m4l28.pdf Beams Continue 02

Prévia do material em texto

Version 2 CE IIT, Kharagpur 
 
 
 
 
 
 
 
 
 
Module 
4 
 
Analysis of Statically 
Indeterminate 
Structures by the Direct 
Stiffness Method 
 
Version 2 CE IIT, Kharagpur 
 
 
 
 
 
 
 
 
Lesson 
29 
The Direct Stiffness 
Method: Beams 
(Continued) 
 
 
Version 2 CE IIT, Kharagpur 
 
Instructional Objectives 
After reading this chapter the student will be able to 
1. Compute moments developed in the continuous beam due to support 
settlements. 
2. Compute moments developed in statically indeterminate beams due to 
temperature changes. 
3. Analyse continuous beam subjected to temperature changes and support 
settlements. 
 
 
29.1 Introduction 
In the last two lessons, the analysis of continuous beam by direct stiffness matrix 
method is discussed. It is assumed in the analysis that the supports are 
unyielding and the temperature is maintained constant. However, support 
settlements can never be prevented altogether and hence it is necessary to 
make provisions in design for future unequal vertical settlements of supports and 
probable rotations of fixed supports. The effect of temperature changes and 
support settlements can easily be incorporated in the direct stiffness method and 
is discussed in this lesson. Both temperature changes and support settlements 
induce fixed end actions in the restrained beams. These fixed end forces are 
handled in the same way as those due to loads on the members in the analysis. 
In other words, the global load vector is formulated by considering fixed end 
actions due to both support settlements and external loads. At the end, a few 
problems are solved to illustrate the procedure. 
 
 
29.2 Support settlements 
Consider continuous beam ABC as shown in Fig. 29.1a. Assume that the flexural 
rigidity of the continuous beam is constant throughout. Let the support B settles 
by an amount Δ as shown in the figure. The fixed end actions due to loads are 
shown in Fig. 29.1b. The support settlements also induce fixed end actions and 
are shown in Fig. 29.1c. In Fig. 29.1d, the equivalent joint loads are shown. Since 
the beam is restrained against displacement in Fig. 29.1b and Fig. 29.1c, the 
displacements produced in the beam by the joint loads in Fig. 29.1d must be 
equal to the displacement produced in the beam by the actual loads in Fig. 
29.1a. Thus to incorporate the effect of support settlement in the analysis it is 
required to modify the load vector by considering the negative of the fixed end 
actions acting on the restrained beam. 
 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Version 2 CE IIT, Kharagpur 
 
29.3 Effect of temperature change 
The effect of temperature on the statically indeterminate beams has already been 
discussed in lesson 9 of module 2 in connection with the flexibility matrix method. 
Consider the continuous beam ABC as shown in Fig. 29.2a, in which span BC is 
subjected to a differential temperature 1T at top and 2T at the bottom of the beam. 
Let temperature in span AB be constant. Let d be the depth of beam and EI 
be the flexural rigidity. As the cross section of the member remains plane after 
bending, the relative angle of rotation θd between two cross sections at a 
distance dxapart is given by 
 ( )dx
d
TTd 21 −= αθ (29.1) 
 
where α is the co-efficient of the thermal expansion of the material. When beam 
is restrained, the temperature change induces fixed end moments in the beam as 
shown in Fig. 29.2b. The fixed end moments developed are, 
 ( )
d
TTEIMM TT 2121
−=−= α (29.2) 
 
Corresponding to the above fixed end moments; the equivalent joint loads can 
easily be constructed. Also due to differential temperatures there will not be any 
vertical forces/reactions in the beam. 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Example 29.1 
Calculate support reactions in the continuous beam ABC (vide Fig. 29.3a) having 
constant flexural rigidityEI , throughout due to vertical settlement of supportB , by 
mm5 as shown in the figure. Assume GPaE 200= and 44104 mI −×= . 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Version 2 CE IIT, Kharagpur 
 
The continuous beam considered is divided into two beam elements. The 
numbering of the joints and members are shown in Fig. 29.3b. The possible 
global degrees of freedom are also shown in the figure. A typical beam element 
with two degrees of freedom at each node is also shown in the figure. For this 
problem, the unconstrained degrees of freedom are 1u and 2u . The fixed end 
actions due to support settlement are, 
 
2
6 96 kN.m; 96 kN.mF FAB BA
EIM M
L
Δ= = = 
 
96 kN.m ; 96 kN.mF FBC CBM M= − = − (1) 
 
The fixed-end moments due to support settlements are shown in Fig. 29.3c. 
 
The equivalent joint loads due to support settlement are shown in Fig. 29.3d. In 
the next step, let us construct member stiffness matrix for each member. 
 
Member 1: mL 5= , node points 1-2. 
 
