Post on 05-Oct-2019
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September 18, 2002 Ahmed Elgamal
Stiffness Coefficients for a Flexural Element
Ahmed Elgamal
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Ahmed ElgamalSeptember 18, 2002
Stiffness coefficients for a flexural element (neglecting axial deformations), Appendix 1, Ch. 1 Dynamics of Structures by Chopra.
u1u3u2 u4
4 degrees of freedom
Positive directionsk11k21
k31k41
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September 18, 2002 Ahmed Elgamal
u1 1.0
L
To obtain k coefficients in 1st column of stiffness matrix, move u1 = 1, u2 = u3 = u4 = 0, and find forces and moments needed to maintain this shape.
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September 18, 2002 Ahmed Elgamal
2L6EI
2L6EI
3L12EI
3L12EI
Σ M = 3L12EI - 3L
12EI= 0
Positive directionsk11k21
k31k41
Note that Σ Forces = 0Σ Moments = 0
3L12EIi.e. remember , and you
can find other forces & moments
These are (see structures textbook)
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September 18, 2002 Ahmed Elgamal
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
44434241
34333231
24232221
14131211
kkkkkkkkkkkkkkkk
k
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
=
6L6L12
12
LEIk 3
, where i is row numberand j is column number
kkij =
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September 18, 2002 Ahmed Elgamal
u21.0
L
u2 = 1, u1 = u3 = u4 = 0
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September 18, 2002 Ahmed Elgamal
2L6EI
2L6EI
3L12EI
3L12EI
u2 = 1
Positive directionsk12k22
k32k42⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
6L6L1212-
LEIk 3
Σ M = 0, Σ F = 0
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September 18, 2002 Ahmed Elgamal
u3 = 1, u1 = u2 = u4 = 0
u3 = 11.0
L
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September 18, 2002 Ahmed Elgamal
2L6EI
2L6EI
L2EI
L4EI
Positive directionsk13k23
k33k43⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
L2EIL
4EIL6EIL6EI-
k2
2
Σ M = 0, Σ F = 0
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September 18, 2002 Ahmed Elgamal
u4 = 1, u1 = u2 = u3 = 0
u4 = 1 1.0
L
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September 18, 2002 Ahmed Elgamal
2L6EI
2L6EI
L2EI
L4EI
Positive directionsk14k24
k34k44⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
L4EIL
2EIL
6EIL6EI-
k2
2
Σ M = 0, Σ F = 0
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September 18, 2002 Ahmed Elgamal
u1u2
ug
m
L
m is lumped at a point & does not contribute in rotation
Example: Water Tank
u2 above was u3 in the earlier section of these notes
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September 18, 2002 Ahmed Elgamal
Example: Water Tank (continued)
u1u2
ug
m
L
k11k21
k12k22
g2
1
2
23
2
1 u0m
uu
L4EI
L6EI
L6EI
L12EI
uu
0m
&&&&
&&⎥⎦
⎤⎢⎣
⎡−=⎥
⎦
⎤⎢⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
+⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
“Note Symmetry”Rotational(used to be u3)
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September 18, 2002 Ahmed Elgamal
Example: Water Tank (continued)
Static Condensation:
Way to solve a smaller system of equations by eliminating degrees of freedom with zero mass.
e.g., in the above, the 2nd equation gives
0uL
4EIuL6EI
212 =+−
or
11122 u2L3u
4L6u
4EIL
L6EIu === -----*
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September 18, 2002 Ahmed Elgamal
Example: Water Tank (continued)Substitute * into Equation 1
g1231 umu2L3
L6EI
L12EIum &&&& −=⎟
⎠⎞
⎜⎝⎛ −+
or,
g131 umu2L
18EI24EIum &&&& −=⎟⎠⎞
⎜⎝⎛ −
+
or,
g131 umuL
3EIum &&&& −=⎟⎠⎞
⎜⎝⎛+
Now, solve for u1 and u2 can be evaluated from Equation * above.
Static condensation can be applied to large MDOF systems of equations, the same way as shown above.
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September 18, 2002 Ahmed Elgamal
Example: Water Tank (continued)
if EIb = 0
Ib
or of column
g131 umuL
3EIum &&&& −=⎟⎠⎞
⎜⎝⎛+
k of water tank as we were given earlier.
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September 18, 2002 Ahmed Elgamal
Mandatory Reading
Example 9.4 page 362-364
Example 9.8 page 368-369
Sample Exercises: 9.5, 9.8, & 9.9
bendingbeam
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September 18, 2002 Ahmed Elgamal
(a)
(b)
Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-086973-2
Example
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September 18, 2002 Ahmed Elgamal
(c)
(d)
Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-086973-2
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September 18, 2002 Ahmed Elgamal
(e)
(f)
Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-086973-2
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September 18, 2002 Ahmed Elgamal
(g)
(h)
Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-086973-2
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September 18, 2002 Ahmed Elgamal
(i)
(j)
Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-086973-2
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September 18, 2002 Ahmed Elgamal
,⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
00
mm
M 2
1
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
−−−
=
22
223
8L2L06L2L4L6L6L06L24126L6L1212
LEIk
Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-086973-2
therefore,
Sample Exercise: For the above cantilever system, write equation of motion and perform static condensation to obtain a 2 DOF system.
