Hibbeler R.C. Structural Analysis
Подождите немного. Документ загружается.
EXAMPLE 14.2
Determine the structure stiffness matrix for the truss shown in
Fig. 14–8a. AE is constant.
SOLUTION
Although the truss is statically indeterminate to the first degree, this
will present no difficulty for obtaining the structure stiffness matrix.
Each joint and member are arbitrarily identified numerically, and the
near and far ends are indicated by the arrows along the members.
As shown in Fig. 14–8b, the unconstrained displacements are code
numbered first. There are eight degrees of freedom for the truss, and
so K will be an matrix. In order to keep all the joint coordinates
positive, the origin of the global coordinates is chosen at ①. Equations
14–5, 14–6, and 14–16 will now be applied to each member.
Member 1. Here so that
Member 2. Here so that
Member 3. Here so that
k
3
= AE
1234
D
0000
0 0.1 0 -0.1
0000
0 -0.1 0 0.1
T
1
2
3
4
l
x
=
0 - 0
10
= 0 l
y
=
10 - 0
10
= 1
L = 10 ft,
k
2
= AE
1278
D
0.035 0.035 -0.035 -0.035
0.035 0.035 -0.035 -0.035
-0.035 -0.035 0.035 0.035
-0.035 -0.035 0.035 0.035
T
1
2
7
8
l
x
=
10 - 0
1022
= 0.707 l
y
=
10 - 0
1022
= 0.707
L = 1022 ft,
k
1
= AE
12 65
D
0.1 0 -0.1 0
0000
-0.1 0 0.1 0
0000
T
1
2
6
5
l
x
=
10 - 0
10
= 1 l
y
=
0 - 0
10
= 0
L = 10 ft,
8 * 8
550
CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD
14
Fig. 14–8
10 ft
10 ft
(a)
10 ft
10 ft
4
y
3
4
4
8
3
6
2
5
5
x
6
2
1
1
2
3
(b)
1
7
6
14
Member 4. Here so that
Member 5. Here so that
Member 6. Here so that
Structure Stiffness Matrix. The foregoing six matrices can now be
assembled into the matrix by algebraically adding their
corresponding elements. For example, since
then,
and so on. The final result
is thus,
AE10.1352,K
11
= AE10.1 + 0.0352=
1k
11
2
3
= 1k
11
2
4
= 1k
11
2
5
= 1k
11
2
6
= 0,1k
11
2
2
= AE10.0352,
1k
11
2
1
= AE10.12,
8 * 8 K
k
6
= AE
6578
D
0000
0 0.1 0 -0.1
0000
0 -0.1 0 0.1
T
6
5
7
8
l
x
=
10 - 10
10
= 0
l
y
=
10 - 0
10
= 1
L = 10 ft,
k
5
= AE
34 65
D
0.035 -0.035 -0.035 0.035
-0.035 0.035 0.035 -0.035
-0.035 0.035 0.035 -0.035
0.035 -0.035 -0.035 0.035
T
3
4
6
5
l
x
=
10 - 0
1022
= 0.707
l
y
=
0 - 10
1022
=-0.707
L = 1022
ft,
k
4
= AE
34 78
D
0.1 0 -0.1 0
0000
-0.1 0 0.1 0
0000
T
3
4
7
8
l
x
=
10 - 0
10
= 1
l
y
=
10 - 10
10
= 0
L = 10 ft,
K = AE
1 2345678
H
0.135 0.035 0 0 0 -0.1 -0.035 -0.035
0.035 0.135 0 -0.1 0 0 -0.035 -0.035
0 0 0.135 -0.035 0.035 -0.035 -0.1 0
0 -0.1 -0.035 0.135 -0.035 0.035 0 0
0 0 0.035 -0.035 0.135 -0.035 0 -0.1
-0.1 0 -0.035 0.035 -0.035 0.135 0 0
-0.035 -0.035 -0.1 0 0 0 0.135 0.035
-0.035 -0.035 0 0 -0.1 0 0.035 0.135
X
1
2
3
4
5
6
7
8
Ans.
