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Seismic Loads
ECP (201)

Forces Resulting From Earthquake

Z Fb (z)
X
Fb (x)
Y

Fb (y)

Characteristics of Earthquake Resistant Buildings (ECP 201):

•Cases of Neglecting earthquake effect

1-Wall bearing buildings ‫مباني الخوائط الحاملة‬
2-For residential buildings satisfying the following conditions: ‫المباني السكنية التي‬
‫تحقق الشروط التالية‬
- Height from foundation level < 10 m (zone 1)
< 8 m (zone 2)
- Columns and shear walls extend from foundation to top storey level
- Columns with adequate rigidity in both directions
- Perimeter columns and staircase columns are connected to beams with
width ≥ 25 cm
- Structural detailing presented in ECP 203 should be fulfilled to ensure
adequate Ductility.
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Structural Systems Resisting Lateral Loads*

2-D Frames,
Space Frames,
Shear Walls,
Coupled Shear Walls,
Wall-Frame,
Framed Tube

*After Smith.S & Coull A , Jhon willy sons

(1991) ‫ م‬03:54 14/03/2020

Essential Factors for Conceptual design:

•Simplicity of structural system )‫(البساطة االنشائية‬

•Uniformity and symmetry )‫(االنتظام و التماثل‬

•Resistance and rigidity in both directions )‫(مقاومة و جسائة في االتجاهين‬

•Resistance and rigidity against Torsion )‫( مقاومة و جساءة عزوم اللي‬

•Rigid Diaphragm effect in all storeys )‫(تأئير لوحي في منسوب االدوار‬

•Suitable foundation system )‫(اساسات مناسبة‬

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Structural regularity ‫ االنتظام االنشائي‬:

Structures are classified to Regular or Irregular

1-Conditions for Regularity in Plan ‫شروط انتظام المسقط االفقي‬

•Masses and Rigidity are almost symmetrical in 2 perpendicular directions

•Plan is regular (setback area ≤ 5% regular floor )
•Slab is rigid enough to distribute loads on vertical elements (columns & shear walls)
•Plan dimensions ratio Lx/Ly ˃ 4 ) Lx ˃ Ly (
•Center of mass and center of rigidity should fulfill:

eox / Lx ˃ 0.15
eox
Ly
eoy / Ly ˃ 0.15 eoy

Lx

Center of Mass
Center of Rigidity ‫ م‬03:54 14/03/2020

2 -Conditions for Regularity in Elevation ‫شروط انتظام المسقط الرأسي‬

•All vertical structural elements should continue from foundation level to top of structure
(or setback level)
•Horizontal Rigidity should be kept constant or regularly decreased not less than 75% of
the above floor
•Mass of each floor not differ +- 50% of next floor
•Uniform setback not more than 20% previous floor area (refer to figure 8-3)

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Basic Representation of Earthquake Action (2-4-8)

• Earthquake motion at a given point on the earth surface is represented by an

elastic ground acceleration response spectrum, called
“Elastic Response Spectrum". ‫طيف التجاوب المرن‬

• The shape and forces resulting form the elastic response spectrum is taken to
ensure:
i- No-collapse requirement (‫)عدم االنهيار‬
ii- Damage limitation requirement. )‫(الحد من التصدعات‬
iii- Increase earthquake safety. )‫(زيادة االمان الزلزالي طبقا الهمية المبني‬

• The horizontal seismic action is described by two perpendicular components

assumed as being independent and represented by the same response
spectrum.

• All structures in A.R.E. should be designed to resist seismic forces calculated

according to response spectrum curve
TYPE1 (‫(لجميع مناطق الجمهورية‬
TYPE2 (‫ كم بمحاذاة الساحل‬40 ‫(للماطق الساحلية علي البحر المتوسط لمسافة‬
‫ م‬03:54 14/03/2020

WHAT IS RESPONSE SPECTRUM ?

Response spectrum is a plot of the maximum response ( maximum displacement,
velocity, acceleration or any other quantity of interest) to a specified load function
( earth quake load function) for all possible Single degree of freedom systems

mass
d (deformation),
v (velocity),
a ( acceleration)

F(t)

Earthquake

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Single Degree Un-damped Harmonic Excitation (READ ONLY)

δ δ δ δ
δ٫ δ᾽٫ δ’’
K W
m
E,I H
δ
Kδ mδ’’
1st Mode Shape
(Dynamically one degree of freedom system)
Un-damped Free Oscillator , Free Body diagram
Summing up forces

mδ’’ + k δ = 0

m = W/g mass of body δ= displacement

K = Spring stiffness δ’’= acceleration
Solution of above equation is:
δ(t) = A cos ωt + B sin ωt ( A, B are constants refer to Dynamic Analysis course )

ω = Fundamental Frequency cycles/sec ( in case of one degree of freedom)

Mode Shape =( shape of deflection corresponding to each frequency)

T = ω / 2π Fundamental cyclic period ( for single degree of freedom system)
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RESPONSE SPECTRUM Chart ?

