CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction For 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 moment ii) Beams and frames subjected to shear force and bending moment iii) Beams and frames subjected to shear force, bending moment and axial forces
Transcript
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
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.
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
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
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
Partitioning the above equation into known and unknown portions as shown below:
W
W
K K
K Kk
u
u
k
11 12
21 22
WhereWk = known loads
Wu = unknown loads
u = 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)
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
L
EI
L
EIL
EI
L
EI
kT42
24
Case-II Beams subjected to Shear Forces & Bending Moment
3322
3322
22
22
121266
121266
6642
6624
L
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EI
kT
Case-IIIFor frame element subjected to axial force, shear force and bending
moment.
L
lAE
L
lAEm
L
AEm
L
AEL
lAE
L
lAEm
L
AEm
L
AEL
mEI
L
mEI
L
lEI
L
lEI
L
EI
L
EIL
mEI
L
mEI
L
lEI
L
lEI
L
EI
L
EIL
mEI
L
mEI
L
lEI
L
lEI
L
EI
L
EIL
mEI
L
mEI
L
lEI
L
lEI
L
EI
L
EI
kT m
..00
..00
.12.12.12.1266
.12.12.12.1266
.6.6.6.642
.6.6.6.624
333322
333322
2222
2222
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.
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
39
39
5.37
5.37
5.12
5.12
' 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
39
5.1
25
5.12
Moments End FixedNet W
4
3
2
1
F
F
F
F
W
W
W
W
F
39
Net fixed end moment
-12.5 -25 -1.5
1 32
39
5.1
25
5.12
4
3
2
1
E
E
E
E
W
W
W
W
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
Structure stiffness matrices are:
2
1
2
1
1
2 1 2 1
4.02.0
2.04.0EI
42
24
10
EIK
3
2
3
2
2
3 2 3 2
267.013.0
13.0267.0EI
42
24
15
EIK
4
3
4
3
3
4 3 4 3
34.0167.0
167.034.0EI
42
24
12
EIK
Forming structure stiffness matrix of the entire structureUsing relation [K] = [K]1 + [K]2 + [K]3 structure stiffness matrix of the entire structure is:
4
3
2
1
33.0167.000
167.034.0267.013.00
013.0267.04.02.0
002.04.0
4 3 2 1
EIK
4
3
2
1
33.0167.000
167.0597.013.00
013.0667.02.0
002.04.0
EIK
4 3 2 1
Finding unknown deformations Unknown deformations are obtained by using the following equation
] = [K]-1 [W]
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
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
Actual forces on the structure are obtained by superimposing the fixed end reactions on above calculated forces.
0
1.50
1.50
2.26
2.26
0
08.39
27.10
67.12
325.11
7.13
5.12
39
39
5.37
5.37
5.12
5.12
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
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
0
1.50
1.50
2.26
2.26
0
08.39
27.10
67.12
325.11
7.13
5.12
39
39
5.37
5.37
5.12
5.12
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
E6
E5
E4
E3
E2
E1
F6
F5
F4
F3
F2
F1
6
5
4
3
2
1
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 1
2 2 w ,3 3
w ,4 4
1
2
Structure forces and deformations
Element forces and deformations
20'
Numbering element forces and deformations
0
0
67.66
67.66
'
4
3
2
1
F
F
F
F
w
w
w
w
sFEM
-16.67k'
Equivalent joint moment
66.67k'
Net fixed end moment
2kip/ft
Finding Fixed end moments and equivalent structure load
0
67.66
67.66
moments end fixedNet
3
2
1
F
F
F
F
W
W
W
W
Equivalent Joint Loads
67.161 EW
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 1
2 10 1 3
Structure stiffness matrices of both elements are:
1
2
2.01.0
1.02.0
1
2
42
24
201
EI
EIK
3
1
4.02.0
2.04.0
3
1
42
24
102
EI
EIK
2 1 2 1
1 3 1 3
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
k
u
u
k
KK
KK
W
W
2221
1211
EIEI
W
W
E
E1
0
0
4.002.0
02.01.0
2.01.06.067.16 1
3
2
EIEI
78.27667.16
6.0
111
[u] = [K11]-1[Wu]
Finding Unknown ReactionsUnknown reactions can be calculated using the following equation.
uW uK 21
556.5
78.2
78.27/12.0
1.0
3
2
3
2
E
E
E
E
W
W
EIEIW
W
W=WF+WE
56.5
45.69
556.5
78.2
0
67.66
3
2
W
W
Calculating element forces(Moments)Using relation
kTw
56.5
11.11
56.5
78.2
1
0
27.78-
0
4.02.00
2.04.00
02.01.0
01.02.0
4
3
2
1
EIEI
w
w
w
w
E
E
E
E
OR
56.5
11.11/1
0
78.27
4.02.0
2.04.
