chapter 5 the analysis of beams & frames 5.1 introduction
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CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction - PowerPoint PPT PresentationTRANSCRIPT
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CHAPTER 5THE ANALYSIS OF BEAMS & FRAMES
5.1 IntroductionFor a two-dimensional frame element each node has the capability of translating in two directions and rotating about one axis. Thus each node of plane frame has three degrees of freedom. Similarly three structure forces (vertical force, shear force and bending moment) act at a node.For a two-dimensional beam element each node has two degrees of freedom (one rotation and one translation). Similarly two structure forces (vertical force and a bending moment) act at a node. However in some structures a node has one degree of freedom either rotation or translation. Therefore they are subjected to a moment or a force as the case may be.The structure stiffness matrices for all these cases have been developed in the previous chapter which can be summarized as under:-
i) Beams and frames subjected to bending momentii) Beams and frames subjected to shear force and bending momentiii) Beams and frames subjected to shear force, bending moment and axial
forces
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Following steps provide a procedure for the determination of unknown deformation, support reactions and element forces (axial forces, shear forces and bending moment) using the force displacement relationship (W=K). The same procedure applies both to determinate and indeterminate structures.
5.2 PROCEDURE TO ANALYSE BEAMS AND FRAMES USING DIRECT STIFFNESS METHOD
5.2.1. Identifying the components of the structural system or labeling the Structures & Elements.As a first step, divide the structure into some finite number of elements by defining nodes or joints. Nodes may be points of supports, points of concentrated loads, corners or bends or the points where the internal forces or displacements are to be determined. Each element extends between the nodes and is identified by arbitrary numbers (1,2,3).
a) Structure Forces and DeformationsAt a node structure forces are assumed to act in their positive direction. The
positive direction of the forces is to the right and upward and positive moments and rotations are clockwise. Start numbering the known forces first and then the unknown forces.
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Structures Forces not acting at the jointsStiffness method is applicable to structures with structure forces acting at
nodes only. However if the structure is subjected to concentrated loads which are not acting at the joints or nodal points or if it is subjected to distributed loads then equivalent joint loads are calculated using the following procedure.
i) All the joints are considered to be fixed. [Figure-5(b)]ii) Fixed End Moments (FEM’s) and Reactions are calculated using the
formulae given in the table as annex-Iiii) If more than one FEM and reactions are present then the net FEM and
Reaction is calculated. This is done by algebric summation. [Figure-5(c)]Equivalent structure forces or loads at the joints/nodes are obtained by
reversing the signs of net FEM’s & Reactions. [Figure-5(d)]orReversing the signs of Net FEM’s or reaction gives the equivalent structure
loads on the joints/nodes.v) Equivalent element forces are calculated from these equivalent structure
loads using equation 5.2, 5.3 and 5.4 as explained in article number 5.vi) Final element forces are obtained by the following equation
w = wE + wF
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wherewE = Equivalent element forceswF = Element forces while considering the elements to be fixed.
FAB
FAB
M
M
Equivalent Joint Loads
Fixed End Moments & Reactions
Fixed End Moments & Reactions
Actual Structure
Fig.5.1
FBA
( + )R
( + )
1R
R1
( M
R2 3
FBC+
R2
M
R3
R4
FCBM
4R
(d)
A B C
FBA
FBA
RR1
( M
FAB M
M+
2
FBC
3R
FBCM
FCBM
R4
(c)
(b)
FCBM
(a)
Net
(wf)
(WE)M
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b) Element ForcesSpecify the near and far end of each element. Draw free body diagram of each
member showing its local co-ordinate and element forces. Arbitrary numbers can identify all the element forces of the structure.
5.2.2. Calculation of Structure Stiffness Matrices of the membersProperties of each element like its length, cross-sectional area, moment of inertia, direction cosines, and numbers identifying the structure forces acting at its near and far ends can be systematically tabulated. Using values of these parameters in equation 4.53 structure stiffness matrix of each member can be formed by applying equation 4.54,4.55 and 4.56 depending upon the situation.
5.2.3. Formation of Structure Stiffness Matrix of the Entire StructureAccording to the procedure discussed in chapter 3 article 3.1.3 stiffness matrix [K] of the entire structure is formed.
