truss by direct stiffness
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DIRECT STIFFNESS METHOD FOR TRUSSES:
3.1 INTRODUCTION
In the previous Lectures the procedure for obtaining thestructure stiffness matrix was discussed. The structure stiffness matrixwas established by the following equation:
[K] = [T]T [kc] [T] -----------(2.21)
However if a large and complicated structure is to be analyzedand if more force components are to be included for an element thenthe size of composite stiffness matrix [kc] and deformationtransformation matrix [T] will be increased. Therefore the procedureoutlined in the preceding chapter for formation of structure stiffnessmatrix appears to be inefficient, furthermore this procedure is notsuitable to automatic computation.
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In this chapter an alternative procedure called the direct stiffnessmethod is introduced. This procedure provides the basis for mostcomputer programs to analyze structures. In this method eachindividual element is treated as a structure and structure stiffnessmatrix is obtained for this element using the relationship:
[K]m = [T]T
m[k]m [T]m -----------(3.1)
Where
[K]m= Structure stiffness matrix of an individual element.
[T]m = Deformation Transformation matrix of an individualelement.
[k]m = Member Stiffness matrix of an individual element.
Total structure stiffness matrix can be obtained by superimposing thestructure stiffness matrices of the individual elements..
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As all members of a truss are not in the same direction i.e.
inclination of the longitudinal axes of the elements varies, thereforestiffness matrices are to be transformed from element coordinatesystem to structure or global coordinate system. When the matrices forall the truss elements have been formed then adding or combiningtogether the stiffness matrices of the individual elements can generatethe structure stiffness matrix [K] for the entire structure, because ofthese considerations two systems of coordinates are required.
i ) Local or member or element coordinate system:
In this coordinate system x-axis is collinear with the longitudinalaxis of the element or member. Element stiffness is calculated with
respect to this axis. This system is illustrated in figure 3.1
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Element 'B'Element 'A'
A B
Structure or
globel axes
Local or member
or element axes for
element B
Local or member
or element axes
for element 'A'
Figure 3.1
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ii) Structure or global coordinate system:
A single coordinate system for the entire structure is chosen, with
respect to which stiffness of all elements must be written.
3.2 PROCEDURE FOR THE FORMATION OF TOTAL STRUCTURESTIFFNESS MATRIX FOR AN ELEMENT USING DIRECT STIFFNESS
METHOD:
Following is the procedure for the formation of structure
stiffness matrix:
i) Formation of the element stiffness matrix using equation 2.16.
11
11
L
AEk m ----------- (2.16)
ii) Formation of the deformation transformation
matrix [T] for a single element:
[]m = [T]m []m -----------(3.2)
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where
[]m= Element or member deformation matrix.[]m= Structure deformation matrix of an element or member[T]m= Element or member deformation transformation matrix.
(iii)Formation of a structure stiffness matrix [K] or an element using the
relation
[K]m = [T]T
m[k]m [T]m -----------(3.1)
The above mentioned procedure is discussed in detail in the subsequent
discussion.
3.2.1 The formation of element sti ffness matr ix in local co-ordinates:
It has already been discussed in the previous chapter. However it
is to be noted that for horizontal members the structure stiffness matrix
and element stiffness matrix are identical because both membercoordinate systems and structure coordinate systems are identical. But
for inclined members deformation transformation matrices are to be
used because member coordinate system and structure coordinate
system are different; therefore their structure stiffness matrix and
element stiffness matrix will also be different.
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3.2.2 The formation of deformation transformation matrix:
As the main difference between the previously discussed method and directstiffness method is the formation of the deformation transformation matrix.In this article deformation transformation matrix for a single element will be
derived.Before the formation of deformation transformation matrix followingconventions are to be established in order to identify joints, members,element and structure deformations.
1) The member is assigned a direction. An arrow is written along the member,with its head directed to the far end of the member.
2) i, j, k, and l, are the x and y structure deformations at near and far (tail& head) ends of the member as shown in figure 3.2. These are positive in theright and upward direction.
3) r and s are the element deformations at near and far (tail & head) ends ofthe member as shown in figure 3.2.
4)The member axis (x-axis of member coordinate system) makes an angle x,with the x-axis of the structure coordinate system as shown in figure 3.2.
