Hibbeler R.C. Structural Analysis
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530 CHAPTER 13 BEAMS AND FRAMES H AVING N ONPRISMATIC M EMBERS
13
Symmetric Beam with Antisymmetric Loading. In the case
of a symmetric beam with antisymmetric loading, we must determine
such that equal rotations occur at the ends of the beam, Fig. 13–6a.To do
this, we first fix end B and apply the moment at A, Fig. 13–6b.
Likewise, application of at end B while end A is fixed is shown in
Fig. 13–6c. Superposition of both cases yields the results of Fig. 13–6a.
Hence,
or, using Eq. 13–1 we have for the absolute stiffness
(13–5)
Substituting the data for a prismatic member, and
yields which is the same as Eq. 12–6.K
A
œ
= 6EI>L,
C
AB
=
1
2
,K
A
= 4EI>L
K
A
œ
= K
A
11 + C
AB
2
1C
BA
K
B
= C
AB
K
A
2,
K
A
œ
= K
A
+ C
BA
K
B
K
B
K
A
K
A
œ
u
A
(1 rad)
K
A
B
C
AB
K
A
(b)
ⴙ
A
K¿
A
A
u
A
(1 rad)
K¿
A
u
B
(1 rad)
(a)
B
ⴝ
A
K
B
u
B
(1 rad)
B
(c)
C
BA
K
B
(
a
)
A
(FEM)
AB
(FEM)
BA
B
L
u
A
__
L
u
B
__
L
ⴝ
A
B
(b)
Fig. 13–6
Fig. 13–7
13.2 MOMENT DISTRIBUTION FOR STRUCTURES HAVING NONPRISMATIC MEMBERS 531
13
Relative Joint Translation of Beam. Fixed-end moments are
developed in a nonprismatic member if it has a relative joint translation
between its ends A and B, Fig. 13–7a. In order to determine these
moments, we proceed as follows. First consider the ends A and B to be
pin connected and allow end B of the beam to be displaced a distance
such that the end rotations are Fig. 13–7b. Second,
assume that B is fixed and apply a moment of to end
A such that it rotates the end Fig. 13–7c. Third, assume that
A is fixed and apply a moment to end B such that it
rotates the end Fig. 13–7d. Since the total sum of these
three operations yields the condition shown in Fig. 13–7a, we have at A
Applying Eq. 13–1 yields
(13–6)
A similar expression can be written for end B. Recall that for a prismatic
member and Thus
which is the same as Eq. 11–5.
If end B is pinned rather than fixed, Fig. 13–8, the fixed-end moment
at A can be determined in a manner similar to that described above. The
result is
(13–7)
Here it is seen that for a prismatic member this equation gives
which is the same as that listed on the inside
back cover.
The following example illustrates application of the moment-distribution
method to structures having nonprismatic members. Once the fixed-end
moments and stiffness and carry-over factors have been determined, and
the stiffness factor modified according to the equations given above, the
procedure for analysis is the same as that discussed in Chapter 12.
1FEM2
AB
œ
=-3EI¢>L
2
,
1FEM2
AB
œ
=-K
A
¢
L
11 - C
AB
C
BA
2
1FEM2
AB
=-6EI¢>L
2
,C
AB
=
1
2
.K
A
= 4EI>L
1FEM2
AB
=-K
A
¢
L
11 + C
AB
2
1C
BA
K
B
= C
AB
K
A
2
1FEM2
AB
=-K
A
¢
L
- C
BA
K
B
¢
L
u
B
=-¢>L,
M
B
œ
=-K
B
1¢>L2
u
A
=-¢>L,
M
A
œ
=-K
A
1¢>L2
u
A
= u
B
=¢>L,
¢
¢
C
AB
K
A
_
_
L
K
A
__
L
u
A
__
L
B
A
(
c
)
ⴙ
ⴙ
A
B
(d)
C
BA
K
B
__
L
K
B
__
L
u
B
__
L
(FEM)¿
AB
B
A
L
Fig. 13–7
Fig. 13–8
532 CHAPTER 13 BEAMS AND FRAMES H AVING N ONPRISMATIC M EMBERS
13
SOLUTION
Since the haunches are parabolic, we will use Table 13–2 to obtain the
moment-distribution properties of the beam.
Span
AB
Entering Table 13–2 with these ratios, we find
Using Eqs. 13–2,
Since the far end of span BA is pinned, we will modify the stiffness
factor of BA using Eq. 13–3. We have
Uniform load, Table 13–2,
1FEM2
BA
= 119.50 k
#
ft
1FEM2
AB
=-10.095621221252
2
=-119.50 k
#
ft
K
BA
œ
= K
BA
11 - C
AB
C
BA
2= 0.171E[1 – 0.61910.6192] = 0.105E
K
AB
= K
BA
=
kEI
C
L
=
6.41E
A
1
12
B
112122
3
25
= 0.171E
k
AB
= k
BA
= 6.41
C
AB
= C
BA
= 0.619
a
A
= a
B
=
5
25
= 0.2 r
A
= r
B
=
4 – 2
2
= 1.0
Determine the internal moments at the supports of the beam shown
in Fig. 13–9a. The beam has a thickness of 1 ft and E is constant.
