CONTINUOUS BEAMS – Statically Indeterminate structures Flexibility Method – Forces are treated as the primary unknowns in the analysis Steps 1. Determine the degree of indeterminacy i of the beam 2. No. of redundant forces = i 3. Choose reaction(s) as redundant(s) 4. Apply releases at redundant (i. e. remove redundant force from acting on structure) to achieve a statically determinate beam with externally applied loads only. (Primary structure) 5. Determine displacement (∆p) in primary structure at the position of the redundant. 6. Apply redundant force to the “complimentary structure” and determine displacement due to redundant (∆R ). ). 7. For compatibility the net displacements (in the primary and complimentary
structures) at the position of the releases must be zero, i.e, + ∆R = 0 ∆p = ∆R
∆p
Example 1
- two-span continuous beam with flexural rigidity EI = constant
P1
P2
A x
LAB
RA
Fig 1
‐
C
B RB
y
LBC
RC
Given Beam (EI constant)
No. of reactions, r=3 No. of hinges, h=0 Equations of statics = 2 (∑Fv = 0; ∑M = 0) Degree of Indeterminacy, I =(r – h – 2) = 3 – 0 -2 = 1 st Beam is statically indeterminate to 1 degree ∑Fv = 0
R A + R B + R C = P1+ P2
∑M = 0 Taking moments about C:
R A (LAB + LBC) + R B LBC = P1( x x + LBC) + P2 y
(1)
(2)
Step 1:Using R B as the redundant, apply release at support B. P1
P2 B
A
C
∆BP
Fig 2a ∆BP
‐
Primary Structure – release at B (EI constant)
= displacement at B in the primary structure due to applied external loads
Step 2:Apply unit redundant at B (R B = 1)
∆B1
A
C
B
1 Fig 2b ∆B1
∆
‐
Complimentary Structure – Unit redundant applied at B
= displacement at B in the complimentary structure due to the redundant force.
Total deflection at B due to RB is
:
∆BR
= RB x ∆B1
Step 3:
For compatibility, the displacements at B must be zero.
∴ Δ BP = Δ BR
Δ BP = Δ BR = R BΔ B1 ∴RB =
Δ BP Δ B1
(3)
Using equations (1), (2) and (3) we can solve for the unknown reactions RA, RB and Rc Bending moment at B is:
M B =R A (LAB ) - P1 ( x)
P1
MB
A B
Fig 2a
‐
Bending Moment Diagram
(4)
P2
C
(Clapeyron, 1857; Mohr, 1860)
Three Moment Theorem
Three Moment Equation – expresses the relationship between the bending moments at three ‐
successive supports of a continuous beam subjected to loads applied on two adjacent spans with or without uneven settlement of supports. Consider two adjacent spans of a continuous beam consisting of N spans, as shown in Fig. 3.1
θ B A
B
RA
C
RB
RC
LAB
LBC
Fig 3.1a Continuous Beam showing tangent to elastic curve at joint B ‐
MB MA
MC RAB
RBA