5)The member axis (x-axis of member coordinate system) makes an angle y,with the y-axis of the structure co-ordinate system as can be seen from figure3.2.
6) The cosines of these angles are used in subsequent discussion Letters l andm
represent these respectively.
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8Figure 3.2
Near
end
Far
end
Memberdirection
x
y
(b)
r
s
i
j
k
Nearend
Farend
Memberdirection
(a)
x
y
l
i
j
k
r
s
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2122
12
1212
YYXX
XX
L
XX
= Cosxand m = Cos
y
= Cos
x=
m = Cosy=
2
12
2
12
1212
YYXX
YY
L
YY
The algebraic signs ofs will be automatically accounted for the members
which are oriented in other quadrants of X-Y plan.
In order to form deformation transformation matrix, once again considerthe member of a truss shown in figure 3.1.
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s=0
r
j
i
s=0
j
r=1
(b)(a)
=1
j
rjc
osy
=mr
cos
xl
x
i
i
i
y
x
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r=0
r=0
y
x
Figure 3.3
(d)(c)
=1k
=1
s1cos
y=m
1
l
kk
s
kco
sx=
1
s
s
x
y
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Following four cases are considered to form deformation transformation matrix.
Case-1: Introduction of horizontal deformation to the structure i = 1, while farend of the member is hinged (restrained against movement). From the geometry
of figure 3.3(a)
r = i Cosx = 1. Cosx = Cosx = lr =l ---------- (3.3)s = 0 ---------- (3.4)
Case-2: I ntroduction of vertical deformation to the structurej = 1, while nearend of the member is hinged (restrained against movement). F rom the geometry of
f igure 3.3 (b) r =j Sinx =j Cosy = 1. Cosy = mr = m ---------- (3.5)s = 0 ---------- (3.6)
Case-3:Int roduc t ion of hor izontal deformat ion to the structurek = 1, whi lenear end o f the member is hinged (restrained against m ovement).From th e
geometry of f igure 3.3(c)
s =k Cosx = 1. Cosx = Cosx = lr = 0 ---------- (3.7)s = l ---------- (3.8)
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Case-4: Introduction of vertical deformation to the structure L = 1, whilenear end is hinged
From the geometry of figure 3.3(d)
L = 1r = 0 ---------- (3.9)s = LCosy = m ---------- (3.10)
Effect of four structure deformations and two member deformations can bewritten as
r =i Cosx +j Cosy +k .0 + L .0s =i .0 +j .0 +k Cosx + L Cosy
r= l.
i+ m.
j+ 0.
k+ 0.
L ----------- (3.11)
s= 0.
i+ 0.
j+ l.
k+ m.
l----------- (3.12)
Arranging equations 3.11 and 3.12 in matrix from
L
k
j
i
s
r
mm
0000
----------- (3.13)
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Comparing this equation with the following equation
[]m = [T]m []m ---------------(3.2)After comparing equation (3.14) and (3.2) following deformation
transformation matrix is obtained.
mm
T m
0000 - ---------- (3.14)
This matrix [T]m transforms four structure deformation (i, j, k, L )into two element deformation (r and s).3.2.3 Formation of structure stiffness matrix:
Structure Stiffness matrix of an individual member is first to be
transformed from member to structure coordinates. This can be done by
using equation 3.1.
[K] = [T]Tm [k]m [T]m --------(3.1)
ml
mlmT
00
00 ----- (3.15)
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1111
L
AEk m ------(2.16)
Lkji
m
m
m
m
L
k
j
i
Km
00
00
11
11
0
0
0
0
Lkji
mm
mm
L
AE
m
m
L
k
ji
Km
0
0
00
L
k
j
iLkji
mmmm
mm
mmmm
mm
L
AEKm
22
22
22
22
--------------------------------(3.16)
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3.2.4 ASSEMBLING OF THE INDI VIDUAL STRUCTURE ELEMENTSTI FFNESS MATRICES TO FORM TOTAL STRUCTURESTIFFNESS MATRIX
Combining the stiffness matrix of the individual members can generatethe stiffness matrix of a structure However the combining processshould be carried out by identifying the truss joint so that matrixelements associated at particular member stiffness matrices arecombined. The procedure of formation of structure stiffness matrix isas follows:
Step 1: Form the individual element stiffness matrices for each member.
