Find the slope and deflection under the loads of a simply supported beam of length 6m?subjected to concentrated loads of 63 kN* each at 2m and 4m from the left

CONJUGATE BEAM METHOD

Conjugate beam is an imaginary beam of same

lenge as original beam

, but is supported in

such a manner that , when loaded with diagram

If real beam , the sheal force and Bending moment

at a section in conjugate beau gives slope and

election at corresponding section of original beam.

Real beam support Conjugate beamSupport

hinged/rolleshinged/roller ⑭X T

fixed free end.

⑭ –

fixed end free end #·

InteriorSupport internal hinge —

⑳

InteriorSupport internal hinge — ⑳

Load on Conjugate Beam- > diagram Area of original

Theorems :- Beam.

&Slope at any port on original beam-there force at

that point in conjugate beam.

② Deflection at any port on original beam-Bending moment at

that point in conjugate beam.

SIGN CONVENTIONS :-

–

1) If M/ diagram is positive , loading outs downwards

of Mez diagram is negative , loading

acts upward .

2) If sheal force -> tre -> slockwise slope

– re -> anti-clockwise slope.

if sheal fork ->

3) if BM ->re -> deflection – downward .

If BM -re -> deflection -> upward .

Problems to point load at free end.

1) Cantileve bear subjected

I w

A B – Real Beam

–

wal -> Bending

moment diagram.

– > diagram

|23

B slope at Bow real beau ,

OB = sheal force at B' of conjugate Beam.

SFi' = RB'= xxL=C

deflection at B of real beam = Oc = Bending moment at B'

of conjugate beam = BMB '

= Cavea of loading x Centroidal distance was)

= (* XL)xzx

= 2) Cantilever Beam subjected to uniformly

distributed load

Wim

mmm – Real beam –

G -> Bending

moment diagram (B . M . D)

w -> diagram

↑

-conjugate beamI

slope at Bow real beau , OB = sheal force at B' of

conjugate Beam.

OB = SFi' = RB'=xx =3

deflection at B of real beam = Oc = Bending moment at B'

of conjugate beam = BMB '

& B = BRB' = =Xxx(x)=

–

③Simply supported beam AB spanh subjected to point loud W at midepan . find slopes at 1 and B and

deflection at midspar , C .

—

W

Y2 X

42 ->

real beam .

B Do T

↑ RA ↑RB

WL F

id -> B . M .D

F —

diagram 42 42

↓ B1 -> conjugate bear

↑ Rig' Ral

Hope at a o real beam-shear force at Alie

RAY of

conjugate beam.

< Fy = 0 => RA'= RB = = Y -xw

↑ 4E1

equilibrium Ra + R= ( Equations

44MA = 0 = L** E) – RBX) = 0

RB'= : Ral=-WTEI//

Ther OA = RA= (2)

OB = RB = E G

= Banding moment at s of conjugate

beam.

= Rsx E – * xEx W * (5 +E)

=- 5r = (

⑦ find Hope and deflection at support and midspan

of simply supported beam subjected to udl w/m.

DA= RAI *b

-> w/m ↳ A = 2munmum Sh

↑ g – Tk L-

↑A+ Rp = EXLX ↑

-> BMD Ra'+ Rs=

we ?

– EMA' = O

k 1- [Rbx4) +) = 0

↑

#) -> digram

R

=-O

At 1- B Ral =We

& c = BM,

↑

= Oc = Ent/

Q) find maximum slope and deflection of following beam using conjugate

beam method?

50 kN may slope at supports A & B

A 2m am " – real

beam man deflection

at

& * ↑

midspan , i. e -> C.

50X4 = 50 kNm

i& -> B-M .D

idiagram

·I i – conjugate beam –

&

&

c &↑ – ↑R C

OA = RA or SEAl and OB = SFil = RB'

RA' + RB = Ex4x = EMAl = 0

LEXivx-4xRp =ER==/

Ra = E – : Slope at A =A= (2)

OB=(

&c = BMC = (2) -x)=

2 find slopes at A & B and deflection at C and D

of following beam using conjugate beau method ?

↓SOKNCork B

A " i F ↑m am am

To analyse the beam , loads will be considered separately.

and the surre gives slope and deflection.

30KN 2 KN Real

Real t beam 2

Y B beam 1 / "4m = *

Om D Do

A T ↑m

L

B 2m T

Wob =2xx – = 30kNm

gram * I T

↑

I

' ↓ – A 6m .

Y 2m B

Mdia . –

~El

↑

I· – 6m- –

= –

there M/c values at D andC are calculated being

Interpolation]

conjugate beam of 30K Conjugatbeam of 20 .

KN

Q

S ~ I

②

"30A M B E 11 1 ↑ –

=>>4m ->↑ 4m

/ D' I "Atra"< 8m- RB RA

p

to conjugate beam theory , sumoy shear forces at

According A and Al will give

the shope at A of real beam. To

find that , calculate RA , RA" , Ris' and RB" .

① RA' + RB) = = X8X

=> RA' + Ri = 2 EMAl = 0

Ex *Xx(E) – 8XR = 0 = Ris =1

due to symmetry RA =

240- A " B & ⑳

k G7b 7

& RA" + Ris"=X8X Distance of C-G

from A =>

EMA" = o (Ea

& x8Xx (A)SXRD"=O R =E

Ra" =

10- –

Now slope at A of Real beam ,

is OA =

shear force at A of conjugate beam = Ra' + Ra"

–

Slope at B of real beam , in OB = shear force at B

of conjugate beam = RB' + Rs

=>2+

DEFLECTION at C

-6 ! I El El

. B/ #Im ↑ A,-

– Ris' RB 1(

Defection at C of real beamie Oc = Bending moment at

of the conjugate beam

= [(RA'x 4) -(e) + ((2" x4) – (tx4x)) um

Total S. F Total S .F

A- c A"->

I

=(x)-()+ 4) – 5 &

= DEFLECTIONAT D

30

& El 30 -43 -X-

k- I El –" ! &

2m B D'

-At R 8m- RB

RB

deflection at D of real beam , is Op = Bending moment

at c of the conjugate beam-

= ((Risx2) – (tx2xx] + ((Ri"x2) -(xx2xx)] = ((x2) – ( ***) + (C +2) – (Ex2xx)

=