5.2.4. Calculation of Unknown Structure Forces and DisplacementsFollowing relation expresses the force-displacement relationship of the structure in the global coordinate system:[W] = [K] []
Where[W] is the structure load vector[K] is the structure stiffness matrix[] is the displacement vector
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Partitioning the above equation into known and unknown portions as shown below:
WW
K KK K
k
u
u
k
11 12
21 22
WhereWk = known loadsWu = unknown loadsu = unknown
deformationk = known
deformationExpansion of the above leads to the following equation.[Wk] = [K11] [u] + [K12] [k] --------------------- (A)[Wu] = [K21] [u] + K22 k --------------------- (B)As k = 0So, unknown structure displacement [u] can be calculated by
solving the relation (A), which takes the following form. [u] = [K11]-1 [Wk] ---------------------- (C)
Unknown structure force i.e. reactions can be calculated by solving equation B which takes the following form
Wu = [K21] [u] ---------------------- (D)
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5.2.5. Calculation of element forces:Finally element forces at the end of the member are computed using the following equation (E).w = k = Tw = kT --------------- (E)
where [w] is the element force vector[kT] is the product of [k] and [T] matrices of the element
where [w] is the element force vector[kT] is the product of [k] and [T] matrices of the element[] is the structure displacement vector for the element.Following are the [kT] matrices for different elements used in the subsequent examples.
Case-I Beam/frame subjected to bending moment only
LEI
LEI
LEI
LEI
kT 42
24
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Case-II Beams subjected to Shear Forces & Bending Moment
3322
3322
22
22
121266
121266
6642
6624
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
kT
Case-IIIFor frame element subjected to axial force, shear force and bending
moment.
LlAE
LlAEm
LAEm
LAE
LlAE
LlAEm
LAEm
LAE
LmEI
LmEI
LlEI
LlEI
LEI
LEI
LmEI
LmEI
LlEI
LlEI
LEI
LEI
LmEI
LmEI
LlEI
LlEI
LEI
LEI
LmEI
LmEI
LlEI
LlEI
LEI
LEI
kT m
..00
..00
.12.12.12.1266
.12.12.12.1266
.6.6.6.642
.6.6.6.624
333322
333322
2222
2222
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Plotting bending moment and shearing force diagrams:
Bending moment and shearing force diagrams of the structure are plotted using the element forces calculated in step-5. Examples on the next pages have been solved using the above-mentioned procedure.
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5.3 ILLUSTRATIVE EXAMPLESEXAMPLE 5.1 Solve the beam shown in the figure using stiffness method. Solution
6'
w ,
W , W , W ,
Structure loads and deformations
w ,w ,1 1
1
2 2
11
w , w ,4 4
Element forces and deformation
3 3
2
5
2 2 3 3
5'
10k
5'
2kip/ft
15' 6'
10k
W ,
w ,
3
5 6 6
4 4
Numbering element and structure forces
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3939
5.375.375.125.12
' sFEF10k
2k/ft 2k/ft10k
Fixed end moments
-12.5k'
-37.5k'
-3912.5k'
37.5k'
39
Calculating fixed end moments and equivalent joint loads
395.1
255.12
Moments End FixedNet W
4
3
2
1
F
F
F
F
WWWW
F
39
Net fixed end moment
-12.5 -25 -1.5
1 32
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395.1
255.12
4
3
2
1
E
E
E
E
WWWW
W K
12.5 25 1.5
1 2 3
Equivalent joint Moments
-39
Equivalent joint loads are :
Calculating structure stiffness matrices of elementFollowing table lists the properties needed to form structure stiffness
matrices of elements.
Member Length (ft) I J
1 10 1 2
2 15 2 3
3 12 3 4
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Structure stiffness matrices are:
2
1
2
1
1
2 1 2 1
4.02.02.04.0
EI 4224
10
EIK
3
2
3
2
2
3 2 3 2
267.013.013.0267.0
EI 4224
15
EIK
4
3
4
3
3
4 3 4 3
34.0167.0
167.034.0EI
4224
12EIK
Forming structure stiffness matrix of the entire structureUsing relation [K] = [K]1 + [K]2 + [K]3 structure stiffness matrix of the entire structure is:
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4321
33.0167.000167.034.0267.013.00
013.0267.04.02.0002.04.04 3 2 1
EIK
4321
33.0167.000167.0597.013.00013.0667.02.0002.04.0
EIK
4 3 2 1
Finding unknown deformations Unknown deformations are obtained by using the following equation
] = [K]-1 [W]
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736.132
35
85.24
8.18
1
39
5.1
25
5.12
56.305.125.0123.0
05.108.249.024.0
25.049.088.194.0
123.024.094.097.2
1
39
5.1
25
5.12
33.0167.000
167.0597.013.00
013.0667.02.0
002.04.0
1
4
3
2
1
4
3
2
1
1
4
3
2
1
EI
EI
EI
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Calculating element forcesUsing relation [w]=[kT][] we get
88.39
27.10
67.12
325.11
7.13
5.12
1
132.74-
35
24.85
18.8
34.0167.000
167.034.000
0267.013.00
013.0267.00
004.02.0
002.04.0
6
5
4
3
2
1
EIEI
w
w
w
w
w
w
E
E
E
E
E
E
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Actual forces on the structure are obtained by superimposing the fixed end reactions on above calculated forces.