EXAMPLE 13.1
3 ft
30 k
2 k/ft
5 ft
C
25 ft
5 ft 15 ft 5 ft 5 ft
10 ft
(a)
5 ft
4 ft
A
B
2 ft
4 ft
2 ft
Fig. 13–9
13.2 MOMENT DISTRIBUTION FOR STRUCTURES HAVING NONPRISMATIC MEMBERS 533
13
Joint AB C
Member
AB BA BC CB
K 0.171E 0.105E 0.875E 1.031E
FEM
6.72 56.05
56.73 22.77
7.91
66.06
43.78
51.59
兺M
0 178.84 178.84 72.60
DF 1 0.107 0.893 0
COF 0.619 0.619 0.781 0.664
119.50
119.50 119.50
73.97
(b)
Dist.
CO
Dist.
CO
Span
BC
From Table 13–2 we find
Thus, from Eqs. 13–2,
Concentrated load,
Using the foregoing values for the stiffness factors, the distribution
factors are computed and entered in the table, Fig. 13–9b. The moment
distribution follows the same procedure outlined in Chapter 12. The
results in are shown on the last line of the table.k
#
ft
1FEM2
CB
= 0.075913021102= 22.77 k
#
ft
1FEM2
BC
=-0.189113021102=-56.73 k
#
ft
b =
3
10
= 0.3
K
CB
=
kEI
C
L
=
15.47E
A
1
12
B
112122
3
10
= 1.031E
K
BC
=
kEI
C
L
=
13.12E
A
1
12
B
112122
3
10
= 0.875E
k
BC
= 13.12
k
CB
= 15.47
C
BC
= 0.781
C
CB
= 0.664
r
C
=
5 – 2
2
= 1.5
a
B
= a
C
=
5
10
= 0.5
r
B
=
4 – 2
2
= 1.0
Fig. 13–9
13.3 Slope-Deflection Equations
for Nonprismatic Members
The slope-deflection equations for prismatic members were developed in
Chapter 11. In this section we will generalize the form of these equations
so that they apply as well to nonprismatic members. To do this, we will use
the results of the previous section and proceed to formulate the equations
in the same manner discussed in Chapter 11, that is, considering the
effects caused by the loads, relative joint displacement, and each joint
rotation separately, and then superimposing the results.
Loads. Loads are specified by the fixed-end moments and
acting at the ends A and B of the span. Positive moments act
clockwise.
Relative Joint Translation. When a relative displacement between
the joints occurs, the induced moments are determined from Eq. 13–6. At
end A this moment is and at end B it is
Rotation at
A
. If end A rotates the required moment in the span
at A is Also, this induces a moment of at
end B.
Rotation at
B
. If end B rotates a moment of must act at end
B, and the moment induced at end A is
The total end moments caused by these effects yield the generalized
slope-deflection equations, which can therefore be written as
Since these two equations are similar, we can express them as a single
equation. Referring to one end of the span as the near end (N) and the
other end as the far end (F ), and representing the member rotation as
we have
(13–8)
Here
internal moment at the near end of the span; this moment is
positive clockwise when acting on the span.
absolute stiffness of the near end determined from tables or by
calculation.
K
N
=
M
N
=
M
N
= K
N
1u
N
+ C
N
u
F
- c11 + C
N
22+ 1FEM2
N
c =¢>L,
M
BA
= K
B
B
u
B
+ C
BA
u
A
-
¢
L
11 + C
BA
2
R
+ 1FEM2
BA
M
AB
= K
A
B
u
A
+ C
AB
u
B
-
¢
L
11 + C
AB
2
R
+ 1FEM2
AB
C
BA
K
B
u
B
= C
AB
K
A
u
B
.
K
B
u
B
u
B
,
C
AB
K
A
u
A
= C
BA
K
B
u
A
K
A
u
A
.
u
A
,
-[K
B
¢>L]11 + C
BA
2.
-[K
A
¢>L]11 + C
AB
2
¢
1FEM2
BA
1FEM2
AB
534 CHAPTER 13 BEAMS AND FRAMES H AVING N ONPRISMATIC M EMBERS
13
near- and far-end slopes of the span at the supports; the
angles are measured in radians and are positive clockwise.
span cord rotation due to a linear displacement,
this angle is measured in radians and is positive clockwise.
fixed-end moment at the near-end support; the moment is
positive clockwise when acting on the span and is obtained
from tables or by calculations.
Application of the equation follows the same procedure outlined in
Chapter 11 and therefore will not be discussed here. In particular, note
that Eq. 13–8 reduces to Eq. 11–8 when applied to members that are
prismatic.
1FEM2
N
=
c =¢>L;c =
u
N
, u
F
=
13.3 SLOPE-DEFLECTION EQUATIONS FOR NONPRISMATIC MEMBERS 535
13
A continuous, reinforced-concrete highway bridge.
