Prepared By - bbsbec.edu.inbbsbec.edu.in/wp-content/uploads/2020/01/PPT... · Deflection and Slope...
Transcript of Prepared By - bbsbec.edu.inbbsbec.edu.in/wp-content/uploads/2020/01/PPT... · Deflection and Slope...
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GuljitSingh
CED
BBSBEC, FGS.
Prepared By
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Solving Statically Determinate
Structures
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Static Equilibrium
Asystem of particles is in static equilibrium
when all the particles of the system are at
rest and the total force on each particle is
permanently zero.
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Statically Determinate
A member or structure that can be analyzed and the reactions
and forces determined from the equations of equilibrium.
Statically Indeterminate
A member or structure that cannot be analyzed by the
equations of statics. It contains unknowns in excess of the
number of equilibrium equations available.
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Determinacy
r = 3n , statically determinate
r > 3n , statically indeterminate
where,
n = the total parts of structuremembers
r = the total number of unknown reactive forces and
moment components
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Examples
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RedundantsThe excess members or reactions of an indeterminate structures are called redundants.
Redundant forces are chosen so that the structure is stable and statically determinate when they are removed.
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How do we make an
indeterminate
structure statically
determinate?
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If there is two degrees of indeterminacy, we have to remove two reactive forces, remove three for three degrees and so forth.
By removing excess supports.
By introducing hinges.
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What are the advantages
of statically indeterminate
structures over
determinate structures?
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There are several advantages in designing
indeterminate structures. These include
the design of lighter and more rigid
structures. With added redundancy in
the structural system, there is an increase
in the overall factor of safety.
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Structural Design and Inspection-
Deflection and Slope of Beams
13
By
Guljit Singh
CED
BBSBEC, FGS
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Suggested Readings
14
Reference 1
Chapter 16
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Objective
15
• To obtain slope and deflection of beam and frame
structures using slope-deflection method
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Introduction
16
• 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
17
• Sign convention:
• Moments, slopes, displacements, shear are all in positive direction as shown
• Axial forces are ignored
EI M S x AB ABdx2
d 2v
x2 x3
x2 dv
EIv M AB 2
SAB 6 C1x C2
EI dx
M AB x SAB 2 C1
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Slope-Deflection Method
18
A Adx
@ x 0dv
,v v
dv
EIv M AB 2
SAB 6 EI Ax EIvA
EI dx
M AB x SAB 2 EIA
x2 x3
x2
B Bdx
@ x Ldv
,v v
EIvB M AB 2
SAB 6 EI AL EIvA
EIB M ABL SAB 2 EI A
L2 L3
L2
Solving for MAB and SAB
A B A B AB
LL B A B
ABLL2
AM 2EI
2 3 v v S
6EI 2 v v
LBA L
A B A B M
2EI 23 v v B A B
BALL2
AS 6EI
2 v v
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Slope-Deflection Method
19
• 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
20
• The beam we considered
so far did not have any
external loading from A to B
• In the presence of mid-span loading (commonengineering
problems) the equations become:
F
ABL
A B A B AB
L M
2EI 2
3 v v M F
ABL
B A B AB
v v S
2 6EI
L2 AS
F
LM BA L
A B A B BA 2
3 v v M 2EI F
BAL
B A B BA
v v S
26EI
L2 AS
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Fixed End Moment/Shear
21
• 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
22
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Fixed End Moment/Shear
23
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Example
• Find support reactions.
