Hibbeler R.C. Structural Analysis
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14.9 Space-Truss Analysis
The analysis of both statically determinate and indeterminate space trusses
can be performed by using the same procedure discussed previously. To
account for the three-dimensional aspects of the problem, however,
additional elements must be included in the transformation matrix T.In
this regard, consider the truss member shown in Fig. 14–17. The stiffness
matrix for the member defined in terms of the local coordinate is given
by Eq. 14–4. Furthermore, by inspection of Fig. 14–17, the direction cosines
between the global and local coordinates can be found using equations
analogous to Eqs. 14–5 and 14–6, that is,
(14–31)
(14–32)
(14–33)
As a result of the third dimension, the transformation matrix, Eq. 14–9,
becomes
Substituting this and Eq. 14–4 into Eq. 14–15, yields
k = F
l
x
0
l
y
0
l
z
0
0 l
x
0 l
y
0 l
z
V
AE
L
c
1 -1
-11
dc
l
x
l
y
l
z
000
000l
x
l
y
l
z
d
k = T
T
k¿T,
T = c
l
x
l
y
l
z
000
000l
x
l
y
l
z
d
=
z
F
- z
N
21x
F
- x
N
2
2
+ 1y
F
- y
N
2
2
+ 1z
F
- z
N
2
2
l
z
= cos u
z
=
z
F
- z
N
L
=
y
F
- y
N
21x
F
- x
N
2
2
+ 1y
F
- y
N
2
2
+ 1z
F
- z
N
2
2
l
y
= cos u
y
=
y
F
- y
N
L
=
x
F
- x
N
21x
F
- x
N
2
2
+ 1y
F
- y
N
2
2
+ 1z
F
- z
N
2
2
l
x
= cos u
x
=
x
F
- x
N
L
x¿
570
CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD
14
F
x¿
L
6
5
4
3
2
1
N
x
z
y
y
N
x
N
y
F
x
F
z
F
z
N
u
y
u
x
u
z
Fig. 14–17
Carrying out the matrix multiplication yields the symmetric matrix
k =
AE
L
N
x
N
y
N
z
F
x
F
y
F
z
F
l
x
2
l
x
l
y
l
x
l
z
l
y
l
x
l
y
2
l
y
l
z
l
z
l
x
l
z
l
y
l
z
2
-l
x
2
-l
x
l
y
-l
x
l
z
-l
y
l
x
-l
y
2
-l
y
l
z
-l
z
l
x
-l
z
l
y
-l
z
2
-l
x
2
-l
x
l
y
-l
x
l
z
-l
y
l
x
-l
y
2
-l
y
l
z
-l
z
l
x
-l
z
l
y
-l
z
2
l
x
2
l
x
l
y
l
x
l
z
l
y
l
x
l
y
2
l
y
l
z
l
z
l
x
l
z
l
y
l
z
2
V
N
x
N
y
N
z
F
x
F
y
F
z
14
(14–34)
This equation represents the member stiffness matrix expressed in
global coordinates. The code numbers along the rows and columns reference
the x, y, z directions at the near end, followed by those at
the far end,
For computer programming, it is generally more efficient to use
Eq. 14–34 than to carry out the matrix multiplication for each
member. Computer storage space is saved if the “structure” stiffness
matrix K is first initialized with all zero elements; then as the elements
of each member stiffness matrix are generated, they are placed directly
into their respective positions in K. After the structure stiffness matrix
has been developed, the same procedure outlined in Sec. 14–6 can be
followed to determine the joint displacements, support reactions, and
internal member forces.
T
T
k¿T
F
z
.F
y
,F
x
,
N
z
,N
y
,N
x
,
CHAPTER REVIEW
The stiffness method is the preferred method for analyzing structures using a computer. It first requires identifying the
number of structural elements and their nodes. The global coordinates for the entire structure are then established, and
each member’s local coordinate system is located so that its origin is at a selected near end, such that the positive axis
extends towards the far end.