01.50
1.502.26
2.260
08.3927.10
67.12325.117.135.12
39395.375.37
5.125.12
6
5
4
3
2
1
6
5
4
3
2
1
6
5
4
3
2
1
E
E
E
E
E
E
F
F
F
F
F
F
wwwwww
wwwwww
wwwwww
01.50
1.502.26
2.260
08.3927.10
67.12325.117.135.12
3939
5.375.37
5.125.12
wwwwww
wwwwww
wwwwww
E6
E5
E4
E3
E2
E1
F6
F5
F4
F3
F2
F1
6
5
4
3
2
1
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EXAMPLE 5.2Analyse the frame shown in the figure using stiffness method.
Solution
2kip/ft 50k'
10'
W ,
W ,
W ,
1 1
2 2
3 3
w ,w ,1 12 2 w ,3 3
w ,4 4
1
2
Structure forces and deformations
Element forces and deformations
20'
Numbering element forces and deformations
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0067.6667.66
'
4
3
2
1
F
F
F
F
wwww
sFEM
-16.67k'
Equivalent joint moment
66.67k'
Net fixed end moment
2kip/ft
Finding Fixed end moments and equivalent structure load
067.66
67.66 moments end fixedNet
3
2
1
F
F
F
F
WWW
W
Equivalent Joint Loads
67.161 EW
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Calculating structure stiffness matrices of elements.Following table shows the properties of the elements required to form structure stiffness matrices of elements.
Members Length (ft) i j
1 20 2 12 10 1 3
Structure stiffness matrices of both elements are:
12
2.01.01.02.0
12
4224
201
EIEIK
3
1
4.02.0
2.04.0
3
1
42
24
102
EIEIK
2 1 2 1
1 3 1 3
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Forming structure stiffness matrix for the entire structure.Structure stiffness matrix for the entire frame is obtained using relation[K] = [K]1 + [K]2
3
2
1
4.002.0
02.01.0
2.01.06.0
EIK
finding unknown deformationUnknown deformation can be calculated by using equation[]u = [K11]-1 [W]k
1 2 3
This can be done by partitioning the structure stiffness matrix into known and unknown deformations and forces
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k
u
u
k
KK
KK
W
W
2221
1211
EIEI
WW
E
E1
00
4.002.002.01.02.01.06.067.16 1
3
2
EIEI
78.27667.166.0
111
[u] = [K11]-1[Wu]
Finding Unknown ReactionsUnknown reactions can be calculated using the following equation.
uW uK 21
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556.578.2
78.27/12.01.0
3
2
3
2
E
E
E
E
WW
EIEIWW
W=WF+WE
56.545.69
556.578.2
067.66
3
2
WW
Calculating element forces(Moments)Using relation
kTw
56.511.1156.578.2
1
027.78-
0
4.02.002.04.00
02.01.001.02.0
4
3
2
1
EIEI
wwww
E
E
E
E
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OR
56.511.11
/10
78.274.02.02.04.
56.578.2
/178.27
02.01.01.02.0
4
3
2
1
EIO
EIww
EIEIww
E
E
E
E
Actual forces acting on the structure can be found by superimposing the fixed end reactions on the forces calculated above.
EF www
56.511.11
11.6145.69
00667.66667.66
56.511.1156.578.2
4
3
2
1
wwww
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1
2
69.45 61.11 11.11
5.56
69.45
61.11
11.11
5.56BMDFinal element forces
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EXAMPLE 5.3Analyse the shown beam using direct stiffness method.Beam subjected to shear and moment.
