Cont Beam Mf 2

8/13/2019 Cont Beam Mf 2

1/6

Gh. Asachi Tehnical University of Iai Structural Statics Hyperstatic Structures

Department of Structural Mechanics Cezar Aanici, Dr. Eng.

- 1 -

CONTINUOUS BEAMTHE THREE BENDING MOMENTS EQUATION (THE CLAPEYRONS EQUATION)

Let draw the internal efforts diagrams for the following continuous beam applying the Clapeyron

equation:

B C D

6,3 m 3,15 m 4,2 m

40 kN/m

A

2 m

2I0 I0 4I0 4I0

1. Establishing the static indeterminacy degree:3 ( ) 3 1 (0 6) 3

sn c l r = + = + = - the beam is three times statically indeterminate

2.

Choosing the primary system:

The primary system (statically determinate) is obtained by removing the corresponding continuityinternal links in the fixed end A, and along the beam, over the simple supports B and C, respectively,

introducing hinges and three couples of unknown bending moments as illustrated below:

6.3 m 3.15 m 4.2 m

40 kN/m

X1X2 X2 X3 X3

M =80 kNm4

X1

80 kN

length=0

0 1 2 3 4

I = 801 2I

or

0 I0 4I0

To find the recurrence relationship, a general primary system will be used, as follows:

i-1 i

Xi-1 Xi-1 Xi Xi

i+1

X i+1X i+1

li li+1

For an internal arbitrary node i, along the continuous beam, the relative rotation in this node of theprimary system is different than zero, comparing with the initial situation. Consequently, the compatibility

equation will be obtained from the condition that the sum of all rotations induced by internal efforts Xi , and by

the external loads, respectively, to be null, hence:

0reli =

or, if developed, the above relationship could be written as:

, 1 1 , , , 1 1( " ' ) " ' 0i i i i i i i i i i i i iX X X + ++ + + + + =


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- 2 -

X =1i-1li li+1

i-1 i

Xi-1 Xi-1 Xi Xi

i+1

X i+1X i+1li li+1

X =1i-1

i,i-1 ' i,i-1

li li+1

i,i "i-1,i

X =1i X =1i

i,i ' i+1,i

li li+1

i+1,i+1 "i,i+1

X =1i+1 X =1i+1

irel

li li+1

i i+1 i+1i' '" "

in which

' , " ,3 3 6

k k k

k k k

k k k

l l l

EI EI EI = = =

are the corresponding end rotations of a simply supported beam (equivalent with a double hinged one)

induced by the unit bending momentsXi= 1applied on the primary system, and

11 i+1

1

" " , ' '6 6

i ii i i

i i

l lm m

EI EI ++

+

= =

are the real rotations caused by the external loads applied on the primary system. The load factors m

and, m, respectively, are calculated with the following relationships:

( ) ( 1)1

1

6 " 6 '" , 'Mp i Mp ii i

i i

R Rm ml l

++

+

= =

whereR and,R are the left and right, respectively,reactive forces in the node iproduced by the

reduced bending moment diagrams considered as loads for the primary system.

Replacing the above expressions of the beam end rotations, the compatibility equation becomes:

( ) ( 1)1 1 11 1

1 1 1 1

6 " 6 '0

6 3 3 6 6 6

M i M ii i i i i ii i i

i i i i i i i i

R Rl l l l l lX X X

EI EI EI EI EI l EI l

++ + + +

+ + + +

+ + + + + =

which, if simplified by 1/6E, could be written as follows:


3/6



- 3 -

1 1 1 1 ( ) ( 1)' 2 ( ' ' ) ' 6 " 6 ' 0i i i i i i i Mp i Mp i

l X l l X l X R R + + + ++ + + + + =

which is known as the three bending moment equationor Clapeyrons equation lare the reduced lengthsof

the corresponding spans of the continuous beam, and are calculated with the following relationships:

0'i ii

Il l I= ,

01 1

1

'i ii

Il l I

+ ++

= , unde I0 = min (I1, I2, Ik )

6.3 m 3.15 m 4.2 m

40 kN/m

M =80 kNm4

80 kN

length=0

0 1 2 3 4

I = 801 2Ior

0 I0 4I0

49.6125 kNm pl /8 88.2 kNm2

M =80 kNm4

+

M =80/4=20 kNm4

M =88.2/4=22.05 kNmM =49.6125 kNm

a)

b)

c)

d)

Mp

R'

+R" R'3p R"4pR"2R'1R"1

R'3t R"4t

3p2p

Applying this recurrence equation for the above continuous beam, the following three-diagonal and

band equation system is obtained, with no more than three unknown bending moments in one equation:

