Hibbeler R.C. Structural Analysis
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600 CHAPTER 16 PLANE FRAME ANALYSIS U SING THE STIFFNESS METHOD
16
16.4 Application of the Stiffness Method
for Frame Analysis
Once the member stiffness matrices are established, they may be assembled
into the structure stiffness matrix in the usual manner. By writing the
structure matrix equation, the displacements at the unconstrained nodes
can be determined, followed by the reactions and internal loadings at the
nodes. Lateral loads acting on a member, fabrication errors, temperature
changes, inclined supports, and internal supports are handled in the same
manner as was outlined for trusses and beams.
Procedure for Analysis
The following method provides a means of finding the displacements,
support reactions, and internal loadings for members of statically
determinate and indeterminate frames.
Notation
• Divide the structure into finite elements and arbitrarily identify each
element and its nodes. Elements usually extend between points of
support, points of concentrated loads, corners or joints, or to points
where internal loadings or displacements are to be determined.
• Establish the x,y,z, global coordinate system, usually for convenience
with the origin located at a nodal point on one of the elements and the
axes located such that all the nodes have positive coordinates.
• At each nodal point of the frame, specify numerically the three x,
y, z coding components. In all cases use the lowest code numbers
to identify all the unconstrained degrees of freedom, followed by
the remaining or highest code numbers to identify the constrained
degrees of freedom.
• From the problem, establish the known displacements and
known external loads When establishing be sure to include
any reversed fixed-end loadings if an element supports an
intermediate load.
Structure Stiffness Matrix
• Apply Eq. 16–10 to determine the stiffness matrix for each element
expressed in global coordinates. In particular, the direction cosines
and are determined from the x, y coordinates of the ends of
the element, Eqs. 14–5 and 14–6.
• After each member stiffness matrix is written, and the six rows
and columns are identified with the near and far code numbers,
merge the matrices to form the structure stiffness matrix K. As a
partial check, the element and structure stiffness matrices should
all be symmetric.
l
y
l
x
Q
k
Q
k
.
D
k
16.4 APPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS 601
16
Displacements and Loads
• Partition the stiffness matrix as indicated by Eq. 14–18. Expansion
then leads to
The unknown displacements are determined from the first of
these equations. Using these values, the support reactions are
computed from the second equation. Finally, the internal loadings
q at the ends of the members can be computed from Eq. 16–7,
namely
If the results of any of the unknowns are calculated as negative
quantities, it indicates they act in the negative coordinate
directions.
q = k¿TD
Q
u
D
u
Q
u
= K
21
D
u
+ K
22
D
k
Q
k
= K
11
D
u
+ K
12
D
k
Fig. 16–4
EXAMPLE 16.1
Determine the loadings at the joints of the two-member frame shown
in Fig. 16–4a. Take and for
both members.
SOLUTION
Notation. By inspection, the frame has two elements and three
nodes, which are identified as shown in Fig. 16–4b. The origin of
the global coordinate system is located at ①.The code numbers at the
nodes are specified with the unconstrained degrees of freedom
numbered first. From the constraints at ① and ➂, and the applied
loading, we have
Structure Stiffness Matrix. The following terms are common to both
element stiffness matrices:
12EI
L
3
=
12[29110
3
215002]
[201122]
3
= 12.6 k>in.
AE
L
=
10[29110
3
2]
201122
= 1208.3 k>in.
D
k
= D
0
0
0
0
T
6
7
8
9
Q
k
= E
5
0
0
0
0
U
1
2
3
4
5
E = 29110
3
2 ksiA = 10 in
2
,I = 500 in
4
,
(b)
y
x
5
6
4
2
1
8
1
2
2
3
7
3
5 k
9
1
20 ft
5 k
20 ft
(a)
602 CHAPTER 16 PLANE FRAME ANALYSIS U SING THE STIFFNESS METHOD
16
EXAMPLE 16.1 (Continued)
Member 1:
Substituting the data into Eq. 16–10, we have
The rows and columns of this matrix are identified by the three
x, y, z code numbers, first at the near end and followed by the far end,
that is, 4, 6, 5, 1, 2, 3, respectively, Fig. 16–4b. This is done for later
assembly of the elements.