15'
2kips
10'
5kip/ft (0.41667kip/inch)
W1 W2 W3
w1
W5W4
w7
w5
w8
w6
W6
w4w3
w2
1 2
Structure forces and deformations
Element forces and deformations
STEP-1 Numbering the forces and deformations
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5.375.37
1125112511
3030
"
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
wwwwwwww
sFEF
37.5k
Fixed end forces
30k"
37.5k
5kip/ft (0.41667kip/inch)
1k 1k
1125k"
2kips 30k"
-1125k"
Finding fixed end forces and equivalent joint loads
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30k" 1095k"
Net fixed end forces
[W]F=Net fixed end moments and forces =
5.375.38
111251095
30
6
5
4
3
2
1
F
F
F
F
F
F
WWWWWW
Equivalent joint loads
30k" 1095k"
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Equivalent joint loads =
1095
30
2
1
E
E
WW
Calculating Structure Stiffness Matrices of ElementsFollowing table shows the properties of the elements required to form structure stiffness matrices of elements
ML I J K L
1 120 1 2 4 52 180 2 3 5 6
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From structural elementsMember-1
1 2 4 5
5421
00000694.000000694.000041667.000041667.000000694.000000694.000041667.000041667.0
00041667.000041667.0033333.0016667.000041667.000041667.0016667.00333.0
1
EIK
Member-2 2 3 5 6
6532
00000206.000000206.0000185.0000185.000000206.000000206.0000185.0000185.0
000185.0000185.00222.00111.0000185.0000185.00111.00222.0
2
EIK
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Forming structure stiffness matrix for the entire structureStructure stiffness matrix for the entire beam is obtained using the relation[K] = [K]1 + [K]2
654321
00000206.00000206.00000185.000018519.0000000206.0000009.000000694.0000185.000023148.000041667.0
0000000694.000000694.0000041667.000041667.0000185.0000185.000222.00111.00
00018519.000023148.000041667.00111.00555.0016667.0000041667.000041667.00016667.00333.0
6 5 4 3 2 1
EIK
Finding unknown deformationsUnknown deformation can be calculated using equation[u] = [K11]-1 [Wk]This can be done by partitioning the structure stiffness matrix into known and unknown deformations and forces.
k
u
u
k
KKKK
WW
2221
1211
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6
5
4
3
2
1
6
5
4
3
2
1
00000206.00000206.00000185.000018519.0000000206.0000009.000000694.0000185.000023148.000041667.0
0000000694.000000694.0000041667.000041667.0000185.0000185.000222.00111.00
00018519.000023148.000041667.00111.00555.0016667.0000041667.000041667.00016667.00333.06 5 4 3 2 1
EI
WWWWWW
E
E
E
E
E
E
Using Equation [u] = [K11]-1 [Wk]
5886.22870
2946.105351
1095
30
0555.0016667.0
016667.00333.011
2
1
EIEIFinding Unknown ReactionsUnknown reactions can be calculated using the following equation[W]u = [K]21 []u
235.4
91.0
14.5
86.253
1
5886.22870
2946.10535
00018519.00
00023148.000041667.0
00041667.000041667.0
0111.00
6
5
4
3
EIEI
W
W
W
W
E
E
E
E
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W = WE + WF
735.41
41.39
14.4
86.1378
5.37
5.38
1
1125
235.4
91.0
14.5
86.253
6
5
4
3
W
W
W
W
Since all the deformations are known to this point we can find the element forces in each member using the relation [w]m = [kT]m []m
Member-1:
1397.51397.57647.586
30
0058863.22870
2946.105351
00000694.000000694.000041667.000041667.000000694.000000694.000041667.0000416667.0
00041667.000041667.0033333.0016667.00001667.000041667.0016667.00333.0
4
3
2
1
EIEI
wwww
E
E
E
E
Superimposing the fixed end forces for member-1 on the above w’s we get
w = wE + wF
![Page 35: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/35.jpg)
1397.61397.4764.616
0
11
3030
1397.51397.57647.586
30
4
3
2
1
wwww
73529.412647.33
117.1379764.616
5.375.37
11251125
23529.423529.41176.2542353.508
8
7
6
5
wwww
23529.423529.41176.2542353.508
000
5886.22870
1
00000206.000000206.0000185.0000185.000000206.000000206.0000185.0000185.0
