Hibbeler R.C. Structural Analysis
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Intermediate Loadings. For application, it is important that the
elements of the beam be free of loading along its length. This is necessary
since the stiffness matrix for each element was developed for loadings
applied only at its ends. (See Fig. 15–4.) Oftentimes, however, beams will
support a distributed loading, and this condition will require modification
in order to perform the matrix analysis.
To handle this case, we will use the principle of superposition in a
manner similar to that used for trusses discussed in Sec. 14–8. To show
its application, consider the beam element of length L in Fig. 15–7a,
which is subjected to the uniform distributed load w. First we will apply
fixed-end moments and reactions to the element, which will be used in
the stiffness method, Fig. 15–7b. We will refer to these loadings as a
column matrix Then the distributed loading and its reactions are
applied, Fig. 15–7c. The actual loading within the beam is determined
by adding these two results. The fixed-end reactions for other cases of
loading are given on the inside back cover. In addition to solving problems
involving lateral loadings such as this, we can also use this method to solve
problems involving temperature changes or fabrication errors.
Member Forces. The shear and moment at the ends of each beam
element can be determined using Eq. 15–2 and adding on any fixed-end
reactions if the element is subjected to an intermediate loading.
We have
(15–5)
If the results are negative, it indicates that the loading acts in the opposite
direction to that shown in Fig. 15–4.
q = kd + q
0
q
0
q
0
-q
0
.
580
CHAPTER 15 BEAM ANALYSIS U SING THE STIFFNESS METHOD
15
L
w
actual loading
(a)
ⴝ
fixed-end
element loading
on joints
(b)
wL
___
2
wL
___
2 wL
___
2
wL
___
2
wL
2
____
12
wL
2
____
12
wL
2
____
12
wL
2
____
12
w
actual loading and
reactions on fixed-
supported element
(c)
ⴙ
Fig. 15–7
15.4 APPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 581
15
Procedure for Analysis
The following method provides a means of determining the
displacements, support reactions, and internal loadings for the
members or finite elements of a statically determinate or statically
indeterminate beam.
Notation
• Divide the beam into finite elements and arbitrarily identify each
element and its nodes. Use a number written in a circle for a node
and a number written in a square for a member. Usually an element
extends between points of support, points of concentrated loads,
and joints, or to points where internal loadings or displacements are
to be determined. Also, E and I for the elements must be constants.
• Specify the near and far ends of each element symbolically by
directing an arrow along the element, with the head directed
toward the far end.
• At each nodal point specify numerically the y and z code
numbers. In all cases use the lowest code numbers to identify all
the unconstrained degrees of freedom, followed by the remaining
or highest numbers to identify the degrees of freedom that are
constrained.
• From the problem, establish the known displacements and
known external loads Include any reversed fixed-end loadings
if an element supports an intermediate load.
Structure Stiffness Matrix
• Apply Eq. 15–1 to determine the stiffness matrix for each element
expressed in global coordinates.
• After each member stiffness matrix is determined, and the rows
and columns are identified with the appropriate code numbers,
assemble the matrices to determine the structure stiffness matrix K.
As a partial check, the member and structure stiffness matrices
should all be symmetric.
Displacements and Loads
• Partition the structure stiffness equation and carry out the
matrix multiplication in order to determine the unknown
displacements and support reactions
• The internal shear and moment q at the ends of each beam
element can be determined from Eq. 15–5, accounting for the
additional fixed-end loadings.
Q
u
.D
u
Q
k
.
D
k
EXAMPLE 15.1
582 CHAPTER 15 BEAM ANALYSIS U SING THE STIFFNESS METHOD
15
k
1
= EI
6453
D
1.5 1.5 -1.5 1.5
1.5 2 -1.5 1
-1.5 -1.5 1.5 -1.5
1.5 1 -1.5 2
T
6
4
5
3
k
2
= EI
5321
D
1.5 1.5 -1.5 1.5
1.5 2 -1.5 1
-1.5 -1.5 1.5 -1.5
1.5 1 -1.5 2
T
5
3
2
1
Determine the reactions at the supports of the beam shown in
Fig. 15–8a. EI is constant.
(
a
)
5 kN
2 m 2 m
SOLUTION
Notation. The beam has two elements and three nodes, which are
identified in Fig. 15–8b. The code numbers 1 through 6 are indicated
such that the lowest numbers 1–4 identify the unconstrained degrees of
freedom.
