Hibbeler R.C. Structural Analysis
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470 CHAPTER 11 DISPLACEMENT METHOD OF A NALYSIS: SLOPE-DEFLECTION EQUATIONS
11
Determine the moments at each joint of the frame shown in Fig. 11–16a.
EI is constant.
SOLUTION
Slope-Deflection Equations. Three spans must be considered in
this problem: AB, BC, and CD. Since the spans are fixed supported at
A and D, Eq. 11–8 applies for the solution.
From the table on the inside back cover, the FEMs for BC are
Note that and since no sidesway
will occur.
Applying Eq. 11–8, we have
(1)
(2)
(3)
(4)
(5)
(6) M
DC
= 0.1667EIu
C
M
DC
= 2Ea
I
12
b[2102+ u
C
- 3102] + 0
M
CD
= 0.333EIu
C
M
CD
= 2Ea
I
12
b[2u
C
+ 0 - 3102] + 0
M
CB
= 0.5EIu
C
+ 0.25EIu
B
+ 80
M
CB
= 2Ea
I
8
b[2u
C
+ u
B
- 3102] + 80
M
BC
= 0.5EIu
B
+ 0.25EIu
C
- 80
M
BC
= 2Ea
I
8
b[2u
B
+ u
C
- 3102] - 80
M
BA
= 0.333EIu
B
M
BA
= 2Ea
I
12
b[2u
B
+ 0 - 3102] + 0
M
AB
= 0.1667EIu
B
M
AB
= 2Ea
I
12
b[2102+ u
B
- 3102] + 0
M
N
= 2Ek12u
N
+ u
F
- 3c2+ 1FEM2
N
c
AB
= c
BC
= c
CD
= 0,u
A
= u
D
= 0
1FEM2
CB
=
5wL
2
96
=
51242182
2
96
= 80 kN
#
m
1FEM2
BC
=-
5wL
2
96
=-
51242182
2
96
=-80 kN
#
m
EXAMPLE 11.5
B
24 kN/m
C
A
D
8 m
12 m
(a)
Fig. 11–16
11.4 ANALYSIS OF FRAMES: NO SIDESWAY 471
11
Equilibrium Equations. The preceding six equations contain eight
unknowns. The remaining two equilibrium equations come from
moment equilibrium at joints B and C, Fig. 11–16b. We have
(7)
(8)
To solve these eight equations, substitute Eqs. (2) and (3) into
Eq. (7) and substitute Eqs. (4) and (5) into Eq. (8). We get
Solving simultaneously yields
which conforms with the way the frame deflects as shown in
Fig. 11–16a. Substituting into Eqs. (1)–(6), we get
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Using these results, the reactions at the ends of each member can
be determined from the equations of equilibrium, and the moment
diagram for the frame can be drawn, Fig. 11–16c.
M
DC
=-22.9 kN
#
m
M
CD
=-45.7 kN
#
m
M
CB
= 45.7 kN
#
m
M
BC
=-45.7 kN
#
m
M
BA
= 45.7 kN
#
m
M
AB
= 22.9 kN
#
m
u
B
=-u
C
=
137.1
EI
0.833EIu
C
+ 0.25EIu
B
=-80
0.833EIu
B
+ 0.25EIu
C
= 80
M
CB
+ M
CD
= 0
M
BA
+ M
BC
= 0
BB
M
BC
M
BC
M
BA
M
BA
CC
M
CB
M
CB
M
CD
M
CD
(
b
)(
b
)
82.3 kNm
45.7 kNm
45.7 kNm
22.9 kNm
45.7 kNm
(
c
)
22.9 kNm
472 CHAPTER 11 DISPLACEMENT METHOD OF A NALYSIS: SLOPE-DEFLECTION EQUATIONS
11
Determine the internal moments at each joint of the frame shown in
Fig. 11–17a. The moment of inertia for each member is given in the
figure. Take E = 29110
3
2 ksi.
EXAMPLE 11.6
SOLUTION
Slope-Deflection Equations. Four spans must be considered in this
problem. Equation 11–8 applies to spans AB and BC, and Eq. 11–10
will be applied to CD and CE, because the ends at D and E are pinned.
