sol2 - drsekaran

Transcript

sol2 - drsekaran

Università degli studi di Napoli Federico II
Facoltà di Ingegneria - a.a. 2007/2008
Corso di TECNICA DELLE COSTRUZIONI II
prof. G. Serino (allievi Civili, Civili_SS, STReGA)
Lecture by: Dr (Eng) Chandrasekaran
Solution 2: Two dimensional analysis for beams and frames
1. What do you understand by degree-of-freedom? Distinguish this from kinematic
indeterminacy.
By definition, degree-of-freedom is the number of independent coordinates required to
fully describe the displacement of a structural element subjected to any arbitrary loading
with respect to any standard reference coordinate system. Whereas kinematic indeterminacy
refers, in particular, to the number of unrestrained joint displacements (rotations and
translations) of the structure. For a give beam element, knowing the nature of support (roller,
hinged or fixed), the number of unrestrained joint displacements can be identified (for ex,
refer to Table 1 of Lesson 1). Hence they become the unknowns to be solved for a given
external loading. On the other hand, degree-of-freedom is the number of independent
displacements required to specify the geometric position of the structure under action of any
given loading. At this point of time, it should be appreciated that they have only a marginal
difference between them and normally leads to a confusion.
2. Define stiffness coefficient.
Stiffness coefficient, kij is defined as the force required (generated or developed) in ith
degree-of-freedom by giving unit displacement (may be rotation or translation) in jth degreeof-freedom by restraining all other degrees-of-freedom (except that of i and j under
consideration) from any type of displacement (neither translational nor rotational). It is to be
appreciated here that the jth shall be always subjected to unit translation (either unit rotation
or unit displacement). Therefore stiffness coefficient can also be understood as force
generated (developed) by giving unit displacement to the structure at appropriate degree-offreedom.
3. Derive [K] for a 2d beam element, considering and neglecting the axial deformation.
Before we derive the stiffness coefficients of the beam element, let us understand some basic
equations related to fixed beam. Though these are extended fundamentals of the classical
indeterminate structural analysis classes, for completion sake they are repeated here.
a) Let us consider a beam of length ‘L’ and flexural rigidity ‘EI’, fixed at both ends as
shown in Fig. 2.3. The end B is given a rotation amounting θ.
MA
MB
RA
θ
RB
Fig. 2.3 Fixed beam subjected to joint rotation
+
MB
RB
The basic equations of compatibility are
θm + θr = θ
∂m +∂r =0
( 2.1)
(2.2)
by substituting in Eq. (2.1), we get
pag. 1/14
− M B L R B L2
+
=θ
EI
2 EI
2EI θ
∴− 2M B + R B L =
L
Also,
(2.3)
(2.4)
− M B L2
R L3
+ B
=0
2EI
3EI
∴− 3M B + 2R B L = 0 ⇒ R B L =
(2.5)
3
MB
2
Substituting Eq. (2.6) in Eq. (2.4), we get M B =
(2.6)
− 4EI θ
L
(-ve sign is due to the fact that moment
is applied in the direction opposite to rotation.
Now M A = M B − R B L
(2.7)
3
2EI θ
= MB − MB =
2
L
(2.8)
Further by substituting Eq. (2.8) in Eq. (2.6), we get R B =
− 6 EI θ
L2
Finally,
4EI θ
L
2EI θ
L
6 EI θ
6 EI θ
2
L2
L
Pl note that the
moment is in the
same direction as
that of the rotation
given
Fig. 2.4 Fixed beam subjected to joint rotation with final values
b) Let us now consider a beam of length ‘L’ and flexural rigidity ‘EI’, fixed at both
ends as shown in Fig. 2.5 with support B settles by Λ.
MB
MA
▲
RA
RB
Fig. 2.5 Fixed beam subjected to settlement of support
+
MB
RB
-RBL/EI
- MB/EI
Fig. 2.6 conjugate beam loaded with M/EI diagram
pag. 2/14
From the classical structural mechanics, it is known that the rotation at point B in the real
beam is same as that of shear force at B in conjugate beam; and deflection at point B in the
real beam is same as that of moment at the same point in the conjugate beam. Hence the
equations for rotation and deflections at point B from the two conjugate beams loaded with
equivalent M/EI diagrams is given by:
− M BL
− M B L2
θM =
and δ M =
EI
2EI
(2.9)
 R L2   2L  R B L3
R L2
1R L

