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Home Work 4
Applications for 3D Structures
Requirements for the Work no. 4
Obtain the seismic response effect according to Earthquake Engineering Code
P100-2006 for a 3D frame structure considering plan frames working together at
floor levels. Horizontal seismic forces due to translation and torsional effects will
be calculated for some vertical structural elements.
The structures to be solved are shown in Appendix IVa.
Notes:
- The data for this work is normally the same as from Work no. 2 and
Work no. 3. Other values might be imposed or added by the course
coordinator.
- Always use the international system of units (i.e. N, kg, m, s). In the case
that there is data given in different units, please transform it into the
international system.
- More or different structures might be imposed by your classroom
coordinator.
- The work is usually given during the eighth week of the semester. The
time to solve the work is six weeks.
4
3
2
1
A B C
i
n
1
y
x
z
Figure 4.1. Example of a 3D framed structure
Data for each student:
The same as during Work no. 2 and Work no. 3 (changes could be applied).
General remarks:
Be a 3D framed structure as that shown in the Figure 4.1. A horizontal seismic
force (calculated conforming to P100-2006 for example) acting on x direction at
level i, in the k-th mode of vibration is also presented in Figure 4.2
A
B
C
1 2 3 4
Fi,k,x
Figure 4.2. Seismic force acting on x direction at level i, in the k-th mode of vibration
The way to interpret the 3D work of the structure is shown in Figure 4.3. The
seismic forces are acting laterally. The floors are working together through simple
infinite stiff horizontal supports (double hinged links) that models the floor’s
infinite in its plan stiffness hypothesis.
1 2 3 4 1 2 3 4 1 2 3 4
Frame A
Fi,k,x
Fn,k,x
F1,k,x
Frame B Frame C
i
n
1
Figure 4.3. Model for frames' inter-connections and seismic action on x direction
The (lateral) stiffnes matrix on x direction of the whole structure will be
xCLxBLxALxL KKKK ,,,,,,,
where xALK ,, , xBLK ,, and xCLK ,, are the (lateral) stiffness matrices of frames A, B
and C respectively.
Knowing the mass of each floor, mi, and of lateral stiffness matrix xLK , the values
of periods of vibrations Tk,x and of the modal shapes si,k,x can be computed. This is
done solving the corresponding eigenproblem.
On the other horizontal direction y, Figure 4.4 (similar to Figure 4.2) is showing
the horizontal action on a current floor. Then, in Figure 4.5, the frames connected
at the floor levels and moving on y direction are shown.
A
B
C
1 2 3 4
Fi,k,y
Figure 4.4. Seismic force acting on Y direction at level i, in the k-th mode of vibration
A B C A B C A B C A B C
Frame 1
Fi,k,y
Fn,k,y
F1,k,y
i
n
1
Frame 2 Frame 3 Frame 4
Figure 4.5. Model for frames' inter-connections and seismic action on y direction
The (lateral) stiffnes matrix on y direction of the whole structure will be
yLyLyLyLyL KKKKK ,4,,3,,2,,1,, where yLK ,1, , yLK ,2, , yLK ,3, and yLK ,4, are the (lateral) stiffness matrices of frames
1, 2, 3 and 4 respectively.
As in the case of x direction, knowing the mass of each floor, mi, and of lateral
stiffness matrix yLK , the values of periods of vibrations Tk,y and of the modal
shapes si,k,y can be computed. This is done solving the corresponding eigenproblem.
Calculation of Seismic forces conforming to P100-2006:
Please follow the steps from the Work no. 3, with the values of periods (Tk,x and Tk,y)
and the modal shapes (si,k,x and si,k,y) calculated as shown before.
For each floor, the mass center (MC) and the stiffness center (SC) must be
established.
p
j
ji
p
j
jxj
i
MC
m
dm
x
1
1,
p
j
ji
p
j
jyj
i
MC
m
dm
y
1
1
where
j – the current number of a column
p – the total number of columns j
im – the lamped mass on floor i at the joint with for the column j
jxd – the distance on x axis of the center of the column j
jyd – the distance on y axis of the center of the column j
p
j
jix
p
j
jxj
ix
SC
K
dK
x
1
1,
p
j
jiy
p
j
jxj
iy
SC
K
dK
y
1
1
where j
ixK – the level relative stiffness of the coumn j, on x direction
3
12
i
j
iyj
ixh
EIK
jiyI – the inertia moment of the column j at the level i on y direction
2ih – the height of the level i
jiyK – the level relative stiffness of the coumn j, on y direction
3
12
i
j
ixj
iyh
EIK
jixI – the inertia moment of the column j at the level i on x direction.