000185.0000185.00222.00111.0000185.0000185.00111.00222.0
8
7
6
5
EIEI
wwww
E
E
E
E
FE www
Member 2 :
Superimposing the fixed end forces
![Page 36: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/36.jpg)
OR
EI
EI
wwwwwwww
w
E
E
E
E
E
E
E
E
E1
0000
5886.228702946.10535
00000206.000000206.00000185.0000185.0000000206.000000206.00000185.0000185.00
000185.0000185.000222.00111.00000185.0000185.000111.00222.00
0000000694.000000694.0000041667.000041667.0000000694.000000694.0000041667.000041667.0000041667.000041667.000333.001667.0000041667.000041667.0001667.00333.0
8
7
6
5
4
3
2
1
24.424.421.25442.508
14.514.553.586
30
8
7
6
5
4
3
2
1
E
E
E
E
E
E
E
E
E
wwwwwwww
w
w = wE +wF
![Page 37: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/37.jpg)
5.375.37
11251125113030
24.424.421.25442.508
14.514.553.586
30
8
7
6
5
4
3
2
1
wwwwwwww
w
74.4126.33
2.137958.616
14.614.453.616
0
8
7
6
5
4
3
2
1
wwwwwwww
w
![Page 38: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/38.jpg)
Shear Force Diagram Bending Moment Diagram
-4.139-6.139
33.2647
-41.735-248.34
-616.7647
711.432
-1379.117
1 2-4.139
0 616.7647k" 1379.177kip"-616.7647k"
-6.139 33.26 41.735
- - - -++
![Page 39: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/39.jpg)
10'
5'
5'
3k/ft
4k
A
B C
D
6'
EXAMPLE 5.4Analyse the frame by STIFFNESS METHOD
![Page 40: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/40.jpg)
W11,11
W10,10
W12,12
W3,3
W2,2 W5,5
W6,6W1,1 W4,4
W7,7W9,9
W8,8
Structure Forcesand Deformation
![Page 41: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/41.jpg)
w11,11
w9,9 w10,10
w8,8w7,7
w12,12
w3,3
w1,1
w5,5
w6,6
w4,4
w2,2
w17,17
w15,15
w13,13
w14,14w16,16
w18,18
Element Forcesand Deformations
13
2
![Page 42: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/42.jpg)
3k/ft
4k
2k
2k
60k"
60k"
300k" 300k"
15k 15k
1
2
3
00000000
151530030022006060
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
wwwwwwwwwwwwwwwwww
sFEM'
Fixed End Moments and reactions
![Page 43: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/43.jpg)
20600000
15300
215240
12
11
10
9
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
F
F
F
F
WWWWWWWWWWWW
2k240 K"
15k
0
15k
300 K"
Net fixed end moments
[W]F=
![Page 44: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/44.jpg)
2k
240 K"
15k
0
15k
300 K"
0
Equivalent joint loads
[W]E=
0015300215
240
7
6
5
4
3
2
1
E
E
E
E
E
E
E
WWWWWWW
![Page 45: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/45.jpg)
Finding the structure stiffness matrices of elements.Next we will calculate the structure stiffness matrices for each element using the properties of members tabulated in below where E = 29000 ksi I = 100 inch4 and A = 5 inch2 (same for all members):
Member Length(in.)
l m I J K L M N
1 120 0 1 10 1 11 2 12 3
2 120 1 0 1 4 2 5 3 6
3 72 0 -1 4 7 5 8 6 9
![Page 46: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/46.jpg)
3122
111
10
13889.2013889.2000333.1208333.120813889.2013889.2000333.1208333.1208
00333.1208333.12080000333.1208333.120800
333.1208333.120800667.96666333.48333333.1208333.120800333.48333667.96666
1
K
For member-1 we have
10 1 11 2 12 3
![Page 47: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/47.jpg)
For member-2
635241
333.1208333.12080000333.1208333.12080000
0013889.2013889.20333.1208333.12080013889.2013889.20333.1208333.120800333.1208333.1208667.96666333.4833300333.1208333.1208333.48333667.96666
2
K
1 4 2 5 3 6
![Page 48: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/48.jpg)
For member-3 4 7 5 8 6
9
968574
2356.932356.9300481.3356481.33562356.932356.9300481.3356481.3356
00889.2013889.20130000889.2013889.201300
481.3356481.335600111.161111556.80555481.3356481.335600556.80555111.161111
3
K
Finding structure stiffness matrix for the entire frame.Using relation [K] = [K]1+[K]2+[K]3 we get the following structure stiffness matrix.