Member 2:
Substituting the data into Eq. 16–10 yields
As usual, column and row identification is referenced by the three code
numbers in x, y, z sequence for the near and far ends, respectively, that
is, 1, 2, 3, then 7, 8, 9, Fig. 16–4b.
k
2
=
12 3 7 8 9
F
12.6 0 1510.4 -12.6 0 1510.4
0 1208.3 0 0 -1208.3 0
1510.4 0 241.7110
3
2 -1510.4 0 120.83110
3
2
-12.6 0 -1510.4 12.6 0 -1510.4
0 -1208.3 0 0 1208.3 0
1510.4 0 120.83110
3
2 -1510.4 0 241.7110
3
2
V
1
2
3
7
8
9
l
x
=
20 - 20
20
= 0
l
y
=
-20 - 0
20
=-1
6 * 6
k
1
=
46 5 12 3
F
1208.3 0 0 -1208.3 0 0
0 12.6 1510.4 0 -12.6 1510.4
0 1510.4 241.7110
3
2 0 -1510.4 120.83110
3
2
-1208.3 0 0 1208.3 0 0
0 -12.6 -1510.4 0 12.6 -1510.4
0 1510.4 120.83110
3
2 0 -1510.4 241.7110
3
2
V
4
6
5
1
2
3
l
x
=
20 - 0
20
= 1
l
y
=
0 - 0
20
= 0
2EI
L
=
2[29110
3
215002]
201122
= 120.83110
3
2 k
#
in.
4EI
L
=
4[29110
3
215002]
201122
= 241.7110
3
2 k
#
in.
6EI
L
2
=
6[29110
3
215002]
[201122]
2
= 1510.4 k
16.4 APPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS 603
16
The structure stiffness matrix is determined by assembling and
The result, shown partitioned, as is
Displacements and Loads. Expanding to determine the displacements
yields
Solving, we obtain
Using these results, the support reactions are determined from Eq. (1)
as follows:
Ans.
D
Q
6
Q
7
Q
8
Q
9
T=
12 345
E
0 -12.6 1510.4 0 1510.4
-12.6 0 -1510.4 0 0
0 -1208.3 0 0 0
1510.4 0 120.83110
3
2 00
U
E
0.696
-1.55110
-3
2
-2.488110
-3
2
0.696
1.234110
-3
2
U
+ D
0
0
0
0
T
= D
-1.87 k
-5.00 k
1.87 k
750 k
#
in.
T
E
D
1
D
2
D
3
D
4
D
5
U= E
0.696 in.
-1.55110
-3
2 in.
-2.488110
-3
2 rad
0.696 in.
1.234110
-3
2 rad
U
E
5
0
0
0
0
U= E
1220.9 0 1510.4 -1208.3 0
0 1220.9 -1510.4 0 -1510.4
1510.4 -1510.4 483.3110
3
2 0 120.83110
3
2
-1208.3 0 0 1208.3 0
0 -1510.4 120.83110
3
2 0 241.7110
3
2
UE
D
1
D
2
D
3
D
4
D
5
U+ E
0
0
0
0
0
U
1
2
3
1220.9
5
0
1510.4
0
0
0
0
0
Q
6
Q
7
Q
8
Q
9
1220.9
1510.4
1510.4
1510.4
483.3(10
3
)
1208.3
0 0
0
1510.4
120.83(10
3
)
0
12.6 1510.4
12.6
0 1510.4
0
1208.3
0
1510.4
0
120.83(10
3
)
45 6 7
1208.3 0 0
12.6
0
1510.4
12.6 0
0
120.83(10
3
)
1510.4 1510.4
1208.3
000
0
241.7(10
3
)
1510.4
0
0
1510.4 12.6 0
0
0
0 12.6
0
0
0
0
0
001510.4
89
0
1510.4
1208.3 0
0 120.83(10
3
)
00
0 0 (1)
00
0 1510.4
1208.3 0
0 241.7(10
3
)
D
1
D
2
D
3
D
4
D
5
0
0
0
0
Q = KD,k
2
.