e y
MC AP
djx xj
ex
Lx
Fy
Fx y SC
y j
SC
Ly
djy
j
xSC
x
y Legend
= structural vertical
element (column)
e0x e 0
y
Figure 4.6 Mass center (MC), stiffness center (SC) must and aplication point (AP) for a
current floor. A current j vertical structural element (column) is also figured out
Then an aplication point (AP) of the seismic forces must be set (conforming the
Code P100-2006), see Figure 4.6. The next relations are available
ixixix eee 10 , iyiyiy eee 10
where
ixe – distance on x axis from the application point (AP) to the stiffness Center (SC),
at level i
ixe0 – distance on x axis from the mass center (MC) to the stiffness Center (SC), at
level i
ixe1 – accidental excentricity on x direction at level i
ixix Le 05.01
ixL – length of the structure on x axis
iye – distance on y axis from the application point (AP) to the stiffness Center (SC),
at level i
iye0 – distance on y axis from the mass center (MC) to the stiffness Center (SC), at
level i
iye1 – accidental excentricity on y direction at level i
iyiy Le 05.01
iyL – length of the structure on y axis
MC AP
SC
x
y
Fix
Δrel,ix
SC
MC AP
Mt
x
y
φt
e iy
Figure 4.7. The translation and torsion effects of an horizontal seismic force
The forces from seismic action on a column j at the level i and on x/y direction will
be composed from the influence from translation and torsion as shown by the next
equations
These forces are used for dimensioning/check of the vertical structural members
(columns).
An example of solving a n d.o.f structure is presented next.
Example of solving a 3D structure
In Figure 4.8, a 3D view of a structure that is the example for this part is
presented. The data for this example was chosen mainly for academic proposes.
Data: E = 2.1·106 daN/cm
2 = 2.1·10
11 N/m
2
I = 105 cm
4 = 10
-3 m
4
m1 = 105 kg, m2 = 2·m1
γI = 1, q = 5
TC = 0.7 s
λ = 1.
Position and values of lamped masses, geometric and elastic characteristics
of each frame are shown in Figures 4.9 (showing the Frames 1,2,3 and 4), 4.10
(Frame A), 4.11 (Frame B) and 4.12 (Frame C).
Figure 4.8. 3D view of the structure chosen for example of solving
Also, in Figures 4.13 and 4.14 there are shown the plans for the first and
for the second floor. The values of lamped masses, elastic and geometric
characteristics are indicated, too. Position and values of the vertical elements
(columns) are mentioned in order to better determine the floors’ stiffness centers.
4 m
4 m
4
m
4m
A B C
5 m
5 m
5 m
1
2
3
4
3D View
x
y
z
Figure 4.9. Frames 1, 2 3 and 4. Masses, geometric and elastic characteristics
To establish the mass center (MC) and stiffness center positions (SC), the
floors’ plans are useful (Figure 4.13 and 4.14).
First floor’s mass center determination, conforming relations shown at the
beginnig of this part (the mass m2 is reduced):
meters
meters
Figure 4.10. Frame A. Masses, geometric and elastic characteristics
2m2
2m1
4 m
4
m
5 m
3EI
2EI
3EI
3EI
m2
m1
2m2
2m1
m2
m1
3EI
2EI 2EI 2EI
EI EI EI EI
5 m 5 m
Frame A
1 2 3 4
3EI 3EI
3m2 (2m2)
3m1 (2m1)
4 m
4 m
4 m
2EI
3EI
2EI
2EI
2m2
2m1
m2
m1
2EI 2EI
(1.5EI)
EI 2EI 1.5EI
(EI)
4 m
Frames 2 and (3)
A B C
2EI
2m2
2m1
4 m
4 m
4 m
2EI
2EI
2EI
2EI
m2
m1
m2
m1
2EI 2EI
EI EI EI
4 m
Frames 1 and 4
A B C
2EI
Second floor’s mass center determination, conforming relations shown at
the beginnig of this part (the mass m1 is reduced):
meters
