Hibbeler R.C. Structural Analysis
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280 CHAPTER 7APPROXIMATE ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES
7
Probs. 7–21/7–22
7 ft
500 lb
BA
EG
HF
DC
6 ft
1.5 ft
1.5 ft 1.5 ft
Probs. 7–23/7–24
8 ft
2 k
8 ft
6 ft
12 ft
AB
C
D
F
E
G
1 k
Prob. 7–25
2 m 2 m
1.5 m
5 m
A
B
D
G
F
C
E
4 kN
8 kN
Prob. 7–26
2 m 2 m
1.5 m
5 m
A
B
D
G
F
C
E
4 kN
8 kN
7–21. Draw (approximately) the moment diagram for
member ACE of the portal constructed with a rigid member
EG and knee braces CF and DH. Assume that all points of
connection are pins. Also determine the force in the knee
brace CF.
7–22. Solve Prob. 7–21 if the supports at A and B are fixed
instead of pinned.
7–25. Draw (approximately) the moment diagram for col-
umn AGF of the portal. Assume all truss members and the
columns to be pin connected at their ends. Also determine
the force in all the truss members.
7–23. Determine (approximately) the force in each truss
member of the portal frame. Also find the reactions at the
fixed column supports A and B. Assume all members of the
truss to be pin connected at their ends.
*7–24. Solve Prob. 7–23 if the supports at A and B are
pinned instead of fixed.
7–26. Draw (approximately) the moment diagram for col-
umn AGF of the portal. Assume all the members of the
truss to be pin connected at their ends. The columns are
fixed at A and B. Also determine the force in all the truss
members.
7.4 PORTAL FRAMES AND TRUSSES 281
7
7–27. Determine (approximately) the force in each truss
member of the portal frame. Also find the reactions at the
fixed column supports A and B. Assume all members of the
truss to be pin connected at their ends.
*7–28. Solve Prob. 7–27 if the supports at A and B are
pinned instead of fixed.
7–31. Draw (approximately) the moment diagram for
column ACD of the portal. Assume all truss members and
the columns to be pin connected at their ends. Also
determine the force in members FG, FH, and EH.
*7–32. Solve Prob. 7–31 if the supports at A and B are
fixed instead of pinned.
7–29. Determine (approximately) the force in members
GF, GK, and JK of the portal frame. Also find the reactions
at the fixed column supports A and B. Assume all members
of the truss to be connected at their ends.
7–30. Solve Prob. 7–29 if the supports at A and B are pin
connected instead of fixed.
7–33. Draw (approximately) the moment diagram for
column AJI of the portal.Assume all truss members and the
columns to be pin connected at their ends. Also determine
the force in members HG, HL, and KL.
7–34. Solve Prob. 7–33 if the supports at A and B are fixed
instead of pinned.
6 m6 m
1.5 m1.5 m
3 m3 m
8 kN8 kN
AA
FF
IIGG
BB
2 m2 m
1.5 m1.5 m
HHCC
DDEE
3 m3 m 3 m3 m
12 kN12 kN
Probs. 7–27/7–28
D
C
E
H
I
K
F
LJ
G
8 ft 8 ft 8 ft 8 ft
12 ft
3 ft
6 ft
A
4 k
B
Probs. 7–29/7–30
D
C
E
HI
K
F
L
J
G
8 ft 8 ft 8 ft 8 ft
12 ft
3 ft
6 ft
6 ft
A
4 k
B
Probs. 7–31/7–32
4 kN
2 kN
OK LMN
AB
C
D
E
F
G
J
I
H
1.5 m
4 m
1 m
6 @ 1.5 m 9 m
Probs. 7–33/7–34
282 CHAPTER 7APPROXIMATE ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES
7
7.5 Lateral Loads on Building Frames:
Portal Method
In Sec. 7–4 we discussed the action of lateral loads on portal frames and
found that for a frame fixed supported at its base, points of inflection
occur at approximately the center of each girder and column and the
columns carry equal shear loads, Fig. 7–8. A building bent deflects in
the same way as a portal frame, Fig. 7–12a, and therefore it would be
appropriate to assume inflection points occur at the center of the columns
and girders. If we consider each bent of the frame to be composed of a
series of portals, Fig. 7–12b, then as a further assumption, the interior
columns would represent the effect of two portal columns and would
therefore carry twice the shear V as the two exterior columns.