56.5
78.2/1
78.27
0
2.01.0
1.02.0
4
3
2
1
EIO
EIw
w
EIEIw
w
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
56.5
11.11
11.61
45.69
0
0
667.66
667.66
56.5
11.11
56.5
78.2
4
3
2
1
w
w
w
w
1
2
69.45 61.11 11.11
5.56
69.45
61.11
11.11
5.56
BMDFinal element forces
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
5.37
5.37
1125
1125
1
1
30
30
"
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
w
w
w
w
w
w
w
w
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
30k" 1095k"
Net fixed end forces
[W]F=Net fixed end moments and forces =
5.37
5.38
1
1125
1095
30
6
5
4
3
2
1
F
F
F
F
F
F
W
W
W
W
W
W
Equivalent joint loads
30k" 1095k"
Equivalent joint loads =
1095
30
2
1
E
E
W
W
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 5
2 180 2 3 5 6
From structural elements
Member-1 1 2 4 5
5
4
2
1
00000694.000000694.000041667.000041667.0
00000694.000000694.000041667.000041667.0
00041667.000041667.0033333.0016667.0
00041667.000041667.0016667.00333.0
1
EIK
Member-2 2 3 5 6
6
5
3
2
00000206.000000206.0000185.0000185.0
00000206.000000206.0000185.0000185.0
000185.0000185.00222.00111.0
000185.0000185.00111.00222.0
2
EIK
Forming structure stiffness matrix for the entire structureStructure stiffness matrix for the entire beam is obtained using the relation[K] = [K]1 + [K]2
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
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:
1397.5
1397.5
7647.586
30
0
0
58863.22870
2946.10535
1
00000694.000000694.000041667.000041667.0
00000694.000000694.000041667.0000416667.0
00041667.000041667.0033333.0016667.0
0001667.000041667.0016667.00333.0
4
3
2
1
EIEI
w
w
w
w
E
E
E
E
Superimposing the fixed end forces for member-1 on the above w’s we get
w = wE + wF
1397.6
1397.4
764.616
0
1
1
30
30
1397.5
1397.5
7647.586
30
4
3
2
1
w
w
w
w
73529.41
2647.33
117.1379
764.616
5.37
5.37
1125
1125
23529.4
23529.4
1176.254
2353.508
8
7
6
5
w
w
w
w
23529.4
23529.4
1176.254
2353.508
0
0
0
5886.22870
1
00000206.000000206.0000185.0000185.0
00000206.000000206.0000185.0000185.0
000185.0000185.00222.00111.0
000185.0000185.00111.00222.0
8
7
6
5
EIEI
w
w
w
w
E
E
E
E
FE www
Member 2 :
Superimposing the fixed end forces
OR
EI
EI
w
w
w
w
w
w
w
w
w
E
E
E
E
E
E
E
E
E
1
0
0
0
0
5886.22870
2946.10535
00000206.000000206.00000185.0000185.00
00000206.000000206.00000185.0000185.00
000185.0000185.000222.00111.00
000185.0000185.000111.00222.00
0000000694.000000694.0000041667.000041667.0
000000694.000000694.0000041667.000041667.0
000041667.000041667.000333.001667.0
000041667.000041667.0001667.00333.0
8
7
6
5
4
3
2
1
24.4
24.4
21.254
42.508
14.5
14.5
53.586
30
8
7
6
5
4
3
2
1
E
E
E
E
E
E
E
E
E
w
w
w
w
w
w
w
w
w
w = wE +wF
5.37
5.37
1125
1125
1
1
30
30
24.4
24.4
21.254
42.508
14.5
14.5
53.586
30
8
7
6
5
4
3
2
1
w
w
w
w
w
w
w
w
w
74.41
26.33
2.1379
58.616
14.6
14.4
53.616
0
8
7
6
5
4
3
2
1
w
w
w
w
w
w
w
w
w
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
- - - -++
10'
5'
5'
3k/ft
4k
A
BC
D
6'
EXAMPLE 5.4Analyse the frame by STIFFNESS METHOD
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
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
3k/ft
4k
2k
2k
60k"
60k"
300k" 300k"
15k 15k
1
2
3
0
0
0
0
0
