We can use the local-to-global DOF relationship in the global DOF table and place the member stiffness components directly into the global stiffness matrix.. For example, component 1,3 o
Trang 1P =
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
3 3 2 2 1 1
y x y x y x
P
P
P
P
P
P
=
1
2 2 1 1
0
0
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
y x y x
F F F F
+
2 3 3 2 2
0 0
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
y x y x
F F F
F
+
3 3 3
1 1
0 0
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
y x
y x
F F
F F
(14)
where the subscript outside of each vector on the RHS indicates the member number Each of the vectors at the RHS, however, can be expressed in terms of their respective
nodal displacement vector using Eq.12, with the nodal forces and displacements referring
to the global nodal force and displacement representation:
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
2
2
1
1
y
x
y
x
F
F
F
F
=
1 44 43 42 41
34 33 32 31
24 23 22 21
14 13 12 11
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
k k k k
k k k k
k k k k
k k k k
⎪
⎪
⎭
⎪⎪
⎬
⎫
⎪
⎪
⎩
⎪⎪
⎨
⎧
2 2 1 1
v u v u
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
3
3
2
2
y
x
y
x
F
F
F
F
=
2 44 43 42 41
34 33 32 31
24 23 22 21
14 13 12 11
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
k k k k
k k k k
k k k k
k k k k
⎪
⎪
⎭
⎪⎪
⎬
⎫
⎪
⎪
⎩
⎪⎪
⎨
⎧
3 3 2 2
v u v u
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
3
3
1
1
y
x
y
x
F
F
F
F
=
3 44 43 42 41
34 33 32 31
24 23 22 21
14 13 12 11
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
k k k k
k k k k
k k k k
k k k k
⎪
⎪
⎭
⎪⎪
⎬
⎫
⎪
⎪
⎩
⎪⎪
⎨
⎧
3 3 1 1
v u v u
Each of the above equations can be expanded to fit the form of Eq 14:
1
2
2
1
1
0
0
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
y
x
y
x
F
F
F
F
=
1
44 43 42 41
34 33 32 31
24 23 22 21
14 13 12 11
0 0 0 0 0 0
0 0 0 0 0 0
0 0
0 0
0 0
0 0
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
k k k k
k k k k
k k k k
k k k k
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
3 3 2 2 1 1
v u v u v u
Trang 22 3
3
2
2
0
0
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
y
x
y
x
F
F
F
F
=
2 44 43 42 41
34 33 32 31
24 23 22 21
14 13 12 11
0 0
0 0
0 0
0 0
0 0 0 0 0 0
0 0 0 0 0 0
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
k k k k
k k k k
k k k k
k k k k
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
3 3 2 2 1 1
v u v u v u
3 3
3
1
1
0
0
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
y
x
y
x
F
F
F
F
=
3 44 43 42
41
34 33 32
31
24 23 22
21
14 13 12
11
0 0
0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0
0 0
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
k k k
k
k k k
k
k k k
k
k k k
k
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
3 3 2 2 1 1
v u v u v u
When each of the RHS vectors in Eq 14 is replaced by the RHS of the above three
equations, the resulting equation is the unconstrained global stiffness equation:
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
66 65 64 63 62 61
56 55 54 53 52 51
46 45 44 43 42 41
36 35 34 33 32 31
26 25 24 23 22 21
16 15 14 13 12 11
K K K K K K
K K K K K K
K K K K K K
K K K K K K
K K K K K K
K K K K K K
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
3 3 2 2 1 1
v u v u v u
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
3 3 2 2 1 1
y x y x y x
P P P P P P
(15)
where the components of the unconstrained global stiffness matrix, K ij, is the
superposition of the corresponding components in each of the three expanded stiffness matrices in the equations above
In actual computation, it is not necessary to expand the stiffness equation in Eq 12 into
the 6-equation form as we did earlier That was necessary only for the understanding of how the results are derived We can use the local-to-global DOF relationship in the global DOF table and place the member stiffness components directly into the global stiffness matrix For example, component (1,3) of the member-2 stiffness matrix is added