RB = RBA + RBC
θ BA
A
RCB
RBC
B
Fig 3.1b = Free body diagrams of Span
θ BC
C
Fig 3.1c End rotations due to applied loads ‐
MB MA
MC A
α
B
β
C
Fig 3.1d End rotations due to reactant (redundant) moments ‐
For compatibility of joint B
(θBA + α ) + (θBC + β ) = 0
(1)
Applying the flexibility coefficients:
1
f 11 1
f 12 2
i
⎛ L ⎞ ⎟ ⎝ 3EI ⎠
⎛ L ⎞ ⎟ ⎝ 63EI ⎠
f11i = ⎜
(
f 21
)
f12i = ⎜
(
1
f 22 i
1
⎛ L ⎞ ⎟ ⎝ 6EI ⎠
i =⎜ f 21
2
⎛ L ⎞ ⎟ ⎝ 3EI ⎠
i =⎜ f 22
)
θ BA + M A f 12i + M Bf 22i + θ BC + M Cf 21j + M Bf 11j = 0
(2)
M A f 12i + M B (f 11j + f 22i ) + M Cf 21j = − (θ BA + θ BC )
(3)
Inserting flexibility coefficients in eqn. (3) gives;
⎧⎛ L ⎞ ⎛ L ⎞ ⎛ L ⎞ ⎫ ⎛ L ⎞ ⎟ + M B ⎨⎜ ⎟ +⎜ ⎟ ⎬ + MC ⎜ ⎟ = − (θ BA + θ BC ) 6EI 3EI 3EI 6EI ⎝ ⎠AB ⎠ AB ⎝ ⎠BC ⎭ ⎝ ⎠ AB ⎩⎝
MA ⎜
(4)
Rewriting;
⎧⎛ L ⎞ ⎛ L ⎞ ⎛ L ⎞ ⎫ ⎛ L ⎞ + + 2M + M ⎨ ⎬ B ⎜ C⎜ ⎟ ⎟ ⎜ ⎟ ⎟ = − (θ BA + θ BC ) ⎝ 6EI ⎠AB ⎝ 6EI ⎠ AB ⎩⎝ 6EI ⎠ AB ⎝ 6EI ⎠ BC ⎭
MA ⎜
(5)
or
⎧⎛ L ⎞ ⎛L⎞ ⎛L⎞ ⎫ ⎛L⎞ ⎟ + 2M B ⎨⎜ ⎟ + ⎜ ⎟ ⎬ + M C ⎜ ⎟ = −6 (θ BA + θ BC ) ⎝ EI ⎠AB ⎝ EI ⎠AB ⎩⎝ EI ⎠ AB ⎝ EI ⎠BC ⎭
MA ⎜
Equation (6) is the Three Moment Equation. ‐
(6)
Settlement of Supports
Let ∆B = settlement of support B, as shown in Fig. 3.3 P1
A
C
∆B
B
Fig 2a
‐
Settlement of support B
Rotations due to settlement at either side of B are given by:
( B− A)
L AB
,
( B− C)
L BC
The Three moment Equation becomes: ‐
⎡⎧ ⎧⎛ L ⎞ ( B −A ) ⎫ ⎧ ( B −C ) ⎫⎤ ⎛L⎞ ⎛L⎞ ⎫ ⎛L⎞ 2M + M 6 θ θ + + = − − + − ⎢ ⎨ BA ⎬ ⎨ BC ⎬⎥ B ⎨⎜ C⎜ ⎟ ⎟ ⎜ ⎟ ⎬ ⎟ L AB ⎭ ⎩ L BC ⎭⎦⎥ ⎝ EI ⎠AB ⎝ EI ⎠ AB ⎩⎝ EI ⎠ AB ⎝ EI ⎠ BC ⎭ ⎣⎢ ⎩
MA ⎜
Flexibility Coefficients , Equations for Slopes and Deflections
(Check Ref. 3: Ghali, A., Neville, A.M. and Brown, T.G. (2003) Structural Analysis – A unified classical and matrix: Appendix B – D)
Example 1
200 kN
5 kN/m B A
C 2m 2m
Fig E1
‐
Given Beam (EI constant)
3m
(5)
Example 2
100 kN
20 kN/m
30 kN
Ao B
A 5m
Fig E2
‐
D
C
3m
2m
5m
Given Beam (EI constant)
Using Three moment equation ‐
⎧⎛ L ⎞ ⎛L⎞ ⎛L⎞ ⎫ ⎛L⎞ 2M + M + + ⎨ ⎬ B ⎜ C⎜ ⎟ ⎟ ⎜ ⎟ ⎟ = −6 (θ BA + θ BC ) ⎝ EI ⎠ AB ⎝ EI ⎠ AB ⎩⎝ EI ⎠ AB ⎝ EI ⎠ BC ⎭
MA ⎜
Creating a fictitious span Ao A with infinite flexural rigidity (EI = ) ∞
Spans Ao A AB ‐
⎛ ⎧⎛ L ⎞ ⎛ 5 ⎞ ⎫ 20(5)3 ⎞ ⎛L⎞ ⎛ 5 ⎞ M AO ⎜ ⎟ + 2M A ⎨⎜ ⎟ + ⎜ ⎟ ⎬ + M B ⎜ ⎟ = −6 ⎜ 0 + ⎟ 24EI ⎠ ⎝∞⎠ ⎝ EI ⎠ ⎩⎝ ∞ ⎠ ⎝ EI ⎠ ⎭ ⎝ ∴10M A +5M B = − 625 Spans AB BC
(1)
MC = (30 x 2) = 60 kNm
‐
‐
‐
⎛ 20(5)3 (100)(2)(52 − 22 ) ⎞ ⎧⎛ 5 ⎞ ⎛ 5 ⎞⎫ ⎛ 5⎞ ⎛ 5⎞ + M A ⎜ ⎟ + 2M B ⎨⎜ ⎟ + ⎜ ⎟⎬ + M C ⎜ ⎟ = −6 ⎜ ⎟ 24EI (6)(5) ⎝ EI ⎠ ⎝ EI ⎠ ⎩⎝ EI ⎠ ⎝ EI ⎠⎭ ⎝ ⎠ ∴ 5M A +20MB + 5(−60) = − 6 (104.167 + 140) ∴ 5M A +20MB = − 1165