0
0
0
15
15
300
300
2
2
0
0
60
60
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
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
sFEM'
Fixed End Moments and reactions
2
0
60
0
0
0
0
15
300
2
15
240
12
11
10
9
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
F
F
F
F
W
W
W
W
W
W
W
W
W
W
W
W
2k240 K"
15k
0
15k
300 K"
Net fixed end moments
[W]F=
2k
240 K"
15k
0
15k
300 K"
0
Equivalent joint loads
[W]E=
0
0
15
300
2
15
240
7
6
5
4
3
2
1
E
E
E
E
E
E
E
W
W
W
W
W
W
W
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
3
12
2
11
1
10
13889.2013889.2000333.1208333.1208
13889.2013889.2000333.1208333.1208
00333.1208333.120800
00333.1208333.120800
333.1208333.120800667.96666333.48333
333.1208333.120800333.48333667.96666
1
K
For member-1 we have
10 1 11 2 12 3
For member-2
6
3
5
2
4
1
333.1208333.12080000
333.1208333.12080000
0013889.2013889.20333.1208333.1208
0013889.2013889.20333.1208333.1208
00333.1208333.1208667.96666333.48333
00333.1208333.1208333.48333667.96666
2
K
1 4 2 5 3 6
For member-3 4 7 5 8 6
9
9
6
8
5
7
4
2356.932356.9300481.3356481.3356
2356.932356.9300481.3356481.3356
00889.2013889.201300
00889.2013889.201300
481.3356481.335600111.161111556.80555
481.3356481.335600556.80555111.161111
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.
Unknown reactions can be calculated using the following equation:
001535.
037023.
007682.
001527.
039859.
01202.
00184.
7
6
5
4
3
2
1
42062.1
5242.14
7704.40
42501.3
4707.15
001535.0
037023.0
007682.0
001527.0
039859.0
01202.0
00184.0
000013889.200333.1208
00000333.12080
0000333.12080333.48333
481.33562356.930481.3356000
00889.20130000
12
11
10
9
8
21
E
E
E
E
E
uu
W
W
W
W
W
KW
Now we will superimpose the fixed end reactions on the above calculated structure forces. [W]=[W]E+[W]F
58.0
52.14
23.19
42.3
47.15
2
0
60
0
0
42.1
52.14
77.40
42.3
47.15
12
11
10
9
8
W
W
W
W
W
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
039859.0
0
01202.0
0
00184.0
0
00333.1208333.120800
00333.12083330.120800
13889.2013889.2000333.1208333.1208
13889.2013889.2000333.1208333.1208
333.1208333.120800667.96666333.48333
333.1208333.120800333.48333667.96666
4
3
6
5
2
1
w
w
w
w
w
w
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
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:
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
KK
KK
W
W
Unknown deformations can be calculated using the equation ; [u] = [K11]-1 [Wk]
0
60
4800
1715.1494804.818064.4272
804.818195.893854.3516
064.4272854.3516333.9252031
3
2
1
Solving above equation we get 1 = 0.00669 c.w. 2 = - 0.223186 downward3 = 0.141442 Rightward
4546.85
75539.5
927.976
4417.85
2432.54
716.728
141442.0
223186.0
00669.0
167.60400
015.321.755
021.75533.120833
0045.890804.818064.4272
804.8180045.890064.4272
064.4272064.427233.341768
9
8
7
6
5
4
21
E
E
E
E
E
E
uu
W
W
W
W
W
W
KW
Now we will superimpose the fixed end forces on the above calculated equivalent forces.
W=WE+WF
4546.85
75539.65
927.5776
4417.85
2432.54
716.728
0
60
4800
0
0
0
4546.85
75539.5
927.976
4417.85
2432.54
716.728
9
8
7
6
5
4
W
W
W
W
W
W
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
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.