θR =  B  L = B
and δ R =  B
=
 2 EI   3 
2  EI 
2 EI
3EI


The compatibility conditions are: θ M + θ R = 0 and δ M + δ R = − ∆
(2.10)
(2.11)
where rotations and displacements due to moments carry suffix M and that due to reactions
carry suffix R.
θ M + θ R = 0 will yield
− M B L R B L2
+
=0
EI
2 EI
− 2M B + R B L = 0 ⇒ R B =
δ M + δ R = − ∆ will yield
(2.12)
2M B
L
(2.13)
− M B L2
R L3
+ B
= −∆
2EI
3EI
(2.14)
substituting the value from Eq. (2.13), we now get M B =
− 6 EI ∆
(2.15)
L2
Similarly, for getting MA, we use the compatibility condition, δ M + δ R = ∆ yielding
MA =
+ 6 EI ∆
(2.16)
L2
By substituting Eq. (2.15) in Eq. 2.13), we get
RB =
− 12 EI ∆
L3
and R A =
12 EI ∆
(2.17)
L3
6EI ▲/L2
▲
12EI ▲/L3
6EI ▲/L2
12EI ▲/L3
Pl note that the
moment is applied to
counter act the
settlement. i.e. in the
direction to lift the
beam which is settled
Fig. 2.7 Fixed beam subjected to support settlement with final values
c) Let us now consider a beam of length ‘L’ and flexural rigidity ‘EI’, fixed at both ends as
shown in Fig. 2.8 with axial deformation taken into account
Fig. 2.8 Fixed beam considered for the analysis
pag. 3/14
The degrees-of-freedom are to be marked at both the supports. The common confusion starts here
with a fundamental question whether to consider the un-restrained degree-of-freedom or restrained
degree-of-freedom (or both). In fact, if you consider a fixed beam, you actually do not have any unrestrained degree-of-freedom and while you think of marking only the un-restrained degree-offreedom, as required for stiffness method of analysis, then you cannot really start the derivation.
You should always consider the beam as a general case and consider all possible degrees-offreedom I every support. The restrained and un-restrained degrees-of-freedom will be identified
actually when this beam becomes a part of any structure or a frame. The order of numbering
degrees-of-freedom is very important in this method of analysis as this shall enable to write a
computer program very easily and systematically. Let the beam has structural properties as A, E, I,
L which are known before hand. One may get a confusion here that usually structural design, which
actually gives some of these values (for ex. b, d etc) following the analysis and therefore how they
are considered as known values in the analysis. It is important to understand that structural design is
an iterative process wherein dimensions are initially assumed (usually known as preliminary values
in the design) and analysis is done. On the basis of the results obtained from the analysis, the stress
values at critical cross-sections are checked with respect to relevant design code in practice and
revised, if required.
The order of degrees-of-freedom can be remembered as below:
1.
First start from the left hand support and number the rotation at this support
Important to
as 1 and that at right hand support as 2.
remember
Always follow an unique sign convention. As this generally
causes confusion, the author strictly advise to assume anti-clockwise
rotations and upward reactions at all supports. It is very important to note that these
conventions do not have any sign attached to it (neither +ve nor –ve).
2.
Next number the reactions at left and right supports as 3 and 4 respectively. Remember to
mark the directions of these degrees-of-freedom as upwards. (They do not have sign
attached to them)
3.
Next number axial degrees-of-freedom in left and right support as 5 and 6 respectively.
Remember to mark them acting from left to right at both the supports. (They do not have
any sign attached to them)
Fig. 2.9 shows the beam considered with all degrees-of-freedom marked as explained above.
1
2
6
5
3
4
Fig. 2.9 Fixed beam with degrees-of-freedom marked
Axis of analysis
It is now important to understand that the axis under consideration in such 2-d analysis of beams.
Imagine a X-Y plane of which is the beam is lying such that the length of the beam is towards the
positive side of X axis. Therefore, depth of the beam shall be normal to the X axis (parallel to XZ
plane) whereas breadth of the beam shall be measured along Y axis (parallel to XY plane) and
length of the beam is measured normal to XZ plane as shown in the Fig. 2.10
pag. 4/14
d
x
y
b
A, E, I and length
Fig. 2.10 Fixed Beam shown with the axis considered for the analysis
Please note that the moments marked in Fig. 2.9 are about y axis and their values shall be taken as
I yy =
bd 3
. Coefficients of [K] shall be derived by giving unit displacement in jth degree-of-freedom
12
and finding the forces in ith degree-of-freedom, by keeping all other degrees-of-freedom as
restrained.
a) by giving unit rotation in 1st degree-of-freedom and obtaining the forces
1
2
3
4
θ =1
6EI yyθ
l2
4EI yyθ
l
= k 31
= k11
First column of K
could be
24/07/2007
 4EI / l 
 2EI / l 