 
[ ]
1
3
5
6
80.024.040.024.0
24.0096.024.0096.0
40.024.080.024.0
24.0096.024.0096.0
'
1356..
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
= zzEIk
fodGlobal
 (2) 
 
 
Member 2: mL 5= , node points 2-3. 
 
 [ ]
2
4
1
3
80.024.040.024.0
24.0096.024.0096.0
40.024.080.024.0
24.0096.024.0096.0
2413..
2
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
= zzEIk
fodGlobal
 (3) 
 
On the member stiffness matrix, the corresponding global degrees of freedom 
are indicated to facilitate assembling. The assembled global stiffness matrix is of 
order 66× . Assembled stiffness matrix [ ]K is given by, 
 
 
 
 
Version 2 CE IIT, Kharagpur 
 
[ ]
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−
−−−
−−−
−
−
= zzEIK (4) 
 
Thus the global load vector corresponding to unconstrained degrees of freedom 
is, 
 
{ } ⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=
96
0
2
1
p
p
pk (5) 
 
Thus the load displacement relation for the entire continuous beam is, 
 
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−
−−−
−−−
−
−
=
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
6
5
4
3
2
1
6
5
4
3
096.024.00096.0024.0
24.08.0024.004.0
00096.0096.024.024.0
096.024.0096.0192.024.00
0024.024.08.04.0
24.04.024.004.06.1
96
0
u
u
u
u
u
u
EI
p
p
p
p
zz
 (6) 
 
Since, 06543 ==== uuuu due to support conditions. We get, 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
2
1
8.04.0
4.06.1
96
0
u
u
EI zz 
 
Thus solving for unknowns 1u and 2u , 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−
−=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
96
0
6.14.0
4.08.0
12.1
1
2
1
zzEIu
u
 
 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
×
×−=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−=
−
−
3
3
10714.1
10429.0
14.137
285.341
zzEI
 
 
radians 10714.1 ;radians 10429.0 32
3
1
−− ×=×−= uu (7) 
 
Now, unknown joint loads are calculated by, 
 
Version 2 CE IIT, Kharagpur⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−−=
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
14.137
285.341
024.0
04.0
24.024.0
24.00
6
5
4
3
zz
zz EI
EI
p
p
p
p
 (8) 
 
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
−
−
−=
23.8
71.13
68.24
91.32
 
 
Now the actual support reactions 543 ,, RRR and 6R must include the fixed end 
support reactions. Thus, 
 
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧−
=
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
−
−
−+
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧−
=
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
17.30
29.82
72.13
88.43
23.8
71.13
68.24
91.32
4.38
96
4.38
8.76
6
5
4
3
R
R
R
R
 (9) 
 
 kN 17.30 kN.m; 29.82 kN; 72.13 kN; 88.43 6543 ===−= RRRR (10) 
 
Example 29.2 
A continuous beam ABCD is carrying a uniformly distributed load of mkN /5 as 
shown in Fig. 29.4a. Compute reactions due to following support settlements. 
 
Support B m005.0 vertically downwards. 
 
Support C m010.0 vertically downwards. 
 
Assume GPaE 200= and 44104 mI −×= . 
 
 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Version 2 CE IIT, Kharagpur 
 
Solution 
The node and member numbering are shown in Fig. 29.4(b), wherein the 
continuous beam is divided into three beam elements. It is observed from the 
figure that the unconstrained degrees of freedom are 1u and 2u . The fixed end 
actions due to support settlements are shown in Fig. 29.4(c). and fixed end 
moments due to external loads are shown in Fig. 29.4(d). The equivalent joint 
loads due to support settlement and external loading are shown in Fig. 29.4(e). 
The fixed end actions due to support settlement are, 
 
( )ψ
L
EIM FA
6−= where ψ is the chord rotation and is taken ve+ if the 
rotation is counterclockwise. 
 
 
Substituting the appropriate values in the above equation, 
 
9 4
3
6 200 10 4 10 0.005 96 kN.m.
5 10 5
F
AM
−× × × × ⎛ ⎞= − − =⎜ ⎟× ⎝ ⎠ 
 
96 96 192 kN.m.FBM = + = 
 
 
96 192 96 kN.m.FCM = − = − 
 
 
192 kN.m.FDM = − (1) 
 
The vertical reactions are calculated from equations of equilibrium. The fixed end 
actions due to external loading are, 
 
2
10.42 kN.m.
12
F
A
w LM = = 
 
10.42 10.42 0 kN.m.FBM = − = 
 
0FCM = 
 
10.42 kN.m.FDM = − (2) 
 
In the next step, construct member stiffness matrix for each member. 
 
Member 1, mL 5= , node points 1-2. 
 
