3. OBJECTIVES
• 1- Rigidity center definition, how to locate it?
• 2- Lateral Load Resisting Systems In Each Direction.
• 3-How To Distribute Lateral Loads Between Lateral
Load Resisting Systems In Each Lateral Direction.
(Relative rigidity method)
4. General rules
• A system or more should be provided in each
direction of the principal directions.
• These systems should be able to resist the
lateral loads.
Check: (Drift-moment-Shear-Stability)
13. الجساءة مركز تحديدXo , Yo
•اتجاه لكل المقاومة عناصر تحديد يتم.
•حساب يتمIeffectiveألخذبالقطاعات التشريخ تاثيرالخرسانية:
How to find the inertia of the lateral resisting system
•محور حول الجساءات عزوم باستخدام الجساءة مركز تحديد يتمX &Y
اآلتية المعادالت طريق عن مكان أس في فرضهم يتم و:
14. محور حول الجساءات عزوم باستخدامX &Yاآلتية المعادالت طريق عن:
محور اتجاه في الجانبية المقاومة لعناصرX:
محور اتجاه في الجانبية المقاومة لعناصرY:
nx Number of lateral load resisting systems in X direction
ny Number of lateral load resisting systems in X direction
25. Lateral load resisting systems
• X اتجاه في الجانبية األحمال لمقاومة االنشائي النظام
• a- The frames on axis (1-1), (2-2) and (3-3).
• Y اتجاه فى االنشائي النظام يتكون و
• a- The frames on axis (A-A), (B-B) and (C-C).
• تأثير نتيجة االنشائية العناصر جميع نصيب احسبالحمل هذا.
26. The sections of C1 600mm x800mm
The sections of C2 800mm x1000mm and
The section of C3 is 1000mm x 1000mm.
4
3
y:1 0179.0
12
0.8x0.6
x0.7ICfor m
Find effective inertia-1
4
3
y 0467.0
12
1.0x0.8
x0.7I:C2For m
4
3
y3 0583.0
12
1.0x1.0
x0.7I:CFor m
27. 2-Find the Centre of Rigidity:
Because the lateral load resisting systems
are symmetric in X and Y directions, the
center of rigidity is located in the center of
the two systems.
- Transfer axes system to the center of rigidity
29. 3 - Load distribution:
In this example, a load in X-direction =Fx is acting at the mid-point of the floor.
We need to find the share of each resisting system of this load
using the following equations. Because ey=0.0, then:
"For X-dir. resisting element"
nx
j 1
yj
yi
xxi
I
I
F=F
30. System in Section Iy Yi
Fx Fx/frame
X-direction (m x m) (m4) (m)
Frame (1-1)
.8*.6 0.0179 3 0.05652
0.2601.0*0.80
0.0467
3 0.147458
.8*.6 0.0179 3 0.05652
Frame (2-2)
1.0*0.80
0.0467
0 0.147458
0.481.0*1.0
0.0583
0 0.184086
1.0*0.80
0.0467
0 0.147458
Frame (3-3)
.8*.6 0.0179 -3 0.05652
0.2601.0*0.80
0.0467
-3 0.147458
.8*.6 0.0179 -3 0.05652
SUM 0.3167 1 1
33. Data of the problem:
The sections of C1 600mm x800mm
The sections of C2 800mm x1000mm and
The section of C3 1000mm x 1000mm.
X اتجاه في الجانبية األحمال لمقاومة االنشائي النظام
a- The frames on axis (2-2) and (3-3).
Y اتجاه فى االنشائي النظام يتكون و
a- The frames on axis (A-A), and (C-C).
FX kN تأث نتيجة االنشائية العناصر جميع نصيب احسبالحمل هذا ير.
35. Frame #2
Frame #1
C2C2 B1
X
Y
2CC1 C1
C3Force
Frame#4
Frame#3
1C1C
C2C2
X
Y
C1 C1
Fx
X اتجاه في الجانبية األحمال لمقاومة االنشائي النظام Yاتجاه في الجانبية األحمال لمقاومة االنشائي النظام
36. Find the Centre of Rigidity:-2
System in
Col.