(See next slide)
![Page 49: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/49.jpg)
139.203.1208139.203.1208
3.1208
3.120867.9666633.12083.48333
048.33562356.9348.3356
0189.20130089.20130
48.33561.16111148.3356056.80555
2356.9348.3356568.130148.33563.1208
00003.20343.1208139.203.1208
48.3356056.8055548.33563.120878.2577773.12083.48333
139.203.12083.1208003.1208
3.12080138.203.120847.12283.1208
3.12083.483333.12083.483333.12083.193333
121110987654321
12 11 10 9 8 7 6 5 4 3 2 1
000000000000000003.12080
000000000002356.930000000000000000000000000089.201300000
0000047.1228000000
0000033.1208
K
![Page 50: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/50.jpg)
Finding unknown deformationsUnknown deformation can be calculated using equation [u] = [K11]–1[Wk]This can be done by partitioning the structure stiffness matrix into known
and unknown deformations and forces.
Unknown deformations can be calculated using the equation ;[]u=[K11]-1[W]kSolving the above equation we get ,
k
u
u
k
2221
1211
KKKK
WW
0015300215
240
111.161111481.33560556.80555000481.3356568.130104814.3356333.120800
000278.2034333.1208013889.20333.1208556.80555481.3356333.1208778.2577770333.1208333.48333
0333.120800722.12280333.12080013889.20333.120804718.1228333.120800333.1208333.48333333.1208333.1208333.193333
7654321
1
![Page 51: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/51.jpg)
Finding Unknown reactionsUnknown reactions can be calculated using the following equation:
001535.037023.007682.001527.
039859.01202.
00184.
7654321
42062.15242.147704.4042501.34707.15
001535.0037023.0007682.0001527.0
039859.001202.0
00184.0
000013889.200333.120800000333.120800000333.12080333.48333
481.33562356.930481.335600000889.20130000
12
11
10
9
8
21
E
E
E
E
E
uu
WWWWW
KW
![Page 52: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/52.jpg)
Now we will superimpose the fixed end reactions on the above calculated structure forces. [W]=[W]E+[W]F
58.052.1423.1942.347.15
206000
42.152.1477.4042.347.15
12
11
10
9
8
WWWWW
Finding the unknown element forces.Up to this point all the deformations are known to us, we can find the element forces in each element using relation [w]m = [kT]m[]m For member-1
039859.0001202.00
00184.00
00333.1208333.12080000333.12083330.12080013889.2013889.2000333.1208333.1208
13889.2013889.2000333.1208333.1208333.1208333.120800667.96666333.48333333.1208333.120800333.48333667.96666
4
3
6
5
2
1
wwwwww
![Page 53: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/53.jpg)
Equation givesw1 = 40.77 kips-inch c.w.w2 = 129.704 kips-inch c.w.w5 = –1.421 kips Downwardw6 = 1.421 kips Upwardw3 =14.524 kips Rightward w4 = –14.524 kips LeftwardSuperimposing the fixed end reactions in their actual direction we get w1 = 41.0269 – 60 = -18.973 kips-inch c.c.w.w2 = 130.2173 + 60 = 190.2173 kips-inch c.w.w5 = 1.427036 – 2 = -3.1427036 kips Rightward w6 = -1.427036 – 2 =0.57296 kips Rightwardw3 = 14.5289 kips Upwardw4 = -14.5289 kips Downward
![Page 54: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/54.jpg)
For member-2
0370233.00398595.00076822.0012024.0001528.0
001845.0
333.1208333.12080000333.1208333.12080000
0013889.2013889.20333.1208333.12080013889.2013889.20333.1208333.120800333.1208333.1208667.96666333.4833300333.1208333.1208333.48333667.96666
12
11
10
9
8
7
wwwwww
Above equation givesw7 = 109.782 kips-inch c.w.w8 = -53.25353 kips-inch c.c.w.w9 = - 0.47048 kips Downwardw10 = 0.47048 kips Upwardw11 = 3.427 kips Rightward w12 = -3.427 kips Leftward
![Page 55: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/55.jpg)
Similarly for member-2 we have to superimpose the fixed end reactionsw7 = 109.782 – 300 = -190.218 kips-inch c.c.w.w8 = –53.25353 + 300 = 246.746 kips-inch c.w.w9 = –0.471071+15 = 14.5289 kips Upwardw10 = 0.471071 +15 = 15.471 kips Upwardw11 = 3.427 kips Rightward w12 = –3.427 kips LeftwardAnd finally for member-3 we get
00370233.0
000768219.0
00153523.00015278.0
00889.2013889.20130000889.2013889.2013002356.932356.9300481.3356481.33562356.932356.9300481.3356481.3356481.3356481.335600111.161111556.80555481.3356481.335600556.80555111.161111
18
17
16
15
14
13
wwwwww
![Page 56: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/56.jpg)
Solving the above equation we get w13 = -246.74227 kips-inch c.c.w.w14 = 0 kips-inchw15 = 3.427kips Upwardw16 = -3.427 kips Downwardw17 = 15.4711 kips Rightward w18 = -15.4711kips Leftward
PLOTTING THE BENDING MOMENT AND SHEARING FORCE DIAGRAM.