meters
Figure 4.11. Frame B. Masses, geometric and elastic characteristics
Figure 4.12. Frame C. Masses, geometric and elastic characteristics
m2
m1
4 m
4
m
5 m
3EI
2EI
3EI
3EI
m2
m1
m2
m1
m2
m1
3EI
2EI 1.5EI 2EI
EI 1.5EI EI EI
5 m 5 m
Frame C
1 2 3 4
3EI 3EI
3m2
3m1
4 m
4 m
5 m
3EI
3EI
3EI
3EI
2m2
2m1
2m2
2m1
2m2
2m1
3EI
2EI 3EI 2EI
EI 2EI 2EI EI
5 m 5 m
Frame B
1 2 3 4
3EI 3EI
Figure 4.13. First floor plan. Masses, geometric and elastic characteristics
Figure 4.14. Second floor plan. Masses, geometric and elastic characteristics
3m1
2m1
4 m
4
m
5 m
3EI
2EI
3EI
3EI 2m1
m1
2m1
2m1
2m1
m1 3EI
2EI 2EI 2EI
2EI 2EI 2EI 2EI
5 m 5 m
Second Floor Plan
1 2 3 4
3EI 3EI
m1 3EI m1 m1 m1 3EI 3EI
A
B
C x
y
EI
EI
EI
2EI
EI
2EI
EI
1.5EI
EI
EI
EI
EI
3m2
2m2
4 m
4 m
5 m
3EI
2EI
3EI
3EI 2m2
m2
2m2
2m2
2m2
m2 3EI
2EI 2EI 2EI
2EI 2EI 2EI 2EI
5 m 5 m
First Floor Plan
1 2 3 4
3EI 3EI
m2 3EI m2 m2 m2 3EI 3EI
A
B
C x
y
2EI
2EI
2EI
3EI
2EI
3EI
2EI
2EI
1.5EI
2EI
2EI
2EI
First floor’s stiffness center determination, conforming relations shown at
the beginnig of this part ( is reduced):
meters
meters
Figure 4.15. A representation of the mass center, stiffness center and applications points of
seismic forces for the first floor
Second floor’s stiffness center determination, conforming relations shown
at the beginnig of this part ( is reduced):
meters
meters
Distances between the mass centers and stiffness centers for each floor and
directions:
meters
meters
meters
meters
Accidental excentricities calculations (for the first and for the second
floor):
0.7
4
AP2
SC
xSC=7.45
0.83
Lx=15.00
y SC=
4.0
8
MC
Ly=
8.0
0
xMC=7.37
x
y
y MC=
4.4
2
AP1
AP4 AP3 0.0
6
0.67
0.75 0.75
0.4
0
0.4
0
First Floor
meters
meters
Figure 4.16. A representation of the mass center, stiffness center and applications points of
seismic forces for the second floor
Total excentricities:
meters
meters
meters
meters
In Figures 4.15 and 4.16, positions of mass centers, stiffness centers and
applications points of seismical forces for the first and for the second floors are
represeted.
0.5
3
AP2 SC
xSC=7.41
0.79
Lx=15.00
y SC=
4.5
5
MC
Ly=
8.0
0
xMC=7.37
x
y
y MC=
4.4
2
AP1
AP4 AP3
0.2
7
0.71
0.75 0.75
0.4
0
0.4
0
Second Floor
Using the computer program shown in Appendix IV, the next was obtained on the
console of Scilab environment, for each different frame of the example:
Frames 1 and 4 ******************************************* *** CALCULATION OF THE STIFFNESS MATRIX *** ******************************************* Number of levels = 2 Number of spans = 2 E*I = 2.1e8 ** LEVEL HEIGHTS: Height of the level no.1= 4
Height of the level no.2= 4 ** SPAN LENGTHS: Length of the span no.1= 4 Length of the span no.2= 4 ** E*I ratio for COLUMNS: * E*I ratio for columns at LEVEL no.1 Level 1, column 1, E*I ratio = 2 Level 1, column 2, E*I ratio = 2 Level 1, column 3, E*I ratio = 2 * E*I ratio for columns at LEVEL no.2 Level 2, column 1, E*I ratio = 1 Level 2, column 2, E*I ratio = 1 Level 2, column 3, E*I ratio = 1
** E*I ratio for BEAMS: * E*I ratio for beams at FLOOR no.1 Floor 1, span 1, E*I ratio = 2 Floor 1, span 2, E*I ratio = 2 * E*I ratio for beams at FLOOR no.2 Floor 2, span 1, E*I ratio = 2 Floor 2, span 2, E*I ratio = 2 maximum of the difference between simmetrical elements is 0 K = 1.0D+07 * 31.648049 - 11.098314 - 11.098314 8.7372349 Execution done.