Section
Ix Xi
Ix . Xi
Y-direction (m x m) (m4) (m)
Frame (A-A)
0.8x0.60 0.01 0 0
1.0x0.80 0.03 0 0
0.8x0.60 0.0583 0 0
Frame (C-C)
0.8x0.60 0.01 6 0.06
1.0x0.80 0.03 6 0.18
0.8x0.60 0.0583 6 0.3498
SUM 0.1966 0.5898
A- Find Xo
m
I
XI
y
y
n
i
ix
n
i
ixi
0.3
1966.0
5898.
=X
1
1
o
Assume that the X axis is axis (3-3) and the Y axis is axis (A-A)
37. System in
Col.
Section
Iy Yi
Iy . Yi
X-direction (m x m) (m4) (m)
Frame (1-1)
0.8x0.60 0.0179 0 0
1.0x0.80 0.0467 0 0
0.8x0.60 0.0179 0 0
Frame (2-2)
1.0x0.80 0.0467 3 0.1401
1.0x1.00 0.0583 3 0.1749
1.0x0.80 0.0467 3 0.1401
SUM 0.2342 0.4551
m
I
YI
x
x
n
i
yi
n
i
iyi
1.94
233.0
455.0
=Y
1
1
o
B- Find Yo
38. 3C
1C1C C2
B12C 2C
C1 C11C
Frame #1
Frame #2Frame#3
Frame#3
C.R.
Force
Y
X
ex=0.0 and ey = 3.0 – 1.94 = 1.06 m
- M = Fx . ey = Fx . (1.06)
39. •-اتجاه في الجانبية القوة حالة فيXوF x
• For X-direction resisting elements:
• For Y-direction resisting elements:
)y*(I)x*(I
.YI
.eF
I
I
F=F
2
j
1
yj
2
j
1
xj
iyi
yx
1
yj
yi
xxi
nx
j
ny
j
nx
j
)y*(I)x*(I
.xI
.eF=F
2
j
1
yj
2
j
1
xj
ixi
yxyi
nx
j
ny
j
40. System
in
Section Iy Yi
Iy . Yi
2 Iy . Yi Fx' Fx"
Fx/
column
Fx/
frameX-
directio
n
(m x m) (m4) (m)
Frame
(1-1)
0.8x0.60 0.0179 -1.94 0.0674 -0.0347 0.0764 -0.027 0.050
0.231.0x0.80 0.0467 -1.94 0.1758 -0.0906 0.1994 -0.070 0.130
0.8x0.60 0.0179 -1.94 0.0674 -0.0347 0.0764 -0.027 0.050
Frame
(2-2)
1.0x0.80 0.0467 1.06 0.0525 0.0495 0.1994 0.038 0.237
0.771.0x1.00 0.0583 1.06 0.0655 0.0618 0.2489 0.047 0.296
1.0x0.80 0.0467 1.06 0.0525 0.0495 0.1994 0.038 0.237
SUM 0.2342 0.4809 1.0000 0.00 1.00 1.00
3 – find the values of [Iy. Yi
2 ] in the X - direction
41. System in Section Ix Xi
Ix. Xi
2 Ix. Xi Fy' Fy"
Fy/
column
Fy/
Frame
Y-direction (m x m) (m4) (m)
Frame
(A-A)
0.8x0.60 0.01008 -3 0.0907 -0.0302 0.0000 0.023 0.023
0.121.0x0.8 0.02987 -3 0.2688 -0.0896 0.0000 0.069 0.069
0.8x0.6 0.01008 -3 0.0907 -0.0302 0.0000 0.023 0.023
Frame
(C-C)
0.8x0.6 0.01008 3 0.0907 0.0302 0.0000 -0.023 -0.023
-0.121.0x0.8 0.02987 3 0.2688 0.0896 0.0000 -0.069 -0.069
0.8x0.6 0.01008 3 0.0907 0.0302 0.0000 -0.023 -0.023
SUM 0.1001 0.9005 0.0000 0.0000 0.0000 0.0000 0.0000
4– Calculate the values of [Ix. Xi
2 ] in the Y - direction
42. - The share of Frame (1-1) = 0.23 Fx
- The share of Frame (2-2) = 0.77 Fx
- The share of Frame (A-A) = 0.12 Fx
- The share of Frame (C-C = -0.12 Fx