According to the forces calculated above bending moment and shearing force diagrams are plotted below:
![Page 57: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/57.jpg)
-3.427k
14.528k
-15.471k
3.427k
+
+
+ -
Shearing Force Diagram
231.959k"
-246.747k"
-190.218k"
-18.973
15.407k"
Bending moment diagram
3.1427k
+
- -
-
![Page 58: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/58.jpg)
EXAMPLE 5.5 To analyse the frame shown in the figure using direct stiffness method.
40' 10'
10'
3k/ft
![Page 59: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/59.jpg)
W 1 , 1
W 3 , 3
W 2 , 2
W 5 , 5
W 6 , 6 W 8 , 8
W 9 , 9
W 7 , 7
W 4 , 4
Structure Forces and Deformation
w 1 , 1
w 2 , 2
w 5 , 5
w 6 , 6
w 3 , 3
w 4 , 4
Element Forces and Deformation
w 7 , 7
w 11 , 11
w 9 , 9
w 8 ,
w 12 , 12
w 10 , 10
![Page 60: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/60.jpg)
006060
48004800000000
12
11
10
9
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
F
F
F
F
wwwwwwwwwwww
Fixed end moments
060
48000000
604800
9
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
F
WWWWWWWWW
4800k"
0
60
4800k"
0
60
[W]F=Net fixed end forces=
![Page 61: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/61.jpg)
Equivalent joint loads=
060
4800
3
2
1
E
E
E
WWW
The properties of each member are shown in the table below.E = 29 x 103 ksi , I = 1000 inch4 and A = 10 inch2 are same for all members.
Member Length l m I J K L M N
1 169.7056 inch
0.707 0.707 4 1 5 2 6 3
2 480 inch 1 0 1 7 2 8 3 9
![Page 62: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/62.jpg)
Using the structure stiffness matrix for frame element in general form we get structure stiffness matrix for member-1.
362514
0045.8900045.890804.818804.818064.4272064.42720045.8900045.890804.818804.818064.4272064.4272804.818804.818045.8900045.890064.4272064.4272804.818804..8180045.8900045.890064.4272064.4272064.4272064.4272064.4272064.4272666.68353633.341768064.4272064.4272064.4272064.427233.341768666.683536
1
K
Similarly for member-2 we get,
938271
167.604167.6040000167.604167.6040000
0015.315.321.75521.7550015.315.321.75521.7550021.75521.755667.24166633.1208330021.75521.75533.120833667.241666
2
K
![Page 63: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/63.jpg)
Calculating the structure stiffness matrix for the entire frame.After getting the structure stiffness matrices for each element we can find the structure stiffness matrix for the whole structure using following relation:
[K]1+[K]2 = [K]
1 2 3 4 5 6 7 8 9
167.60400000167.60400015.321.755000015.321.755021.75566.241666000021.75533.1208330000045.890804.818064.42720045.890804.818064.4272000804.8180045.890064.4272804.8180045.890064.4272000064.4272064.427266.683536064.4272064.427233.341768
167.604000045.890804.818064.42721715.1494804.818064.4272015.321.755804.8180045.890064.4272804.818195.893854.3516021.75533.120833064.427206.427233.341678064.4272854.351632.925203
987654321
K
![Page 64: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/64.jpg)
Finding unknown deformationsUnknown deformation can be calculated using equation [u] = [K11]–1[Wk]This can be done by partitioning the structure stiffness matrix into known and unknown deformations and forces.
k
u
u
k
2221
1211
KKKK
WW
Unknown deformations can be calculated using the equation ; [u] = [K11]-1 [Wk]
060
4800
1715.1494804.818064.4272804.818195.893854.3516
064.4272854.3516333.925203 1
3
2
1
Solving above equation we get 1 = 0.00669 c.w. 2 = - 0.223186 downward3 = 0.141442 Rightward
![Page 65: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/65.jpg)
4546.8575539.5
927.9764417.852432.54
716.728
141442.0223186.000669.0
167.60400015.321.755021.75533.1208330045.890804.818064.4272804.8180045.890064.4272064.4272064.427233.341768
9
8
7
6
5
4
21
E
E
E
E
E
E
uu
WWWWWW
KW
![Page 66: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/66.jpg)
Now we will superimpose the fixed end forces on the above calculated equivalent forces.