2
 6EI / l 

2
− 6EI / l 


0




0
2EI yyθ
l
Dr. Chandrasekaran
= k 21
6EI yyθ
l2
= k 41
8
Fig. 2.11 Obtaining first column of [K]
pag. 5/14
b) by giving unit rotation in 2nd degree-of-freedom and obtaining the forces
1
2
3
4EI yyθ
l
6EI yyθ
l2
2EI yyθ
= k 22
θ =1
= k 32
= k12
l
4
second column of
K could be
24/07/2007
6EI yyθ
 2EI / l 
 4EI / l 


 6EI / l 2 

2
− 6EI / l 


0




0
l2
= k 42
9
Dr. Chandrasekaran
Fig. 2.12 Obtaining second column of [K]
c) by giving unit displacement along 3rd degree-of-freedom and obtaining the forces
1
2
3
4
∆ =1
12EI yy ∆
6EI yy ∆
l
2
= k13
l3
= k 33
third column of K
could be
24/07/2007
6EI yy ∆
 6EI / l 2 

2 
 6EI / l 
 12EI / l 3 

3
−12EI / l 


0




0
l2
= k 23
Dr. Chandrasekaran
12EI yy ∆
l3
= k 43
10
Fig. 2.13 Obtaining third column of [K]
pag. 6/14
d) by giving unit displacement along 4th degree-of-freedom and obtaining the forces
1
2
3
4
∆ =1
12EI yy ∆
6EI yy ∆
l
2
l
3
6EI yy ∆
= k 34
 − 6EI / l 

2
 − 6EI / l 
−12EI / l 3 

3 
 12EI / l 


0




0
2
= k14
fourth column of K
could be
24/07/2007
l2
= k 24
12EI yy ∆
l3
= k 44
11
Dr. Chandrasekaran
Fig. 2.14 Obtaining fourth column of [K]
e) by giving unit axial displacement along 5th degree-of-freedom and obtaining the forces
5
6
1
2
3
4
∆ =1
5
6
AE ∆
= k 55
l
fifth column of K
could be
24/07/2007
 0 
 0 


 0 


 0 
 AE / l 


− AE / l 
Dr. Chandrasekaran
AE ∆
= k 65
l
12
Fig. 2.15 Obtaining fifth column of [K]
pag. 7/14
f) by giving unit axial displacement along 6th degree-of-freedom and obtaining the forces
5
6
1
2
3
4
∆ =1
5
6
AE ∆
= k 56
l
AE ∆
= k 66
l
 0 
 0 


 0 


 0 
− AE / l 


 AE / l 
sixth column of K
could be
24/07/2007
13
Dr. Chandrasekaran
Fig. 2.16 Obtaining sixth column of [K]
Thus the complete stiffness matrix of the fixed beam is given as:
 4EI
 l

 2EI
 l
 6EI
 2
 l
 − 6EI
 2
 l
 0


 0

6EI
− 6EI
2
l
6EI
2
l
− 6EI
l2
12EI
l2
− 12EI
l2
− 6EI
l3
− 12EI
l3
12EI
l2
l3
l3
0
0
0
0
0
0
2EI
l
4EI
l
6EI
0
0
0
0
AE
l
− AE
l