Acceleration mass x acc. = Force

mass a

Structure Single Degree of T

freedom Natural (Fundamental)Frequency (T1)

The X-axis is the Fundamental Period of the structure (T),

The Y-axis is the maximum response required ( acceleration).
To determine the response from any available spectral chart for a specified earth
quake , it is required to know only the Natural frequency (FUNDAMENTAL PERIOD)of
the system.

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Elastic Response Spectrum

‫طيف التجاوب المرن‬

R =1
Horizontal Elastic Table ( 8 - ‫)أ‬ Vertical Elastic Response
Response Spectrum Spectrum
‫طيف التجاوب االفقي المرن‬ ‫طيف التجاوب الرأسي المرن‬

Design Response Spectrum

‫طيف التجاوب التصميمي المرن‬

R˃1
Horizontal Design Elastic For all structures Vertical Design Elastic
Response Spectrum except Water Tanks
Response Spectrum
Table ( 10 - 1 )
‫طيف التجاوب االفقي المرن‬ ‫طيف التجاوب الرأسي المرن‬

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TYPE1
‫لجميع مناطق الجمهورية‬

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TYPE2
‫ كم‬40 ‫للمناطق الساحلية علي البحر المتوسط لمسافة‬
‫بمحاذاة الساحل‬

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Factors affecting Seismic forces:

• Soil Classification: Table (8-1)

• Earthquake Zones: Table (8-2)

• Soil Factor S: Table (8-3)

• Damping Factor : Table (8-4)

• Building Importance Factor :Table(8-9)

• Force Reduction factor R: Appendix (8—‫) أ‬

(R is used in Design Spectrum for elastic analysis)

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Factor Symbol Code ref.

Soil Classification A (Rock), Table (8-1)

(Nspt, Cu, Vs,30) B (Sand, Gravel) ,…
C, D, and E
Earthquake Zone Zone 1 ag = 0.10 g Table (8-2A)
Zone 2 ag = 0.125 g Table (8-2B)
Zone 3 ag = 0.15 g
Zone 4 ag = 0.20 g
Zone 5A ag = 0.25 g
Zone 5B ag = 0.30 g

Soil Factor S S=1 to 1.8 Table (8-3A)

(according to soil classification Table (8-3B)
A,B,…)

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Factor Symbol Code ref.

Damping Factor  =0.95 to 1.20 Table(8-4)

(Type of building) v =0.65 to 1.0
‫معامل االضمحالل‬ Reinforced concrete (=1.0 , v=0.70)
Pre stressed conc. (=1.05, v=0.75)
Building Importance I ( Hospitals, Power stations …..) I =1.4 Table(8-9)
Factor : I II (Schools, Halls, Mosques,.. ) I =1.2
III (Ordinary buildings…) I =1.0
IV (low importance building, Agricultural) I
=0.8
Force Reduction Bearing walls carry vertical loads , shear Appendix
walls carry horizontal load
Factor : R (8-A)
R=4.5
3-D frame system and Shear Walls R=5
Frames with limited Ductility R=5
Frames with enough Ductility R=7, ‫ م‬03:54 14/03/2020

……..

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Methods of Analysis

1- Simplified Modal Response Spectrum

)‫طريقة طيف التجاوب المبسطة ( طريقة الحمل االستاتيكي المكافئ‬
(will be considered in this course)

2- Multi –Modal Response Spectrum Method

)‫طريقة طيف التجاوب المركب ( متعدد االنماط‬

3- Time History Method ‫طريقةالتحليل الديناميكى الزمنى‬

NOTE:
All seismic forces calculated are ULTIMATE loads.
For using these forces in working cases divide by (1.4)

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Simplified Modal Response Spectrum

)‫طريقة طيف التجاوب المبسطة ( طريقة الحمل االستاتيكي المكافئ‬

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Seismic Loads ECP (201)

*Simplified Modal Response Spectrum
)‫طريقة طيف التجاوب المبسطة ( طريقة الحمل االستاتيكي المكافئ‬
Dynamic Forces. Static Forces
Horizontal Seismic forces can be calculated by either:

a. Elastic Horizontal Response spectrum. ‫طيف التجاوب االفقي المرن‬

b. Elastic Vertical Response spectrum. ‫طيف التجاوب الرأسي المرن‬
c. Horizontal Design Spectrum for Elastic analysis. ‫طيف التجاوب التصميمي االفقي‬
d. Vertical Design Spectrum for Elastic analysis. ‫طيف التجاوب التصميمي الرأسي‬
Stress