24
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Solution
25
• 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/deflections
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Solution
26
F F
AB BA8
61.0M M 0.75kNm F F
BC CB8
101.0M M 1.25kNm F F
CD MDC 1.00kNm2
12
121.0M
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Solution
27
F
ijL
i j i j ij
L v v M
3M 2
2EI
F
Lj i j
ji
ji L
i v v M32EI
M 2
vi=0 and vj=0 for all
cases
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Solution
• Equilibrium moments at the joints
MCB MCD
M AB 0
MBA MBC
MB 0
MBA MBC 0
MC 0
MCBMCD 0
M DC 0
28
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Solution
29
• 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:
30
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Solution
31
1.0AB
RBA 6 1.85 4.15
R
• Now support reactions can easily be calculated as
6 10
1.0BC
RCB 10 4.75 5.25
60.51.15
1.85 R
12
1.0
120.51.4 7.4
CD
RDC 12 7.4 4.6
100.51.15 1.4
4.75 R
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Solution
1.85
32
4.15 4.75 5.25 7.4 4.6
1.85 8.9 12.65 4.60
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Solution
33
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Example 2
34
• 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
35
F
ijL
i j i j ij
L v v M
3M 2
2EI F
L
ji L
i j i j ji
32EI v v MM 2
A
4kN/m
5mm
B
F
A B A B ABAB v M
LL
3 vM 2EI
2
12
3 4000 L22EI
M AB 0 L
2A 0 L 0.005 0
LA
3
224EI 4
9.3310 rad
0.0015 4000 L
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Solution
36
4kN/m
B
j iiijL
j ij
F
L v v M
3M 2
2EI j i
Lj
ji
F
ji L i v v M
32EI M 2
5mm 9.33 10 4 radA
A
F
B A A B BABA v M
LL M
2EI 2
3 v
12 4
3 0.005 0
4000 L2
LL
0 9.33 10 2EI
MB A
MB A 18705.3N.m
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Case study
37
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Case study
t=3.552 mmb=7.792 mm
F
ijL
i j i j ij
L v v M
3M 2
2EI
38
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Case study
Rotation and
displacement
can be obtained from the
FEM 39
Wire in compositebeam
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Case study
Rotation and
displacement
can be obtained from the
FEM 40
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Q1
• Determine the support reactions in the beam
shown below.
41
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Q2
• Calculate the support reactions in the beam
shown below.
42
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Q3
• Determine the end moments in the members of
the portal frame shown. The second moment of
areaof the vertical members is 2.5I while that of
the horizontal members is I.
43
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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.
44
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Outline of thepresentation
• Introduction to moment distributionmethod.
• Important terms.
• Signconventions.
• Fixed end moments (FEM)
• Examples;
• (A) example of simply supportedbeam
• (B) example of fixed supported beam with sinking of
support.
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Introduction
• The moment distribution method was first introduced
by Prof. Hardy Cross of Illinois University in 1930.
• This method provides a convinient means of analysing
statically indeterminate beams and rigid frames.
• It is used when number ofreduntants are large and
when other method becomes verytedious.
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Important terms
1. Stiffness
The moment required to produce a unit rotation(slope)
at a simply supported end of a member is called
Stiffness. It is denoted by'K'.
A) Stiffness when both ends arehinged.
B)Stiffness when both ends are fixed.
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Cont...
A) Beam hinged at both ends:
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Cont..
B) Beam hinged at near end and fixed at far end:
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Cont..
Carry over factor(C.O.F):
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Cont..
Distribution factor (D.F.)
● The factor by which the applied moment isdistributed to the member is known as thedistribution factor.
- far-end pined (DF = 1)
● Figure:
●
- far-end fixed (DF = 0)
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Cont..
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Cont....
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SignConventions
A)Support moments :
clockwise moment = +ve
anticlockwise moment =-ve
B)Rotation (slope):
clockwise moment = +ve
anticlockwise moment =-ve
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Cont...
C) Sinking (settlement)
● The settlement will be taken as +ve, if it rotates the
beam as a whole in clockwise direction.
● The settlement will be taken as -ve, if it rotates the
beam as a whole in anti-clockwise direction.
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Fixed EndMoments
● The fixed end moments for the various load casesis
as shown in figure;
● a) for centric loading;
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Cont..
b) for eccentric loading, udl,rotation,sinking of
supports & uvl
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Cont..
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Cont...
● Fixed end moment for sinking of supports:
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Example 1
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Cont..
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Example3
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Structural Analysis -II
Approximate
Methods
Guljit Singh
CED
BBSBEC,FGS.