Light-weight metal buildings are often designed using
frame members having variable moments of inertia.
536 CHAPTER 13 BEAMS AND FRAMES H AVING N ONPRISMATIC M EMBERS
13
13–1. Determine the moments at A, B, and C by the
moment-distribution method. Assume the supports at A
and C are fixed and a roller support at B is on a rigid base.
The girder has a thickness of 4 ft. Use Table 13–1. E is
constant. The haunches are straight.
13–2. Solve Prob. 13–1 using the slope-deflection equations.
13–5. Use the moment-distribution method to determine
the moment at each joint of the symmetric bridge frame.
Supports at F and E are fixed and B and C are fixed
connected. Use Table 13–2. Assume E is constant and the
members are each 1 ft thick.
13–6. Solve Prob. 13–5 using the slope-deflection equations.
PROBLEMS
13–3. Apply the moment-distribution method to determine
the moment at each joint of the parabolic haunched frame.
Supports A and B are fixed. Use Table 13–2. The members
are each 1 ft thick. E is constant.
*13–4. Solve Prob.13–3 using the slope-deflection equations.
13–7. Apply the moment-distribution method to determine
the moment at each joint of the symmetric parabolic
haunched frame. Supports A and D are fixed. Use Table 13–2.
The members are each 1 ft thick. E is constant.
*13–8. Solve Prob.13–7 using the slope-deflection equations.
6 ft
4 ft
4 ft
2 ft
A
C
4 ft
4 ft
4 ft
6 ft
B
20 ft
20 ft
8 k/ft
9 ft
6 ft
8 ft
2 ft 4 ft
4 ft
25 ft
2 ft
A
B
C
D
F
E
30 ft 40 ft 30 ft
4 k/ft
5 ft
40 ft
40 ft
2 ft
2 ft
8 ft
12 ft
C
B
A
1.5 k/ft
4 ft
3.2 ft
3.2 ft
15 ft
5 ft
2.5 ft
2 k
8 ft 8 ft
A
D
B
C
3 ft 3 ft
Probs. 13–1/13–2
Probs. 13–3/13–4
Probs. 13–5/13–6
Probs. 13–7/13–8
CHAPTER REVIEW 537
13
20 ft
2.5 ft
6 ft
1 ft
6 ft
18 ft
2.5 ft
1 ft 1 ft
A
B
C
D
500 lb/ft
4 ft
3 ft
2 ft
2 ft
30 ft
30 ft
30 ft
30 ft
3 ft
D
C
B
EF
A
2 k/ft
40 ft40 ft 40 ft
12 ft
13–9. Use the moment-distribution method to determine
the moment at each joint of the frame. The supports at A
and C are pinned and the joints at B and D are fixed
connected. Assume that E is constant and the members
have a thickness of 1 ft. The haunches are straight so use
Table 13–1.
13–10. Solve Prob.13–9 using the slope-deflection equations.
13–11. Use the moment-distribution method to determine
the moment at each joint of the symmetric bridge frame.
Supports F and E are fixed and B and C are fixed connected.
The haunches are straight so use Table 13–2. Assume E is
constant and the members are each 1 ft thick.
*13–12. Solve Prob. 13–11 using the slope-deflection
equations.
Non-prismatic members having a variable moment of inertia are often used on long-span bridges and building frames to
save material.
A structural analysis using non-prismatic members can be performed using either the slope-deflection equations or moment
distribution. If this is done, it then becomes necessary to obtain the fixed-end moments, stiffness factors, and carry-over factors
for the member. One way to obtain these values is to use the conjugate beam method, although the work is somewhat tedious.
It is also possible to obtain these values from tabulated data, such as published by the Portland Cement Association.
If the moment distribution method is used, then the process can be simplified if the stiffness of some of the members is
modified.
CHAPTER REVIEW
Probs. 13–9/13–10
Probs. 13–11/13–12
The space-truss analysis of electrical transmission towers can be performed
using the stiffness method.
14
539
In this chapter we will explain the basic fundamentals of using the
stiffness method for analyzing structures. It will be shown that this
method, although tedious to do by hand, is quite suited for use on
a computer. Examples of specific applications to planar trusses will
be given. The method will then be expanded to include space-truss
analysis. Beams and framed structures will be discussed in the next
chapters.
14.1 Fundamentals of the Stiffness Method
There are essentially two ways in which structures can be analyzed using
matrix methods. The stiffness method, to be used in this and the following
chapters, is a displacement method of analysis. A force method, called the
flexibility method, as outlined in Sec. 9–1, can also be used to analyze
structures; however, this method will not be presented in this text.
There are several reasons for this. Most important, the stiffness method
can be used to analyze both statically determinate and indeterminate
structures, whereas the flexibility method requires a different procedure
for each of these two cases. Also, the stiffness method yields the dis-
placements and forces directly, whereas with the flexibility method the
displacements are not obtained directly. Furthermore, it is generally much
easier to formulate the necessary matrices for the computer operations
using the stiffness method; and once this is done, the computer
calculations can be performed efficiently.
Truss Analysis Using
the Stiffness Method