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8-5 Conjugate-Beam method
The basis for the method comes from the similarity of eqn 4.1 & 4.2 to eqn 8.2 & 8.4
To show this similarity, we can write these eqn as shown
dx w
dV = w
dx M
d =
2 2
EI M dx dθ =
EI M dx
v d =
2 2
8-5 Conjugate-Beam method
Or intergrating
wdx
V =∫ M =∫
[ ]
wdxdxEI dx M ⎟
⎠
⎜ ⎞
⎝
∫⎛
θ= dx dx
EI
v M ⎥
⎦
⎢ ⎤
⎣
⎡ ⎟
⎠
⎜ ⎞
⎝
∫ ⎛
=
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8-5 Conjugate-Beam method
Here the shear V compares with the slope , the moment M compares with the disp v & 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
“conjugate beam”, Fig 8.22
θ
8-5 Conjugate-Beam method
Fig 8.22
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8-5 Conjugate-Beam method
The conjugate beam is loaded with the M/EI diagram derived from the load w on the real beam
From the above comparisons, we can state 2 theorems related to the conjugate beam
Theorem 1
{ The slope at a point in the real beam is numerically equal to the shear at the corresponding point in the conjugate beam
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8-5 Conjugate-Beam method
Theorem 2
{ The disp. 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 & disp of the real beam at its supports
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8-5 Conjugate-Beam method
For e.g, as shown in Table 8.2, a pin or roller support at the end of the real beam provides zero disp. 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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8-5 Conjugate-Beam method
Corresponding real & conjugate beam supports for other cases are listed in the table
Table 8.2
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Example 8.12
Determine the slope & deflection at point B of the steel beam shown in Fig 8.24(a)
The reactions have been computed
Take
{ E = 200GPa
{ I = 475(106)mm4
Example 8.12
Fig 8.24
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Example 8.12 - solution
The conjugate beam is shown in Fig 8.24(b)
The supports at A’ and B’ correspond to supports A and B on the real beam
The M/EI diagram is –ve, so the distributed load acts downward, away from the beam
Since θBand ∆Bare to be determined, we must compute VB’and MB’in the conjugate beam, Fig 8.24(c)
Example 8.12 - solution
m m
kN
kNm EI V kNm
EI V kNm
F
B B
B y
] ) 10 )(
10 ( 475 ][
/ ) 10 ( 200 [
250 250 250 0
0
4 12 6 2
6
2 2 '
' 2
−
=
= −
−
=
=
=
−
−
=
∑
↑ +
−
θ
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Example 8.12 - solution
mm m
m m
kN
kNm EI M kNm
M EI m
kNm
M
B B
B
B
9 . 21 0219 . 0
] ) 10 )(
10 ( 475 ][
/ ) 10 ( 200 [
2083 2083
0 )
33 . 8 250 (
0 ve,
as moments ise
anticlockw With
4 12 6 2
6
3 3 '
' 2
'
−
=
−
=
= −
−
=
=
∆
= +
=
∑ +
−
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Example 8.12 - solution
The –ve signs indicate the slope of the beam is measured clockwise & the disp is
downward, Fig 8.24(d)
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Example 8.13
Determine the max deflection of the steel beam shown in Fig 8.25(a)
The reactions have been computed
Take
{ E = 200GPa
{ I = 60(106)mm4
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Example 8.13
Fig 8.25
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Example 8.13 - solution
The conjugate beam loaded with the M/EI diagram is shown in Fig 8.25(b)
Since M/EI diagram is +ve, the distributed load acts upward
The external reactions on the conjugate beam are determined first and are indicated on the free-body diagram in Fig 8.25(c)
Max deflection of the real beam occurs at the point where the slope of the beam is zero
Assuming this point acts within the region 0≤x≤9m from A’ we can isolate the section
Example 8.13 - solution
Note that the peak of the distributed loading was determined from proportional triangles
OK m x m x
EI x x EI
F V
EI x
w
y
) 9 0 ( 71 . 6
2 0 2 1 45
0 0 '
9 / ) / 18 ( /
≤
≤
=
⎟ =
⎠
⎜ ⎞
⎝ + ⎛
−
=
∑
↑ +
= =
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Example 8.13 - solution
Using this value for x, the max deflection in the real beam corresponds to the moment M’
Hence,
0 ' ) 71 . 6 3( 71 1 . ) 6 71 . 6 ( 2 2 ) 1 71 . 6 45(
0 ve, as moments ise
anticlockw With
=
⎥ +
⎦
⎢ ⎤
⎣
⎡ ⎟
⎠
⎜ ⎞
⎝
− ⎛
=
∑ +
EI M EI
M
Example 8.13 - solution
The –ve sign indicates the deflection is downward
mm m
mm m
mm m
kN
kNm EI
M kNm
8 . 16 0168 . 0
)]
) 10 /(
1 ( ) 10 ( 60 ][
/ ) 10 ( 200 [
2 . 201 2 . ' 201
4 4 3 4 4 6 2 6
3 3 max
−
=
−
=
= −
−
=
=
∆