Formulation of the method first requires that each member stiffness matrix be constructed. It relates the loads at the
ends of the member, q, to their displacements, d, where . Then, using the transformation matrix T, the local
displacements d are related to the global displacements D, where . Also, the local forces q are transformed
into the global forces Q using the transformation matrix T, i.e., . When these matrices are combined, one
obtains the member’s stiffness matrix K in global coordinates, . Assembling all the member stiffness matrices
yields the stiffness matrix K for the entire structure.
The displacements and loads on the structure are then obtained by partitioning , such that the unknown
displacements are obtained from , provided the supports do not displace. Finally, the support reactions
are obtained from , and each member force is found from .q = k¿TDQ
u
= K
21
, D
u
D
u
= 3K
11
4
-1
Q
k
Q = KD
k = T
T
k¿T
Q = T
T
q
d = TD
q = k¿d
k¿
x¿
The structural framework of this aircraft
hangar is constructed entirely of trusses in
order to reduce significantly the weight of
the structure. (Courtesy of Bethlehem Steel
Corporation)
CHAPTER REVIEW 571
572 CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD
14
*14–4. Determine the stiffness matrix K for the truss.Take
ksi.
14–5. Determine the horizontal displacement of joint ①
and the force in member . Take ,
ksi.
14–6. Determine the force in member if its temperature
is increased by . Take ,
.a = 6.5(10
-6
)>°F
E = 29(10
3
) ksi,A = 0.75 in
2
100°F
ƒ
2
ƒ
E = 29(10
3
)
A = 0.75 in
2
ƒ
2
ƒ
E = 29(10
3
)A = 0.75 in
2
,
14–9. Determine the stiffness matrix K for the truss. Take
and GPa for each member.
14–10. Determine the force in member . Take
and GPa for each member.
14–11. Determine the vertical displacement of node
② if member was 10 mm too long before it was fitted
into the truss. For the solution, remove the 20-k load. Take
and GPa for each member.E = 200A = 0.0015 m
2
ƒ
6
ƒ
E = 200A = 0.0015 m
2
ƒ
5
ƒ
E = 200A = 0.0015 m
2
14–7. Determine the stiffness matrix K for the truss. Take
and GPa for each member.
*14–8. Determine the vertical displacement at joint ②
and the force in member . Take and
GPa.E = 200
A = 0.0015 m
2
ƒ
5
ƒ
E = 200A = 0.0015 m
2
14–1. Determine the stiffness matrix K for the assembly.
Take and ksi for each member.
14–2. Determine the horizontal and vertical displacements
at joint ③ of the assembly in Prob. 14–1.
14–3. Determine the force in each member of the assem-
bly in Prob. 14–1.
E = 29(10
3
)A = 0.5 in
2
PROBLEMS
Probs. 14–4/14–5/14–6
Probs. 14–7/14–8
Probs. 14–9/14–10/14–11
Probs. 14–1/14–2/14–3
10
9
3
3
6
5
4
4
3 m
4 m 4 m
20 kN
4
7
6
1
2
2
1
5
3
5
7
8
1 2
2
4 ft 3 ft
4 ft
1
2
500 lb
8
7
6
4
5
3
2
3
4
1
1
3
7
2
1
3
4 ft 6 ft
3 ft
3 ft
4 k
3
3
5
8
1
6
4
24
1
2
30 kN
10
9
3
3
6
5
4
4
2 m
2 m2 m
4
6
1
1
2
5
3
5
7
1
2
2
8
14
14–15. Determine the stiffness matrix K for the truss. AE
is constant.
*14–16. Determine the vertical displacement of joint ②
and the support reactions. AE is constant.
*14–12. Determine the stiffness matrix K for the truss.
Take ksi.
14–13. Determine the horizontal displacement of joint ②
and the force in member . Take
ksi. Neglect the short link at
➁.
14–14. Determine the force in member if this member
was 0.025 in. too short before it was fitted onto the truss. Take
ksi. Neglect the short link at
➁.E = 29(10
3
)A = 2 in
2
.
ƒ
3
ƒ
E = 29(10
3
)A = 2 in
2
,
ƒ
5
ƒ
E = 29(10
3
)A = 2 in
2
,
14–17. Use a computer program and determine the
reactions on the truss and the force in each member. AE
is constant.