The known load and displacement matrices are
Q
k
= D
0
-5
0
0
T
1
2
3
4
D
k
= c
0
0
d
5
6
(b)
1 2
6
5
2
1
3
4
3
1
5 kN
2
Member Stiffness Matrices. Each of the two member stiffness
matrices is determined from Eq. 15–1. Note carefully how the code
numbers for each column and row are established.
Fig. 15–8
15
15.4 A
PPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 583
Displacements and Loads. We can now assemble these elements into
the structure stiffness matrix. For example, element
etc.Thus,
The matrices are partitioned as shown. Carrying out the multiplication
for the first four rows, we have
Solving,
Using these results, and multiplying the last two rows, gives
Ans.
Ans.=-5 kN
Q
6
= 0 + 0 + 1.5EIa-
6.67
EI
b+ 1.5EIa
3.33
EI
b
= 10 kN
Q
5
= 1.5EIa-
16.67
EI
b- 1.5EIa-
26.67
EI
b+ 0 - 1.5EIa
3.33
EI
b
D
4
=
3.33
EI
D
3
=-
6.67
EI
D
2
=-
26.67
EI
D
1
=-
16.67
EI
0 = 0 + 0 + D
3
+ 2D
4
0 = D
1
- 1.5D
2
+ 4D
3
+ D
4
-
5
EI
=-1.5D
1
+ 1.5D
2
- 1.5D
3
+ 0
0 = 2D
1
- 1.5D
2
+ D
3
+ 0
F
0
-5
0
0
Q
5
Q
6
V
= EI
12 345 6
F
2 -1.5 1 0 1.5 0
-1.5 1.5 -1.5 0 -1.5 0
1 -1.5 4 1 0 1.5
0012-1.5 1.5
1.5 -1.5 0 -1.5 3 -1.5
0 0 1.5 1.5 -1.5 1.5
VF
D
1
D
2
D
3
D
4
0
0
V
Q = KD
K
55
= 1.5 + 1.5 = 3,
K
11
= 0 + 2 = 2,
EXAMPLE 15.2
Determine the internal shear and moment in member 1 of the
compound beam shown in Fig. 15–9a. EI is constant.
SOLUTION
Notation. When the beam deflects, the internal pin will allow a
single deflection, however, the slope of each connected member will
be different. Also, a slope at the roller will occur.These four unknown
degrees of freedom are labeled with the code numbers 1, 2, 3, and 4,
Fig. 15–9b.
Member Stiffness Matrices. Applying Eq. 15–1 to each member,
in accordance with the code numbers shown in Fig. 15–9b, we have
Q
k
= ≥
0
0
0
-M
0
¥
1
2
3
4
D
k
= C
0
0
0
S
5
6
7
584
CHAPTER 15 BEAM ANALYSIS U SING THE STIFFNESS METHOD
15
k
2
= EI
3
2
5
4
H
12
L
3
6
L
2
-
12
L
3
6
L
2
6
L
2
4
L
-
6
L
2
2
L
-
12
L
3
-
6
L
2
12
L
3
-
6
L
2
6
L
2
2
L
-
6
L
2
4
L
X
3
2
5
4
k
1
= EI
6
7
3
1
H
12
L
3
6
L
2
-
12
L
3
6
L
2
6
L
2
4
L
-
6
L
2
2
L
-
12
L
3
-
6
L
2
12
L
3
-
6
L
2
6
L
2
2
L
-
6
L
2
4
L
X
6
7
3
1
Displacements and Loads. The structure stiffness matrix is formed
by assembling the elements of the member stiffness matrices.Applying
the structure matrix equation, we have
Q = KD
Fig. 15–9
1
2
L L
M
0
(b)
1
2
L L
63
5
4
2
1
7
2
3
1
M
0
(a)
15
15.4 A
PPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 585
Multiplying the first four rows to determine the displacement yields
So that
Using these results, the reaction is obtained from the multiplication
of the fifth row.
Ans.
This result can be easily checked by statics applied to member 2 .