Step 2: Form a square matrix, whose order should be equal to that ofstructure deformations.
Step 3: Place the elements of each individual element stiffness matrix
framed into structure in the corresponding rows and columns ofstructure stiffness matrix of step-2.
Step 4: If more than one element are to be placed in the same location ofthe structure stiffness matrix then those elements will be added.
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1
2 3
2
3
2
1
Structurestiffness
matrix formember-2
Structurestiffness
matrix formember-1
Structure
stiffness matrix
Foll owing f igure shows the above mentioned process.
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2
121
kk
kkK
11
11
1
3
2
32
kk
kkK
22
22
2
21 KKK
3
2
1
kk
kkkk
kk
K
321
22
2211
11
0
0
In the above matrix the element in the second row and
second column is k1+k2 where k1 is for member 1 and k2
is for member 2.This is because both k1 and k2 have
structure coordinates 2-2.
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Illustrative Examples Regarding the Formation of [K] Matrix:
L/2
(x1,y1) = (0,0)
(x1,y1) = (L/2,2L/3)(x2,y2) = (L/2,2L/3)
(x2,y2) = (L,0)
(x1,y1) = (L,0)(x2,y2) = (0,0)
(a)
(b)
(c)
(Structure forces and deformations)
(Free body diagram indicating structure forces
and coordinates)
2L/31 1
2 2
6 4
5 3
6 4
5 3
1
5
6
3
4
2
12
11.5
1
1
2
23
3
L/2
Example 3.1: Form the structure stiffness matrix for the following
truss by direct stiffness method.
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Member Length Near
End
X1 Y1
Far End
X2, Y2
i j k L l=Cosx =X2-X1/
Length
m=CosyY2-Y1/
Length
1 5L/6 0 0 L/2 2L/3 5 6 1 2 0.6 0.8
2 5L/6 L/
2
2L/3 L 0 1 2 3 4 0.6 -0.8
3 L L 0 0 0 3 4 5 6 -1 0
l
k
ji
mlmmlm
lmllmlmlmmlmlmllml
lkji
L
AEK m
22
22
22
22
According to eq. (3.16) we get
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2165
768.0576.0768.0576.0576.0432.0576.0432.0768.0576.0768.0576.0576.0432.0576.0432.0
2165
1
L
AE
K
432
1
768.0576.0768.0576.0576.0432.0576.0432.0 768.0576.0768.0576.0
576.0432.0576.0432.0
4321
2
LAEK
6
5
43
0000
000.1000.1
0000000.1000.1
6543
L
AE3K
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6
5
4
3
2
1
768.0576.0000.0000.0768.0576.0
576.0432.1000.0000.1576.0432.0
000.0000.0768.0576.0768.0576.0
000.0000.1576.0432.1576.0432.0
768.0576.0768.0576.0536.1000.0
576.0432.0576.0432.0000.0864.0
654321
L
AEK
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Example 3.2: Form the structure stif fness matrix of the following truss.
L L
P
1
2
3
4
5
6
7
8
(L,L)
(0,0)
(L,0)
(2L,0)
(L,L)
(L,L)
y
x
1
2
32 2
1
1
3
4
65
7
8
1
2
Free body diagram showing structure forces andcoordinates
Structure forces and deformations
Structure to be analysed
L
1
2
3
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Member Near End Far End Length =Cosx m=Cosy i j k L
X1 Y1 X2 Y2
1 0 0 L L 2L 0.707 0.707 7 8 1 2
2 L L L 0 L 0 -1 1 2 5 6
3 L L 2L 0 2L 0.707 -0.707 1 2 3 4
22
22
22
22
mlmmlm
lmllml
mlmmlm
lmllml
L
AEmK
So
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25
2
1
8
7
0.50.5.5.5
0.50.5.5.5
.5.50.50.5
.5.50.50.5
2187
1
2L
AEK
6
5
2
1
1010
0000
1010
0000
6521
2
L
AE
K 4
3
2
1
.5.5.5.5
.5.5.5.5
.5.5.5.5
.5.5.5.5
3
4321
2L
AE
K
Now the matrix can be written as [k]2:
6
5
2
1
414.10414.10
0000
414.10414.10
0000
6521
22
L
AEK
2
1
8
7
5.5.5.5.
5.5.5.5.
5.5.5.5.