 
Version 2 CE IIT, Kharagpur 
 
 
[ ]
1
3
5
6
80.024.040.024.0
24.0096.024.0096.0
40.024.080.024.0
24.0096.024.0096.0
'
1356..
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
= zzEIk
fodGlobal
 (3) 
 
 
Member 2, mL 5= , node points 2-3. 
 
 
[ ]
2
4
1
3
80.024.040.024.0
24.0096.024.0096.0
40.024.080.024.0
24.0096.024.0096.0
2413..
2
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
= zzEIk
fodGlobal
 (4) 
 
 
Member 3, mL 5= , node points 3-4. 
 
 
[ ]
7
8
2
4
80.024.040.024.0
24.0096.024.0096.0
40.024.080.024.0
24.0096.024.0096.0
7824..
3
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
= zzEIk
fodGlobal
 (5) 
 
 
On the member stiffness matrix, the corresponding global degrees of freedom 
are indicated to facilitate assembling. The assembled global stiffness matrix is of 
the order 88× . Assembled stiffness matrix [ ]K is, 
 
 
 
Version 2 CE IIT, Kharagpur 
 
[ ]
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−−−
−
−
−
−−−
−−−
−
−
=
096.024.000096.0024.00
24.080.00024.0040.00
00096.024.00096.0024.0
0024.080.0024.0040.0
096.024.000192.0096.0024.0
00096.024.0096.0192.024.00
24.040.000024.060.140.0
0024.040.024.00.040.060.1
zzEIK
 (6) 
 
 
The global load vector corresponding to unconstrained degree of freedom is, 
 
{ } ⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=
96
192
2
1
p
p
pk (7) 
 
 
Writing the load displacement relation for the entire continuous beam, 
 
 
 
⎪⎪
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−−−
−
−
−
−−−
−−−
−
−
=
⎪⎪
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪⎪
⎪
⎨
⎧−
8
7
6
5
4
3
2
1
8
7
6
5
4
3
096.024.000096.0024.00
24.080.00024.0040.00
00096.024.00096.0024.0
0024.080.0024.0040.0
096.024.000192.0096.0024.0
00096.024.0096.0192.024.00
24.040.000024.060.140.0
00375.040.024.00.040.060.1
96
192
u
u
u
u
u
u
u
u
EI
p
p
p
p
p
p
zz
 
 
 (8) 
We know that 0876543 ====== uuuuuu . Thus solving for unknowns 
displacements 1u and 2u from equation, 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−
2
1
60.140.0
40.060.1
96
192
u
u
EI zz (9) 
 
 
 
Version 2 CE IIT, Kharagpur 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−
−
×=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
96
192
60.140.0
40.060.1
)1080(4.2
1
3
2
1
u
u
 
 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
×
×−=
−
−
3
3
1020.1
1080.1
 (10) 
 
 
radians 1020.1 ;radians 1080.1 32
3
1
−− ×=×−= uu (11) 
 
The unknown joint loads are calculated as, 
 
 
( )
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
×
×−
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−
×=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
3
3
3
8
7
6
5
4
3
1020.1
1080.1
24.00
40.00
024.0
040.0
024.0
24.00
1080
p
p
p
p
p
p
 
 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
−=
04.23
40.38
56.34
60.57
56.34
04.23
 (12) 
 
Now the actual support reactions 76543 ,,,, RRRRR and 8R must include the fixed 
end support reactions. Thus, 
 
 
Version 2 CE IIT, Kharagpur 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
−+
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
+
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
26.66
02.164
34.16
82.48
64.55
04.48
04.23
40.38
56.34
60.57
56.34
04.23
3.89
42.202
9.50
42.106
2.90
25
8
7
6
5
4
3
8
7
6
5
4
3
8
7
6
5
4
3
p
p
p
p
p
p
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
 (13) 
 
 kN.m; 82.48 kN; 64.55 kN; 04.48 543 =−== RRR 
 kN 2666 kN.m; 02.164 kN; 34.16 876 .RRR =−== (14) 
 
 
SummaryThe effect of temperature changes and support settlements can easily be 
incorporated in the direct stiffness method and is discussed in the present 
lesson. Both temperature changes and support settlements induce fixed end 
actions in the restrained beams. These fixed end forces are handled in the same 
way as those due to loads on the members in the analysis. In other words, the 
global load vector is formulated by considering fixed end actions due to both 
support settlements and external loads. At the end, a few problems are solved to 
illustrate the procedure. 
 
	Analysis of StaticallyIndeterminateStructures by the DirectStiffness Method
	The Direct StiffnessMethod: Beams(Continued)
	Instructional Objectives
	29.1 Introduction
	29.2 Support settlements
	29.3 Effect of temperature change
	Summary