Frame 2 ******************************************* *** CALCULATION OF THE STIFFNESS MATRIX *** ******************************************* Number of levels = 2 Number of spans = 2 E*I = 2.1e8 ** LEVEL HEIGHTS: Height of the level no.1= 4 Height of the level no.2= 4 ** SPAN LENGTHS: Length of the span no.1= 4 Length of the span no.2= 4
** E*I ratio for COLUMNS: * E*I ratio for columns at LEVEL no.1 Level 1, column 1, E*I ratio = 2 Level 1, column 2, E*I ratio = 3 Level 1, column 3, E*I ratio = 2 * E*I ratio for columns at LEVEL no.2 Level 2, column 1, E*I ratio = 1
Level 2, column 2, E*I ratio = 2 Level 2, column 3, E*I ratio = 1.5 ** E*I ratio for BEAMS: * E*I ratio for beams at FLOOR no.1 Floor 1, span 1, E*I ratio = 2 Floor 1, span 2, E*I ratio = 2 * E*I ratio for beams at FLOOR no.2 Floor 2, span 1, E*I ratio = 2 Floor 2, span 2, E*I ratio = 2 maximum of the difference between simmetrical elements is 0 K = 1.0D+11 * 4.0210637 - 1.5511824 - 1.5511824 1.2062802
Execution done.
Frame 3 ******************************************* *** CALCULATION OF THE STIFFNESS MATRIX *** ******************************************* Number of levels = 2 Number of spans = 2 E*I = 2.1e8 ** LEVEL HEIGHTS: Height of the level no.1= 4 Height of the level no.2= 4 ** SPAN LENGTHS:
Length of the span no.1= 4 Length of the span no.2= 4 ** E*I ratio for COLUMNS: * E*I ratio for columns at LEVEL no.1 Level 1, column 1, E*I ratio = 2 Level 1, column 2, E*I ratio = 3 Level 1, column 3, E*I ratio = 1.5 * E*I ratio for columns at LEVEL no.2 Level 2, column 1, E*I ratio = 1 Level 2, column 2, E*I ratio = 2 Level 2, column 3, E*I ratio = 1 ** E*I ratio for BEAMS: * E*I ratio for beams at FLOOR no.1 Floor 1, span 1, E*I ratio = 2
Floor 1, span 2, E*I ratio = 2 * E*I ratio for beams at FLOOR no.2 Floor 2, span 1, E*I ratio = 2 Floor 2, span 2, E*I ratio = 2 maximum of the difference between simmetrical elements is 0 K = 1.0D+08 * 3.7101287 - 1.4173245 - 1.4173245 1.1167863 Execution done.
Frame A ******************************************* *** CALCULATION OF THE STIFFNESS MATRIX *** ******************************************* Number of levels = 2 Number of spans = 3 E*I = 2.1e8 ** LEVEL HEIGHTS:
Height of the level no.1= 4 Height of the level no.2= 4 ** SPAN LENGTHS: Length of the span no.1= 5 Length of the span no.2= 5 Length of the span no.3= 5 ** E*I ratio for COLUMNS: * E*I ratio for columns at LEVEL no.1 Level 1, column 1, E*I ratio = 2 Level 1, column 2, E*I ratio = 2 Level 1, column 3, E*I ratio = 2 Level 1, column 4, E*I ratio = 2 * E*I ratio for columns at LEVEL no.2 Level 2, column 1, E*I ratio = 1
Level 2, column 2, E*I ratio = 1 Level 2, column 3, E*I ratio = 1 Level 2, column 4, E*I ratio = 1 ** E*I ratio for BEAMS: * E*I ratio for beams at FLOOR no.1 Floor 1, span 1, E*I ratio = 3 Floor 1, span 2, E*I ratio = 3 Floor 1, span 3, E*I ratio = 3 * E*I ratio for beams at FLOOR no.2 Floor 2, span 1, E*I ratio = 3 Floor 2, span 2, E*I ratio = 3 Floor 2, span 3, E*I ratio = 3 maximum of the difference between simmetrical elements is 0 K =
1.0D+08 * 4.3294327 - 1.513303 - 1.513303 1.2411297 Execution done.