Computing the member stiffnesses, we have
The FEMs due to the loadings are
Applying Eqs. 11–8 and 11–10 to the frame and noting that
since no sidesway occurs, we have
(1) M
AB
= 10740.7u
B
M
AB
= 2[29110
3
21122
2
]10.0012862[2102+ u
B
- 3102] + 0
M
N
= 2Ek12u
N
+ u
F
- 3c2+ 1FEM2
N
c
AB
= c
BC
= c
CD
= c
CE
= 0
u
A
= 0,
1FEM2
CE
=-
wL
2
8
=-
31122
2
8
=-54 k
#
ft
1FEM2
CB
=
PL
8
=
61162
8
= 12 k
#
ft
1FEM2
BC
=-
PL
8
=-
61162
8
=-12 k
#
ft
k
BC
=
800
161122
4
= 0.002411 ft
3
k
CE
=
650
121122
4
= 0.002612 ft
3
k
AB
=
400
151122
4
= 0.001286 ft
3
k
CD
=
200
151122
4
= 0.000643 ft
3
3 k/ft
6 k
8 ft
8 ft
15 ft
B
A
C
D
E
400 in
4
200 in
4
800 in
4
650 in
4
(a)
12 ft
Fig. 11–17
11.4 ANALYSIS OF FRAMES: NO SIDESWAY 473
11
(2)
(3)
(4)
(5)
(6)
Equations of Equilibrium. These six equations contain eight
unknowns. Two moment equilibrium equations can be written for
joints B and C, Fig. 11–17b. We have
(7)
(8)
In order to solve, substitute Eqs. (2) and (3) into Eq. (7), and Eqs. (4)–(6)
into Eq. (8). This gives
Solving these equations simultaneously yields
These values, being clockwise, tend to distort the frame as shown in
Fig. 11–17a. Substituting these values into Eqs. (1)–(6) and solving,
we get
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.M
CE
=-37.3 k
#
ft
M
CD
= 4.12 k
#
ft
M
CB
= 33.1 k
#
ft
M
BC
=-0.592 k
#
ft
M
BA
= 0.592 k
#
ft
M
AB
= 0.296 k
#
ft
u
B
= 2.758110
-5
2 rad u
C
= 5.113110
-4
2 rad
20 138.9u
B
+ 81 059.0u
C
= 42
61 759.3u
B
+ 20 138.9u
C
= 12
M
CB
+ M
CD
+ M
CE
= 0
M
BA
+ M
BC
= 0
M
CE
= 32 725.7u
C
- 54
M
CE
= 3[29110
3
21122
2
]10.0026122[u
C
- 0] - 54
M
CD
= 8055.6u
C
M
CD
= 3[29110
3
21122
2
]10.0006432[u
C
- 0] + 0
M
N
= 3Ek1u
N
- c2+ 1FEM2
N
M
CB
= 20 138.9u
B
+ 40 277.8u
C
+ 12
M
CB
= 2[29110
3
21122
2
]10.0024112[2u
C
+ u
B
- 3102] + 12
M
BC
= 40 277.8u
B
+ 20 138.9u
C
- 12
M
BC
= 2[29110
3
21122
2
]10.0024112[2u
B
+ u
C
- 3102] - 12
M
BA
= 21 481.5u
B
M
BA
= 2[29110
3
21122
2
]10.0012862[2u
B
+ 0 - 3102] + 0
M
BC
M
BC
M
BA
M
BA
BB
CC
M
CE
M
CE
M
CD
M
CD
M
CB
M
CB
(
b
)(
b
)
474 CHAPTER 11 DISPLACEMENT METHOD OF A NALYSIS: SLOPE-DEFLECTION EQUATIONS
11
11.5 Analysis of Frames: Sidesway
A frame will sidesway, or be displaced to the side, when it or the loading
acting on it is nonsymmetric. To illustrate this effect, consider the frame
shown in Fig. 11–18. Here the loading P causes unequal moments
and at the joints B and C, respectively. tends to displace joint B
to the right, whereas tends to displace joint C to the left. Since
is larger than the net result is a sidesway of both joints B and C to
the right, as shown in the figure.* When applying the slope-deflection
equation to each column of this frame, we must therefore consider the
column rotation (since ) as unknown in the equation. As a
result an extra equilibrium equation must be included for the solution. In
the previous sections it was shown that unknown angular displacements
were related by joint moment equilibrium equations. In a similar manner,
when unknown joint linear displacements (or span rotations ) occur,
we must write force equilibrium equations in order to obtain the complete
solution. The unknowns in these equations, however, must only involve
the internal moments acting at the ends of the columns, since the slope-
deflection equations involve these moments. The technique for solving
problems for frames with sidesway is best illustrated by examples.
c¢
u
c =¢>Lc
¢M
CB
,
M
BC
M
CB
M
BC
M
CB
M
BC
AD
P
B
C
M
BC
M
CB
L
Determine the moments at each joint of the frame shown in Fig. 11–19a.