W=WE+WF
4546.8575539.65
927.57764417.852432.54
716.728
060
4800000
4546.8575539.5
927.9764417.852432.54
716.728
9
8
7
6
5
4
WWWWWW
![Page 67: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/67.jpg)
Calculating the unknown element forcesAll the deformations are known up to this point, therefore we can calculate the forces in all elements using relation [w] =[kT][]For member-1 we get [w]1=[kT]1[]1
14144.00
22319.00
321.1208321.1208321.1208321.120800321.1208321.1208321.1208321.120800
34676.5034676.5034676.5034676.50668.6041668.604134676.5034676.5034676.5034676.50668.6041668.6041
063.4272063.4272063.4272063.4272665.683536332.341768063.4272063.4272063.4272063.4272332.341768665.683536
0.006690
6
5
4
3
2
1
E
E
E
E
E
E
wwwwww
Solving the above equation we get following values of w’s for the equivalent loading condition provided by member-1 w1E= 729.0578 kip-inch c.w.w2E= 3015.829 kip-inch c..w.w3E= –22.066 kip Downwardw4E= 22.0669 kip Upwardw5E= 98.773 kip Rightwardw6E= –98.773 kip Leftwardw = wE+wF
![Page 68: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/68.jpg)
As wF are zero therefore w’s will be having the same values as those of wE ‘sFor member-2 we get [w]2=[kT]2[]2
0141442.0
0223186.00
00669.0
16667.60416667.604000016667.60416667.6040000001467.31467.3208.755208.755001467.31467.3208.755208.75500208.755208.755667.241666333.12083300208.755208.755333.120833667.241666
12
11
10
9
8
7
E
E
E
E
E
E
wwwwww
solving the above equation we get the structure forces for the equivalent structure loads provided by member-2 w7E= 1785.543 kip-inch c.w.w8E= 977.047 kip-inch c..w.w9E= –5.7553 kip Downwardw10E= 5.7553 kip Upwardw11E= 85.454 kip Rightwardw12E= –85.454 kip Leftward
![Page 69: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/69.jpg)
Superimposing the fixed end forces in their actual direction we get the element forces for the actual loading condition.
w7= 1785.543 – 4800 = -3014.698 kip-inch c.c.w.w8= 977.047 + 4800 = 5776.927 kip-inch c..w.w9= –5.7553 +60 = 54.244 kip Upwardw10= 5.7553 + 60 = 65.754 kip Upwardw11 = 85.454 kip Rightwardw12= – 85.454 kip Leftward
Plotting the bending moment and shearing force diagrams.Bending moment and the shearing force diagram can be drawn according to the element forces calculated above.
![Page 70: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/70.jpg)
-3014.698k" 5776.927k"
54.244k 65.754k
-85.454k85.454k
![Page 71: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/71.jpg)
Shearing force Diagram Bending Moment Diagram
-22.059kip
54.244 kip
-65.7541 kips
263.016 " 729.0578 kip-inch
-3015.829 kip-inch
2870.376 kip-inch
-5797.047 kip-inch
- - --+
+
+
![Page 72: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/72.jpg)
a
L
L/2
L
L
A
A
bM
B
L/2wB
A B
L
L
L
A
A
A
L/2w
w B
B
w B
a
L/2
L
MAA
b
L/2
P
P
BMB
MA MB
MA MB
MA MB
MA MB
MA MB
MA MB
2
2
B2
2
AL
bPa M L
Pab M
2
2
B2
2
AL
bPa M L
Pab M
12wL
- M2
A 12
wL M2
B
2A wL
1925 - M 2
B wL19211
M
30wL - M
2
A 20
wL M
2
B
965wL - M
2
A 96
5wL M2
B
1 -
L3a
LMb - MA
1 -
L3b
LMa M B
![Page 73: CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction](https://reader033.vdocuments.mx/reader033/viewer/2022061502/56815bcd550346895dc9c39f/html5/thumbnails/73.jpg)
TheThe End End