5.5.5.5.2187
21
L
AEK
So [k] = [k]1+ [k]
2+ [k]
3
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8
7
6
54
3
2
1
5.5.00005.5.
5.5.00005.5.
00414.1000414.10
0000000000005.5.5.5.
00005.5.5.5.
5.5.414.105.5.414.20
5.5.005.05.01
87654321
2
L
AEK
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3.4 Force transformation matrix:
The axial force (wm) in the members of a truss can befound by using the relationship between member forces
and member deformations [equation (2.15)] and betweenelement and structure deformations [equation (3.1)]
[w]m= [k]m []m ------------ (2.15)[]m = [T]m []m ------------ (3.3)
Substituting value of []m from equation 3.1 into equation2.15
[w]m = [k]m [T]m []m ------------ (3.18)Substituting value of [k]m and [T]m from equation 2.16 and
3.15 respectively
L
k
j
i
mmmm
L
AEmw 11
11
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Lk
j
i
mmmm
L
AE
wsrw
In above Equations, wr= Force at near end
See figure below:
ws
= Force at far end
As wr = -ws so:
l
k
j
i
smlml
L
AEw -------------- (3.19)
Sign conventions
wr
ws
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If
wris positive then member is in compression.
wsis negative then member is in compression.
wris negative then member is in tension.
wsis positive then member is in tension.
3.5 ANALYSIS OF TRUSSES USING DIRECT STI FFNESS METHOD:
Basing on the derivations in the preceding sections of this chaptera truss can be completely analyzed. The analysis comprises ofdetermining.
i) Joint deformations.
ii) Support reactions.
iii) Internal member forces.
As the first step in the analysis is the determination of unknown jointdeformation. Using the equations can do this.
[W] = [K] []The matrices [W], [K] and [] can be divided in submatrices in the
following form
k
u
u
k
KK
KK
W
W
2221
1211 ----------- (3.20)
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Where
Wk = Known values of loads at joints.
Wu = Unknown support reaction.
u = Unknown joint deformation.k = Known deformations, generally zero due to supportconditions.
K11, K12, K21, K22are the sub-matrices of [K]
Expanding equation 3.20
Wk = K11u + K12
k ----------- (3.21)
Wu = K21u + K22k ----------- (3.22)If the supports do not move, then k = 0 therefore equation 3.21
& 3.22 can be written as
Wk = K11u ----------- (3.23)Wu = K21u ----------- (3.24)
By pre-multiplying equation 3.23 by [K11]-1 following equation isobtained
[K11]-1 [WK]=[K11]-1 [K11][U][u] = [K11]-1 [Wk] ----------- (3.25)
Substituting value u from equation 3.25 into equation 3.24[Wu] = [K21] [K11]-1 [Wk] ----------- (3.26)
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Using equation 3.25, 3.26 and 3.19, joint deformations, support reactionsand internal member forces can be determined respectively.
As this method does not depend upon degree of indeterminacy so it canbe used for determinate as well as indeterminate structures.
Using the basic concepts as discussed in the previous pages, followingare the necessary steps for the analysis of the truss using stiffnessmethod.
1-Identify the separate elements of the structure numerically and specifynear end and far end of the member by directing an arrow along thelength of the member with head directed to the far end as shown in thefig. 3.2
2-Establish the x,y structure co-ordinate system. Origin be located at oneof the joints. Identify all nodal co-ordinates by numbers and specifytwo different numbers for each joint (one for x and one for y). Firstnumber the joints with unknown displacements.
3-Form structure stiffness matrix for each element using equation 3.16.
4-Form the total structure stiffness matrix by superposition of the element
stiffness matrices.5-Get values of unknown displacements using equation 3.25.
6-Determine support reaction using equation 3.26.
7-Compute element or member forces using equation 3.19.
3 6 Ill t ti E l R di C l t A l i f T
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3.6- Illustrative Examples Regarding Complete Analysis of Trusses:
Example 3.3: Solve truss in example 3.1 to find member forces.