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Module
III
Approximate Methods of Analysis of Multi-storey Frames
• Analysis for vertical loads - Substitute frames-Loading conditions for maximum positive and negative bending moments in beams and maximum bending moment incolumns
• Analysis for lateral loads - Portal method–Cantilever method– Factor method.
2
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Why approximateanalysis?
• Rapid check on computer aided analysis
• Preliminary dimensioning before exact
analysis
Advantage?
• Faste
r
Disadvantage?
• Results are
approximate
•Approximate methods are particularly useful for
multi-storey frames taller than 3 storeys.
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Approximate analysis for Vertical
Loads
SUBSTITUTE FRAME METHOD
• Analyse only a part of the frame – substitute
frame
• Carry out a two-cycle moment distribution
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Substitute
frame
Actual frame
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• Analysis done for:
• Beam span moments
• Beam support moments
• Column moments
• Liveload positioning for the worst condition
• For the same frame, liveload positions for maximum
span moments, support moments and column moments
may be different
• For maximum moments at different points,
liveload positions may de different
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LL positions for maximum positive span moment at
B
B
Influence line forMB
Deadloads
Liveloads
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LL positions for maximum negative support moment at
A
AInfluence line forMA
Live loads
Deadloads
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LL positions for maximum column moment M1 at
C
C
M1
Liveloads
Deadloads
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LL positions for maximum column moment M2 at
D
M2
D
Liveloads
Deadloads
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Problem 1: Total dead load is 12 kN/m. Total live
load is 20kN/m. Analyse the frame for midspan positive moment onBC.
6 m 6 m 6 m
B C DA
4m
4m
11
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12+20kN/m12
kN/m
12kN/m
B DA 6 m 6 m C 6
m
wl2 1262
FEMAB
36kNm B
A
FEM 36kNm
Fixed end
moments
12
1
23262FEM
96kNm12FEMBC
96kNm
CB
FEMCD FEMDC 36kNm
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Distribution
factors
AB DC
K1 4 EI
6
DF
0.25
DFK1 K2 K3 4EI 64EI 44EI
4
4
BA
K1 4 EI
6DF
0.2K1 K2 K3 K 4EI 44EI 64EI 44EI
6
DFBC DFCD DFCB DFBA 0.2
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A B C D
0.25 0.2 0.2 0.2 0.2 0.25
*
*
*
*
*
*
*
*
*
*
*
** *
*
*
*
*
*
*
DFs
FE
M
Dis
t
CO
Dist
Final
Moments
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A B C D
0.20.
2-36 36 -96 96 -36 36
FEM
0.20.
2
0.25
0.25DF
s9 12 12 -12 -12 -9
Dist
6 4.5 -6 6 -4.5 -6
CO2.25 0.3 0.3 -0.3 -0.3 -2.25 Dist
-18.75 52.8 -89.7 89.7 52.8 18.75 Final
Moments
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89.7kNm 89.7kNmA B32kN m
Midspan positive moment onBC,
32
2
326ME 89.7 32
2
3 54.3
kNm
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Problem 2: Analyse the frame for beam negative moment at B. Moment of inertia of beams is 1.5 times that of columns. Total dead load is 14 kN/m and total live load is 9 kN/m.
