Example 1 – Design Seismic Forces Including Missing Mass
Correction
Problem Statement
Consider a four-storey reinforced concrete office building located in Shillong (seismic zone V). The
soil conditions are medium stiff and the entire building is supported on a raft foundation. The R. C.
frames are infilled with brick-masonry. The lumped weight due to dead loads is 12 kN/m2 on floors and
10 kN/m2 on the roof. The floors are to cater for a live load of 4 kN/m2 on floors and 1.5 kN/m2 on the
roof. This building is taken from Example 1 of IITGN-WB-EQ3 V3.0 [1].
Now, for this building, shear walls of 350 mm thickness and M30 grade concrete are provided at ground
storey in X-direction. Ground floor plan and elevation of the building is shown in Figure 1.1. Due to
these shear walls, the ground becomes very stiff, and it is quite likely that this building will have high
frequency modes with substantial modal mass. The objective of this example is to obtain design seismic
forces in X-direction by Response Spectrum Analysis incorporating the effect of missing mass.
y
(2)
(1)
(4)
(3)
(5)
(A)
3@5m
(B)
(C)
(D)
x
4@5m
Ground Floor PLAN
z
3.2 m
3.2 m
3.2 m
4.2 m
x
5m
5m
5m
5m
Elevation XZ
Figure 1.1 ̶ Plan and Elevation of building
1
In order to perform free vibration analysis of the
building, 3-D model is prepared in SAP
software. Beams and columns are modeled
using frame element with effective stiffness, as
per Clause 6.4.3.1 of IS 1893. The effective
stiffness of column is taken as 0.7EI and that of
beam is taken as 0.35EI, rigid diaphragm is
provided to model in plane rigidity of slab.
Shear walls are modeled using shell elements.
There is no reduction in shear wall stiffness and
gross stiffness is used in the model. 3D line
model of the building is shown in Figure 1.2.
All the structural elements are considered mass
less, and mass corresponding to seismic weight
of each floor is lumped at the respective floor
level. Free vibration properties of building in Xdirection are shown in Table 1.1.
Solution
In this building, due to presence of shear walls
at the ground floor, the structure becomes very
stiff in X direction. Seismic weight of all other
floors are same as in Example 1 of IITGN-WBEQ3 V3.0 [1], but the seismic weight of ground
floor is re-calculated including the weight of
shear walls. Thus, revised floor weights are:
W1 = 2920 kN
W2 = 4300 kN
W3 = 4300 kN
W4 = 5800 kN
Figure 1.2 ̶ 3D model of building with shear wall
2
Table 1.1 ̶ Free vibration properties of building in X-direction
Natural Period
(sec)
Natural Freq (Hz)
Modal Mass (%)
Mode 1
1.153
0.87
59.4
Roof
3rd Floor
2nd Floor
1st Floor
1.000
0.752
0.342
0.001
Mode 2
0.356
2.81
7.6
Mode Shape
1.000
-0.491
-0.912
-0.005
Mode 3
0.203
4.93
1.9
Mode 4
0.026
38.46
10.3
Mode 5
0.022
45.45
19.8
1.000
-1.648
1.634
0.013
1.000
1.158
-13.368
1206.789
1.000
-2.750
8.714
-1133.500
procedure for calculating seismic forces using
response spectrum analysis and missing mass
correction is given here.
It is to be noted that frequency of fourth and
fifth modes are more than 33Hz, and in these
two modes about 30% of modal mass gets
excited. Thus, without including these modes,
more than 90% mass does not participate.
Hence, as per cl. 7.7.5.2 of IS 1893 Part 1, their
effect in seismic response should be considered
using missing mass correction. The procedure
for missing mass correction is adopted from
Regulatory Guide 1.92 of U.S. Nuclear
Regulatory commission [2]. The step by step
Step 1: Determine seismic forces using RSA by
considering only those modes whose frequency
is less than 33 Hz. (Cl 7.7.5.4 of IS 1893, Part1,
2016). In the present example, first three modes
have frequency less than 33Hz. The details of
seismic forces obtained using first three modes
are given in Table 1.2.