EI is constant.
SOLUTION
Slope-Deflection Equations. Since the ends A and D are fixed,
Eq. 11–8 applies for all three spans of the frame. Sidesway occurs here
since both the applied loading and the geometry of the frame are non-
symmetric. Here the load is applied directly to joint B and therefore
no FEMs act at the joints. As shown in Fig. 11–19a, both joints B and
C are assumed to be displaced an equal amount Consequently,
and Both terms are positive since the cords
of members AB and CD “rotate” clockwise. Relating to we
have Applying Eq. 11–8 to the frame, we havec
AB
= 118>122c
DC
.
c
DC
,c
AB
c
DC
=¢>18.c
AB
=¢>12
¢.
EXAMPLE 11.7
(1)
(2)
(3)M
BC
= 2Ea
I
15
b[2u
B
+ u
C
- 3102] + 0 = EI10.267u
B
+ 0.133u
C
2
M
BA
= 2Ea
I
12
bc2u
B
+ 0 - 3a
18
12
c
DC
bd+ 0 = EI10.333u
B
- 0.75c
DC
2
M
AB
= 2Ea
I
12
bc2102+ u
B
- 3 a
18
12
c
DC
bd+ 0 = EI10.1667u
B
- 0.75c
DC
2
Fig. 11–18
Fig. 11–19
*Recall that the deformation of all three members due to shear and axial force is neglected.
B
C
A
D
18 ft
40 k
12 ft
15 ft
(a)
11.5 ANALYSIS OF FRAMES: SIDESWAY 475
11
BB
M
BC
M
BC
M
BA
M
BA
CC
M
CB
M
CB
M
CD
M
CD
(b)(b)
40 k 40 k
B
M
BA
M
AB
V
A
12 ft
C
M
CD
18 ft
M
DC
V
D
(c)
(4)
(5)
(6)
Equations of Equilibrium. The six equations contain nine unknowns.
Two moment equilibrium equations for joints B and C, Fig. 11–19b, can
be written, namely,
(7)
(8)
Since a horizontal displacement occurs, we will consider summing
forces on the entire frame in the x direction. This yields
The horizontal reactions or column shears and can be related
to the internal moments by considering the free-body diagram of each
column separately, Fig. 11–19c. We have
Thus,
(9)
In order to solve, substitute Eqs. (2) and (3) into Eq. (7), Eqs. (4)
and (5) into Eq. (8), and Eqs. (1), (2), (5), (6) into Eq. (9). This yields
Solving simultaneously, we have
Finally, using these results and solving Eqs. (1)–(6) yields
Ans.
Ans.
Ans.
Ans.
Ans.
Ans. M
DC
=-110 k
#
ft
M
CD
=-94.8 k
#
ft
M
CB
= 94.8 k
#
ft
M
BC
= 135 k
#
ft
M
BA
=-135 k
#
ft
M
AB
=-208 k
#
ft
EIu
B
= 438.81
EIu
C
= 136.18
EIc
DC
= 375.26
0.5u
B
+ 0.222u
C
- 1.944c
DC
=-
480
EI
0.133u
B
+ 0.489u
C
- 0.333c
DC
= 0
0.6u
B
+ 0.133u
C
- 0.75c
DC
= 0
40 +
M
AB
+ M
BA
12
+
M
DC
+ M
CD
18
= 0
V
D
=-
M
DC
+ M
CD
18
©M
C
= 0;
V
A
=-
M
AB
+ M
BA
12
©M
B
= 0;
V
D
V
A
40 - V
A
- V
D
= 0
:
+
©F
x
= 0;
¢
M
CB
+ M
CD
= 0
M
BA
+ M
BC
= 0
M
DC
= 2Ea
I
18
b[2102+ u
C
- 3c
DC
] + 0 = EI10.111u
C
- 0.333c
DC
2
M
CD
= 2Ea
I
18
b[2u
C
+ 0 - 3c
DC
] + 0 = EI10.222u
C
- 0.333c
DC
2
M
CB
= 2Ea
I
15
b[2u
C
+ u
B
- 3102] + 0 = EI10.267u
C
+ 0.133u
B
2
476 CHAPTER 11 DISPLACEMENT METHOD OF A NALYSIS: SLOPE-DEFLECTION EQUATIONS
11
Determine the moments at each joint of the frame shown in Fig. 11–20a.