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September 18, 2002 Ahmed Elgamal
m
L
bendingbeam
3L3EIk = , m
k=ω
beam with EIbeam = 0
⎟⎠⎞
⎜⎝⎛= 3L
3EI2k⇒
Column Stiffness (lateral vibration)
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September 18, 2002 Ahmed Elgamal
u
3L12EI
k =L ⎟⎠⎞
⎜⎝⎛= 3L12EI2k
Case ofEIb = ∞
h1 h2
(rigid roof)
32
2231
11
hI12E
hI12E
k +=
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September 18, 2002 Ahmed Elgamal
h
L = 2h
3c
7h96EIk = if cb EIEI =
beam column
, & EEE bc ==412ρ112ρ
h24EIk 3
c
++
=c
b
4IIρ =
(See example 1.1 in Dynamics of Structures by Chopra)
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September 18, 2002 Ahmed Elgamal
u1
θ1θ2
fs
call it u2
call it u3
Obtained by “static condensation” of 3x3 system
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
00f
uuu S
3
2
1
force
Use to represent u2 and u3 in terms of u1 & plug back into
Technique can also be used for large systems of equationsand get fs = ku1
(See example 1.1 in Dynamics of Structures by Chopra)
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September 18, 2002 Ahmed Elgamal
u1
u2u3
IcIc
Ib
Neglect axial deformationDraft Example
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September 18, 2002 Ahmed Elgamal
L
L
2c
L6EI
( ) ⎟⎠⎞
⎜⎝⎛
3c
L12EI2
2c
L6EI
u1 = 1
u2 = u3 = 0
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September 18, 2002 Ahmed Elgamal
u2 = 1
u1 = u3 = 0
2c
L6EI
L4EIc
L4EIb
L2EIb
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September 18, 2002 Ahmed Elgamal
u3 = 1
u1 = u2 = 0
2c
L6EI
L4EIc
L4EIb
L2EIb
( )( ) ⎥
⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
++=
2cb
2bc
2b
2cbc
ccc
3
LII4L2IL6IL2ILII4L6IL6IL6I24I
LEk
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September 18, 2002 Ahmed Elgamal
If frame is subjected to lateral force fs
Then (for simplicity, let Ic = Ib = I)
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
00f
uuu
8L2L6L2L8L6L6L6L24
LEI s
3
2
1
22
223
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September 18, 2002 Ahmed Elgamal
Static condensation:
From 2nd and 3rd equations,
1
1
22
22
3
2 u6L6L
8L2L2L8L
uu
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−=⎥
⎦
⎤⎢⎣
⎡−
1
1
22
22
44 u6L6L
8L2L-2L-8L
4L64L1
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−−
=−
1u11
10L6
⎥⎦
⎤⎢⎣
⎡−=
⎥⎦
⎤⎢⎣
⎡−
−−
=⎥⎦
⎤⎢⎣
⎡−
acbd
cbad1
dcba 1
Note matrix inverse:
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September 18, 2002 Ahmed Elgamal
Substitute into 1st equation
13s13 u10L
168EIfu1036
103624
LEI
==⎥⎦⎤
⎢⎣⎡ −−
or
310L168EIk = (check this result)
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September 18, 2002 Ahmed Elgamal
2L
L
Draft Example 2
u1
u2u3
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September 18, 2002 Ahmed Elgamal
2c
L6EI
( ) ⎟⎠⎞
⎜⎝⎛
3c
L12EI2
2c
L6EI
u1 = 1
u2 = u3 = 0
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September 18, 2002 Ahmed Elgamal
2c
L6EI
L4EIc
2L4EIb
2L2EIb
u2 = 1
u1 = u3 = 0
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September 18, 2002 Ahmed Elgamal
2c
L6EI
L4EIc
2L4EIb
2L2EIb
u3 = 1
u1 = u2 = 0
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September 18, 2002 Ahmed Elgamal
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +=
2c
b2bc
2b
2c
bc
ccc
3
LI2I4LIL6I
LILI2I4L6I
L6IL6I24I
LEk
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
00f
uuu
6LL6LL6L6L6L6L24
LEI s
3
2
1
22
223
For simplicity, let Ib = Ic
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September 18, 2002 Ahmed Elgamal
Static condensation:
1
1
22
22
3
2 u6L6L
6LLL6L
uu
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−=⎥
⎦
⎤⎢⎣
⎡−
1
1
22
22
44 u6L6L
6LL-L-6L
L36L1
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
−=
−
11 u11
7L6u
11
35L03
⎥⎦
⎤⎢⎣
⎡−=⎥
⎦
⎤⎢⎣
⎡−=
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September 18, 2002 Ahmed Elgamal
Substitute in 1st Equation
s13 fu7
367
3624LEI
=⎥⎦⎤
⎢⎣⎡ −−
13s uLEI
796f =
37L96EIk =
or,
or, Same as in Example 1.1, Dynamics of Structures by Chopra
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September 18, 2002 Ahmed Elgamal
Sample Exercise
1.1 Derive stiffness matrix fork
EIc2
EIc1
EIb
h2
h1
L
1.2 For the special case of Ic1 = Ic2 = Ib, h1 = h2 = h and L = 2h,find lateral stiffness k of the frame.
1)
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September 18, 2002 Ahmed Elgamal
Sample Exercise
2) Derive equation of motion for:
m
2m
EIb
EIb
EIc
EIcEIc
EIch
h
2h
use 600 lb/ft
Flexural rigidity of beams and columns
E = 29,000 ksi, Columns W8x24 sections
with Ic = 82.4 in4
h = 12 ftIb = ½ Ic
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September 18, 2002 Ahmed Elgamal
Sample Exercise (Optional)
3) Derive lateral k of system (need to use computer to invert3x3 matrix)
Ib Ib
IcIcIch = 12 ft
24 ft 24 ft
E = 29,000 ksi,
W8x24 sectionsIc = 82.4 in4
Ib = ½ Ic