Distribution of lateral loads using the relative rigidity method. This is a simple and easy method to distribute lateral loads

1. Lateral load Resisting Systems
Dr. Ahmed Tarabia
High rise concrete building, 3rd Civil
Spring 2015

2. ‫الجانبية‬ ‫لالحمال‬ ‫المقاومة‬ ‫االنظمة‬ ‫انواع‬
•‫المقاومة‬ ‫االطارات‬‫للعزوم‬:
‫الكمرات‬ ‫و‬ ‫األعمدة‬ ‫من‬ ‫تتكون‬‫الخرسانية‬.
•‫حوائط‬‫الحوائط‬ ‫أو‬ ‫القص‬‫االنشائية‬.
•‫المركبة‬ ‫االنظمة‬:
‫االطارات‬ ‫من‬ ‫مجموعة‬ ‫من‬ ‫تتكون‬+‫القص‬ ‫حوائط‬.

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)

5. Moment resisting frames

6. Moment resisting frames

7. Shear wall system

8. Shear wall system
   

9. Dual system:(Shear walls + Frames)

10. ‫الجساءة‬ ‫مركز‬Center of Rigidity
•‫المستوي‬ ‫في‬ ‫جسئ‬ ‫كسطح‬ ‫تتحرك‬ ‫دور‬ ‫كل‬ ‫بالطة‬ ‫أن‬ ‫افتراض‬
•‫مركز‬‫أثر‬ ‫اذا‬ ‫الدور‬ ‫مستوى‬ ‫فى‬ ‫افتراضية‬ ‫نقطة‬ ‫هو‬ ‫الجساءة‬
‫انتقالية‬ ‫حركة‬ ‫جسئ‬ ‫كجسم‬ ‫المستوي‬ ‫يتحرك‬ ‫فيها‬ ‫الجانبية‬ ‫القوة‬
‫المؤثرة‬ ‫القوة‬ ‫اتجاه‬ ‫في‬‫بدون‬‫حدوث‬‫دوران‬.
•‫انتقالية‬ ‫حركة‬ ‫يحدث‬ ‫آخر‬ ‫مكان‬ ‫اي‬ ‫في‬ ‫القوة‬ ‫أثرت‬ ‫اذا‬ ‫و‬
‫التواء‬ ‫عزم‬ ‫لوجود‬ ‫نتيجة‬ ‫دوران‬ ‫حدوث‬ ‫مع‬ ‫للمستوي‬.

11. ‫األولي‬ ‫الحالة‬:‫الجساءة‬ ‫مركز‬ ‫في‬ ‫تؤثر‬ ‫القوة‬

12. ‫الثانية‬ ‫الحالة‬:‫الجساءة‬ ‫مركز‬ ‫خارج‬ ‫تؤثر‬ ‫القوة‬

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

15. Y
C.R.Xo
Yo

16. Simple examples: #1

17. Simple examples: #2

18. In the case of symmetry of lateral load resisting system
Ey=0.0
Force 3C
C.R.
1C1C C2
Y
X
A B C
1
2
3 3
2
1
CBA
B12C 2C
2CC1 C1

19. Step 2:
• Transfer axes systems to the rigidity center.
Y
C.R.

20. Distribution of lateral forces
•-‫الجانبية‬ ‫القوة‬ ‫حالة‬ ‫في‬FY‫اتجاه‬ ‫في‬Y:
• For each Y-direction resisting elements:
• For each X-direction resisting elements:
)y*(I)x*(I
.XI
.eF
I
I
F=F
2
j
1
yj
2
j
1
xj
ixi
xy
1
xj
xi
yyi



 nx
j
ny
j
ny
j
)y*(I)x*(I
.yI
.eF=F
2
j
1
yj
2
j
1
xj
iyi
xyxi



 nx
j
ny
j

21. -‫الجانبية‬ ‫القوة‬ ‫حالة‬ ‫في‬FY‫اتجاه‬ ‫في‬Y:
Then, For Y-direction resisting elements:


ny
1j
xj
xi
yyi
I
I
.F=F
0.0eif y 

22. •-‫الجانبية‬ ‫القوة‬ ‫حالة‬ ‫في‬F x‫اتجاه‬ ‫في‬X:
• 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

23. -‫الجانبية‬ ‫القوة‬ ‫حالة‬ ‫في‬F x‫اتجاه‬ ‫في‬:X
For X-direction resisting elements:


nx
j
yj
1
yi
xxi
I
I
.F=F
0.0eif x


24. Example #1
Force 3C
C.R.
1C1C C2
Y
X
A B C
1
2
3 3
2
1
CBA
B12C 2C
2CC1 C1

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

28. 1C
A
3
Fx
2
2C
1
A
C1
X
3
2
1
23C
C.R.
C2
Y
B
B
B1
2C
1C
C
C
C
C1

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

31. Distribution of lateral loads on frames

32. Frame#3
Frame #2
Frame #1
C1 1C1C
C2C2 B1
A B C
1
2
33
2
1
CBA
X
Y
2CC1 C1
C3Fx
Frame#4
Example #2

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.

34. Find effective inertia-1
4
3
x
:
4
3
y:1
010.0
12
0.6x0.8
x0.7I
0179.0
12
0.8x0.6
x0.7ICfor
m
m


4
3
x
4
3
y
029.0
12
0.8x1.0
x0.7I
0467.0
12
1.0x0.8
x0.7I:C2For
m
m


4
3
x
4
3
y3
0583.0
12
1.0x1.0
x0.7I
0583.0
12
1.0x1.0
x0.7I:CFor
m
m



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

43. .12Fx
.12Fx
C1
.23 Fx
.77 Fx
Fx
3C
1C1C C2
Y
X
B12C 2C
Frame #1
Frame #2
Distribution of lateral loads on frames