Post on 20-Feb-2017
1
Structural Design and Inspection-Deflection and Slope of Beams
By
Dr. Mahdi Damghani
2016-2017
2
Suggested Readings
Reference 1Chapter 16
3
Objective
• To obtain slope and deflection of beam and frame structures using slope-deflection method
4
Introduction
• Structural analysis method for beams and frames introduced in 1914 by George A. Maney
• This method was later replaced by moment distribution method which is more advanced and useful (students are encouraged) to study this separately
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Slope-Deflection Method
• Sign convention:• Moments, slopes,
displacements, shear are all in positive direction as shown• Axial forces are ignored
xSMdxvdEI ABAB2
2
21
32
1
2
62
2
CxCxSxMEIv
CxSxMdxdvEI
ABAB
ABAB
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Slope-Deflection Method
AA vvdxdvx ,0@
AAABAB
AABAB
EIvxEIxSxMEIv
EIxSxMdxdvEI
62
232
2
BB vvdxdvLx ,@
AAABABB
AABABB
EIvLEILSLMEIv
EILSLMEI
62
232
2
Solving for MAB and SAB
BABAAB vv
LLEIM 322
BABAAB vv
LLEIS 262
BABABA vv
LLEIM 322
BABABA vv
LLEIS 262
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Slope-Deflection Method
• Last 4 equations obtained in previous slide are called slope-deflection equations
• They establish force-displacement relationship
• This method can find exact solution to indeterminate structures
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Slope-Deflection Method
• The beam we considered so far did not have any external loading from A to B
• In the presence of mid-span loading (common engineering problems) the equations become:
FABBABAAB Mvv
LLEIM
322 FABBABAAB Svv
LLEIS
262
FBABABABA Mvv
LLEIM
322 FBABABABA Svv
LLEIS
262
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Fixed End Moment/Shear
• MABF, MBA
F are fixed end moments at nodes A and B, respectively.
• Moments at two ends of beam when beam is clamped at both ends under external loading (see next slides)
• SABF, SBA
F are fixed end shears at nodes A and B, respectively.
• Shears at two ends of beam when beam is clamped at both ends under external loading (see next slides)
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Fixed End Moment/Shear
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Fixed End Moment/Shear
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Example
• Find support reactions.
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Solution
• This beam has 2 degrees of indeterminacy
1- Assume all beams are fixed & calculate FEM
2- Establish Slope-Deflection
equations
3- Enforce boundary
conditions & equilibrium
conditions at joints
4- Solve simultaneous
equations to get slopes/deflection
s
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Solution
kNmMM FBA
FAB 75.0
80.16 kNmMM F
CBFBC 25.1
80.110 kNmMM F
DCFCD 00.1
120.112 2
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Solution
Fijjijiij Mvv
LLEIM
322
Fjijijiji Mvv
LLEIM
322
vi=0 and vj=0 for all cases
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Solution
• Equilibrium moments at the joints
MBA MBCMCB MCD
0ABM
0
0
BCBA
B
MM
M
0
0
CDCB
C
MM
M 0DCM
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Solution
• Substitution into slope deflection equations gives 4 equations and 4 unknown slopes.
• By simultaneously solving the equations
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Solution
• Simply operation of substitution:
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Solution
• Now support reactions can easily be calculated as
15.485.16
85.10.115.15.06
BA
AB
R
R
6 10
25.575.410
75.40.1
4.115.15.010
CB
BC
R
R
12
6.44.712
4.70.1
4.15.012
DC
CD
R
R
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Solution
1.85 4.15 4.75 5.25 7.4 4.6
1.85 8.9 12.65 4.60
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Solution
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Example 2
• Obtain moment reaction at the clamped support B for 6m long beam if support A settles down by 5mm. EI=17×1012 Nmm2
A
4kN/m
B
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Solution
Fijjijiij Mvv
LLEIM
322 Fjijijiji Mvv
LLEIM
322
A
4kN/m5mm
B
F
ABBABAAB MvvLL
EIM 322
1240000005.030220
2LLL
EIM AAB
radLL
EIA
4
3
1033.92
0015.0244000
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Solution
A
4kN/m5mm
B
Fijjijiij Mvv
LLEIM
322 Fjijijiji Mvv
LLEIM
322
radA41033.9
F
BABAABBA MvvLL
EIM 322
1240000005.031033.902 2
4 LLL
EIM BA
mNM BA .3.18705
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Case study
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Case study
t=3.552 mmb=7.792 mm
Fijjijiij Mvv
LLEIM
322
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Case study
Rotation and displacement can be obtained from the
FEM
Wire in composite beam
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Case study
Rotation and displacement can be obtained from the
FEM
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Q1
• Determine the support reactions in the beam shown below.
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Q2
• Calculate the support reactions in the beam shown below.
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Q3
• Determine the end moments in the members of the portal frame shown. The second moment of area of the vertical members is 2.5I while that of the horizontal members is I.
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Q4
• Analyze two span continuous beam ABC by slope deflection method. Then draw Bending moment & Shear force diagram. Take EI constant.