Ahsanullah University of Science and Technology
ARC 4229
Structure And Architecture VI
(2.0 Credit Hrs/week)
Course Teacher
Md. Nahid Hossain
Lecturer
Department of Civil Engineering
Ahsanullah University of Science and Technology
EARTHQUAKE LOAD
Definitions
Base: The level at which the earthquake motions are considered
to be imparted to the structure.
Base Shear: Total design lateral force or shear at the base of a
structure.
Building Frame System: An essentially complete space frame
which provides support for gravity loads.
Essential Facilities: Buildings and structures which are
necessary to remain functional during an emergency or a post
disaster period.
Moment Resisting Frame: A frame in which members and joints
are capable of resisting forces primarily by flexure.
Definitions
Shear Wall: A wall designed to resist lateral forces parallel to the
plane of the wall.
Story: The space between floor levels. Story-x is the story below
level-x.
Story Shear: The summation of design lateral forces above the
story under consideration.
Structural Irregularities
Regular Structures
Regular structures are structures having no significant
physical discontinuities in plan or vertical configuration or in
their lateral force resisting.
Irregular Structures
Irregular structures are structures having significant physical
discontinuities in configuration or in their lateral force
resisting systems (See Table 1 for detailed description of
such structures).
Table 1: Vertical Structural Irregularities
Irregularity Type and Definition How to Deal with
A. Stiffness Irregularity (Soft Story): A
soft story is one in which the lateral
stiffness is less than 70 percent of that Use the dynamic lateral
in the story above or less than 80 force procedure.
percent of the average stiffness of the
three stories above.
Table 1: Vertical Structural Irregularities
Irregularity Type and Definition How to Deal with
B. Mass Irregularity: The seismic weight
of any story is more than twice of that of Use the dynamic lateral
its adjacent story. This irregularity need force procedure.
not be considered in case of roofs.
2
Table 1: Vertical Structural Irregularities
Irregularity Type and Definition How to Deal with
C. Vertical Geometric Irregularity: Vertical
geometric irregularity shall be considered
to exist where the horizontal dimension of Use the dynamic lateral
the lateral force-resisting system in any force procedure.
story is more than 130 percent of that in an
adjacent story.
Wall
Table 1: Vertical Structural Irregularities
Irregularity Type and Definition How to Deal with
D. In-Plane Discontinuity in Vertical The Structure is to be
Lateral Force-resisting Element: An in- designed to resist the
plane offset of the lateral load overturning effects caused by
resisting elements greater than the seismic forces, down to the
length of these elements. foundations level.
Table 1: Vertical Structural Irregularities
Irregularity Type and Definition How to Deal with
E. Discontinuity in Capacity (Weak Story):
Structures are not to be
A weak story is one in which the story
over two stories or 9 m in
strength is less than 80 percent of that in
height where the weak
the story above. The story strength is the
story has calculated
total strength of all seismic-resisting
strength of less than 65 %
elements sharing the story shear for the
of the story above.
direction under consideration.
Problem 01
Calculate the vertical distribution of earthquake forces on a 45 m high 15-storied
residential building with story height 3 m each and located in Dhaka town. The