k
1
604 CHAPTER 16 PLANE FRAME ANALYSIS U SING THE STIFFNESS METHOD
16
EXAMPLE 16.1 (Continued)
The internal loadings at node ➁ can be determined by applying
Eq. 16–7 to member 1. Here is defined by Eq. 16–1 and by
Eq. 16–3. Thus,
Note the appropriate arrangement of the elements in the matrices as
indicated by the code numbers alongside the columns and rows.
Solving yields
Ans.
The above results are shown in Fig. 16–4c. The directions of these
vectors are in accordance with the positive directions defined in
Fig. 16–1. Furthermore, the origin of the local axes is at the
near end of the member. In a similar manner, the free-body diagram
of member 2 is shown in Fig. 16–4d.
z¿y¿,x¿,
F
q
4
q
6
q
5
q
1
q
2
q
3
V= F
0
-1.87 k
0
0
1.87 k
-450 k
#
in.
V
1208.3
4
1208.3
0
0
0
0
0
6
0
12.6
1510.4
12.6
1510.4
0
5
0
1510.4
241.7(10
3
)
1510.4
120.83(10
3
)
1208.3
1
1208.3
0
0
0
0
0
2
0
12.6
1510.4
12.6
1510.4
0
3
0
1510.4
120.83(10
3
)
1510.4
241.7(10
3
)
1
0
0
0
0
0
0
0
1
0
0
0
4
1
6
5
2
3
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0.696
0.696
0
1.234(10
3
)
1.55(10
3
)
2.488(10
3
)
q
1
k
1
T
1
D
T
1
k
1
œ
y¿
1.87 k
1.87 k
450 kin
.
x¿
(c)
1.87 k
450 kin.
5 k
1.87 k
750 kin.
5 k
(d)
y¿
x¿
Fig. 16–4
16
16.4 A
PPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS 605
Determine the loadings at the ends of each member of the frame shown
in Fig. 16–5a. Take and for
each member.
SOLUTION
Notation. To perform a matrix analysis, the distributed loading
acting on the horizontal member will be replaced by equivalent end
moments and shears computed from statics and the table listed on the
inside back cover. (Note that no external force of 30 k or moment of
is placed at ➂ since the reactions at code numbers 8 and 9
are to be unknowns in the load matrix.) Then using superposition, the
results obtained for the frame in Fig. 16–5b will be modified for this
member by the loads shown in Fig. 16–5c.
As shown in Fig. 16–5b, the nodes and members are numbered and
the origin of the global coordinate system is placed at node ①.As
usual, the code numbers are specified with numbers assigned first to
the unconstrained degrees of freedom. Thus,
Structure Stiffness Matrix
Member 1:
l
x
=
20 - 0
25
= 0.8
l
y
=
15 - 0
25
= 0.6
2EI
L
=
2[29110
3
2]600
251122
= 116110
3
2 k
#
in.
4EI
L
=
4[29110
3
2]600
251122
= 232110
3
2 k
#
in.
6EI
L
2
=
6[29110
3
)]600
[25112)]
2
= 1160 k
12EI
L
3
=
12[29110
3
2]600
[251122]
3
= 7.73 k>in.
AE
L
=
12[29110
3
2]
251122
= 1160 k>in.
D
k
= F
0
0
0
0
0
0
V
4
5
6
7
8
9
Q
k
= C
0
-30
-1200
S
1
2
3
1200 k
#
in.