0 

0 


0 


0 

− AE 
l 
AE 

l 
The partition lines above show the 4x4 stiffness matrix of the fixed beam if axial deformation
degrees-of-freedom are not considered.
pag. 8/14
4. In your opinion, why a fixed beam should be considered as a general case for derivation.?
Does it have any significant advantage?
The fixed beam is the most probable case of a structural element which is a subset of the structural
frame. In the analysis, if beams with other types of supports are encountered, only the degrees-offreedom has to be changed from restrained to un-restrained and so on. This makes the analysis very
simple and systematic. It has a major advantage because we do not have to derive the stiffness
matrix for each beam in relationship with its type of support and just add or delete the respective
rows and columns from the original matrix (appending) to generate a new stiffness matrix suitable
for the specific beam under consideration. This also saves time (during hand calculation) and
optimizes the memory allocation in computer program.
5. What are the important characteristics of [K] of a beam element
The important characteristics of [K] can be listed as:
It is a square matrix, the size of being the degrees-of-freedom considered
The matrix will be always a symmetrical matrix
[K] is written for a member and NOT FOR A JOINT
all leading diagonal elements of this matrix are positive
It is diagonally dominant
The elements of [K] are functions of E, I, l and ▲and not a function of applied force and
the member response.
IF ANY OF THE ABOVE CHARACTERISTICS ARE NOT SEEN IN THE DERIVED
[K], THEN IT IS DERIVED FOR A SPECIAL MEMBER.
Axial
deformation is
not considered
6. Derive [K] for the following
3
2
1
A
4m, EI
4
B
6m, EI
C
5
6
Fig. 2.17 Continuous beam ABC – Problem statement
The degrees-of-freedom are to be labelled for the given problem, which is the most important step
in the analysis. For the continuous beam shown in Fig. 2.17, starting from left support A, let us
mark the un-restrained degrees-of-freedom in each support. These labels are shown in grey shade
and please notice that they are continuously numbered. Then start labelling restrained degrees-offreedom in each support (again starting from left support A). These labels are shown without shade
as 4,5 and 6. Now this gives us an idea that [K] for the beam ABC will of size 6x6 (as there are 6
degrees-of-freedom in total. The stiffness matrix derived for beam AB and beam BC are given
below:
pag. 9/14
1
2
5
1
0. 5
0. 375
- 0. 375
1
0. 5
1
0. 375
- 0. 375
2
- 0. 1875
4
5
KAB = EI
0. 375
0. 375
0. 1875
- 0. 375
- 0. 375
- 0. 1875
0. 1875
2
3
5
6
1. 5
0. 75
0. 375
- 0. 375
2
0. 75
1. 5
0. 375
- 0. 375
3
0. 375
0. 375
0. 125
- 0. 375
- 0. 375
- 0. 125
KBC = EI
KABC = EI
- 0. 125
5
0. 125
6
1
2
3
4
1.0
0. 5
0. 0
0. 375
0. 75
0. 375
1. 50
0. 5
2. 5
0.0
0.75
5
6
0.0
1
0.0
-0. 375
2
0.0
0. 375
-0. 375
3
-0. 375
0. 375
0. 375
0.0
0. 1875
0. 1875
0.0
4
0. 375
0.0
0. 375
- 0. 1875
0. 3125
- 0. 125
5
0.0
0. 375
0.0
- 0. 125
0. 125
6
- 0. 375
pag. 10/14
The above matrix can also be re-written in a general form as below:
S
K ABC = EI  UU
S RU
S UR 