14.5 TRUSS STIFFNESS MATRIX 551
14.6 Application of the Stiffness Method
for Truss Analysis
Once the structure stiffness matrix is formed, the global force
components Q acting on the truss can then be related to its global
displacements D using
(14–17)
This equation is referred to as the structure stiffness equation. Since
we have always assigned the lowest code numbers to identify the
unconstrained degrees of freedom, this will allow us now to partition this
equation in the following form*:
(14–18)
Here
known external loads and displacements; the loads here exist
on the truss as part of the problem, and the displacements are
generally specified as zero due to support constraints such as
pins or rollers.
unknown loads and displacements; the loads here represent
the unknown support reactions, and the displacements are at
joints where motion is unconstrained in a particular direction.
structure stiffness matrix, which is partitioned to be compati-
ble with the partitioning of Q and D.
Expanding Eq. 14–18 yields
(14–19)
(14–20)
Most often since the supports are not displaced. Provided this is
the case, Eq. 14–19 becomes
Since the elements in the partitioned matrix represent the total
resistance at a truss joint to a unit displacement in either the x or y
direction, then the above equation symbolizes the collection of all the
force equilibrium equations applied to the joints where the external loads
are zero or have a known value Solving for we have
(14–21)
From this equation we can obtain a direct solution for all the unknown
joint displacements; then using Eq. 14–20 with yields
(14–22)
from which the unknown support reactions can be determined. The
member forces can be determined using Eq. 14–13, namely
q = k¿TD
Q
u
= K
21
D
u
D
k
= 0
D
u
= [K
11
]
-1
Q
k
D
u
,1Q
k
2.
K
11
Q
k
= K
11
D
u
D
k
= 0
Q
u
= K
21
D
u
+ K
22
D
k
Q
k
= K
11
D
u
+ K
12
D
k
K =
D
u
=Q
u
,
D
k
=Q
k
,
c
Q
k
Q
u
d= c
K
11
K
12
K
21
K
22
dc
D
u
D
k
d
Q = KD
552
CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD
14
*This partitioning scheme will become obvious in the numerical examples that follow.
14.6 APPLICATION OF THE STIFFNESS METHOD FOR TRUSS ANALYSIS 553
14
Expanding this equation yields
Since for equilibrium, only one of the forces has to be found.
Here we will determine the one that exerts tension in the member,
Fig. 14–2c.
(14–23)
In particular, if the computed result using this equation is negative, the
member is then in compression.
q
F
=
AE
L
[-l
x
-l
y
l
x
l
y
]D
D
Nx
D
Ny
D
Fx
D
Fy
T
q
F
,
q
N
=-q
F
c
q
N
q
F
d=
AE
L
c
1 -1
-11
dc
l
x
l
y
00
00l
x
l
y
dD
D
Nx
D
Ny
D
Fx
D
Fy
T
Procedure for Analysis
The following method provides a means for determining the unknown
displacements and support reactions for a truss using the stiffness method.
Notation
• Establish the x, y global coordinate system. The origin is usually located
at the joint for which the coordinates for all the other joints are positive.
• Identify each joint and member numerically, and arbitrarily specify the
near and far ends of each member symbolically by directing an arrow
along the member with the head directed towards the far end.
• Specify the two code numbers at each joint, using the lowest numbers to
identify unconstrained degrees of freedom, followed by the highest
numbers to identify the constrained degrees of freedom.
• From the problem, establish and
Structure Stiffness Matrix
• For each member determine and and the member stiffness matrix
using Eq. 14–16.
• Assemble these matrices to form the stiffness matrix for the entire truss
as explained in Sec. 14–5. As a partial check of the calculations, the
member and structure stiffness matrices should be symmetric.
Displacements and Loads
• Partition the structure stiffness matrix as indicated by Eq. 14–18.