Probs. 14–12/14–13/14–14
Probs. 14–15/14–16
Prob. 14–17
Prob. 14–18
14–18. Use a computer program and determine the
reactions on the truss and the force in each member. AE
is constant.
5
1
6
5
3
6 ft
1
2
4
3
2
1
7
8
3 k
4
3
2
4
6
8 ft
E
C
D
8 ft8 ft8 ft
6 ft
6 ft
6 ft
B
A
2 k
F
1.5 k
3
2
1
3 m
4 m
6
5
2
1
3
4
2
1
3
3 kN
45
4 kN
F
ED
C
B
A
60
2 m
2 m
2 m
PROBLEMS 573
The statically indeterminate loading in bridge girders that are continuous
over their piers can be determined using the stiffness method.
15
575
The concepts presented in the previous chapter will be extended here
and applied to the analysis of beams. It will be shown that once the
member stiffness matrix and the transformation matrix have been
developed, the procedure for application is exactly the same as that
for trusses. Special consideration will be given to cases of differential
settlement and temperature.
15.1 Preliminary Remarks
Before we show how the stiffness method applies to beams, we will first
discuss some preliminary concepts and definitions related to these
members.
Member and Node Identification. In order to apply the
stiffness method to beams, we must first determine how to subdivide the
beam into its component finite elements. In general, each element must
be free from load and have a prismatic cross section. For this reason the
nodes of each element are located at a support or at points where
members are connected together, where an external force is applied,
where the cross-sectional area suddenly changes, or where the vertical or
rotational displacement at a point is to be determined. For example,
consider the beam in Fig. 15–1a. Using the same scheme as that for
trusses, four nodes are specified numerically within a circle, and the three
elements are identified numerically within a square. Also, notice that the
“near” and “far” ends of each element are identified by the arrows written
alongside each element.
Beam Analysis Using
the Stiffness Method
(a)
x
y
2 3
1
1
3
5
6
2
4
7
1
8
3
P
2
4
Fig. 15–1
Global and Member Coordinates. The global coordinate
system will be identified using x, y, z axes that generally have their origin
at a node and are positioned so that the nodes at other points on the
beam all have positive coordinates, Fig. 15–1a. The local or member
coordinates have their origin at the “near” end of each element,
and the positive axis is directed towards the “far” end. Figure 15–1b
shows these coordinates for element 2. In both cases we have used a
right-handed coordinate system, so that if the fingers of the right hand
are curled from the axis towards the axis, the thumb points
in the positive direction of the axis, which is directed out of the
page. Notice that for each beam element the x and axes will be
collinear and the global and member coordinates will all be parallel.
Therefore, unlike the case for trusses, here we will not need to develop
transformation matrices between these coordinate systems.
Kinematic Indeterminacy. Once the elements and nodes have
been identified, and the global coordinate system has been established, the
degrees of freedom for the beam and its kinematic determinacy can be
determined. If we consider the effects of both bending and shear, then each
node on a beam can have two degrees of freedom, namely, a vertical
displacement and a rotation. As in the case of trusses, these linear and
rotational displacements will be identified by code numbers. The lowest
code numbers will be used to identify the unknown displacements
(unconstrained degrees of freedom), and the highest numbers are used to
identify the known displacements (constrained degrees of freedom). Recall
that the reason for choosing this method of identification has to do with the
convenience of later partitioning the structure stiffness matrix, so that the
unknown displacements can be found in the most direct manner.
To show an example of code-number labeling, consider again the con-
tinuous beam in Fig. 15–1a. Here the beam is kinematically indeterminate
to the fourth degree. There are eight degrees of freedom, for which code
numbers 1 through 4 represent the unknown displacements, and numbers
5 through 8 represent the known displacements, which in this case are all
zero. As another example, the beam in Fig. 15–2a can be subdivided into
three elements and four nodes. In particular, notice that the internal hinge
at node 3 deflects the same for both elements 2 and 3; however, the
rotation at the end of each element is different. For this reason three code
numbers are used to show these deflections. Here there are nine degrees
of freedom, five of which are unknown, as shown in Fig. 15–2b, and four
known; again they are all zero. Finally, consider the slider mechanism
used on the beam in Fig. 15–3a. Here the deflection of the beam is shown
in Fig. 15–3b, and so there are five unknown deflection components
labeled with the lowest code numbers. The beam is kinematically
indeterminate to the fifth degree.
