

Lx

w unit /m

X

X



Applying boundary conditions:-

at x =L, dy/dx=0 from equation(1)

0=-wL3/6 + C1 C1 = wL3/6therefore,

EI dy/dx=-wx3/6+wL

3/6---------(2)

Integrating once again,

EI y=-wx4/24 + wL3.x/6 +C2 ---------------- (3)

where C2 is 2nd constant of integration

Applying boundary condition;

at x=L, y=0



0=-wL4/24+wL

4/6+C2

Therefore,

C2=wL4/24-wL

4/6

C2=-wL4/8.

Therefore, equation (3) becomes,

EI(y)=-wx4/24 + wL

3.x/6 –wL

4/8--------(4)

Maximum deflectionIt occurs at free end where x= 0

From (4),EIy=-0+0-wL4/8



ymax=-wL4/8EI

similarly maximum slope occurs at free end where x=0

from (2),

EI (dy/dx) =-0+wL

3

/6

(dy/dx )max=wL3/6EI



Case 3:- When simply supported beam is subjected

to a single concentrated load at mid-span.

L/2 L/2

P

P

C

A B

R A=P/2

RB=P/2

x X

X

Mx = (P/2) x

EI d2y/dx2=(P/2)x

EI dy/dx=(P/2)x2/2 + C1

Due to symmetry slope at x = L/2 is zero

C1 = -PL2/16

EI dy/dx=(P/2)x2/2 -PL2/16

Integrating again we get

EIY = (P/2)x3

/6 – (PL2

/16) x + C2

At x=0 , Y = 0

C2 = 0



Hence EIY = (Px3/12) – (PL2/16) x

Deflection at mid span i.e. at x = L/2 is

Y = -PL3/48EI

= PL3/48EI (downward)

Slope at support, is obtained by puttingx = 0, in slope equation

EI WLWL

EI dxdy

x

A

16161

22

0

q



Case4:- Simply supported beam of span L carrying a

uniformly distributed load of intensity w per unit run over the

whole span.

RB=WL/2

A

x

W unit / run

B

R A=WL/2

X

X

1

32

2

2

2

2

64

22

22

C

xw xwL

dx

dy

EI

xw x

wL

xd

yd EI

xw

x

wL

Mx



Due to symmetry dy/dx = 0 at x = L/2

24

3

1

wL

C Integrating both side w.r.t. x, we get

2

343

242412

C xwl wxwLx

Y EI

At x = 0, y = 0

C2=0

xwl wxwLx

Y EI 242412

343

Hence

Maximum deflection yc which occurs at centre C is

obtained by substituting x = L/2 in the above equation



Due to symmetry dy/dx = 0 at x = L/2

24

3

1

wL

C Integrating both side w.r.t. x, we get

2

343

242412 C x

wl wxwLx

Y EI

At x = 0, y = 0

C2=0

xwl wxwLx

Y EI 242412

343

Hence

Maximum deflection yc

which occurs at centre C is

obtained by substituting x =

L/2 in the above equation.

EI

wL

dx

dyend at Slope

downward EI wl Yc

EI

wl Yc

x

A

24

)(3845

384

5

3

0

4

4

q



MACAULAY‟S METHOD

For a general case of loading on the beam, though the

expression for B.M. varies from region to region, the

constants of integration remain the same for all regions.

Macaulay recognized this fact and proposed a method

which is known as the Macaulay‟s method to obtain the

slope and deflection equations. It is essentially modified

method of double integration of the B.M. expression but

following a set of rules given below:-



(1). Assuming origin of the beam at extreme left end, take a

section in the last region of the beam at a distance x from the

origin and write the expression for B.M. at that section

considering all the force on the left of section.(2). Integrate the term of the form (x-a)

n using the formula

∫(x-a)n dx=(x-a)

n+1 /n+1

where a=distance of load from origin.