Strain
Elastic region Plastic region
Note:
Forces from Design Spectrum < Forces from Elastic Spectrum due to the capability
of structure to resist seismic forces in plastic region

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*Simplified Modal Response Spectrum

)‫طريقة طيف التجاوب المبسطة ( طريقة الحمل االستاتيكي المكافئ‬
Conditions for application of method:

1. The structure can be modeled by 2 planer perpendicular models ‫(مستويين‬

)‫متعامدين‬

2. The structure fulfills conditions of uniformity in

HL. Plans (8-6-3-2)& in VL. Plans (8-6-3-3)
)‫(يحقق شروط انتظام المسقط االفقي و المسقط الرأسي‬

3. The Fundamental Period T1 of the structure in each direction

:‫(ان يكون الزمن الدوري االساسي في كل من االتجاهين أقل من او يساوي التالي‬

T1 ≤ 2 seconds T1≤4 Tc (Table 8-3)

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Simplified Modal Response Spectrum

)‫طريقة طيف التجاوب المبسطة (طريقة الحمل االستاتيكي المكافئ‬
Z

X
Ultimate Base Shear Force Fb : Fb (x)
Y

Fb = Sd (T1) λ W / g Fb (y)

Sd(T1) = calculate according to item (8-4-2-5) OR item (8-2-4-6) for fundamental

period T1
W = Total Design Load of structure above foundation level
λ = correction factor ‫معامل تصحيح‬
λ=0.85 for T1 ≤ 2 Tc
or ‫وعدد االدوار اكثر من دورين‬
λ=1.0 for T1 ˃ 2 Tc
Fundamental Period of Buildings: T1 Appendix(8-B)

T1  Ct  H 3/ 4 ‫الطول الموجي االساسي‬

Ct =0.085 Steel Space Frames

0.075 Concrete Space Frames
0.05 other types
H = Height of building from top of foundation in METERS
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*Horizontal Design Spectrum (8-4-2-5)

)‫طريقة طيف التجاوب النصميمي االفقي ( للتحليل االنشائي المرن‬

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Simplified Modal Response Spectrum

A- Determine the following parameters:

1- Type & profile of Soil Table(8-1) (A,B,C and D)

2- Building Location Table (8-2 – ‫( & )أ‬8-2- ‫ (ب‬Design ground acceleration (ag)

3- Soil parameter Table (8-3) (‫ )أ‬or (‫)ب‬ S, TB , TC , & TD

4- Damping Factor  Table (8-4)

5- Importance Factor I Table (8-9)

6- Fundamental Period of building T Calculate T1  Ct  H 3 / 4

7- Reduction Factor R Table (8- ‫)أ‬

B- Calculate Sd ( choose equation according to T1 Value )

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C- Calculate Base Shear (Fb)

W
Calculate for each X- Direction & Y-Direction
W
Fb  S d (T )   
g
=0.85 if T1 ≤ 2 Tc
 =1.0 if T1 > 2 Tc

W=Total Load of Building item[8-7-1 (4)]

Fb
TABLE (8-7) ( % of Live Load)

W =Total Dead Load + Ψ x L.L

Ψ = 1.0 Water tanks , Storage areas, ,…

= 0.5 Schools, Hospitals,…..
= 0.25 Residential
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D- Distribution of Base Shear (Fb) along stories

If w is equal for all storys:

W W
Wi Wi
Fi
zi Fi zi

Fbx Fby
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Accidental Torsion effects

Li
Mti = ei x Fi
ei = ± 0.05 x L i

ei = Minimum additional
Li
eccentricity Fi x eyi
exi
Fiy

Combination of Components of Earthquake actions

Fy
ET = [ E2(Fx) + E2(Fy) ]½
0.3Fy
OR
Fx 0.3Fx
ET = E(Fx) + 0.30 E(Fy)
E.Q. in X E.Q. in Y
ET = 0.30 E(Fx) + E(Fy) Direction Direction

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Seismic Loads ECP (201)

Displacement Analysis

dS = 0.70 R de ds = Storey Displacement

Note: 0.70 to consider working load instead of Ultimate load

Requirements for Earthquake Separation

de = (dx2 + dy2)1/2 de ≤ ΔS Δs

ΔS = 0.7 (0.70 R de) ‫تساوي مناسيب االدوار‬

ΔS = 0.70 R de ‫عدم تساوي مناسيب االدوار‬

ΔS ≥4 cm Δs
‫ حائط قص علي االقل‬2 ‫في حالة وجود حوائط قص علي محيط المباني منهم‬
‫متعامدين علي اتجاة الفاصل بكامل ارتفاع المبني‬