Frame B ******************************************* *** CALCULATION OF THE STIFFNESS MATRIX *** ******************************************* Number of levels = 2 Number of spans = 3 E*I = 2.1e8 ** LEVEL HEIGHTS:
Height of the level no.1= 4 Height of the level no.2= 4 ** SPAN LENGTHS: Length of the span no.1= 5 Length of the span no.2= 5 Length of the span no.3= 5 ** E*I ratio for COLUMNS: * E*I ratio for columns at LEVEL no.1 Level 1, column 1, E*I ratio = 2 Level 1, column 2, E*I ratio = 3 Level 1, column 3, E*I ratio = 3 Level 1, column 4, E*I ratio = 2 * E*I ratio for columns at LEVEL no.2
Level 2, column 1, E*I ratio = 1 Level 2, column 2, E*I ratio = 2 Level 2, column 3, E*I ratio = 2 Level 2, column 4, E*I ratio = 1 ** E*I ratio for BEAMS: * E*I ratio for beams at FLOOR no.1 Floor 1, span 1, E*I ratio = 3
Floor 1, span 2, E*I ratio = 3 Floor 1, span 3, E*I ratio = 3 * E*I ratio for beams at FLOOR no.2 Floor 2, span 1, E*I ratio = 3 Floor 2, span 2, E*I ratio = 3 Floor 2, span 3, E*I ratio = 3 maximum of the difference between simmetrical elements is 0 K = 1.0D+08 * 5.7267139 - 2.1716949 - 2.1716949 1.7502779 Execution done.
Frame C ******************************************* *** CALCULATION OF THE STIFFNESS MATRIX *** ******************************************* Number of levels = 2 Number of spans = 3 E*I = 2.1e8 ** LEVEL HEIGHTS: Height of the level no.1= 4 Height of the level no.2= 4 ** SPAN LENGTHS: Length of the span no.1= 5 Length of the span no.2= 5 Length of the span no.3= 5
** E*I ratio for COLUMNS: * E*I ratio for columns at LEVEL no.1 Level 1, column 1, E*I ratio = 2 Level 1, column 2, E*I ratio = 2 Level 1, column 3, E*I ratio = 1.5 Level 1, column 4, E*I ratio = 2 * E*I ratio for columns at LEVEL no.2 Level 2, column 1, E*I ratio = 1 Level 2, column 2, E*I ratio = 1.5 Level 2, column 3, E*I ratio = 1 Level 2, column 4, E*I ratio = 1 ** E*I ratio for BEAMS: * E*I ratio for beams at FLOOR no.1 Floor 1, span 1, E*I ratio = 3
Floor 1, span 2, E*I ratio = 3 Floor 1, span 3, E*I ratio = 3 * E*I ratio for beams at FLOOR no.2 Floor 2, span 1, E*I ratio = 3 Floor 2, span 2, E*I ratio = 3 Floor 2, span 3, E*I ratio = 3 maximum of the difference between simmetrical elements is 0 K = 1.0D+08 * 4.3275675 - 1.6536029 - 1.6536029 1.3718141 Execution done.
Therefore, the next stiffness matrices were obtained:
N/m
N/m
N/m
N/m
N/m
N/m
The stiffness matrix on direction x and y is calculated:
N/m
N/m
Then, knowing the mass matrix (the same on both directions):
kg
the eigenvalues and eigenvectors lead to:
- On x direction:
rad/s, rad/s,
s, s, Hz, Hz,
, kg, kg,
m/s2, m/s
2,
N, N,
N, N,
N, N.
- On y direction:
rad/s, rad/s,
s, s, Hz, Hz,
, kg, kg,
m/s2, m/s
2,
N, N,
N, N,
N, N.
For this example, only one column will be taken into account: that one
located at the intersection of axes “2” and “B”. For any other column the
calculations are very similar. Also, in order to simplify, only the eccentricities
e1x=0.77 m, e1y=0.74 m, e2x=0.71 m, e2y=0.27 m will be considered.
Position of the column “2B” is x2B = 5 m on x axis and y2B = 4 m on y axis.
On both directions the moments of inertia are 3EI at the first level and 2EI at the
second level.
Forces for each direction of seismic action and for each mode of vibration
on the column “2B” at each level will be calculated. In calculations, or
is reduced.
- On x direction, the generic relation is (see the beginning of this part):
Mode 1: N, N.
- Level 2:
e2y=0.27 m, m, m
N
- Level 1:
e1y=0.74 m, m, m
N
Mode 2: N, N.
- Level 2:
e2y=0.27 m, m, m
N
- Level 1:
e1y=0.74 m, m, m
N
- On y direction, the generic relation is (see the beginning of this part):
Mode 1: N, N.
- Level 2:
e2x=0.71 m, m, m
N
- Level 1:
e1x=0.77 m, m, m
N
Mode 2: N, N.
- Level 2:
e2x=0.71 m, m, m
N
- Level 1:
e1x=0.77 m, m, m
N
Figure 4.17. Application of seismic forces on the 2B column
h
h
3EI
2EI
x direction
2
3EI
2EI
B
Mode 1
Mode 2
3EI
2EI
y direction
2
3EI
2EI
B
Mode 1
Mode 2
