statically indeterminate beam & frame -...
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Statically Indeterminate Beam & Frame
STRUCTURAL ANALYSISBFC 21403
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Introductionwww.uthm.edu.my With Wisdom, We Explore
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Analysis for indeterminate structure of beam and frame:
1. Slope-deflection method
2. Moment distribution method
3. Modified stiffness
Shear force
Moment
The main aim is to calculate
reaction forces
Beam Frame (non-sway/sway)
Displacement
Methods
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The main problem of indeterminate beam is todeterminate support reactions.
A static equilibrium is not enough to solve this problem.
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In reinforced concrete building:
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In steel building:
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In frame:
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In bridge:
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Two method that can be used:
Slope deflection method Moment-distribution method
1. Fixed end moment
2. Moment result deflection and moment deposit (support shift)
3. Slope of support
4. Moment of support
1. Stiffness member
2. Distribution factor
3. Cary over factor
4. Fixed end moment
5. Distribution process
6. Moment of support
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For indeterminate structure, moment of member happenfrom:
i. Fixed end moment
ii. Deflection slope or rotation
iii. Support shift (support settlement)
To form equation of state, member must have uniformand homogeneous among two support.
Redundancy create the value of unknown that related toforce method, e.g. flexibility method.
Deformation can also contribute to the value of unknownthat related to deformation method, e.g. slope-deflectionmethod.
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Consider typical beam BC from continuous beam:
Moment resultant at the end B and C can be identified as:
1. Fixed end moment
(FEM)
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2. Moment of Slope
(MS)
3. Moment of Support Displacement
(MSD)
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FEM: moment resultant at end to end outside taxincidence member that imposed to stated member whenboth supports are assumed as fixed, therefore therotation is zero.
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MS: Moment that occur at different direction of rotation.End gradient member would be positive if the rotation isclockwise. MS can be phased as following:
- End gradient B if C’s end control/fixed
- End gradient C if B’s end control/fixed
∴ The results are:
L
2EIθMS
2
1MS ;
L
4EIθMS
BBCCB
BBC
L
2EIθMS
2
1MS ;
L
4EIθMS
CCBBC
CCB
L
θ2θ2EI
L
2EIθ
L
4EIθMS
L
θ2θ2EI
L
2EIθ
L
4EIθMS
BCBCcerCB
CBCBcerBC
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MSD: Moment that occur due to deformation(displacement) at one end to another end, e.g. B’s endand C constrained.
- Moment of support displacement is generally calculated
using:
where
2DCBC
L
6EIΔMSDMSD
L
Δδ
L
6EIδMSDMSD DCBC
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Moment resultant for beam BC is:
3δθ2θL
2EI
L
PabM
L
6EIδθ2θ
L
EI2
L
PabM
MSDMSFEMM
CB2
2
BC
CB2
2
BC
BCBCBCBC
3δθ2θL
2EI
L
PabM
L
6EIδθ2θ
L
EI2
L
PabM
MSDMSFEMM
BC2
2
CB
BC2
2
CB
CBCBCBCB
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Principle of analysis:
- Equilibrium equation: - Boundary condition:
Fixed end slope, ∆=0
0MM
0MMM
0MM
CBC
BCBAB
ABA
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Example 4.1:
Determine the moment value and shear force to eachsupport and draw shear force diagram (SFD) andbending moment diagram (BMD) for structure beambelow. Assume EI is constant.
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Procedure:
1. Calculate fixed end moment (FEM)2. Determine boundary conditions at support3. Calculate moment resultant for both ends4. Calculate rotation based on equilibrium equation5. Calculate end moments, MAB and MBA
6. Calculate reaction forces7. Calculate shear forces8. Draw shear forces diagram9. Calculate moment from end moments and shear forces
diagram10. Draw bending moment diagram
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E
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Example 4.2:
Determine the moment value and shear force to eachsupport and draw shear force diagram (SFD) andbending moment diagram (BMD) for structure beambelow. Assume EI is constant.
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E
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Tutorial 4.1
Determine the reaction to each support for thecontinuous beam in the figure below. The support at Caccidently constructed 10mm below its intendedposition. Given E=210e6kN/m4 and I=180e-6m4.
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Tutorial 4.2
Determine the moment value and shear force to eachsupport and draw shear force diagram (SFD) andbending moment diagram (BMD) for structure beambelow. Assume EI is constant