Table 1.2 ̶ Calculation of modal mass and modal participation factor and seismic forces
Mode 1
Floor
Wi(kN)
2920
4
4300
3
4300
2
5800
1
17320

Mik(tons)
% Mass
Pk
T(sec)
Sa/g
Ak
i
1.000
0.752
0.342
0.001
i
1.000
0.752
0.342
0.001
Floor
4
3
2
1
2
2
Wii
2920
3236
1471
8
7634
1013
Mode 2
Wii 2
2920
2435
503
0
5858
i
1.000
-0.491
-0.912
-0.005
Wii
2920
-2111
-3923
-27
-3141
134
Mode 3
Wii 2
2920
1036
3579
0
7536
i
1.000
-1.648
1.634
0.013
Wii
2920
-7087
7025
75
2933
34
59.4
1.303
1.153
7.6
-0.417
0.356
1.9
0.112
0.203
1.180
0.042
2.500
0.09
2.500
0.09
Qik
162
179
81
0
Vik
162
341
422
422
Shear Forces
Qik
i
1.000
-110
-0.491
79
-0.912
147
-0.005
1
Vik
-110
-30
117
118
i
1.000
-1.648
1.634
0.013
Qik
30
-72
71
1
Wii 2
2920
11680
11476
1
26078
Vik
30
-42
29
30
V2 = [(422)2+ (117)2+ (29)2]1/2 = 439 kN
V1 = [(422)2+ (118)2+ (30)2]1/2 = 440 kN
(Clause 7.7.5.3b of IS: 1893 Part 1)
2 1/
V4 = [(162) + (-110) + (30) ] = 197 kN
V3 = [(341)2+ (-30)2+ (-42)2]1/2 = 345 kN
3
The Storey forces are then obtained as:
𝑍𝑃𝐴 = (
Q4 = V4 = 197 kN
Q3 = V3 – V4 = 345 – 197 = 147 kN
Q2 = V2 – V3 = 439 – 345 = 94 kN
Q1 = V1 – V2 = 6440 – 439 = 1 kN
(Clause 7.7.5.4f of IS: 1893 Part 1)
𝑃𝑖 = (
𝑍 𝐼
× × 𝑔) × (𝑀𝑖 ) × (𝑒𝑖 )
2 𝑅
𝑃𝑖 = (0.3532) × (𝑀𝑖 ) × (𝑒𝑖 )
Step 2: For each degree of freedom (DoF), i.e.
for each floor, determine the fraction of DoF
mass included in the summation of all modes
considered in Step 1. This fraction, di for each
DoF i is given by:
The seismic forces obtained as per Steps 2, 3
and 4 are shown in Table 1.3.
Table 1.3 ̶ Forces (kN) corresponding to
missing mass correction
Storey
di
ei
Pi (kN)
Level
Storey 4 0.99899
0.00101
0.11
Storey 3 0.99994
0.00006
0.01
Storey 2 1.00970 -0.00970
-1.50
Storey 1 0.00519
0.99481
207.72
𝑁
𝑑𝑖 = ∑ 𝑃𝑘 × 𝜙𝑖𝑘
𝑘=1
Where,
k = mode number (1, 2, 3….)
N = number of modes included in step 1
ik = mode shape value for mode k and DOF i
Pk = participation factor for mode k
Step 3: Determine the fraction of DoF mass not
included in the summation of modes from step
1. This is given by the following equation.
Step 5: Forces (Qi) obtained from RSA using
flexible modes (< 33 Hz), and forces (Pi)
obtained from missing mass correction (rigid
modes > 33 Hz), are combined using SRSS
combination [3]. This combination is shown in
Table 1.4.
Table 1.4 Combination of seismic forces
(SRSS) (kN)
𝑒𝑖 = 1 − 𝑑𝑖
Step 4: Determine the force for fraction of
missing mass of ith floor. This force (Pi) is given
the following equation.
𝑃𝑖 = 𝑍PA × (𝑀𝑖 ) × (𝑒𝑖 )
Storey 4
Storey 3
Storey 2
Storey 1
Here, Mi = Seismic mass associated with ith
floor and ZPA is zero period acceleration, given
by
𝑍𝑃𝐴 = (
0.36 1
× × 1 × 9.81)
2
5
= 0.3532 𝑚/𝑠 2
𝑍 𝐼
𝑆𝑎
× × ( ) × 𝑔)
2 𝑅
𝑔
Design seismic forces
Due to
Due to Combined
forces
Modes
modes
less than more than
√𝑸𝟐𝒊 + 𝑷𝟐𝒊
33Hz (Qi) 33Hz (Pi)
197
0.11
197
147
0.01
147
94
-1.50
94
1
207.72
208
These combined forces will be used in various
load combinations to get the response of the
building. Thus, a simple procedure to include
effect of missing mass in the RSA of IS 1893 is
described in this example. This method can be
used even if more than one modes have
frequency higher than 33Hz.
In the present case, since we are using dynamic
analysis, hence, as per Fig. 2B of IS1893 Part 1
at zero period, value of Sa/g is unity.
For this building, as per Example 1, Z = 0.36, R
= 5 and I = 1, Hence,
Observations
4
analysis,” Rev. 2, U.S. Nuclear
Regulatory Commission, February,
2005.
3. R.
Morante
and
Y.
Wang,
“Reevaluation of Regulatory Guidance
on Modal Response Combination
Methods for Seismic Response
Spectrum Analysis,” NUREG/CR6645, U.S. Nuclear Regulatory
Commission,
Washington,
DC,
December 1999.
From the present example, it appears that
buildings with basements having RC walls,
become a potential structure to have high
frequency modes with substantial modal
masses. For such buildings, missing mass
correction will be required.
References
1. Jain, S.K., Jaiswal, O.R., Ingle, R.K.,
Roy, D., Explanatory Examples on
Indian Seismic Code IS 1893 (Part I),
Document No. IITGN-WB-EQ3 V3.0.
2. Regulatory Guide 1.92, “Combining
modal
responses
and
spatial
components in seismic response
5
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