(a)
P
inflection point
(b)
V V V V
Fig. 7–12
7.5 LATERAL LOADS ON BUILDING FRAMES: PORTAL METHOD 283
7
In summary, the portal method for analyzing fixed-supported building
frames requires the following assumptions:
1. A hinge is placed at the center of each girder, since this is assumed
to be a point of zero moment.
2. A hinge is placed at the center of each column, since this is assumed
to be a point of zero moment.
3. At a given floor level the shear at the interior column hinges is twice
that at the exterior column hinges, since the frame is considered to
be a superposition of portals.
These assumptions provide an adequate reduction of the frame to one
that is statically determinate yet stable under loading.
By comparison with the more exact statically indeterminate analysis,
the portal method is most suitable for buildings having low elevation and
uniform framing. The reason for this has to do with the structure’s action
under load. In this regard, consider the frame as acting like a cantilevered
beam that is fixed to the ground. Recall from mechanics of materials that
shear resistance becomes more important in the design of short beams,
whereas bending is more important if the beam is long. (See Sec. 7–6.)
The portal method is based on the assumption related to shear as stated
in item 3 above.
The following examples illustrate how to apply the portal method to
analyze a building bent.
The portal method of analysis can be used to (approximately) perform a lateral-load
analysis of this single-story frame.
284 CHAPTER 7APPROXIMATE ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES
7
Determine (approximately) the reactions at the base of the columns
of the frame shown in Fig. 7–13a. Use the portal method of analysis.
EXAMPLE 7.5
SOLUTION
Applying the first two assumptions of the portal method, we place
hinges at the centers of the girders and columns of the frame,
Fig. 7–13a. A section through the column hinges at I, J, K, L yields
the free-body diagram shown in Fig. 7–13b. Here the third assumption
regarding the column shears applies. We require
Using this result, we can now proceed to dismember the frame at
the hinges and determine their reactions. As a general rule, always
start this analysis at the corner or joint where the horizontal load is
applied. Hence, the free-body diagram of segment IBM is shown in
Fig. 7–13c. The three reaction components at the hinges and
are determined by applying
respectively. The adjacent segment MJN is analyzed next, Fig. 7–13d,
followed by segment NKO, Fig. 7–13e, and finally segment OGL,
Fig. 7–13f. Using these results, the free-body diagrams of the columns
with their support reactions are shown in Fig. 7–13g.
©F
y
= 0,©F
x
= 0,©M
M
= 0,M
y
M
x
,I
y
,
:
+
©F
x
= 0;
1200 - 6V = 0
V = 200 lb
BM DN F O G
IJKL
AC EH
1200 lb
12 ft
16 ft 16 ft 16 ft
(
a
)
Fig. 7–13
1200 lb
V
I
y
I
2V
J
y
J
2V
K
y
K
V
L
y
L
(b)
7.5 LATERAL LOADS ON BUILDING FRAMES: PORTAL METHOD 285
7
If the horizontal segments of the girders in Figs. 7–13c, d, e and f are
considered, show that the moment diagram for the girder looks like
that shown in Fig. 7–13h.
J
y
0
150 lb
J
8 ft
400 lb
N
x
= 600 lb
N
y
150 lb
8 ft
1000 lb
6 ft
(d)
M
N
K
y
0
8 ft
400 lb
O
O
x
= 200 lb
O
y
= 150 lb
8 ft
600 lb
150 lb
6 ft
(e)
N
K
O
G
L
L
y
150 lb
8 ft
6 ft
200 lb
200 lb
150 lb
(
f
)
I
1200 lb
I
y
150 lb
8 ft
6 ft
200 lb
M
M
x
1000 lb
M
y
150 lb
(c)
B
L
K
J
I
A
x
200 lb
6 ft
150 lb
(g)
200 lb
M
A
1200 lbft
A
y
150 lb
C
x
400 lb
6 ft
400 lb
M
C
2400 lbft
E
x
400 lb
6 ft
400 lb
M
E
2400 lbft
H
x
200 lb
6 ft
150 lb
200 lb
M
H
1200 lbft
H
y
150 lb
(
h
)
1.2
1.2 1.2
1.2 1.2 1.2
M (kft)
816243240 48
x (ft)
286 CHAPTER 7APPROXIMATE ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES
7
Determine (approximately) the reactions at the base of the columns
of the frame shown in Fig. 7–14a. Use the portal method of analysis.
