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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

wdx

dV= w

dx

Md=

2

2

EI

M

dx

d=

θEI

M

dx

vd=

2

2

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8-5 Conjugate-Beam method

Or intergrating

wdxV ∫= [ ]dxwdxM ∫=

dxEI

M⎟⎠⎞

⎜⎝⎛∫=θ dxdx

EI

Mv ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∫=

2

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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

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8-5 Conjugate-Beam method

Fig 8.22

3

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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

4

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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

5

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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

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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 θB and ∆B are to be determined, we must compute VB’ and MB’ in the conjugate beam, Fig 8.24(c)

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Example 8.12 - solution

rad

mmkN

kNmEI

kNmV

VEI

kNm

F

BB

B

y

00263.0

])10)(10(475][/)10(200[

250

250

0250

0

412626

2

2

'

'

2

−=

−=

−==

=−−

=∑↑+

−

θ

7

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Example 8.12 - solution

mmm

mmkN

kNmEI

kNmM

MmEI

kNm

M

BB

B

B

9.210219.0

])10)(10(475][/)10(200[

2083

2083

0)33.8(250

0 ve, as moments iseanticlockwWith

412626

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 upwardThe 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 zeroAssuming this point acts within the region 0≤x≤9m from A’ we can isolate the section

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Example 8.13 - solution

Note that the peak of the distributed loading was determined from proportional triangles

OKmxmx

xEI

x

EI

F

VEIxw

y

)90( 71.6

02

2

1450

0'9/)/18(/

≤≤=

=⎟⎠⎞

⎜⎝⎛+−

=∑↑+

==

10

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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

171.6

)71.6(2

2

1)71.6(

45

0 ve, as moments iseanticlockwWith

=+⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

=∑+

MEIEI

M

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Example 8.13 - solution

The –ve sign indicates the deflection is downward

mmm

mmmmmmkN

kNmEI

kNmM

8.160168.0

)])10/(1()10(60][/)10(200[

2.201

2.201'

44344626

3

3

max

−=−=

−=

−==∆