to component (3,5) of the global stiffness matrix This simple way of assembling the global stiffness matrix is called the Direct Stiffness Method
To carry out the above procedures numerically, we need to use the dimension and
member property given at the beginning of this section to arrive at the stiffness matrix for each of the three members:
Trang 31 2 2
2 2
2 2
2 2
1
)
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
S CS S
CS
CS C
CS C
S CS S
CS
CS C
CS C
L
EA
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
8 12 6 9 8 12 6 9
6 9 2 7 6 9 2 7
8 12 6 9 8 12 6 9
6 9 2 7 6 9 2 7
(kG)2=(
2 2 2
2 2
2 2
2 2
2
)
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
S CS S
CS
CS C
CS C
S CS S
CS
CS C
CS C
L
EA
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
−
−
8 12 6 9 8 12 6 9
6 9 2 7 6 9 2 7
8 12 6 9 8 12 6 9
6 9 2 7 6 9 2 7
(kG)3=(
3 2 2
2 2
2 2
2 2
3
)
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
S CS S
CS
CS C
CS C
S CS S
CS
CS C
CS C
L
EA
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
0 6 9 0 0
0 7 16 0 7 16
0 0 0
0
0 7 16 0 7 16
When the three member stiffness matrices are assembled according to the Direct Stiffness Method, the unconstrained global stiffness equation given at the beginning of this section
is obtained For example, the unconstrained global stiffness matrix component k34 is the
superposition of (k34)1 of member 1 and (k12)2 of member 2 Note that the unconstrained global stiffness matrix has the same features as the member stiffness matrix: symmetric and singular, etc
Example 2 Now consider the same three-bar truss as shown before with E=70 GPa,
A=1,430 mm2 for each member but with the support and loading conditions added
A constrained and loaded truss in a global coordinate system.
x
y
1
2
4m
2m
3 3m
3
1.0 MN 0.5 MN
Trang 4Solution. The support conditions are: u1=0, v1=0, and v3=0 The loading conditions are:
Px2=0.5 MN, Py2= −1.0 MN, and Px3=0 The stiffness equation given at the beginning of the last section now becomes
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
8 12 6 9 8 12 6 9 0 0
6 19 9 23 6 9 2 7 0 6
16
8 12 6 9 6 25 0 8
12 6
9
6 9 2 7 0 4
14 6 9 2
7
0 0
8 12 6 9 8 12 6
9
0 6
16 6 9 2 7 6 9 9
23
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
0
0 0
3 2 2
u v
u
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
−
3
1 1
0
0 1
5 0
y
y x
P
P P
Note that there are exactly six unknown in the six equations The solution of the six unknowns is obtained in two steps In the first step, we notice that the three equations, 3rd through 5th , are independent from the other three and can be dealt with separately
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
− 9 23 6 9 2
7
6 9 6 25 0
2 7 0 4
14
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
3 2 2
u v
u
=
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
− 0
0 1
5 0
(16)
Eq 16 is the constrained stiffness equation of the loaded truss The constrained 3x3 stiffness matrix is symmetric but not singular The solution of Eq 16 is: u2=0.053 m,
v2= −0.053 m, and u3= 0.037 m In the second step, the reactions are obtained from the direct substitution of the displacement values into the other three equations, 1st, 2nd and
6th :
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
8 12 6 9 8 12 6 9 0 0
0 0 8
12 6 9 8 12 6
9
0 6 16 6 9 2 7 6 9 9
23
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
− 0
037 0
053 0
053 0 0 0
=
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧−
83 0
17 0
5 0
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
3 1 1
y y x
P P P
or
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
3
1
1
y
y
x
P
P
P
=
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧−
83 0
16 0
5 0 MN
The member deformation represented by the member elongation can be computed by the
member deformation equation, Eq 3:
Trang 5Member 1:
∆1 = ⎣−C −S C S⎦1
⎪
⎪
⎭
⎪⎪
⎬
⎫
⎪
⎪
⎩
⎪⎪
⎨
⎧
2 2 1 1
v u v u
= ⎣−0.6 −0.8 0.6 0.8⎦1
⎪
⎪
⎭
⎪⎪
⎬
⎫
⎪
⎪
⎩
⎪⎪
⎨
⎧
− 0530
053 0 0 0
= -0.011m
For member 2 and member 3, the elongations are ∆2= −0.052m, and ∆3=0.037m
The member forces are computed using Eq 1.