6 m 4 m
4m3.5
m
BA
3.5
m
DC
3.5
m3.5
m
17
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14+9kN/m
14+9kN/m 14
kN/m
I
B DA
I
6 m 1.5I
4 m 1.5I
C 4 m 1.5I
wl2 2362
FEMAB
69kNm B
A
FEM 69kNm
Fixed end
moments
12
1
2 2342
30.67kNm
FEMBC FEMCB12
1442
36kNm12FEMCD FEMDC
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Distribution
factors
AB
K1 4 E1.5I 6
DF
0.304K1 K2 K3 4E 1.5I 64EI 3.54EI
3.5
K1
4
1.5I
6BADF
0.209K1 K2 K3
K
1.5I 6 I 3.5 I 3.5 1.5I
4
4
BC
K1 1.5I
4DF
0.313K1 K2 K3
K
1.5I 6 I 3.5 I 3.5 1.5I
4
DFCB 0.284, DFCD 0.284, DFDC
0.396
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A B C D
0.304 0.209 0.313 0.284 0.284 0.396
*
*
*
*
*
*
*
*
*
*
*
*
*
* *
DFs
FE
M
Dis
t
CO
Dist
Final
Moments
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A B C D
0.2090.31
3-69 69 -30.67 30.67 -18.67
FEM
0.304
0.284 0.284 0.396DFs
20.98 -8.01 -12 -3.41
Dist
10.49 -1.71
CO-1.84 -2.75 Dist
69.64 -47.13 Final
Moments
B 69.64kNm47.13kNm
Max. beam negative moment at B 69.64
kNm
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Approximate analysis for Horizontal
Loads
1. Portal method
2. Cantilever
method
3. Factor method
22
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PORTAL METHOD
Assumptions
1. The points of contraflexure in all the members lie at
their midpoints.
2. Horizontal shear taken by each interior column is
double that taken by each exterior column.
Horizontal forces are assumed to act only at thejoints.
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B C DAP1
P
P
2P
2P
2P
2P
P
P
FE GH
P2
Q 2Q 2Q Q
Q 2Q 2Q Q
J K LI
24
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1P P2P2P
P6 P
P1
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P1P2Q2Q2QQ6
Q P1P2
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Problem 3: Analyse the frame using portal
method.
B C DA120 kN
7 m 3.5m
5 m
FE G
3.5
m
180kN H
3.5
m
J K LI
27
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Horizontal shears:
16For the top storey, P P
2P 2P P P 120
20kNFor the bottom storey, Q P1
P2
120180
50kN6 6
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120kNMoments:
35kNm
A3.5m
1.7
5m 35kN
m10k
N
20k
N
35kNm 35kNm
B
10k
N
70kNm10kN
40kN
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Beammoments:
B C D35
A35
35
35
35 35
FE G122.5
122.5
122.5
122.5
122.5
H 122.5
J K LI
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Columnmoments:
C DB 70A 35 kNm
70
35kNm
FE GH
3587.
5
87.5
175
3570 7017
5
J KI87.5
L87.517
5175
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Beam and Columnmoments:
35
D35
35B 3
570
3535
357
0
35
87.5
175
70
3570 17
5
122.5
122.5
122.5
35
87.5
122.5
122.5
122.5
87.5
87.5
175
175
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Homework
40kN
B CA
F
3.5
m
80kN
5m
I
5 m
D
7.5 m
E
G H
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CANTILEVER
METHOD
• Frame considered as a vertical cantilever
Assumptions
1. The points of contraflexure in all the members lie
at their midpoints.
2. The direct stresses (axial stresses) in the columns
are directly proportional to their distance from the
centroidal vertical axis of the frame.
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P1
y1
y2 y3
y4
P2
A1 A2A3 A4
Centroidal verticalaxis of the frame
Area of
cross
section
y A1d1 A2d2 A3d3 A4d4
To locate centroidal verticalaxis of the frame,
A1 A2 A3 A4
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V1 V2 V3 V4
xI
My
M
Iis constant at a given height (of the ‘verticalcantilever’).
1 2
3 4
y1 y2 y3 y4
V1 A1
V2 1y1
A2 V3 A3
V4
A4
y2 y3
y4
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P1
m1
m2
H1 H2 H3 H4
h
2 B
V1 V2
V3
V4
B 1 2 1 1 2 2 3 3 4
4
M P hVm V m V m V
m
2
From 1and 2, V1,V2 ,V3,V4 can be
found.
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P1
H1 H4H2
H3
P1 H1 H2 H3 H4
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Problem 4: Analyse the frame using cantilever
method, if allthe columns have the same area of crosssection.