Development of the stiffness method for beams follows a similar
procedure as that used for trusses. First we must establish the stiffness
matrix for each element, and then these matrices are combined to
form the beam or structure stiffness matrix. Using the structure
x¿
z 1z¿2
y 1y¿2x 1x¿2
x¿
z¿y¿,
x¿,
576
CHAPTER 15 BEAM ANALYSIS U SING THE STIFFNESS METHOD
15
2
x¿
y¿
(b)
2
3
1 2
3
3
2
4
P
5
8
4
1
7
6
2
3
(a)
1
9
D
1
D
2
D
4
D
5
D
3
(
b
)
1
2
3
1
3
2
4
P
4
6
5
7
9
3
8
12
(
a
)
D
4
D
2
D
3
D
3
D
1
D
5
(
b
)
Fig. 15–2
Fig. 15–3
Fig. 15–1
15.2 BEAM-MEMBER STIFFNESS MATRIX 577
15
matrix equation, we can then proceed to determine the unknown
displacements at the nodes and from this determine the reactions on
the beam and the internal shear and moment at the nodes.
15.2 Beam-Member Stiffness Matrix
In this section we will develop the stiffness matrix for a beam element or
member having a constant cross-sectional area and referenced from the
local coordinate system, Fig. 15–4. The origin of the coordinates
is placed at the “near” end N, and the positive axis extends toward the
“far” end F. There are two reactions at each end of the element,
consisting of shear forces and and bending moments and
These loadings all act in the positive coordinate directions.
In particular, the moments and are positive counterclockwise,
since by the right-hand rule the moment vectors are then directed along
the positive axis, which is out of the page.
Linear and angular displacements associated with these loadings also
follow this same positive sign convention. We will now impose each of
these displacements separately and then determine the loadings acting
on the member caused by each displacement.
z¿
q
Fz¿
q
Nz¿
q
Fz¿
.
q
Nz¿
q
Fy¿
q
Ny¿
x¿
z¿y¿,x¿,
x¿
y¿
N
q
Ny¿
d
Ny¿
q
Nz¿
d
Nz¿
F
q
Fy¿
d
Fy¿
positive sign convention
q
Fz¿
d
Fz¿
Displacements. When a positive displacement is imposed
while other possible displacements are prevented, the resulting shear
forces and bending moments that are created are shown in Fig. 15–5a.
In particular, the moment has been developed in Sec. 11–2 as Eq. 11–5.
Likewise, when is imposed, the required shear forces and bending
moments are given in Fig. 15–5b.
d
Fy¿
d
Ny¿
y
œ
y¿
x¿
(a)
y¿ displacements
y¿
x¿
(b)
d
Fy¿
12EI
q
Ny¿
_____
d
Ny¿
L
3
12EI
q
Fy¿
_____
d
Ny¿
L
3
12EI
q
Ny¿
_____
d
Fy¿
L
3
12EI
q
Fy¿
_____
d
Fy¿
L
3
6EI
q
Nz¿
____
d
Ny¿
L
2
6EI
q
Nz¿
____
d
Fy¿
L
2
6EI
q
Fz¿
____
d
Fy¿
L
2
6EI
q
Fz¿
____
d
Ny¿
L
2
d
Ny¿
Fig. 15–4
Fig. 15–5
Rotations. If a positive rotation is imposed while all other
possible displacements are prevented, the required shear forces and
moments necessary for the deformation are shown in Fig. 15–6a.
In particular, the moment results have been developed in Sec. 11–2 as
Eqs. 11–1 and 11–2. Likewise, when is imposed, the resultant
loadings are shown in Fig. 15–6b.