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Seismic Loads ECP (201)

Limitations of Inter-storey Drift (dr)

).... ‫منشات بها عناصر غير انشائية قاصفة ( مباني الطوب‬

drv = dr* v ≤ 0.005 * h

drv = dr* v ≤ 0.0075*h
drv = dr * v ≤ 0.01 * h
ds3
dr = ex. (dr= ds3 - ds2)
h
ds2

V = factor depending on building importance

= 0.4, 0.5 (Table 8-10)

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EXAMPLE
Data:
Office Building; Plan(30m 40m) , Height 45m (15 story)
Soil C , Location Cairo
Average partitions load =2.0 kN/m2
F.Cover= 2.5 kN/m2 40m
L.L=2.5 kN/m2
Required:
Base Shear
Shear per floor
Check Stability 30m

Simplified Modal Response Spectrum:

Soil C ,
Cairo ---- Type 1 ------- Zone 3
Table (8-3) ---- S=1.5, TB= 0.1, TC=0.25, TD= 1.2
Building Importance -----1=1
Force Reduction Factor R (Shear walls) (Table8-A) -------R=5
T1=ct H3/4 , Ct=0.05
T1=0.05 (45) ¾ = 0.868 ( T1< 2.0 sec) , ( T1 ≤ 4 *Tc) O.K.
Select Equation Tc < T1 < TD (8-13)

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Sd (T1) = 0.15 * g * 1.0 * 1.5 * 2.5/5 * (0.25/0.868)

=0.0324 g
Check
Sd(T1) = 0.0324 g ˃ (0.2 * 0.15 * 1.0 = 0.3 g) (O.K.)

Calculate Base Shear

L.L % =0.5
w = D.L + α * L.L + F. cover + Partitions Load
=(0.25*25) 1.10 + 0.5 * 2.5 + 2.5 + 2.0
= 13.125 kN/m2

W (Total) = 13.125 * 30 * 40 * 15
=236250 kN

Fb = Sd(T1) * λ * W/g
=0.0324g * 1.0 * 236250 /g (T1 ≤ 2Tc . . . λ=1)
=7654.5 kN

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Floor Height Lateral S.F B.M.

No m Load kN. kN.m
kN
Lateral Load on each Story:
15 45 956.81 956.81 0
Q (Story Height) =Fb* Hi(story) / Σ Hi 14 42 893.03 1849.84 2870.4
13 39 829.24 2679.08 8420
Σ Hi = (3+6+9+…..+45) 12 36 765.45 3444.53 16457
=360 11 33 701.66 4146.19 26791
10 30 637.88 4784.06 39229
Q(3) = 7654.5 * 3 / 360 9 27 574.09 5358.15 53582
=63.79kN 8 24 510.30 5868.45 69656
7 21 446.51 6314.96 87261
Q(6) = 7654.5 * 6 / 360 6 18 382.73 6697.69 106206
=127.58 kN 5 15 318.94 7016.63 126299
….. 4 12 255.15 7271.78 147349
Q(24) =7654.5 * 24 / 360 3 9 191.36 7463.14 169164
=510.3 kN 2 6 127.58 7590.71 191554
1 3 63.79 7654.50 214326
Q(45) =7654.5 * 45 / 360 base 0 63.79 7654.5 237290
=956.81 kN

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SFD:
From top of structure
F15 = 956.82 kN
Q15 =956.82 kN
Q14 =Q15+P14
= 956.81 + 893.02
=1849.83 kN
Q13 = Q14+P13
=1849.83 + 829.24
=2679.07 kN
….
BMD:
From Top of structure
Mi = M(i-1) + Q(i-1)*h

M15=0 + 956.82*3.0 =2870.46 kNm

M14=M15 + Q14*h
=2870.46 + 1849.83 * 3.0
= 8420 kN m

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956.81
956.81 15
2870.4375
893.03
1849.84 14
8419.95
2679.08 13 829.24
16457.175

26790.75 3444.53 12 765.45

39229.312 4146.19 11 701.66

5
53581.5 4784.06 10 637.88

69655.95 5358.15 9 574.09

87261.3 5868.45 8 510.30

106206.18
75 6314.96 7 446.51

126299.25 6697.69 6 382.73

147349.12
5 7016.63 5 318.94
169164.45
7271.78 4 255.15
191553.86
25 7463.14 3 191.36
214326
7590.71 2 127.58
237289.5
7654.50 1 63.79

BMD Load
SFD

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Thanks

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