(3). While finding slope and deflection in the form

(x-a)n ,if (x-a) becomes negative on substituting the value of x,

neglect the term containing the factor (x – a)n

(4). If a couple (moment) of magnitude „c‟ is acting at a distance„a‟ from the origin of the beam, then write the BM due to couple

in the form c (x-a)0.

(5). If the beam carries a U.D.L, extend it up to the extreme right

end and superimpose an UDL equal and opposite to that which

has been added while extending the given UDL on the beam.

EXERCISE PROBLEMS



EXERCISE PROBLEMS :

Q.(1) Figure shows a simply supported beam of span 5m carrying

two point loads. Find (1)the deflection at the section of the point

loads. (ii) Slope at A,B,C and D, (iii) maximum deflectionof the beam.

Take E=200GPa, I=7332.9 cm4

Solution:-

R A =34 KN, RB =36 KN A B

R A RB

30KN 40KN1m

5m

1.25m

x

X

X

c D

Mx = R Ax - 30(x-1) - 40(x-3.75)



EI d2y/dx

2= 34 x -30(x-1) -40(x-3.75)

Integrating once,

EI ( dy/dx) =c1 + 34 x2/2 -30(x-1)

2/2 - 40(x-3.75)

2/2 ---(1)

Integrating once again,

---------- (2)

Support conditions:

at x=0,y=0 therefore,C2=0

at x=5m,y=0 therefore, C1=-75.06 ≈ -75.1

6

)75.3(40

6

)1(30

6

34 333

12

x x x

xC C EIy

Thus the equation for slope and deflection will be



Thus the equation for slope and deflection will be

E I (dy / dx) =-75.1+17x2 - 15(x-1)

2-20(x-3.75)

2 -----------(3)

E I (y)= -75.1 x +17x3/3 -5(x-1)3 -20(x-3.75)3/3 ---------(4)

Total deflection at section of point loads

At C x=1m , deflection y= yC (say)

EI yC = -75.1×1 +17(1)3/3 =-69.393

Or, yC =-69.393/EI

=-69.393 /(200×106×7332.9×10-8)=- 4.73× 10-3 m = - 4.73mm



(dy/dx)D =[17 ×(3.75)2-75.06-15 ×(2.75)2] × 1/EI

(when x=3.75m)

=50.525/EI=3.445 radians.

(dy/ dx) at B=1/EI[17 ×(5)2-75.06-15 ×(4)2-20

×(1.25)2]=78.65/EI=5.363 × 10-3rad =0.197 degree

Assuming the deflection to be maximum in the region

CD:

(at point of maximum deflection dy/dx = 0)



say at x=x1 where dy / dx=0

From equation (3),

EI(0)=34/2(x1)2-75.06-30(x1-1)

2/2

=17 x12 -75.06 -15x1

2 +30x1 -15

=2x12+30x1-90.06=0

x1= 2.56m

The assumption that the maximum deflection is withinthe region CD is correct.

EI ymax=34(2.56)3/6-75.06 × 2.56-30(2.56-1)

3/6

Ymax=-116.13/EI=-7.918 × 10-3

m

=-7.918mm



(Q-2)

(A) Obtain the equation for slope and elastic curve for the

beam loaded as shown in figure and find the deflection and

slope at mid-point of beam.

Take EI=15× 103 kNm2

(B) Find the slope at A,C and D

Solution:-

Reactions

R A=1/4[80×3+120]

=90KN( )

RB =80-90=-10kN=10( )

80KN 1m

120kN

mx

X

X

2m1m A

c

DB

Alternatively R =1/4[80×1-120]



Alternatively, RB =1/4[80×1-120]

=-10KN =10 KN( )

EI (dy /dx )= C1 + 90x2

/2 - 80(x-1)2

/2 –120 (x-3) -------(1)

EI(y)= C2+C1(x) + 90 x3/6 -80(x-1)

3/6 -120(x-3)

2/2 ------(2)

