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Lateral load resisting systems

Ahmad T., profile picture
Ahmad T.

The document discusses lateral load resisting systems in high-rise concrete buildings, detailing types such as moment-resisting frames and shear walls, as well as their combinations. It covers key concepts including the definition of the center of rigidity, methods for distributing lateral loads, and the evaluation of structural stability through critical parameters. Practical examples illustrate calculations for effective inertia and load distribution among various structural frames.

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Lateral load Resisting Systems Dr. Ahmed Tarabia High rise concrete building, 3rd Civil Spring 2015
‫الجانبية‬ ‫لالحمال‬ ‫المقاومة‬ ‫االنظمة‬ ‫انواع‬ •‫المقاومة‬ ‫االطارات‬‫للعزوم‬: ‫الكمرات‬ ‫و‬ ‫األعمدة‬ ‫من‬ ‫تتكون‬‫الخرسانية‬. •‫حوائط‬‫الحوائط‬ ‫أو‬ ‫القص‬‫االنشائية‬. •‫المركبة‬ ‫االنظمة‬: ‫االطارات‬ ‫من‬ ‫مجموعة‬ ‫من‬ ‫تتكون‬+‫القص‬ ‫حوائط‬.
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)
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)
Moment resisting frames
Moment resisting frames
Shear wall system
Shear wall system    
Dual system:(Shear walls + Frames)
‫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.
‫األولي‬ ‫الحالة‬:‫الجساءة‬ ‫مركز‬ ‫في‬ ‫تؤثر‬ ‫القوة‬
‫الثانية‬ ‫الحالة‬:‫الجساءة‬ ‫مركز‬ ‫خارج‬ ‫تؤثر‬ ‫القوة‬
‫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:
‫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
Y C.R.Xo Yo
Simple examples: #1
Simple examples: #2
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
Step 2: • Transfer axes systems to the rigidity center. Y C.R.
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
-‫الجانبية‬ ‫القوة‬ ‫حالة‬ ‫في‬FY‫اتجاه‬ ‫في‬Y: Then, For Y-direction resisting elements:   ny 1j xj xi yyi I I .F=F 0.0eif y 
•-‫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
-‫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 
Example #1 Force 3C C.R. 1C1C C2 Y X A B C 1 2 3 3 2 1 CBA B12C 2C 2CC1 C1
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.
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
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
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
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
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
Distribution of lateral loads on frames
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
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.
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  
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
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)
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
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)
•-‫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
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
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
- 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
.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

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Lateral load resisting systems

  • 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)
  • 8. Shear wall system    
  • 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.
  • 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
  • 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
  • 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.
  • 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