The supports at A and D are fixed and joint C is assumed pin connected.
EI is constant for each member.
SOLUTION
Slope-Deflection Equations. We will apply Eq. 11–8 to member
AB since it is fixed connected at both ends. Equation 11–10 can be
applied from B to C and from D to C since the pin at C supports zero
moment. As shown by the deflection diagram, Fig. 11–20b, there is an
unknown linear displacement of the frame and unknown angular
displacement at joint B.* Due to the cord members AB and CD
rotate clockwise, Realizing that
and that there are no FEMs for the members, we have
(1)
(2)
(3)
(4)
Equilibrium Equations. Moment equilibrium of joint B, Fig. 11–20c,
requires
(5)
If forces are summed for the entire frame in the horizontal direction,
we have
(6)
As shown on the free-body diagram of each column, Fig. 11–20d,we
have
V
D
=-
M
DC
4
©M
C
= 0;
V
A
=-
M
AB
+ M
BA
4
©M
B
= 0;
10 - V
A
- V
D
= 0
:
+
©F
x
= 0;
M
BA
+ M
BC
= 0
M
DC
= 3Ea
I
4
b10 - c2+ 0
M
BC
= 3Ea
I
3
b1u
B
- 02+ 0
M
N
= 3Ea
I
L
b1u
N
- c2+ 1FEM2
N
M
BA
= 2Ea
I
4
b12u
B
+ 0 - 3c2+ 0
M
AB
= 2Ea
I
4
b[2102+ u
B
- 3c] + 0
M
N
= 2Ea
I
L
b12u
N
+ u
F
- 3c2+ 1FEM2
N
u
A
= u
D
= 0c = c
AB
= c
DC
=¢>4.
¢,u
B
¢
EXAMPLE 11.8
Fig. 11–20
*The angular displacements and at joint C (pin) are not included in the
analysis since Eq. 11–10 is to be used.
u
CD
u
CB
3 m
10 kN
4 m
A
B
C
D
(a)
u
B
u
B
u
CB
u
CD
c
CD
c
AB
CB
AD
(b)
M
BC
M
BC
10 kN10 kN
M
BA
M
BA
(c)(c)
11.5 ANALYSIS OF FRAMES: SIDESWAY 477
11
Thus, from Eq. (6),
(7)
Substituting the slope-deflection equations into Eqs. (5) and (7) and
simplifying yields
Thus,
Substituting these values into Eqs. (1)–(4), we have
Ans.
Ans.
Using these results, the end reactions on each member can be
determined from the equations of equilibrium, Fig. 11–20e.The
moment diagram for the frame is shown in Fig. 11–20f.
M
DC
=-11.4 kN
#
mM
BC
= 11.4 kN
#
m,
M
BA
=-11.4 kN
#
mM
AB
=-17.1 kN
#
m,
u
B
=
240
21EI
c =
320
21EI
10 +
EI
4
a
3
2
u
B
-
15
4
cb = 0
u
B
=
3
4
c
10 +
M
AB
+ M
BA
4
+
M
DC
4
= 0
11.4
11.4
17.1
11.4
(f)
10 kN10 kN
(e)(e)
7.14 kN7.14 kN
3.81 kN3.81 kN
3.81 kN3.81 kN
3.81 kN3.81 kN
11.4 kNm11.4 kNm
11.4 kNm11.4 kNm
2.86 kN2.86 kN
2.86 kN2.86 kN
3.81 kN3.81 kN 3.81 kN3.81 kN
2.86 kN2.86 kN
2.86 kN2.86 kN
3.81 kN3.81 kN
11.4 kNm11.4 kNm
2.86 kN2.86 kN
3.81 kN3.81 kN
11.4 kNm11.4 kNm
17.1 kNm17.1 kNm
11.4 kNm11.4 kNm
7.14 kN7.14 kN
7.14 kN7.14 kN
3.81 kN3.81 kN
M
BA
M
BA
V
B
V
B
V
A
V
A
M
AB
M
AB
V
C
V
C
V
D
V
D
M
DC
M
DC
(d)(d)
478 CHAPTER 11 DISPLACEMENT METHOD OF A NALYSIS: SLOPE-DEFLECTION EQUATIONS
11
Fig. 11–21
Explain how the moments in each joint of the two-story frame shown
in Fig. 11–21a are determined. EI is constant.