1 2 1

1 1( ) 1 0 0 2 0 6.3 6.3 6 "

2 2a nod X X R

+ + + +

6 'R+ 1

1 2 3 2

0

1 1( ) 2 6.3 2 6.3 3.15 3.15 6 "

2 2b nod X X X R

=

+ + + +

2

2 3 4 3 3

2 3

6 ' 0

1 1( ) 3 3.15 2 3.15 4.2 4.2 6 " 6 ' 0

4 4

sau

1 1( ) 3 3.15 2 3.15 4.2 4.2 0

4 4

p

p p

R

c nod X X M R R

d nod X X

+ =

+ + + + =

+ + +

( )3 3 36 " 6 ' ' 0p p tR R R+ + =


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- 4 -

When the continuous beam is extended with a cantilever, it will be removed, and its effect is replaced

with equivalent loads (shear forces, and bending moment, respectively). In such cases, like in the current

application, there are two options to apply the Clapeyrons equation for the node 3, as below:

- case c) in which the equivalent bending moment corresponding to the replaced cantilever , could be

used in the equation as the fourth unknown, with the initial value (unreduced), the requiredreactive forces being solved only for the parabolic diagram load;

- or case d) in which the concentrated bending moment corresponding to the replaced cantilever isreducedand it is superposed to the parabolic distribution load, with its own sign.

The bending moment diagrams Mp on the primary system consist in two second order parabolicvariations, both having a maximum magnitude ofpl

2/8, anda linear variation on beam 3-4, corresponding to the

concentrated bending moment from the overhang. Applying as loads on the opposite fibre of the primary system

the reduced bending moment diagramsMp , the magnitude of the reactive forcesR andR areas follows:

1 2 2' " " 0R R R= = = ,2

2 32 3 2 3

1 2 1 2' " 49.6125 3.15 52.093 kN

2 3 8 2 3p p

plR R l = = = = ,

2

3 43 4 3 4

1 2 1 2' " 22.05 4.20 30.87 kN2 3 8 2 3

p p

plR R l = = = = ,

3 4 3 4

1 1 1 1' 20 4.20 14 kN

3 2 3 2tR M l = = = ,

4 4 3 4

2 1 2 1" 20 4.20 28 kN

3 2 3 2tR M l = = = ,

and the system of compatibility equations becomes:

1 2

1 2 3

2 3

( ) 1 6.30 3.15 0

( ) 2 3.15 12.6 3.15 6 52.093 0

1( ) 3 3.15 8.40 4.2

4

a nod X X

b nod X X X

c nod X X

+ =

+ + + =

+ 80

2 3

206 52.093 6 30.87 0

or

1( ) 3 3.15 8.40 4.2 0

4d nod X X

+ + =

+ + 6 52.093 6 30.87 6+ + 1

3

1

2 20 4.2 0

=

or

1 2

1 2 3

2 3

6.3 3.15 0

3.15 12.6 3,15 312.558

3,15 8.40 413.778

X X

X X X

X X

+ = + + = + =

with the solution:

1

2

3

7.994

-15.989

-43.263

X

X kNm

X

=

Once solved for the unknowns, a couple of bending moments will be applied at the ends of the

simply supported beam corresponding to each span of the continuous beam, with their correspondingsigns, and the internal efforts diagrams will be drawn separately. Finally, all these diagrams will be

assembled in one, corresponding to the full length continuous beam.

The final reactive forces are obtained from an algebraic sum of the shear forces in thecorresponding outward links of the continuous beam.


5/6



- 5 -

2

6,3 m

1

7,994

3.80 3.80

Q3.80

M

7.994

15.989

15.989

-

2 3

3,15 m

54.34 71.66

15.98940 kN/m

Q

1.3585

M

43.263

20.92

15.989

43.263

54.34

71.66

3 4

4,2 m

40 kN/m

43.263

75.25 172.75

Q

75.25

92.75

1.88

43.623

27.52

M

80

80

80


6/6



- 6 -

FINAL INTERNAL EFFORTS DIAGRAMS

2 3 4

6,3 m 3,15 m 4,2 m

40 kN/m

1

75.25

71.66

3.80

43.263

27.52

20.92

15.989

7.994

Q

M

2 m

92.75

8054.34

80

3.80 58.14 146.91 172.75

The vertical reactive forces in external links are:

13.80V kN=

2 3.80 54.34 58.14V kN= + =

375.25 71.66 146.91V kN= + =

480 92.75 172.25V kN= + =

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Cont Beam Mf 2