S RR 
where SUU, SUR etc are elements of the complete stiffness matrix of the beam ABC. “U” stands for
un-restrained and “R” stands for restrained degree-of-freedom respectively. The reader can easily
identify that size of all the four sub matrices shall be 3x3 (as this problem has 3 un-restrained and 3
restrained degree-of-freedom. The advantage of partitioning the complete structure stiffness matrix
into sub-matrices as above shall be discussed later.
The above matrix KABC has some special characteristics, which are interesting to note. The left
square area of 3x3 show matrix of unrestrained degrees-of-freedom and right square area show 3x3
of restrained degrees-of-freedom. It is arranged in this way as they are labelled in a specific pattern.
Should the labelling pattern changed, the unrestrained and restrained degrees-of-freedom will not
get automatically grouped as obtained now. This grouping has an advantage in inverse of the
matrix, which is required to obtain the complete solution. The elements of KABC are obtained by
collecting the respective terms from KAB and KBC and adding them. For ex, to obtain the element 11, we shall collect elements of 1-1 from KAB and KBC which shall be
(1.0 + 0 as there is no 1-1 in KBC ).
Problem 2:
Note that supports B and C do
not have separate lateral dof .
Axial deformations are
neglected
2
1
4m, EI
B
3
C
3m, EI
4m, EI
4
9
D
5
7
A
8
6
Fig. 2.18 Single storey- single bay frame- Problem statement
For the frame, the unrestrained and restrained degrees-of-freedom are identified in the same manner
as that of the previous problem. It can be easily seen that since there are 4 unrestrained degree-offreedom and 5 restrained degree-of-freedom, the size of [K] shall be 9x9. In this case, there will be
three stiffness matrices, one for each element of the frame, namely KAB, KBC and KDC respectively.
Let us first derive KBC which is a simple beam element and is given as below:
pag. 11/14
1
2
6
1
0. 5
0. 375
- 0. 375
1
0. 5
1
0. 375
- 0. 375
2
- 0. 1875
6
0. 1875
8
8
KBC = EI
0. 375
0. 375
0. 1875
- 0. 375
- 0. 375
- 0. 1875
The column elements AB and DC can be simply viewed as beam elements, by doing the following:
For column element AB, hold the support A and rotate the column clockwise by 90° and for
member DC, hold support D and rotate the member clockwise similarly. As a result, we get
horizontal members AB and DC as shown in the Fig.
5
1
A
4
D
B
4m, EI
7
2
3m, EI
3
9
C
3
Fig. 2.19 Beam elements derived from the frame
5
1
7
1
0. 5
0. 375
- 0. 375
5
0. 5
1
0. 375
- 0. 375
1
- 0. 1875
7
0. 1875
3
3
KAB = EI
0. 375
0. 375
0. 1875
- 0. 375
- 0. 375
- 0. 1875
pag. 12/14
4
2
1.3333
0.6667
0.6667
1.3333
9
0.6667
0.6667
3
-0.6667
4
-0.6667
2
KDC = EI
0.6667
0.6667
0.4445
-0.4445
-0.6667
-0.6667
-0.4445
0.4445
9
3
4
2
2. 00
0. 50
0.50
2. 3333
9
3
- 0. 375
0. 00
- 0. 6667
0. 6667
4
2
SUU = EI
- 0. 375
- 0. 6667
0. 632
-0. 6667
0. 0
0. 6667
- 0. 6667
1. 3333
9
3
pag. 13/14
1
2
3
2.0
0.5
-0.375
0.5
2.3333
-0.6667
0.6667
-0.375
-0.6667
0.632
-0.6667
0.0
0.6667
0.5
4
5
6
7
0.0
0.5
0.375
0.375
-0.375
0.0
0.375
0.0
-0.375
0.6667
-0.375
0.0
-0.1875
0.0
-0.4445
3
-0.6667 1.3333
0.0
0.0
0.0
0.0
0.6667
4
0.0
-0.375
1.0
0.0
0.375
0.0
0.0
5
0.375
0.375
0.0
0.0
0.0
0.1875
0.0
-0.1875
0.0
6
0.375
0.0
-0.1875
0.0
0.375
0.0
0.1875
0.0
0.0
7
-0.375
-0.375
0.0
0.0
0.0
0.1875
0.0
8
0.0
0.6667
0.6667
0.0
0.0
0.0
0.0
-0.4445
0.0
-0.1875
0.0
8
9
0.0
0.4445
pag. 14/14
1
2
9

sol2 - drsekaran

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