• Determine the unknown joint displacements using Eq. 14–21, the
support reactions using Eq. 14–22, and each member force using
Eq. 14–23.
q
F
Q
u
D
u
l
y
l
x
Q
k
.D
k
554 CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD
14
Determine the force in each member of the two-member truss shown
in Fig. 14–9a. AE is constant.
SOLUTION
Notation. The origin of x, y and the numbering of the joints and
members are shown in Fig. 14–9b. Also, the near and far ends of each
member are identified by arrows, and code numbers are used at each
joint. By inspection it is seen that the known external displacements
are Also, the known external loads are
Hence,
Structure Stiffness Matrix. Using the same notation as used here,
this matrix has been developed in Example 14–1.
Displacements and Loads. Writing Eq. 14–17, for the
truss we have
Q = KD,
D
k
= D
0
0
0
0
T
3
4
5
6
Q
k
= c
0
-2
d
1
2
Q
2
=-2 k.Q
1
= 0,
D
3
= D
4
= D
5
= D
6
= 0.
EXAMPLE 14.3
(1)F
0
-2
Q
3
Q
4
Q
5
Q
6
V= AE F
0.405 0.096 -0.333 0 -0.072 -0.096
0.096 0.128 0 0 -0.096 -0.128
-0.333 0 0.333 0 0 0
000000
-0.072 -0.096 0 0 0.072 0.096
-0.096 -0.128 0 0 0.096 0.128
VF
D
1
D
2
0
0
0
0
V
From this equation we can now identify and thereby determine
It is seen that the matrix multiplication, like Eq. 14–19, yields
Here it is easy to solve by a direct expansion,
Physically these equations represent and applied
to joint ②. Solving, we get
D
1
=
4.505
AE
D
2
=
-19.003
AE
©F
y
= 0©F
x
= 0
-2 = AE10.096D
1
+ 0.128D
2
2
0 = AE10.405D
1
+ 0.096D
2
2
c
0
-2
d= AE c
0.405 0.096
0.096 0.128
dc
D
1
D
2
d+ c
0
0
d
D
u
.
K
11
Fig. 14–9
3 ft
4 ft
(a)
2 k
14
By inspection of Fig. 14–9b, one would indeed expect a rightward
and downward displacement to occur at joint ② as indicated by the
positive and negative signs of these answers.
Using these results, the support reactions are now obtained from
Eq. (1), written in the form of Eq. 14–20 (or Eq. 14–22) as
Expanding and solving for the reactions,
The force in each member is found from Eq. 14–23. Using the data
for and in Example 14–1, we have
Member 1:
Ans.
Member 2:
Ans.
These answers can of course be verified by equilibrium, applied at
joint ②.
=
1
5
[-0.614.5052- 0.81-19.0032] = 2.5 k
q
2
=
AE
5
1256
C
-0.6 -0.8 0.6 0.8
D
1
AE
D
4.505
-19.003
0
0
T
1
2
5
6
L = 5 ftl
y
= 0.8,l
x
= 0.6,
=
1
3
[-4.505] =-1.5 k
q
1
=
AE
3
1234
C
-1010
D
1
AE
D
4.505
-19.003
0
0
T
1
2
3
4
L = 3 ftl
y
= 0,l
x
= 1,
l
y
l
x
Q
6
=-0.09614.5052- 0.1281-19.0032= 2.0 k
Q
5
=-0.07214.5052- 0.0961-19.0032= 1.5 k
Q
4
= 0
Q
3
=-0.33314.5052=-1.5 k
D
Q
3
Q
4
Q
5
Q
6
T= AE D
-0.333 0
00
-0.072 -0.096
-0.096 -0.128
T
1
AE
c
4.505
-19.003
d+ D
0
0
0
0
T
2
3
1
6
5
2
1
4
3
1
2
x
y
(
b
)
2 k
14.6 APPLICATION OF THE STIFFNESS METHOD FOR TRUSS ANALYSIS 555
EXAMPLE 14.4
Determine the support reactions and the force in member 2 of the
truss shown in Fig. 14–10a. AE is constant.