B C DA120 kN
7 m 3.5m
5 m
FE G
3.5
m
180kN H
3.5
m
J K LI
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To locate centroidal vertical axis of theframe,
4y
A1 0 A1 7 A1 10.5 A1 15.5
33 8.25mA1 A1 A1 A1
120
8.251.2
52.2
5
7.2
5
180
Also,V1 A1
V2 A2 V3 A3
V4
A4
V1
V2 V3
V48.25 1.25 2.25
7.25
8.25 1.25 2.25
7.25
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1 1 1
2 3 48.2
5
8.2
5
8.2
5
1.25V 2.25V 7.25VV
, V
, V
P1
m1
m2
H1 H2 H3 H4
h
2 O
V1 V2 V3 V4
O 1 2 1 1 2 2 3 3 4
4
M P hVm V m V m V
m
For the top
storey,
1 2 3
42120
3.5 V 15.5 V 8.5 V 5V
0
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12120 3.5
V 15.5 1.25V1
8.52.25V15
Dr.RajeshKNDept. of CE, GCE Kannur
8.25
8.25
V1 13.615kN
28.2
5
V 1.2513.615
2.063kN,
38.2
5
V 2.2513.615
3.713kN,
48.2
5
V 7.2513.615
11.965kN
Check : 13.6152.0633.71311.965
0
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H1 H2 H3 H4
V4
O
V1 V2 V3
For the bottom
storey,
1 2 3
42O
M 1203.53.5180
3.5V 15.5V 8.5V 5V
0
2
Dr.RajeshKNDept. of CE, GCE Kannur
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1 11
3.
5
3.
5 2
1.25V
Dr.RajeshKNDept. of CE, GCE Kannur
2.25V 120
180
V 15.5
8.5
5
3.5
2
8.25
8.25
28.2
5
V1 61.267kN
V 1.2561.267
9.283kN,
38.2
5
V 2.2561.267
16.709kN,
4V
7.2561.267
53.841kN8.2
5Check : 61.2679.28316.70953.841
0
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Moments:
47.652kN
mA120k
N3.5m
1.7
5m 47.652kN
m13.615k
N
27.3kN
13.615k
N
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Beam and Columnmoments:
29.9
D47.6
27.4B 29.
97547.
647.6
27.4 57.3 29.9
74.8
187.8
57.3
29.9
75
143.2
166.8
96
104.7
47.6119.
2
166.8
96
104.7
119.2
74.8
187.8
143.2
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25kN
Homework
A B
6 m
DC50kN
3.5
m3
.5m
FE55kN
4.5
m
HG
47
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FACTOR
METHOD
• More accurate than Portal and Cantilever methods
• Specially useful when moments of inertia of
various members are different.
Basis:
• At any joint the total moment is shared by all the
members in proportion to their stiffnesses
• Half the moment gets carried over to the far end
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Girder and column factors:
• Relative stiffness of amember
k I
L
Girder factor at a joint
g k, of all columns meeting at the
joint
k, of all members meeting at the
joint
Column factor at a jointc 1
g
k, of all beams meeting at the joint
k, of all members meeting at the
joint
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Moment factor for a member
C cmk, for a
columnG gmk, for abeam
where cm chalf of column facor of far
end and gm ghalf of
girder facor of far end
C sum ofcolumn moment factors for a
storey
G sum ofbeam moment factors for a joint
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Problem 5: Analyse the frame using factor
method.
40kN
H IG
5 m 7.5m
FED
3.5
m
80kN
5m
B CA
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Total column moment
above DEF 403.5
140kNm
Total column moment above ABC 408.5805
740kNm
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1 2 3 4 5 6 7 8 9 10 11 12JOINT MEMBER k=I/L Ʃk FACTOR c/2,
g/2
fro
m
far
end
5+6 MOMEN
T
FACTO
R
Tot
al
Col
.
Mom, MT
Col
.