Q
5
=
M
0
L
Q
5
=-
6EI
L
2
a-
M
0
L
6EI
b-
12EI
L
3
a
M
0
L
2
3EI
b-
6EI
L
2
a-
2M
0
L
3EI
b
Q
5
D
4
=-
2M
0
L
3EI
D
3
=
M
0
L
2
3EI
D
2
=-
M
0
L
6EI
D
1
=
M
0
L
2EI
-M
0
=
2
L
D
2
+
6
L
2
D
3
+
4
L
D
4
0 =-
6
L
2
D
1
+
6
L
2
D
2
+
24
L
3
D
3
+
6
L
2
D
4
0 =
4
L
D
2
+
6
L
2
D
3
+
2
L
D
4
0 =
4
L
D
1
-
6
L
2
D
3
= EI
12 34567
4
L
0 -
6
L
2
00
6
L
2
2
L
0
4
L
6
L
2
2
L
-
6
L
2
00
-
6
L
2
6
L
2
24
L
3
6
L
2
-
12
L
3
-
12
L
3
-
6
L
2
0
2
L
6
L
2
4
L
-
6
L
2
00
0 -
6
L
2
-
12
L
3
-
6
L
2
12
L
3
00
6
L
2
0 -
12
L
3
00
12
L
3
6
L
2
2
L
0 -
6
L
2
00
6
L
2
4
L
D
1
D
2
D
3
D
4
0
0
0
1
2
3
4
5
6
7
0
0
0
-M
0
Q
5
Q
6
Q
7
⎡
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎣
⎤
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎦
⎡
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎣
⎤
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎦
⎡
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎣
⎤
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎦
1
2
3
4
5
6
7
The beam in Fig. 15–10a is subjected to the two couple moments. If
the center support ② settles 1.5 mm, determine the reactions at the
supports. Assume the roller supports at ➀ and ③ can pull down or
push up on the beam. Take and I = 22110
-6
2 m
4
.E = 200 GPa
586
CHAPTER 15 BEAM ANALYSIS U SING THE STIFFNESS METHOD
15
Fig. 15–10
EXAMPLE 15.3
SOLUTION
Notation. The beam has two elements and three unknown degrees of
freedom. These are labeled with the lowest code numbers, Fig. 15–10b.
Here the known load and displacement matrices are
Q
k
= C
4
0
-4
S
1
2
3
D
k
= C
0
-0.0015
0
S
4
5
6
4 kNm
1 2
2 m 2 m
4 kNm
3
1
2
Member Stiffness Matrices. The member stiffness matrices are
determined using Eq. 15–1 in accordance with the code numbers and
member directions shown in Fig. 15–10b. We have,
3
2
6
(b)
4 kNm
1 2
2 m 2 m
4 kNm
3
1
5
1
4
2
(a)
15
15.4 A
PPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 587
Displacements and Loads. Assembling the structure stiffness
matrix and writing the stiffness equation for the structure, yields
Solving for the unknown displacements,
Substituting and solving,
Using these results, the support reactions are therefore
D
3
=-0.001580 radD
2
= 0,D
1
= 0.001580 rad,
EI = 200110
6
21222110
-6
2,
-4
EI
= 0D
1
+ 1D
2
+ 2D
3
+ 0 - 1.51-0.00152+ 0
0 = 1D
1
+ 4D
2
+ 1D
3
- 1.5102+ 0 + 0
4
EI
= 2D
1
+ D
2
+ 0D
3
- 1.5102+ 1.51-0.00152+ 0
F
4
0
-4
Q
4
Q
5
Q
6
V
= EI
123 456
F
210-1.5 1.5 0
141-1.5 0 1.5
0120-1.5 1.5
-1.5 -1.5 0 1.5 -1.5 0
1.5 0 -1.5 -1.5 3 -1.5
0 1.5 1.5 0 -1.5 1.5
VF
D
1
D
2
D
3
0
-0.0015
0
V
k
2
= EI
5241
D
1.5 1.5 -1.5 1.5
1.5 2 -1.5 1
-1.5 -1.5 1.5 -1.5
1.5 1 -1.5 2
T
5
2
4
1
k
1
= EI
6 3 5 2
D
1.5 1.5 -1.5 1.5
1.5 2 -1.5 1
-1.5 -1.5 1.5 -1.5
1.5 1 -1.5 2
T
6
3
5
2
Ans.
Ans.
Ans.
Q
6
= 200110
6
222110
-6
2[0 + 1.5102+ 1.51-0.0015802+ 0 - 1.51-0.00152+ 1.5102] =-0.525 kN
Q
5
= 200110
6
222110
-6
2[1.510.0015802+ 0 - 1.51-0.0015802- 1.5102+ 31-0.00152- 1.5102] = 1.05 kN
Q
4
= 200110
6
222110
-6
2[-1.510.0015802- 1.5102+ 0 + 1.5102- 1.51-0.00152+ 0] =-0.525 kN
Determine the moment developed at support A of the beam shown in
Fig. 15–11a. Assume the roller supports can pull down or push up on
the beam. Take
SOLUTION
Notation. Here the beam has two unconstrained degrees of
freedom, identified by the code numbers 1 and 2.
