Congugate beam
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)
=