F=k∆ =
L
EA
∆ F1= −0.20 MN, F2= −1.04 MN, F3= 0.62 MN The results are summarized in the following table
Nodal and Member Solutions
Node
x-direction y-direction x-direction y-direction
Problem 2 Consider the same three-bar truss as that in Example 2, but with a different
numbering system for members Construct the constrained stiffness equation, Eq 16.
Problem 2.
x
y
1
2
4m
2m
3 3m
2
1.0 MN 0.5 MN
Trang 66 Procedures of Truss Analysis
Example 3 Consider the following two truss problems, each with member properties
E=70 GPa and A=1,430 mm2 The only difference is the existence of an additional diagonal member in the second truss It is instructive to see how the analyses and results differ
Two truss problems.
Solution. We will carry out a step-by-step solution procedure for the two problems, referring to the truss at the left and at the right as the first and second truss, respectively
We also define the global coordinate system in both cases as one with the origin at node 1
and its x-and y-direction coincide with the horizontal and vertical directions, respectively.
(1) Number the nodes and members and define the nodal coordinates
Nodal Coordinates Node x (m) y (m)
(2) Define member property, starting and end nodes and compute member data
1 MN 0.5 MN
3 2
4 m
3 m
1
2
3
4 5
1 MN 0.5 MN
3 2
4 m
3 m
1
2
3
4
Trang 7Member Data
Member
* S Node and E Node represent Starting and End nodes.
(3) Compute member stiffness matrices
kG =
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
2 2
2 2
2 2
2 2
S CS S
CS
CS C
CS C
S CS S
CS
CS C
CS C
L
EA
Member 1:
(kG)1=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
− 25 0 25 0
0 0 0 0
25 0 25 0
0 0 0 0
Member 2:
(kG)2=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
0 0
0 0
0 33
33 0 33
33
0 0
0 0
0 33
33 0 33
33
Member 3:
(kG)3=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
− 25 0 25 0
0 0 0 0
25 0 25 0
0 0 0 0
Member 4:
Trang 8⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
0 0
0 0
0 33
33 0 33
33
0 0
0 0
0 33
33 0 33
33
Member 5:
(kG)5=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
8 12 6 9 25 6
9
6 9 2 7 6 9 2 7
25 6
9 8 12 6 9
6 9 2 7 6 9 2 7
Member 6(for the second truss only):
(kG)6=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
8 12 6 9 8 12 6 9
6 9 2 7 6 9 2 7
8 12 6 9 8 12 6 9
6 9 2 7 6 9 2 7
(4) Assemble the unconstrained global stiffness matrix
In order to use the Direct Stiffness Method to assemble the global stiffness matrix, we need the following table which gives the global DOF number corresponding to each local DOF of each member This table is generated using the member data given in the table in subsection (2), namely the starting and end nodes data
Global DOF Number for Each Member
Global DOF Number for Member Local DOF
* For the second truss only
Armed with this table we can easily direct the member stiffness components to the right location in the global stiffness matrix For example, the (2,3) component of (kG)5 will be added to the (4,7) component of the global stiffness matrix The unconstrained global stiffness matrix is obtained after all the assembling is done
For the first truss:
Trang 9K1 =
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
80 37 60 9 00 25 0 80
12 60 9 0 0
60 9 53 40 0 0
60 9 20 7 0 33
33
00 25 0 00
25 0 0
0 0
0
0 0
0 33
33 0 33
33 0 0
8 12 60 9 0 0
80 37 60 9 00 25 0
60 9 20 7 0 33
33 60 9 53 40 0 0
0 0
0 0
00 25 0 00
25 0
0 33
33 0 0
0 0
0 33
33
For the second truss:
K2 =
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
80 37 60 9 00 25 0 80
12 60 9 0 0
60 9 53 40 0 0
60 9 20 7 0 33
33
00 25 0 80
37 60 9 0 0
80 12 60 9
0 0
60 9 52 40 0 33
33 60 9 20 7
8 12 60 9 0 0
80 37 60 9 00 25 0
60 9 20 7 0 33
33 60 9 53 40 0 0
0 0
80 12 60 9 00 25 0 80
37 60 9
0 33
33 60 9 20 7 0 0
60 9 53 40
Note that K2 is obtained by adding (KG)6 to K1 at the proper locations in columns and rows 1, 2, 5, and 6 (enclosed in dashed lines above)
(5) Assemble the constrained global stiffness equation
Once the support and loading conditions are incorporated into the stiffness equations we obtain:
For the first truss:
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
80 37 60 9 00 25 0 80
12 60 9 0 0
60 9 53 40 0 0
60 9 20 7 0 33
.