By superposition, if the above results in Figs. 15–5 and 15–6 are added,
the resulting four load-displacement relations for the member can be
expressed in matrix form as
These equations can also be written in abbreviated form as
(15–2)
The symmetric matrix k in Eq. 15–1 is referred to as the member stiffness
matrix. The 16 influence coefficients that comprise it account for the
shear-force and bending-moment displacements of the member. Physically
these coefficients represent the load on the member when the member
undergoes a specified unit displacement. For example, if Fig. 15–5a,
while all other displacements are zero, the member will be subjected
only to the four loadings indicated in the first column of the k matrix. In
a similar manner, the other columns of the k matrix are the member
loadings for unit displacements identified by the degree-of-freedom code
numbers listed above the columns. From the development, both equilibrium
and compatibility of displacements have been satisfied. Also, it should be
noted that this matrix is the same in both the local and global coordinates
since the axes are parallel to x, y, z and, therefore, transformation
matrices are not needed between the coordinates.
z¿y¿,x¿,
d
Ny¿
= 1,
k
ij
q = kd
H
q
Ny¿
q
Nz¿
q
Fy¿
q
Fz¿
X
=
N
y¿
N
z¿
F
y¿
F
z¿
H
12EI
L
3
6EI
L
2
-
12EI
L
3
6EI
L
2
6EI
L
2
4EI
L
-
6EI
L
2
2EI
L
-
12EI
L
3
-
6EI
L
2
12EI
L
3
-
6EI
L
2
6EI
L
2
2EI
L
-
6EI
L
2
4EI
L
X
H
d
Ny¿
d
Nz¿
d
Fy¿
d
Fz¿
X
d
Fz¿
d
Nz¿
z
œ
578 CHAPTER 15 BEAM ANALYSIS U SING THE STIFFNESS METHOD
15
d
Nz¿
(
a
)
6EI
q
Ny¿
____
d
Nz¿
L
2
6EI
q
Fy¿
____
d
Nz¿
L
2
2EI
q
Fz¿
____
d
Nz¿
L
4EI
q
Nz¿
____
d
Nz¿
L
x¿
y¿
d
Fz¿
6EI
q
Ny¿
____
d
Fz¿
L
2
6EI
q
Fy¿
____
d
Fz¿
L
2
4EI
q
Fz¿
____
d
Fz¿
L
2EI
q
Nz¿
____
d
Fz¿
L
y¿
x¿
Fig. 15–6
(15–1)
15.4 APPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 579
15
15.3 Beam-Structure Stiffness Matrix
Once all the member stiffness matrices have been found, we must
assemble them into the structure stiffness matrix K. This process depends
on first knowing the location of each element in the member stiffness
matrix. Here the rows and columns of each k matrix (Eq. 15–1) are
identified by the two code numbers at the near end of the member
followed by those at the far end Therefore, when
assembling the matrices, each element must be placed in the same location
of the K matrix. In this way, K will have an order that will be equal to the
highest code number assigned to the beam, since this represents the total
number of degrees of freedom. Also, where several members are
connected to a node, their member stiffness influence coefficients will
have the same position in the K matrix and therefore must be
algebraically added together to determine the nodal stiffness influence
coefficient for the structure. This is necessary since each coefficient
represents the nodal resistance of the structure in a particular direction
( or ) when a unit displacement ( or ) occurs either at the same
or at another node. For example, represents the load in the direction
and at the location of code number “2” when a unit displacement occurs
in the direction and at the location of code number “3.”
15.4 Application of the Stiffness Method
for Beam Analysis
After the structure stiffness matrix is determined, the loads at the nodes
of the beam can be related to the displacements using the structure
stiffness equation
Here Q and D are column matrices that represent both the known and
unknown loads and displacements. Partitioning the stiffness matrix into
the known and unknown elements of load and displacement, we have
which when expanded yields the two equations
(15–3)
(15–4)
The unknown displacements are determined from the first of these
equations. Using these values, the support reactions are computed for
the second equation.
Q
u
D
u
Q
u
= K
21
D
u
+ K
22
D
k
Q
k
= K
11
D
u
+ K
12
D
k
c
Q
k
Q
u
d= c
K
11
K
12
K
21
K
22
dc
D
u
D
k
d
Q = KD
K
23
z¿y¿z¿y¿
1F
y¿
, F
z¿
2.1N
y¿
, N
z¿
2