Support reactions

at x = 0 ,y = 0 ,

C2= 0

MX= 90x - 80(x-1) -120(x-3)0

Mx = 90 x – 80 (x-1) – 120 (x-3)0

At 4 0 C 135



At x=4,y=0 ,C1=-135

Equation for slope (dy/dx):-

EI (dy /dx)=-135 +90 x2/2 -40(x-1)2 -120(x-3)

Equation for deflection (y):-

EIy = -135x + 90 x3/6 -80(x-1)

3/6 -120(x-3)

2/2

To find deflection at centre (i.e. x=2m, at mid span ):-EIy=90(2)

3/6 -135(2) -80/6(2-1)

3/6 =-163.33

y=-163.33/(15×103)=-10.89×10

-3 m =-10.89 mm

To find slope at centre (i.e. x=2m, at mid span ):-

EI (dy/dx) =90× (2)2/2-135-80(2-1)

2/2 =+5

3 4



dy/dx =5/(15×103)=3.33×10

-4 radians~ 0.019°

=19.1×10-3

degree ~ 0.02°

(b)θ A(at x=0) =-135/EI=-135/(15×103)

θC ( at x= 1m ) =[45× (1)2-135 ×1)/15 × 10

3=-90/15 ×

103radian

θD (at x=3m )= [45 ×(3)2-135-40(2)

2] =110/15 × 10

3radian



(Q.3) Find maximum slope and maximum deflection of the

beam loaded as shown in fig.

Take E=200KN/mm2, I=60 × 10

6mm4

2m 1m

25kNm15kN

x

X

X

A B

c

EI(d2y/dx

2)=-15x -25(x-2)

0

EI(dy/dx)= C1 -15x2/2 -25(x-2) …………(1)

EI(y)= C2 + C1x -(15/2) x3/3 -25(x-2)

2/2 …(2)



Support conditions:

slope and deflection are zero at fixed support

at x=3m,dy/dx=0 from equation (1),

EI(0)=-15(3)2/2 + C1- 25(3-2)=0

C1

=25+67.5=92.5

At C, y=0, x=3 from (2)

EI(0)=-15 (3)3/6+92.5(3) +C2 -25(3-2)

2/2

C2=12.5-277.5+67.5=-197.5

Now equation (1)

EI(dy/dx)=-15x2/2+92.5 -25(x-2)



Now equation (1)

EI(dy/dx)=-15x2/2 +92.5 -25(x-2) ---------(1)

and

Maximum slope at A when x=0 from (1)

EI(dy/dx) A=92.5

(dy/dx) A

=92.5/EI=7.708×10-3

radian

=0.441degree

EI(y)=-15/2*x3/3+92.5x-197.5 -25(x-2)

2/2 -----(2)

Maximum Deflection at free end when x=0

EI(y) A=-197.5, yA=-197.5/EI=-16.458mm.



Q. 3 Determine the equation for the elastic curve for

the beam loaded as shown in figure. Find the

maximum deflection.

1m1m3m

A C D B2kN/m

Solution: R A=RB = 2x3/2 = 3KN

D A1m1m 3m

C

2kN/m

x

M =3(x) -2(x-1)2/2 +2 (x-4)

2/2



M =3(x) -2(x-1)2/2 +2 (x-4)

2/2

EI (dy/dx)=3/2(x)2 +C1 –(x-1)

3/3 +(x-4)

3/3 -------(1)

EIy=3 x3/6 +C1x +C 2 – (x-1)

4/12 +(x-4)

4/12 ------(2)

S t ti



Support reactions:

at x=0,y=0, C2=0

at x=5m,y=0 , C1=-8.25

Equation for slope:-

EI (dy/dx)=3x2/2-8.25 –(x-1)

3/3 +(x-4)

3/3 ------(3)

Equation for the elastic curve :

EIy=x3/2 -8.25x –(x-1)

4/12 +(x-4)

4/12 ---------(4)

Due to symmetry, deflection is maximum at centre at

x=2.5m,EI ymax=(2.5)

3/2 -8.25x(2.5)-(2.5-1)

4/12=-13.23

ymax = -13.23/EI



Practice problems:-

(Q-1) A cantilever beam of span L carries a udl of

intensity w/unit length for half of its span as shown in

figure.If E is the modulus of elasticity and I is moment of

inertia,determine the following in terms of w,L,E and I.