Mo
m,
MC =
MT
×
C/Ʃ
C
DFB
=G/ƩG
Beam
M
o
=
M
D
F
m
C
×
B
Col Beam Col Beam c
=Ʃk(b
e
ams)/
Ʃ k
g
=Ʃk(
c
olum
n
s)/Ʃk
cm gm C=
cm
×
k
G =
gm
×
k
D
DA 0.2
0.686
0.29 0.5 0.79 0.15
8
740 99.4
DE 0.2 0.71 0.3 1.01 0.202 1 122 .6
DG 0.286 0.29 0.3 0.59 0.16
9
140 23.2
E
ED 0.2
0.819
0.59 0.36 0.95 0.19 0.59 83.2 5
EH 0.286 0.41 0.27 0.68 0.19
4
140 26.6
EF 0.133 0.59 0.4 0.99 0.132 0.41 57.8 5
EB 0.2 0.41 0.5 0.91 0.18
2
740 114.5
F
FE 0.133
0.772
0.79 0.3 1.09 0.145 1 13
3
.0
3
FI 0.286 0.21 0.16 0.37 0.10
6
140 14.6
FC 0.2 0.21 0.5 0.71 0.14
2
740 89.3
GGD 0.286
0.4860.59 0.15 0.74 0.21
2
140 29.1
GH 0.2 0.41 0.23 0.64 0.128 1 29. 1
H
HG 0.2
0.772
0.46 0.21 0.67 0.134 0.559 16. 5
HE 0.286 0.54 0.21 0.75 0.21
5
140 29.5
HI 0.133 0.46 0.34 0.8 0.106 0.441 13. 0
storey,C 0.1580.1820.1420.230.2420.26
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Homework
B CA120 kN
6 m 6 m
4m
FED60kN
6m
H IG
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Summar
y
Approximate Methods of Analysis of Multi-storey Frames
• Analysis for vertical loads - Substitute frames-Loading conditions for maximum positive and negative bending moments in beams and maximum bending moment incolumns
• Analysis for lateral loads - Portal method–Cantilever method– Factor method.
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Influence Line for Reaction , Moment & Shear for
Indeterminate Structure
Guljit Singh
CED
BBSBEC, FGS.
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Influence Lines
Variation of
Reaction, Shear, Moment or
Deflection
at a SPECIFIC POINT
due to a concentrated force moving on
member
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SIGNIFICANCE
Influence lines are important in the design of structures that
resist large live loads.
If a structure is subjected to a live or moving load, the variation in
shear and moment is best described using influence lines.
Once the influence line is drawn, the location of the live load
which will cause the greatest influence on the structure can be
found very quickly
Influence Lines
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Influence Lines
As the car moves across the bridge, the forces
in the truss members change with the position of
the
car and the maximum force in each member will
be at a different car location.
The design of each
member must be based on the
maximum probable load each
member will experience
If a structure is to be safely designed, members
must be proportioned such that the maximum
force produced by dead and live loads is less than
the available section capacity.
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Structural analysis for variable loads
consists of two steps:
1.Determining the positions of
the loads at which the response function is
maximum;
AND
2.Computing the maximum value of the
response function.
Influence Lines
Response Function = support reaction, axial force,
shear force, or bending moment.
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Influence Lines
INFLUENCE LINE VS SFD/BMD
shear and moment diagrams represent the effect of
fixed loads at all points along the member.
Influence lines represent the effect of a moving load
only at a specified point on a member
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TYPES OF INFLUENCE LINES
Reaction I.L.
Shear I.L.
Moment I.L.
Floor Girder I.L.
Truss Bar force I.L.
Influence Lines
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Influence Lines
Structure type
Indeterminate
Determinate
Influence lines
For Determinate Structure
For Indeterminate Structure
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Methods of constructing the shape of Influence Lines
Tabulation Method.
Muller –Breslau principles.
Influence Lines
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Muller-Breslau Principle
Influence Lines
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Influence Lines
Indeterminate VS Determinate
Influence lines for statically determinate structures are always piecewise linear.
For indeterminate structures, the influence line is not straight lines!
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Influence Lines
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Influence Lines