EXAMPLE 7.6
SOLUTION
First hinges are placed at the centers of the girders and columns of the
frame, Fig. 7–14a. A section through the hinges at O, P, Q and J, K, L
yields the free-body diagrams shown in Fig. 7–14b. The column shears
are calculated as follows:
20 + 30 - 4V¿=0
V¿=12.5 kN
:
+
©F
x
= 0;
20 - 4V = 0
V = 5 kN
:
+
©F
x
= 0;
20 kN
V
O
y
2.5 m
2V
P
y
V
Q
y
20 kN
30 kN
GR H S I
DM ENF
OP Q
JKL
AC
5 m
6 m
8 m 8 m
(a)
B
20 kN
5 m
30 kN
3 m
V¿
J
y
2V¿
K
y
V¿
L
y
(b)
Fig. 7–14
7.5 LATERAL LOADS ON BUILDING FRAMES: PORTAL METHOD 287
7
Using these results, we can now proceed to analyze each part of the
frame. The analysis starts with the corner segment OGR, Fig. 7–14c.
The three unknowns and have been calculated using the
equations of equilibrium. With these results segment OJM is analyzed
next, Fig. 7–14d; then segment JA, Fig. 7–14e; RPS, Fig. 7–14f; PMKN,
Fig. 7–14g; and KB, Fig. 7–14h. Complete this example and analyze seg-
ments SIQ, then QNL, and finally LC, and show that
and Also, use the results and
show that the moment diagram for DMENF is given in Fig. 7–14i.
M
C
= 37.5 kN
#
m.C
y
= 15.625 kN,
C
x
= 12.5 kN,
R
y
R
x
,O
y
,
20 kN
O
y
3.125 kN
4 m
2.5 m
5 kN
R
x
15 kN
R
y
3.125 kN
(c)
G
O
R
30 kN
J
y
15.625 kN
4 m
3 m
M
x
22.5 kN
M
y
12.5 kN
(
d
)
2.5 m
3.125 kN
5 kN
O
M
J
12.5 kN
2.5 m
4 m
R
10 kN
4 m
P
S
S
x
5 kN
S
y
3.125 kN
P
y
0
(f)
3.125 kN
15 kN
M
4 m
K
y
0
4 m
2.5 m
3 m
10 kN
22.5 kN
12.5 kN
25 kN
N
y
12.5 kN
N
x
7.5 kN
(g)
P
K
N
(
i
)
50
50
50 50
M (kNm)
4 8 12 16
x (m)
J
A
3 m
15.625 kN
12.5 kN
A
x
12.5 kN
M
A
37.5 kNm
A
y
15.625 kN
(e)
K
B
3 m
25 kN
B
x
25 kN
M
B
75 kNm
B
y
0
(h)
288 CHAPTER 7APPROXIMATE ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES
7
Fig. 7–15
7.6 Lateral Loads on Building Frames:
Cantilever Method
The cantilever method is based on the same action as a long
cantilevered beam subjected to a transverse load. It may be recalled
from mechanics of materials that such a loading causes a bending stress
in the beam that varies linearly from the beam’s neutral axis, Fig. 7–15a.
In a similar manner, the lateral loads on a frame tend to tip the frame
over, or cause a rotation of the frame about a “neutral axis” lying in a
horizontal plane that passes through the columns between each floor.
To counteract this tipping, the axial forces (or stress) in the columns will
be tensile on one side of the neutral axis and compressive on the other
side, Fig. 7–15b. Like the cantilevered beam, it therefore seems reasonable
to assume this axial stress has a linear variation from the centroid of the
column areas or neutral axis. The cantilever method is therefore
appropriate if the frame is tall and slender, or has columns with different
cross-sectional areas.
P
beam
(a)
building frame
(b)
7.6 LATERAL LOADS ON BUILDING FRAMES: CANTILEVER METHOD 289
7
The building framework has rigid connections.A lateral-load analysis can be performed
(approximately) by using the cantilever method of analysis.
In summary, using the cantilever method, the following assumptions
apply to a fixed-supported frame.
1. A hinge is placed at the center of each girder, since this is assumed
to be a point of zero moment.
2. A hinge is placed at the center of each column, since this is assumed
to be a point of zero moment.
3. The axial stress in a column is proportional to its distance from the
centroid of the cross-sectional areas of the columns at a given floor
level. Since stress equals force per area, then in the special case of
the columns having equal cross-sectional areas, the force in a column
is also proportional to its distance from the centroid of the column
areas.
These three assumptions reduce the frame to one that is both stable and
statically determinate.
The following examples illustrate how to apply the cantilever method
to analyze a building bent.