SOLUTION
Notation. The joints and members are numbered and the origin of
the x, y axes is established at ①, Fig. 14–10b. Also, arrows are used to
reference the near and far ends of each member. Using the code
numbers, where the lowest numbers denote the unconstrained degrees
of freedom, Fig. 14–10b, we have
Structure Stiffness Matrix. This matrix has been determined in
Example 14–2 using the same notation as in Fig. 14–10b.
Displacements and Loads. For this problem isQ = KD
D
k
= C
0
0
0
S
6
7
8
Q
k
= E
0
0
2
-4
0
U
1
2
3
4
5
556
CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD
14
(1)H
0
0
2
-4
0
Q
6
Q
7
Q
8
X= AE H
0.135 0.035 0 0 0 -0.1 -0.035 -0.035
0.035 0.135 0 -0.1 0 0 -0.035 -0.035
0 0 0.135 -0.035 0.035 -0.035 -0.1 0
0 -0.1 -0.035 0.135 -0.035 0.035 0 0
0 0 0.035 -0.035 0.135 -0.035 0 -0.1
-0.1 0 -0.035 0.035 -0.035 0.135 0 0
-0.035 -0.035 -0.1 0 0 0 0.135 0.035
-0.035 -0.035 0 0 -0.1 0 0.035 0.135
XH
D
1
D
2
D
3
D
4
D
5
0
0
0
X
4
y
3
4
4
8
3
6
2
5
5
x
2
1
1
2
3
(b)
4 k
2 k
1
7
6
10 ft
10 ft
(a)
2 k
4 k
2
Fig. 14–10
Multiplying so as to formulate the unknown displacement
equation 14–18, we get
E
0
0
2
-4
0
U= AE E
0.135 0.035 0 0 0
0.035 0.135 0 -0.1 0
0 0 0.135 -0.035 0.035
0 -0.1 -0.035 0.135 -0.035
0 0 0.035 -0.035 0.135
UE
D
1
D
2
D
3
D
4
D
5
U+ E
0
0
0
0
0
U
14
Expanding and solving the equations for the displacements yields
Developing Eq. 14–20 from Eq. (1) using the calculated results, we
have
E
D
1
D
2
D
3
D
4
D
5
U=
1
AE
E
17.94
-69.20
-2.06
-87.14
-22.06
U
Expanding and computing the support reactions yields
C
Q
6
Q
7
Q
8
S= AE C
-0.1 0 -0.035 0.035 -0.035
-0.035 -0.035 -0.1 0 0
-0.035 -0.035 0 0 -0.1
S
1
AE
E
17.94
-69.20
-2.06
-87.14
-22.06
U+ C
0
0
0
S
Ans.
Ans.
Ans.
The negative sign for indicates that the rocker support reaction
acts in the negative x direction. The force in member 2 is found from
Eq. 14–23, where from Example 14–2,
Thus,
Ans. = 2.56 k
q
2
=
AE
1022
[-0.707 -0.707 0.707 0.707]
1
AE
D
17.94
-69.20
0
0
T
L = 1012
ft.
l
y
= 0.707,l
x
= 0.707,
Q
6
Q
8
= 4.0 k
Q
7
= 2.0 k
Q
6
=-4.0 k
14.6 APPLICATION OF THE STIFFNESS METHOD FOR TRUSS ANALYSIS 557
EXAMPLE 14.5
Determine the force in member 2 of the assembly in Fig. 14–11a if the
support at joint ① settles downward 25 mm. Take
SOLUTION
Notation. For convenience the origin of the global coordinates in
Fig. 14–11b is established at joint ③, and as usual the lowest code num-
bers are used to reference the unconstrained degrees of freedom.