29110
3
2 ksiE =A = 12 in
2
,I = 600 in
4
,
EXAMPLE 16.2
20 ft
3 k/ft
20 ft
15 ft
(a)
1
30 k
y
x
2
2
3
1
6
5
4
2
3
1
9
8
7
1200 kin.
(b)
3 k/ft
20 ft
30 k30 k
ⴙ
1
_
_
(3)(20)
2
100 kft
12
100 kft
(1200 kin.)
(
c
)
(1200 kin.)
Fig. 16–5
EXAMPLE 16.2 (Continued)
606 CHAPTER 16 PLANE FRAME ANALYSIS U SING THE STIFFNESS METHOD
16
Applying Eq. 16–10, we have
k
1
=
456 123
F
745.18 553.09 -696 -745.18 -553.09 -696
553.09 422.55 928 -553.09 -422.55 928
-696 928 232110
3
2 696 -928 116110
3
2
-745.18 -553.09 696 745.18 553.09 696
-553.09 -422.55 -928 553.09 422.55 -928
-696 928 116110
3
2 696 -928 232110
3
2
V
4
5
6
1
2
3
Member 2:
Thus, Eq. 16–10 becomes
l
x
=
40 - 20
20
= 1
l
y
=
15 - 15
20
= 0
2EI
L
=
2329110
3
)4600
320112)4
= 1.45110
5
2 k
#
in.
4EI
L
=
4[29110
3
2]600
201122
= 2.90110
5
2
k
#
in.
6EI
L
2
=
6[29110
3
2]600
[201122]
2
= 1812.50 k
12EI
L
3
=
12[29110
3
2]600
[201122]
3
= 15.10 k>in.
AE
L
=
12[29110
3
2]
201122
= 1450 k>in.
k
2
=
12378 9
F
1450 0 0 -1450 0 0
0 15.10 1812.50 0 -15.10 1812.50
0 1812.50 290110
3
2 0 -1812.50 145110
3
2
-1450 0 0 1450 0 0
0 -15.10 -1812.50 0 15.10 -1812.50
0 1812.50 145110
3
2 0 -1812.50 290110
3
2
V
1
2
3
7
8
9
16
16.4 A
PPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS 607
The structure stiffness matrix, included in becomesQ = KD,
0
30
1200
Q
4
0
0
0
0
0
0
D
1
D
2
D
3
Q
5
Q
6
Q
7
Q
8
Q
9
2195.18
553.09
696
745.18
553.09
696
1450
0
0
553.09
437.65
884.5
553.09
422.55
928
00
15.10
1812.50
1812.50
145(10
3
)
696
884.5
522(10
3
)
696
928
116(10
3
)
745.18
553.09
696
745.18
553.09
696
553.09
422.55
928
553.09
422.55
928
696
928
116(10
3
)
696
928
232(10
3
)
1450
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1450
15.10
1812.50
0
0
0
0
00
1812.50
145(10
3
)
0
0
0
0
15.10
1812.50
1812.50
290(10
3
)
12 3 4 56 78 9
(1)
Displacements and Loads. Expanding to determine the displacements,
and solving, we have
Using these results, the support reactions are determined from
Eq. (1) as follows:
C
D
1
D
2
D
3
S= C
0.0247 in.
-0.0954 in.
-0.00217 rad
S
C
0
-30
-1200
S= C
2195.18 553.09 696
553.09 437.65 884.5
696 884.5 522110
3
2
SC
D
1
D
2
D
3
S+ C
0
0
0
S
F
Q
4
Q
5
Q
6
Q
7
Q
8
Q
9
V= F
-745.18 -553.09 -696
-553.09 -422.55 928
696 -928 116110
3
2
-1450 0 0
0 -15.10 -1812.50
0 1812.50 145110
3
2
VC
0.0247
-0.0954
-0.00217
S+ F
0
0
0
0
0
0
V= F
35.85 k
24.63 k
-145.99 k
#
in.