33
00 25 0 00
25 0 0
0 0
0
0 0
0 33
33 0 33
33 0 0
8 12 60 9 0 0
80 37 60 9 00 25 0
60 9 20 7 0 33
33 60 9 53 40 0 0
0 0
0 0
00 25 0 00
25 0
0 33
33 0 0
0 0
0 33
.
33
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎨
⎧
0
0 0
4 3 3 2 2
u v u v u
=
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎨
⎧
−
4 0 0 0
0 1
5 0
1 1
y P
P
P
y x
For the second truss:
Trang 10⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
80 37 60 9 00 25 0 80
12 60 9 0 0
60 9 53 40 0 0
60 9 20 7 0 33
.
33
00 25 0 80
37 60 9 0 0
80 12 60
.
9
0 0
60 9 52 40 0 33
33 60 9 20
.
7
8 12 60 9 0 0
80 37 60 9 00 25 0
60 9 20 7 0 33
33 60 9 53 40 0 0
0 0
80 12 60 9 00 25 0 80
37 60
.
9
0 33
33 60 9 20 7 0 0
60 9 53
.
40
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎨
⎧
0
0 0
4 3 3 2 2
u v u v u
=
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎨
⎧
−
4 0 0 0
0 1
5 0
1 1
y P
P
P
y x
(6) Solve the constrained global stiffness equation
The constrained global stiffness equation in either case contains five equations
corresponding to the third through seventh equations (enclosed in dashed lines above) that are independent from the other three equations and can be solved for the five
unknown nodal displacements
For the first truss:
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
53 40 0 0 60
9 20
.
7
0 00
25 0 0
0
0 0
33 33 0 33
.
33
60 9 0 0 80
37 60
.
9
20 7 0 33 33 60 9 53
.
40
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
4 3 3
2 2
u v u v u
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
−
0 0 0
0 1
5 0
For the second truss:
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
53 40 0 0 60
9 20
.
7
0 80
37 60 9 0 0
0 60
9 52 40 0 33
.
33
60 9 0 0 80
37 60
.
9
20 7 0 33 33 60 9 53
.
40
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
4 3 3
2 2
u v u v u
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
−
0 0 0
0 1
5 0
The reactions are computed by direct substitution
For the first truss:
Trang 11⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
80 37 60 9 00 25 0 80 12 60 9 0 0
0 0 0
0 00 25 0 00 25 0
0 33 33 0 0
0 0
0 33
.
33
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎨
⎧
0
0 0
4 3 3 2 2
u v u v u
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
4 1 1
y y x
P P P
For the second truss:
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
−
80 37 60 9 00 25 0 80
12 60 9 0 0
0 0 80
12 60 9 00 25 0 80 37 60
.
9
0 33 33 60 9 20 7 0 0
60 9 53
.
40
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎨
⎧
0
0 0
4 3 3 2 2
u v u v u
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
4 1 1
y y x
P P P
Results will be summarized at the end of the example
(7) Compute the member elongations and forces
For a typical member i:
∆i = ⎣−C −S C S⎦i
i
v u v u
⎪
⎪
⎭
⎪⎪
⎬
⎫
⎪
⎪
⎩
⎪⎪
⎨
⎧
2 2 1 1
F i =(k∆)ι = (
L
EA
∆)ι
(8) Summarizing results
Trang 12Results for the First Truss
Node
x-direction y-direction x-direction y-direction
Results for the Second Truss
Node
x-direction y-direction x-direction y-direction
Note that the reactions at node 1 and 4 are identical in the two cases, but other results are changed by the addition of one more diagonal member
(9) Concluding remarks
If the number of nodes is N and the number of constrained DOF is C, then
(a) the number of simultaneous equations in the unconstrained stiffness equation is
2N.
(b) the number of simultaneous equations for the solution of unknown nodal