Thus,
Structure Stiffness Matrix. Using Eq. 14–16, we have
Member 1: so that
Member 2: so that
Member 3: so that
By assembling these matrices, the structure stiffness matrix becomes
K = AE
1234 5678
H
0.378 0.096 0 0 -0.128 -0.096 -0.25 0
0.096 0.405 0 -0.333 -0.096 -0.072 0 0
00000000
0 -0.333 0 0.333 0 0 0 0
-0.128 -0.096 0 0 0.128 0.096 0 0
-0.096 -0.072 0 0 0.096 0.072 0 0
-0.25 0 0 0 0 0 0.25 0
00000000
X
1
2
3
4
5
6
7
8
k
3
= AE
7812
D
0.25 0 -0.25 0
0000
-0.25 0 0.25 0
0000
T
7
8
1
2
L = 4 m,l
y
= 0,l
x
= 1,
k
2
= AE
1256
D
0.128 0.096 -0.128 -0.096
0.096 0.072 -0.096 -0.072
-0.128 -0.096 0.128 0.096
-0.096 -0.072 0.096 0.072
T
1
2
5
6
L = 5 m,l
y
=-0.6,l
x
=-0.8,
k
1
= AE
3412
D
00 00
0 0.333 0 -0.333
00 00
0 -0.333 0 0.333
T
3
4
1
2
L = 3 m,l
y
= 1,l
x
= 0,
D
k
= F
0
-0.025
0
0
0
0
V
3
4
5
6
7
8
Q
k
= c
0
0
d
1
2
AE = 8110
3
2 kN.
558
CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD
14
7
1
3
2
1
2
1
2
4
3
4
8
6
5
3
(b)
x
y
1
3
2
1
4 m
3 m
(a)
2
4
3
Fig. 14–11
14
Displacements and Loads. Here yields
Developing the solution for the displacements, Eq. 14–19, we have
c
0
0
d= AE c
0.378 0.096
0.096 0.405
dc
D
1
D
2
d+ AE c
00 -0.128 -0.096 -0.25 0
0 -0.333 -0.096 -0.072 0 0
dF
0
-0.025
0
0
0
0
V
H
0
0
Q
3
Q
4
Q
5
Q
6
Q
7
Q
8
X= AE H
0.378 0.096 0 0 -0.128 -0.096 -0.25 0
0.096 0.405 0 -0.333 -0.096 -0.072 0 0
00000000
0 -0.333 0 0.333 0 0 0 0
-0.128 -0.096 0 0 0.128 0.096 0 0
-0.096 -0.072 0 0 0.096 0.072 0 0
-0.25 0 0 0 0 0 0.25 0
00000000
XH
D
1
D
2
0
-0.025
0
0
0
0
X
Q = KD
which yields
Solving these equations simultaneously gives
Although the support reactions do not have to be calculated, if needed
they can be found from the expansion defined by Eq. 14–20. Using
Eq. 14–23 to determine the force in member 2 yields
Member 2: so
that
Ans.
Using the same procedure, show that the force in member 1 is
and in member 3, The results are shown
on the free-body diagram of joint ②, Fig. 14–11c, which can be checked
to be in equilibrium.
q
3
= 11.1 kN.q
1
= 8.34 kN
=
8110
3
2
5
10.00444 - 0.01312=-13.9 kN
q
2
=
8110
3
2
5
[0.8 0.6 -0.8 -0.6]D
0.00556
-0.021875
0
0
T
AE = 8110
3
2 kN,L = 5 m,l
y
=-0.6,l
x
=-0.8,
D
2
=-0.021875 m
D
1
= 0.00556 m
0 = AE[10.096D
1
+ 0.405D
2
2+ 0.00833]
0 = AE[10.378D
1
+ 0.096D
2
2+ 0]
2
11.1 kN
8.34 kN
13.9 kN
3
4
5
(c)
14.6 APPLICATION OF THE STIFFNESS METHOD FOR TRUSS ANALYSIS 559