(i)A expression for slope(dy/dx)at free end

(ii)An expression for deflection( y ) at free end

(iii)The magnitude of upward vertical force to be applied at

free end in order to resume this end to the same horizontal

level as built in end.



[Ans (i)θA=WL3 /48EI (ii) ∂ A =7wL4 /384EI( ) (iii)P=7wL/128]

L/2 L/2

w/m

A B

Q (2 ) f f C



Q- (2 ) Determine the values of deflections at points C,D

and E in the beam as shown in figure.Take E=2*105MPa ; I=

60 *108 mm

4

1m 2m

10kN/m

1m 1m

20kN30kN

A C D E B

Answers:

[ᵟC=0.0603mm(downward), ᵟ D=0.0953mm(downward)

ᵟE=0.0606mm(downward)]



.

Q-(3) Find the position and magnitude of maximum deflection

for the beam loaded as shown in fig.

Take E=200GPa ,I=7500cm4 .

3KN/m20 KN

AD B

X

4m 4m 4mC

X

X

[Answers: ymax at 3.7 m from A=-118/EI=7.99mm

yc=-32/EI=-2.13mm]



DOUBLE INTEGRATION METHOD

50

1.



51



EXAMPLE SIMPLY SUPPORTED BEAM

Consider a simply supported uniform sectionbeam with a single load F at the

centre. The beam will be deflect

symmetrically about the centre line with aslope (dy/dx) at the centre line. It is

convenient to select the origin at the centre

line.

52



53



54



THE DOUBLE INTEGRATION METHOD

The constants of integration are determined byevaluating the functions for slope or

displacement at a particular point on the beam

where the value of the function is known. These values are called boundary conditions.

55



2. THE DOUBLE INTEGRATION METHOD

56

Consider the beam shown in Fig.Here x1 and x2 coordinates are valid within the regions

AB & BC.



3-3 THE DOUBLE INTEGRATION METHOD

Once the functions for the slope and

deflections are obtained, they must give

the same values for slope & deflection at

point B. This is so as for the beam to be physically

continuous.

57



EXAMPLE 3

The cantilevered beam shown in Fig. 8.11(a) issubjected to a couple moment Mo at its end.

Determine the equation of the elastic curve

EI is constant.

58



EXAMPLE 3.3 - SOLUTION

The load tends to deflect the beam asshown in Fig. 8.9(a).

By inspection, the internal moment can be

represented throughout the beam using a

single x coordinate.

From the free-body diagram, with M acting

in +ve direction as shown in Fig. 8.11(b), we

have:

59

o M M




Applying equation 8.4 & integrating twice yields:

Using boundary conditions, dy/dx = 0 at x = 0 & y =0 at x = 0 then C1 = C2 =0

60

21

2

1

2

2

2 C xC x M y EI

C x M dx

dv EI

M dx

vd EI

o

o

o




Substituting these values into earlierequations, we get:

Max slope & displacement occur at A (x = L)

for which

61

EI

x M y

EI

x M

dxdy

oo

2

;

/with

2

q

q

EI L M y

EI L M o

Ao

A2

;2

q



EXAMPLE

The +ve result for q A

indicates counterclockwiserotation & the +ve result for y A indicates that it isupwards.

This agrees with results sketched in Fig. 8.11(a).

In order to obtain some idea to the actual magnitudeof the slope

Consider the beam in Fig. 8.11(a) to:

Have a length of 3.6m

Support a couple moment of 20kNmBe made of steel having Est = 200GPa

62



MOMENT-AREA THEOREMS To develop the theorems, reference is made to Fig. 8.13(a).