The matrix analysis requires that the external loading be applied at
the nodes, and therefore the distributed and concentrated loads are
replaced by their equivalent fixed-end moments, which are determined
from the table on the inside back cover. (See Example 11–2.) Note that
no external loads are placed at ① and no external vertical forces are
placed at ② since the reactions at code numbers 3, 4 and 5 are to be
unknowns in the load matrix. Using superposition, the results of the
matrix analysis for the loading in Fig. 15–11b will later be modified
by the loads shown in Fig. 15–11c. From Fig. 15–11b, the known
displacement and load matrices are
Member Stiffness Matrices. Each of the two member stiffness
matrices is determined from Eq. 15–1.
Member 1:
Member 2:
6EI
L
2
=
61292110
3
215102
[81122]
2
= 9628.91
12EI
L
3
=
121292110
3
215102
[81122]
3
= 200.602
k
1
=
4
3
5
2
D
7.430 1069.9 -7.430 1069.9
1069.9 205 417 -1069.9 102 708
-7.430 -1069.9 7.430 -1069.9
1069.9 102 708 -1069.9 205 417
T
4
3
5
2
2EI
L
=
21292110
3
215102
241122
= 102 708
4EI
L
=
41292110
3
215102
241122
= 205 417
6EI
L
2
=
61292110
3
215102
[241122]
2
= 1069.9
12EI
L
3
=
121292110
3
215102
[241122]
3
= 7.430
D
k
= C
0
0
0
S
4
5
6
Q
k
= c
144
1008
d
1
2
I = 510 in
4
.E = 29110
3
2 ksi,
588
CHAPTER 15 BEAM ANALYSIS U SING THE STIFFNESS METHOD
15
EXAMPLE 15.4
12 k
2 k/ft
A
B
C
24 ft
4 ft 4 ft
(a)
1
1
3
4
96 kft 12 k ft 1008 k in.
12 kft 144 k in.
2
5
2
6
1
2
2
3
beam to be analyzed by stiffness method
(b)
12 k
2 k/ft
24 k
12 kft 144 k in.
96 kft 1152 k in.
beam subjected to actual load and
fixed-supported reactions
(c)
A
B
C
24 k
6 k
6 k
Fig. 15–11
15
15.4 A
PPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 589
Displacements and Loads. We require
Q = KD
k
2
=
5261
D
200.602 9628.91 -200.602 9628.91
9628.91 616 250 -9628.91 308 125
-200.602 -9628.91 200.602 -9628.91
9628.91 308 125 -9628.91 616 250
T
5
2
6
1
2EI
L
=
21292110
3
215102
81122
= 308 125
4EI
L
=
41292110
3
215102
81122
= 616 250
F
144
1008
Q
3
Q
4
Q
5
Q
6
V
=
12 34 5 6
F
616 250 308 125 0 0 9628.91 -9628.91
308 125 821 667 102 708 1069.9 8559.01 -9628.91
0 102 708 205 417 1069.9 -1069.9 0
0 1069.9 1069.9 7.430 -7.430 0
9628.91 8559.01 -1069.9 -7.430 208.03 -200.602
-9628.91 -9628.91 0 0 -200.602 200.602
VF
D
1
D
2
0
0
0
0
V
Solving in the usual manner,
Thus,
The actual moment at A must include the fixed-supported reaction of
shown in Fig. 15–11c, along with the calculated result for
Thus,
Ans.
This result compares with that determined in Example 11–2.
Although not required here, we can determine the internal moment
and shear at B by considering, for example, member 1, node 2,
Fig. 15–11b. The result requires expanding
D
q
4
q
3
q
5
q
2
T
=
43 52
D
7.430 1069.9 -7.430 1069.9
1069.9 205 417 -1069.9 102 708
-7.430 -1069.9 7.430 -1069.9
1069.9 102 708 -1069.9 205 417
TD
0
0
0
1.40203
T
110
-3
2+
D
24
1152
24
-1152
T
q
1
= k
1
d + 1q
0
2
1
M
AB
= 12 k
#
ft + 96 k
#
ft = 108 k
#
ftg
Q
3
.+96 k
#
ft
Q
3
= 0 + 102 70811.402032110
-3
2= 144 k
#
in. = 12 k
#
ft
D
2
= 1.40203110
-3
2 in.
D
1
=-0.4673110
-3
2 in.
1008 = 308 125D
1
+ 821 667D
2
144 = 616 250D
1
+ 308 125D
2