concrete building has a rectangular plan 30 m x 60 m and the basic structural
system is developed. Dead load including partitions = 9 kN/m2 (each floor), live load
= 3 kN/m2 (each floor) and viscous damping ratio of the structure = 5.
Depth (m) SPT – N Depth (m) SPT – N Depth (m) SPT – N
0.0 0 10.5 16 21.0 50
1.5 5 12.0 29 22.5 50
3.0 7 13.5 32 24.0 50
4.5 10 15.0 38 25.5 50
6.0 12 16.5 46 27.0 50
7.5 13 18.0 50 28.5 50
9.0 10 19.5 50 30.0 50
Solution 01
Step 01: Calculation of Site Classification Depth (m) d (m) SPT – N d / SPT – N
0.0 0 0 -
1.5 1.5 5 0.3
∑𝑑
𝑆𝑃𝑇 − 𝑁 𝑎𝑣𝑔 = 3.0 1.5 7 0.21
∑ 𝑆𝑃𝑇𝑑 − 𝑁 4.5 1.5 10 0.15
(Page 3189)
6.0 1.5 12 0.13
30 7.5 1.5 13 0.12
=
1.59 9.0 1.5 10 0.15
10.5 1.5 16 0.09
= 18.88
12.0 1.5 29 0.05
13.5 1.5 32 0.05
From Table 6.2.13, we get 15.0 1.5 38 0.04
(Page 3190) 16.5 1.5 46 0.03
Site Classification = SC 18.0 1.5 50 0.03
19.5 1.5 50 0.03
21.0 1.5 50 0.03
From Table 6.2.16, we get
22.5 1.5 50 0.03
S = 1.15 (Page 3197) 24.0 1.5 50 0.03
TB = 0.2 sec 25.5 1.5 50 0.03
27.0 1.5 50 0.03
TC = 0.6 sec
28.5 1.5 50 0.03
TD = 2 sec 30.0 1.5 50 0.03
Total 30.0 1.59
Solution 01
Step 02: Calculation of zone coefficient and importance factor:
From Table 6.2.15, we get for Dhaka
(Page 3196)
Z = Zone - II = 0.2
From Table 6.1.1 & Table 6.2.17, we get (Page 3061) (Page 3197)
I = Occupancy Category - II = 1
Step 03: Calculation of seismic design category and response reduction factor:
From Table 6.2.18, we get
Seismic design category = C (Page 3198)
From 8.3.2 (Provisions), we get
IMRF (Intermediate Moment Resisting Frame) (Page 3669)
From Table 6.2.19, we get (Page 3202)
R=5
𝐼 1
= = 0.2 (<1) (OK) (Page 3193)
𝑅 5
Solution 01
Step 04: Calculation of structure (building) period
𝑇 =𝐶𝑡 ℎ𝑛 𝑚 From given data
hn = 45 m
(Page 3208) = 0.0466 × 45 0.9
= 1.433 𝑠𝑒𝑐 From Table 6.2.20, we get
Ct = 0.0466 (Page 3209)
m = 0.9
Step 05: Calculation of normalized acceleration response spectrum
𝑇𝐶
𝐶 = 2.5 𝑆η (Page 3193) 10
𝑇 𝜂= (Page 3194)
5+𝜉
0.6
= 2.5 × 1.15 × 1
1.433 10
= 𝜉= 5
= 1.204 5 +5
= 1 > 0.55 (𝑂𝐾)
Solution 01
Step 06: Calculation of design spectral acceleration or lateral seismic force coefficient
2 𝑍𝐼 0.67𝛽𝑍𝐼𝑆 = 0.67 × 0.11 × 0.2 × 1 × 1.15
𝑆𝑎 = 𝐶𝑠 (Page 3193)
3𝑅 = 0.01695 (Page 3193)
2 0.2 × 1
= × × 1.204 0.044 𝑆𝐷𝑆𝐼 = 0.044 × 0.383 × 1
3 5
= 0.01685 (Page 3208)
= 0.03211 > 0.01695 (𝑂𝐾)
Step 07: Calculation of total seismic load
𝐷𝐿 + 0.25 𝐿𝐿 = 9 + 0.25 × 3 = 9.75 𝑘𝑁/𝑚2 (Page 3209)
𝑆𝑒𝑖𝑠𝑚𝑖𝑐 𝑙𝑜𝑎𝑑 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑠𝑡𝑜𝑟𝑦, 𝑤𝑖 = 9.75 × 30 × 60 = 17550 𝑘𝑁 𝑖 = 1~15
𝑤1 = 𝑤2 = 𝑤3 = … … … … … = 𝑤15 = 17550 𝑘𝑁
𝑇𝑜𝑡𝑎𝑙 𝑠𝑒𝑖𝑠𝑚𝑖𝑐 𝑙𝑜𝑎𝑑, 𝑊 = 17550 × 15 = 263250 𝑘𝑁
Step 08: Calculation of seismic design base shear
𝑉 = 𝑆𝑎𝑊 = 0.03211 × 263250 = 8452.96 𝑘𝑁 (Page 3207)
Solution 01
Step 09: Calculation of vertical distribution of lateral force
T (sec) ≤ 0.5 ≥ 2.5
k 1 2
T = 1.433 sec
𝑘 = 0.5𝑇 + 0.75 = 1.4665
Solution 01
Fx Wx (kN) hx (m) Wxhxk Fx (kN)
F1 17550 3 87897.28 24.18
F2 17550 6 242904.71 66.82
F3 17550 9 440224.05 121.10
F4 17550 12 671268.73 184.66
F5 17550 15 931139.93 256.14 87897.28
F1 = 30728403.55 × 8452.96 = 24.18
F6 17550 18 1216562.01 334.66
F7 17550 21 1525147.04 419.55
F8 17550 24 1855055.50 510.30
F9 17550 27 2204814.62 606.51 242904.71
F2 = 30728403.55 × 8452.96 = 66.82
F10 17550 30 2573211.25 707.85
F11 17550 33 2959223.82 814.04
F12 17550 36 3361977.00 924.83
F13 17550 39 3780710.01 1040.02
F14 17550 42 4214753.74 1159.42
F15 17550 45 4663513.86 1282.87
∑ Wxhxk 30728403.55
Solution 01
𝐹15
𝐹14
𝐹13
𝐹12
𝐹11
𝐹10
𝐹9
𝐹8
𝐹7
𝐹6
𝐹5
𝐹4
𝐹3
𝐹2
𝐹1
References
➢ BNBC 2021