If we draw the moment diagram for the beam & then divide itby the flexural rigidity, EI, the “M/EI diagram” shown in Fig.8.13(b) results.

By equation

63

dx

EI

M d

q



3-4 MOMENT-AREA THEOREMS

dq on either side of the element dx = the lighter shadearea under the M/EI diagram.

Integrating from point A on the elastic curve to point B,

Fig 8.13(c), we have

This equation forms the basis for the first moment-

area theorem.

64

/ dx EI

M B

A A B q



MOMENT-AREA THEOREMS

Theorem 1: The change in slope betweenany two points on the elastic curve equals thearea of the M/EI diagram between the twopoints.

The second moment-area theorem is basedon the relative derivation of tangents to theelastic curve.

Shown in Fig 8.12(c) is a greatly exaggeratedview of the vertical deviation dt of thetangents on each side of the differentialelement, dx.

65




Fig 8.12(c)

66



Since slope of elastic curve & its deflectionare assumed to be very small, it issatisfactory to approximate the length ofeach tangent line by x & the arc ds‟ by dt.

Using qs = r dt = xdq Using equation 8.2, dq = (M/EI) dx

The vertical deviation of the tangent at A

with the tangent at B can be found byintegration.

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8.6eqn/ dx EI

M xt

B

A B A


MOMENT AREA THEOREMS



Centroid of an area

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B.&A betweenareatheof centroidthetoA

throughaxisverticalthefromdistance

8.7eqn

dAdA

/

x

dx EI M xt

x x

B

A B A





Theorem 1

The vertical deviation of the tangent at a point

(A) on the elastic curve with the tangent

extended from another point (B) equals the“moment” of the area under the M/EI diagram

between the 2 points (A & B).

This moment is computed about point A where thederivation is to be determined.

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Provided the moment of a +ve M/EI area from A

to B is computed as in Fig. 8.14(b), it indicates

that the tangent at point A is above the tangent

to the curve extended from point B as shown inFig. 8.14(c).

-ve areas indicate that the tangent at A is below

the tangent extended from B.

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71





It is important to realize that the moment-area

theorems can only be used to determine the

angles and deviations between 2 tangents on

the beam‟s elastic curve. In general, they do not give a direct solution for

the slope or displacement at a point.

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3-5 CONJUGATE-BEAM METHOD

Here the shear V compares with the slope ,the moment M compares with the displacement

Y & the external load w compares with the M/EI

diagram. To make use of this comparison we will now

consider a beam having the same length as the

real beam but referred to as the “conjugatebeam” as shown in Fig. 8.22.

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q




74




To show this similarity, we can write theseequation as shown

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wdx

dV w

dx

M d

2

2

EI

M

dx

d

q

EI

M

dx

vd

2

2




Or intergrating

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wdxV dxwdx M

dx EI

M

q dxdx

EI

M v




The conjugate beam is loaded with theM/EI diagram derived from the load w on

the real beam.

From the above comparisons, we can state2 theorems related to the conjugate beam.

Theorem 1

The slope at a point in the real beam isnumerically equal to the shear at the

corresponding point in the conjugate beam.

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Theorem 2The displacement of a point in the real beam is

numerically equal to the moment at the

corresponding point in the conjugate beam.

When drawing the conjugate beam, it is

important that the shear & moment developed

at the supports of the conjugate beam account

for the corresponding slope and displacementof the real beam at its supports.

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3-5 CONJUGATE-BEAM METHOD For example, as shown in Table 8.2, a pin or roller support at

the end of the real beam provides zero displacement but the

beam has a non-zero slope.

Consequently from Theorem 1 & 2, the conjugate beam must

be supported by a pin or roller since this support has zero

moment but has a shear or end reaction.

When the real beam is fixed supported, both beam has a free

end since at this end there is zero shear & moment.

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Corresponding real & conjugate beamsupports for other cases are listed in the

table

Documents

Deflection of Beam- IsM