-35.85 k
5.37 k
-487.60 k
#
in.
V
608 CHAPTER 16 PLANE FRAME ANALYSIS U SING THE STIFFNESS METHOD
16
EXAMPLE 16.2 (Continued)
The internal loadings can be determined from Eq. 16–7 applied to
members 1 and 2. In the case of member 1, where is
determined from Eq. 16–1, and from Eq. 16–3. Thus,T
1
k
œ
1
q = k
œ
1
T
1
D,
45 6 12 3
F
q
4
q
5
q
6
q
1
q
2
q
3
V= F
1160 0 0 -1160 0 0
0 7.73 1160 0 -7.73 1160
0 1160 232110
3
2 0 -1160 116110
3
2
-1160 0 0 1160 0 0
0 -7.73 -1160 0 7.73 -1160
0 1160 116110
3
2 0 -1160 232110
3
2
VF
0.8 0.6 0 0 0 0
-0.6 0.8 0 0 0 0
001000
0 0 0 0.8 0.6 0
000-0.6 0.8 0
000001
VF
0
0
0
0.0247
-0.0954
-0.00217
V
4
5
6
1
2
3
Here the code numbers indicate the rows and columns for the near
and far ends of the member, respectively, that is, 4, 5, 6, then 1, 2, 3,
Fig. 16–5b. Thus,
Ans.
These results are shown in Fig. 16–5d.
A similar analysis is performed for member 2.The results are shown
at the left in Fig. 16–5e. For this member we must superimpose the
loadings of Fig. 16–5c, so that the final results for member 2 are shown
to the right.
F
q
4
q
5
q
6
q
1
q
2
q
3
V= F
43.5 k
-1.81 k
-146 k
#
in.
-43.5 k
1.81 k
-398 k
#
in.
V
3 k/ft
35.85 k
802.3 kin.
5.37 k
5.37 k
35.85 k
35.85 k
487.6 kin.
30 k
1200 kin. 1200 kin.
30 k
3 k/ft
24.6 k
398 kin.
(e)
35.85 k
35.4 k
1688 kin.
ⴙ
ⴝ
y¿
(d)
x¿
43.5 k
146 kin.
1.81 k
1.81 k
398 kin.
43.5 k
Fig. 16–5
16
16.4 A
PPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS 609
16–1. Determine the structure stiffness matrix K for
the frame. Assume
① and ➂ are fixed. Take ,
, for each member.
16–2. Determine the support reactions at the fixed
supports
① and ➂. Take , ,
for each member.A = 10110
3
2 mm
2
I = 300110
6
2 mm
4
E = 200 GPa
A = 10110
3
2 mm
2
I = 300110
6
2 mm
4
E = 200 GPa
*16–4. Determine the support reactions at
① and ➂.
Take , ,
for each member.
A = 21110
3
2 mm
2
I = 300110
6
2 mm
4
E = 200 MPa
PROBLEMS
16–3. Determine the structure stiffness matrix K for
the frame. Assume
➂ is pinned and ① is fixed. Take
, , for
each member.
A = 21110
3
2 mm
2
I = 300110
6
2 mm
4
E = 200 MPa
Prob. 16–3
Prob. 16–5
Probs. 16–1/16–2
8
1
2
3
1
2
2
12 kN/m
1
3
4
6
4 m
9
7
5
2 m
2 m
10 kN
2
1
2
3
2
300 kN m
1
9
5 m
4 m
3
6
4
5
1
8
7
Prob. 16–4
2
1
2
3
2
300 kN m
1
9
5 m
4 m
3
6
4
5
1
8
7
9
1
2
3
1
2
2
1
3
6
4
4 m
2 m
2 m
60 kN
5
8
7
16–5. Determine the structure stiffness matrix K for the frame.
Take , ,
for each member. Joints at
① and ➂ are pins.
A = 15110
3
2 mm
2
I = 350110
6
2 mm
4
E = 200 GPa