L/2
(x1,y1) = (0,0)
(x1,y1) = (L/2,2L/3)(x2,y2) = (L/2,2L/3)
(x2,y2) = (L,0)
(x1,y1) = (L,0)(x2,y2) = (0,0)
(a)
(b)
(c)
(Structure forces and deformations)
(Free body diagram indicating structure forces and
coordinates)
2L/31 1
2 2
6 4
5 3
6 4
5 3
1
5
6
3
4
2
12
11.5
1
1
2
23
3
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The [K] matrix for the truss as formed in example 3.1 is:
6
5
4
3
2
1
768.0576.0000.0000.0768.0576.0
576.0432.1000.0000.1576.0432.0
000.0000.0768.0576.0768.0576.0
000.0000.1576.0432.1576.0432.0
768.0576.0768.0576.0536.1000.0
576.0432.0576.0432.0000.0864.0
654321
L
AEK
Using the relation [W] = [K] [] we get
k
u
u
k
KK
KK
W
W
2221
1211
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6
5
4
3
2
1
654321
6
5
4
3
2
1
768.0576.0000.0000.0768.0576.0
576.0432.1000.0000.1576.0432.0
000.0000.0768.0576.0768.0576.0
000.0000.1576.0432.1576.0432.0
768.0576.0768.0576.0536.1000.0
576.0432.0576.0432.0000.0864.0
L
AE
W
W
W
W
W
W
From equation (3.25) [u] = [K11]-1 [Wk]
2
1
536.1000.0
000.0864.0
2
11
W
W
AE
L
125.11
651.0000.0000.0157.1
2
1AEL
812.7
31.13
2
1
AE
L
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Using the relation:
[Wu] = [K21] [u]
812.7
310.13
768.0576.0
576.0432.0
768.0576.0
576.0432.0
6
5
4
3
W
W
W
W
1.66
1.25
13.66
10.25
6W5W4W3W
Now the member forces can be found using the relation:
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L
k
j
i
mmL
AEwr
09.2
812.7
31.13
00.0
00.0
8.06.08.06.05
61
AE
L
L
AEw kips
09.17
00.0
00.0
812.7
31.13
8.06.08.06.05
62
AE
L
L
AEw kips
00.0
00.0
00.0
00.0
00.0
0.00.10.00.13
AE
L
L
AEw
kips
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11.5k
12k
1.25k 10.25k
13.66k
17.09kip
2.09
kip
s
1.66k
0 kips
Final Resul ts & sketch
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Example 3.4:
L L
P
1
2
3
4
5
6
7
8
(L,L)
(0,0)
(L,0)
(2L,0)
(L,L)
(L,L)
y
x
1
2
32 2
1
1
3
4
65
7
8
1
2
(Structure forces and deformations)
(Structure to be analysed)
L
1
2
3
(Free body diagram showing structure forces and
coordinates)
Solve the truss in example 3.2 to find member forces.
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The [K]matrix for the truss as formed in example 3.2 is as follows:
8
7
65
4
3
2
1
5.5.00005.5.
5.5.00005.5.
00414.1000414.1000000000
00005.5.5.5.
00005.5.5.5.
5.5.414.105.5.414.20
5.5.005.05.01
87654321
2
L
AEK
Now as W = k
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8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
5.5.00005.5.
5.5.00005.5.
00414.1000414.10
0000000000005.5.5.5.
00005.5.5.5.
5.5.414.105.5.414.20
5.5.005.05.01
87654321
2LAE
W
W
W
WW
W
W
W
k
u
u
k
kk
kk
W
W
2221
1211
As k= 0 or 3, 4, 5, 6, 7,8, all are zero due to supports and using[W
k] = [k
11] [
u] we get
W
W
AE
L
1
2
1
22
1 0
0 2 414
.
Which gives
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W1= 0
W2= P
Also using [Wu] = [k21] [u] we have
0
2
1 0
0 2 414
1
2P
AE
L
.
Or
PAE
L 0
414.20
0121
2
1
2 1
2 4142 414 0
0 10L
AE P.
..
Or
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1
2
2
2 414
0
L
AE P.
Or
22
2 414 LP
AE.Or
2 05858 .PL
AE
Now Wu= [K
21] [
u]
W
W
WW
W
W
AE
L
3
4
5
6
7
8
1
22
5 5
5 5
0 00 1414
5 5
5 5
. .
. .
.
. .
. .
W
W
W
W
W
W
P
3
4
5
6
7
8
1
2 414
5 5
5 5
0 0
0 1414
5 5
5 5
0
.
. .
. .
.
. .
