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The forms of the present continuous
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Continuous Beams - Flexibility Method
Qu.1 Sketch the BM diagram for the beam shown in Fig.1. Take E = 200kN/mm 2. 60kN-m
50kN =
A
=
10kN/m B !!!!!!!!!!!!!!!!!! C
3m
I =
3m
60
D
3m
50
4 x 10 6 mm
40
Fig.1 60.0 23.5
A
C
B
D
16.9
25.7
BM diagram in kN-m units
Qu.2 Sketch the BM diagram for the beam shown in Fig.2. Take EI = constant.
80kN A
=
=
15kN/m B !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! C
D
20kN
4m
8m
2m
Fig.2 83.6
40.0
18.3
A
B
C
58.2 @ mid-span
BM diagram in kN-m units
D
Continuous Beams - Flexibility Method
Qu.3 Sketch the BM diagram for the beam shown in Fig.3. Also calculate the support reactions. Take EI const. [R A = -29.9 , R B = 236.8, R C = 381.1kN] 160kN 40kN/m C B !!!!!!!!!!!!!!!!!!!!!!!
A
D 3m
8m
5m
11m
8m
Fig.3 239.2 383.7
A
C
B 293.6 @ mid-span
D
179.8
BM diagram in kN-m units
Qu.4 Sketch the BM diagram for the beam shown in Fig.4. Support B undergoes a 4. settlement of 15mm. Take E = 200kN/mm 2 and I = 1 x 10-4 m
80kN B
A
C 2m
D
6m
40kN 8m
8m
4m
Fig.4
160.0
17.0
A
B 5.8 84.4
BM diagram in kN-m units
C
D
End rotations of simply supported beams Span = L
1
Flexural rigidity = EI
f11
1
f11 =
1
f12
2
L 3EI
f12 =
f22
f21
f21 =
L 6EI
f22 =
L 6EI
2
1
L 3EI
UDL of intensity 1
"
#
#
=
"
=
L3 24EI
1
"
# a
#
=
$
=
a L
L2(1 - $)(2 - $)$ 6EI
"
=
L2(1 - $2) $ 6EI
Flexibility method applied to continuous beams
Example 1.
Two-span continuous beam with the flexural rigidity EI = constant
Q
P A
B
C
X
Y
R A
RB
RC LBC
L AB
Using statics Res. vertically:
R + RB + RC = P + Q A
(1)
Taking mom. about C:
R A x (L AB + LBC) + RB x LBC = P x XC + Q x YC
(2)
The three reactions are the unknowns but only two equations are available. Therefore the problem is statically indeterminate. We must use compatibility of displacements to generate an additional equation in order to complete the solution. One procedure for doing this is illustrated in the figures below:
Q
P A
B X
Step 1
C
%B
Y
Remove prop and calculate the deflection
%B
&B
A
C
B 1 Step 2
Introduce unit force at B and calculate
For compatibility
R B =
&B
%B / &B
(3)
We use these three equations to solve for R A , RB and R C respectively. The bending moment at B
is
M B = R A x L AB - P x XB
(4)
Q
P MB A
C B Bending Moment Diagram
The major disadvantage of this method arises when EI varies from span to span. The deflections cannot be calculated from standard tables which assume that the flexural rigidity EI is constant. The free-body diagram for the individual spans are given below: P
Q
MB
A
C B
R A
RBA
RBC
RC
RB =RBA + RBC
P
MB
A
R A
RBA
MB
Reactant BM P
Free BM
Final BM = Reactant BM + Free BM
A powerful method that does not suffer from this defect is now described. It uses displacement compatibility and leads to the Three Moment Theorem. The method breaks a continuous beam into a series of simply supported spans. The slopes are therefore discontinuous over the supports. Rotations are then introduced via the reactant moments to ‘heal the cuts’.
Three Moment Theorem Consider a continuous beam consisting of N spans.