SOLUTION
Slope-Deflection Equation. Since the supports at A and F are
fixed, Eq. 11–8 applies for all six spans of the frame. No FEMs have to
be calculated, since the applied loading acts at the joints. Here the
loading displaces joints B and E an amount and C and D an
amount As a result, members AB and FE undergo rotations
of and BC and ED undergo rotations of
Applying Eq. 11–8 to the frame yields
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
These 12 equations contain 18 unknowns.
M
EF
= 2Ea
I
5
b[2u
E
+ 0 - 3c
1
] + 0
M
FE
= 2Ea
I
5
b[2102+ u
E
- 3c
1
] + 0
M
DE
= 2Ea
I
5
b[2u
D
+ u
E
- 3c
2
] + 0
M
ED
= 2Ea
I
5
b[2u
E
+ u
D
- 3c
2
] + 0
M
EB
= 2Ea
I
7
b[2u
E
+ u
B
- 3102] + 0
M
BE
= 2Ea
I
7
b[2u
B
+ u
E
- 3102] + 0
M
DC
= 2Ea
I
7
b[2u
D
+ u
C
- 3102] + 0
M
CD
= 2Ea
I
7
b[2u
C
+ u
D
- 3102] + 0
M
CB
= 2Ea
I
5
b[2u
C
+ u
B
- 3c
2
] + 0
M
BC
= 2Ea
I
5
b[2u
B
+ u
C
- 3c
2
] + 0
M
BA
= 2Ea
I
5
b[2u
B
+ 0 - 3c
1
] + 0
M
AB
= 2Ea
I
5
b[2102+ u
B
- 3c
1
] + 0
c
2
=¢
2
>5.c
1
=¢
1
>5
¢
1
+¢
2
.
¢
1
,
A
B
D
E
C
F
40 kN
80 kN
5 m
(a)
5 m
⌬
1
⌬
1
⌬
1
⫹⌬
2
⌬
1
⫹⌬
2
7 m
EXAMPLE 11.9
11.5 ANALYSIS OF FRAMES: SIDESWAY 479
11
Equilibrium Equations. Moment equilibrium of joints B, C, D, and E,
Fig. 11–21b, requires
(13)
(14)
(15)
(16)
As in the preceding examples, the shear at the base of all the columns
for any story must balance the applied horizontal loads, Fig. 11–21c.
This yields
(17)
(18)
Solution requires substituting Eqs. (1)–(12) into Eqs. (13)–(18), which
yields six equations having six unknowns, and
These equations can then be solved simultaneously. The results are
resubstituted into Eqs. (1)–(12), which yields the moments at the
joints.
u
E
.u
D
,u
C
,u
B
,c
2
,c
1
,
120 +
M
AB
+ M
BA
5
+
M
EF
+ M
FE
5
= 0
40 + 80 - V
AB
- V
FE
= 0
:
+
©F
x
= 0;
40 +
M
BC
+ M
CB
5
+
M
ED
+ M
DE
5
= 0
40 - V
BC
- V
ED
= 0
:
+
©F
x
= 0;
M
EF
+ M
EB
+ M
ED
= 0
M
DC
+ M
DE
= 0
M
CB
+ M
CD
= 0
M
BA
+ M
BE
+ M
BC
= 0
M
BC
M
BC
M
BE
M
BE
M
BA
M
BA
BB
CC
M
CB
M
CB
M
CD
M
CD
M
DC
M
DC
M
D
E
M
D
E
DD
M
ED
M
ED
M
EB
M
EB
M
EF
M
EF
EE
(b)(b)
40 kN40 kN
V
BC
V
BC
V
ED
V
ED
40 kN40 kN
80 kN80 kN
V
AB
V
AB
V
FE
V
FE
(
c
)(
c
)