. .
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W
W
W
W
W
W
P
P
P
P
P
3
4
5
6
7
8
1
2 414
0 5
0 5
0
1414
0 5
0 5
.
.
.
.
.
.
W3= 0.207 P
W4= -0.207 P
W5= 0
W6= -0.5858 PW
7= -0.207 P
W8= -0.207 P
Now as
l
k
j
i
s
r
mlml
mlml
L
AE
w
w
So
2
1
8
7
l
k
j
s
i
mlmlL
AEw
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So, for member # 1:
AE
LP
L
AE
wsw5858.0
0
0
0
707.0707.0707.0707.01
Or
ws= 0.707 x 0.5858 P
w1= 0.414 P
Similarl y, for member # 2:
6
5
2
1
0
0
5858.
0
10102
AE
PL
L
AEwws
Or w2= 1 x 0.5858 P
Or w2= 0.5858 P
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For member # 3:
4
3
2
1
l
k
js
i
mlmlL
AEW
0
0
5858.0
707.707.707.707.3 AE
PL
LAEw
Orw3= 0.707 x 0.5858 P
w3= 0.414 P
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P
0.207P 0.59P 0.207P
0.207P
0.207P
0.41
4P0.414P
0.5
86P
w1= 0.414 P
w2= 0.586 P
w3= 0.414 P
Example 3 5 Analyze the truss shown in the f igure
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Example 3.5 Analyze the truss shown in the f igure.
5k2kW1,1
W7,7
W8,8
W5,5W6,
6
W3,3
W4,4
W2,2
12
3
4
5
8' 6'
8'
Truss to be analysedStructure forces and
displacements
A and E are constant for each member.
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Member Length l m i j k l
1 10 -0.6 -0.8 7 8 1 2
2 11.34 -0.7071 0.7071 1 2 3 4
3 8 1 0 3 4 5 6
4 6 1 0 5 6 7 8
5 8 0 1 1 2 5 6
Using the properties given in the above table we can find the structures
stiffness matrices for each element as follows.
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Element forces and displacements
w , 5
w , 4
5
4
w , w ,
2
3 3
w , 9 9
6w ,
3
6w ,
5
w , 10
7
10
7
1
2 2
w ,
w ,
4
8
1
8
1
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50
6
5
4
3
0000
0125.00125.0
0000
0125.00125.0
k
6543
4
3
2
1
044.0044.0044.0044.0
044.0044.0044.0044.0
044.0044.0044.0044.0
044.0044.0044.0044.0
k
4321
21
8
7
064.0048.0064.0048.0048.0036.0048.0036.0
064.0048.0064.0048.0
048.0036.0048.0036.0
k
2187
3
2
1
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51
8
7
4
3
125.00125.00
0000
125.00125.00
0000
k
8743
8
7
6
5
0000
0167.00167.0
0000
0167.00167.0
k
8765
5
4
Using equation [K] = [K1]+[K2]+[K3]+[K4]+[K5] we get the structure
stiffness matrix of (8x8) dimensions. Partitioning this matrix with respectto known and unknown deformations we get [K11] and [K12] portions as
follows.
21
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665.9
983.62
2
5
233.0004.0
004.008.01
2
1
2
1
233.0004.0
004.008.0
21
11
K
8
7
6
5
4
3
064.0048.0
048.0036.0
125.00
00
044.0044.0
044.0004.0
21
12
K
Using equation [Du]=[K11]1[Wk] we get the
Putting the above value in equation [Wu]=[K21][Du]
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405.2803.1
208.1
0197.3
197.3
665.9
983.62
064.0048.0048.0036.0
125.00
00044.0044.0
044.0004.0
6
5
4
3
2
1
WW
W
WW
W
Element forces can be calculated using equation as follows.
0
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54
4.54kips
3.197 K
3.197 K
1.2
08
kips
-3.006
kip
s
2 K
5 K
1.803 K
2.405 K
0 K
1.208 K
kipw
kipw
kipw
kipw
kipw
208.1
0
0
665.9
983.62
10108
1
0
0
0
0
0
01016
1
0
0
0
0
0
01018
1
540.4
0
0
665.9
983.62
707.0707.0707.0707.0314.11
1
006.3
665.9
983.62
08.06.08.06.0
10
1
5
4
3
2
1
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