A
B (a)
I (b)
J
K
(i)
M
(j)
tangent to curve
I
"J
J
RI
K
RJ
RK LJK
LIJ
MI
MJ
MK
J I
K
RIJ
RJI
RJK
RKL
RJ =RJI + RJk
I
J
K
#JK
#JI
Stage A : End rotations due to applied loads
MI
$
MJ J
MK
'
I
K
Stage A : End rotations due to reactant moments
In order to restore continuity at joint J,
"J = #JI - $ = ' - #JK
1
f11
f12
1
f22
f21 2
1 2
1
Flexibility Coefficients: End rotations due to unit couples
Making use of the flexibility coef’ts shown above we may write the compatibility condition as follows:
#JI - MI fi12 - MJ fi22 = MJ fj11 - MK fj21 - #JK or MI fi12 + MJ (fj11 + fi22) + MK fj21 =
#JI + #JK
Note fi12 = (L/6EI) IJ , fi22 = (L/3EI) IJ , fj11 = (L/3EI) JK , fj21 = (L/6EI) JK This is the so-called Three Moment Theorem (due to Clapeyron). Settlement of supports The theorem can be easily extended to deal with settlement of supports. Let
%J
be the settlement of support J. The rotations on each side of support J due to the differential settlement are ( %J - %I)/LIJ and (%J - %K)/LJK respectively. We get
MI fi12 + MJ (fj11 + fi22) + MK fj21 =
(#JI - (%J -
%I)/LIJ) + (#JK - (%J -
%K)/LJK)
Fixed end (zero rotation)
M A
MB B A R AB
RBA
RBC
A fixed end, A in the figure above, does not rotate. The rotation therefore be balanced by the reactant rotations. We have
a a M A f 11 + MB f 21 = # AB - (% A -
%B)/L AB
# AB must
Worked examples
Example 1. Calculate the reactions at the supports of the beam shown in Fig.1. The flexural rigidity EI is the same for both spans. 10kN/m B A !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! C
3m
3m Fig.1
Step 1 Remove prop
10kN/m B A !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! C
%B
&B
Step 2 Apply unit load
1
%B =
5 x 10 x 64/(384EI)
&B =
1 x 63/(48EI)
=
=
168.75/EI
4.5/EI
Let RB denote the reaction at the support B in Fig. 1 then RB &B = % B
which yields
RB = 37.5kN Resolving vertically we get for the beam in Fig. 1, R + RB + RC = 60 A Also Therefore
R = RC A
by symmetry
R = 11.25kN A
Worked examples
Example 2. Calculate the rotations at the supports of the beam shown in Fig.2. The flexural rigidity is EI.
B
A
1
L Fig.2
BM diagram
1
x/L
A
B
x
Deflection diagram
Apply
Integrating
# A
#B
EI d 2y/dx 2
EI dy/dx
= -M = x/L = x 2/(2L) + C
, C is a constant [1]
Integrating again
EI y
= x 3/(6L) + Cx + D [2]
Applying b.c’s : At x = 0 , y = 0
D = 0
At x = L, y = 0
C = -L/6
Substituting in Eq.{1]
from Eq.[2]
EI A # = C = -L/6
or
# = A
-L/(6EI)
EI#B = L/2 + C
or
#B =
L/(3EI)
Note: The difference in signs of the rotations is due to the direction of the rotation. A positive sign denotes an clockwise rotation whilst the negative sign denotes an anti-clockwise rotation.
Worked examples
Example 3. Draw the BM diagram for the beam shown in Fig. 3 below. The flexural rigidity EI is the same for both spans. 200kN 5kN/m B A !!!!!!!!!!!!!!!!!
C
2m 2m
3m
Fig.3
We note that M = 0 and M C = 0 , simple supports. Applying the three-moment A theorem at B we get 2
3 +
3EI
MB
=
+ #BC #BA
=
815/(9EI)
3EI
()
27.2
= 54.33kN-m
54.3 18.1
Reactant BM (kN-m)
2.5 133.3
24.7
Free BM (kN-m)
Final BM (kN-m) 115.2
Worked examples
Example 4. Calculate the bending moments at the supports of the beam in Fig.4. The flexural rigidity EI is the same for all spans. 400kN
200kN 5kN/m
B A !!!!!!!!!!!!!!!!!
C
=
=
D
2m 2m
3m
2m
Fig.4
We note that M A = 0 and M D = 0 , simple supports. Applying the three-moment we get: Joint B
Taking EI = 1,
[2/3 + 3/3] M B +
3/6 M C
=
5/3 + 800/9
10 () + 3 MC = 543.3 Joint C
3/6 M B +
[3/3 + 2/3] M C
[1] =
3 MB + 10 M C = 1266.7
Solving Eqs.[1] and [2] we get
MB = 18.0kN-m
M C = 121.3kN-m
1000/9 + 100
[2]
Worked examples
Example 5. Rework the problem of example 3 taking the support A as being fixed.
200kN 5kN/m
B A !!!!!!!!!!!!!!!!!
C
2m 2m
3m
Fig.3
We note that M C = 0 , simple support. Applying the three-moment theorem at B we get: Joint A
