i s code 1893:2002

Numerical problem of is code 1893: 2002

1. Earthquakes LoadsEarthquakes Loads
As per IS 1893 : 2002As per IS 1893 : 2002

2. BuildingBuilding
Step: 1 Design Horizontal Seismic Coefficient,
A
h
= ZIS
a
/ 2Rg
where, Z = Zone Factor (Table 2)
I = Importance Factor (Table 6)
R = Response Reduction Factor (Table 7)
S
a
/g = Average Response Acceleration Coefficient (Fig. 2)
Step: 2 Calculation of Seismic Weight ,
Step:3 Design Seismic Lateral Force :
V
b
= A
h
. W
W = Seismic weight of building (clause 7.4.2)

3. Design Horizontal Seismic CoefficientDesign Horizontal Seismic Coefficient
It is the factor which is multiplied with Seismic wt. to
determine design lateral force.
A
h
= ZIS
a
/ 2Rg
1) Z = Zone Factor



Seismic Zone Z
II 0.10
III 0.16
IV 0.24
V 0.36

4. MUSE 11BMUSE 11B

5. 2) I = Importance factor
3)R = Response Reduction Factor
Sr. No. Structure I
I
Important services
and community
buildings
1.5
II All other building 1.0
Sr. No. Structure R
I
Ordinary RC moment
resisting frame
3
II
Special RC moment-
resisting frame
5

6. 4) S
a
/ g = Spectral Acceleration Coefficient
Step : 2 Seismic Weight
Sr. No. Floor Load %
I Upto 3 25
II Above 3 50

7. ExampleExample
10m
5m
10m
Consider a single storey reinforced concrete office building which
is located in Ludhiana. The soil conditions are medium stiff . The
lumped weight due to dead loads is 12kN/m
2
on roofs and a live
load of 4 kN/m2 . Determine design seismic load on the structure
as per new code.

8. Data GivenData Given
•Zone factor (Z) = 0.24 ( Ldh in Zone IV)
•Importance factor (I) = 1.0 (Office Building)
•Response reduction factor = 5 (R.C.C. Frame)
•Live Load = 4 KN/m
2
•Dead load = 12KN/m
2

+

9. Design Horizontal Seismic CoefficientDesign Horizontal Seismic Coefficient
•Fundamental period = 2 sec
•Soil Condition = medium stiff
•Hence S
a
/ g = 2.5
A
h
= ZIS
a
/ 2Rg
= 0.24 x 1.0 x 2.5 / 5x2
= 0.06

10. G. N. D..E.C.G. N. D..E.C.
Seismic WeightSeismic Weight
Floor area = 10 X 10 m
2
Live Load = 4 KN/m
2
Only 50% of live load is to be lumped at roofs.
Dead load = 12KN/m
2
Seismic Weight = 100 X (4*0.5+ 12) = 1400KN

11. Seismic Shear forceSeismic Shear force
V
b
= A
h
. W
= 0.06 x 1400
= 84 KN

12. •Problem:
A single storey R.C.C. building is situated at Jalandhar. The
soil below foundation is hard strata. Determining seismic
shear force acting on the building. Compare the seismic
shear force with an identical building in Bangalore. take
fundamental time period as unity and seismic weight of
building = 1500 KN.

13. Case I: Building in JalandharCase I: Building in Jalandhar
•Zone factor (Z) = 0.24 As Jalandhar in Zone IV
•Importance factor (I) = 1.5
•Response reduction factor = 5 (R.C.C. frame building)
•Spectral acceleration S
a
/g = 1 (Hard Strata)
•Seismic weight (W) = 1500 KN

14. Seismic Base ShearSeismic Base Shear
V
b = A
h . W
= ZIS
a / 2Rg x W
= 0.24x 1.5x1x15000/2/5
= 540 KN

15. Case II: Building in BangaloreCase II: Building in Bangalore
•Zone factor (Z) = 0.10 As Bangalore in Zone IV
•Importance factor (I) = 1.5
•Response reduction factor = 5 (R.C.C. frame building)
•Spectral acceleration S
a
/g = 1 (Hard Strata)
•Seismic weight (W) = 1500 KN

16. Seismic Base ShearSeismic Base Shear
V
b
= A
h
. W
= ZIS
a
/ 2Rg x W
= 0.10x 1.5x1x15000/2/5
= 225 KN
Result: Comparison of base shear values in the above two cases
show that seismic design force in Jalandhar is 2.4 times that
of Bangalore for identical type of building $ soil condition.

17. ExampleExample
•Problem:
Considering a four storey building of R.C.C. is located in Bhuj.
The soil conditions are medium stiff and entire building is
located on raft footing. The lumped weight due to dead load is
12 KN/m2 on floors and 10KN/m2 on roof. The floor are cater
to carry a live load of 4KN/ m2 on floor and 1.5 KN/m2 on
roof. Determine design seismic load on structure.

18. Data GivenData Given
•Zone factor (Z) = 0.36 As Bhuj in Zone IV
•Importance factor (I) = 1.0
•Response reduction factor = 5 (R.C.C. frame building)
•Spectral acceleration S
a
/g = 1 (Hard Strata)

19. Seismic weightSeismic weight
The floor area is 15×20=300 sq. m. Since the live load class is
4kN/sq.m, only 50% of the live load is lumped at the floors. At roof,
no live load is to be lumped. Hence, the total seismic weight on the
floors and the roof is:
Floors:
W1=W2 =W3 =300×(12+0.5×4)
= 4,200 KN
Roof:
W4 = 300×10
= 3,000 KN
(clause7.3.1, Table 8 of IS: 1893 Part 1)
Total Seismic weight of the structure,
W = ΣWi = 3×4,200 + 3,000
= 15,600 KN

20. Design Horizontal Seismic CoefficientDesign Horizontal Seismic Coefficient
•Fundamental period = 2 sec
•Soil Condition = medium stiff
•Hence S
a
/ g = 2.5
A
h
= ZIS
a
/ 2Rg
= 0.36 x 1.0 x 2.5 / 5x2
= 0.09

21. Seismic Shear forceSeismic Shear force
V
b
= A
h
. W
= 0.09 x 15600
= 1440 KN

22. Distribution of design force to each floor,
Q
i
= V
b
.w
i
.h
i
2
/ ∑ w
j
.h
j
2
Q
i
= Design lateral force at floor i
w
i
= Seismic weight of floor i
h
i
= height of floor i measured from base
n = number of floors

23. Lateral load distributionLateral load distribution
Storey
Level
W
i
h
i
w
i
.h
i
2
w
i
.h
i
2
/
∑ w
j
.h
j
2
Lateral
Force
4 3000 13.8 571.3 .424 611
3 4200 10.6 471.9 .350 501
2 4200 7.4 230 .171 246
1 4200 4.2 74.1 .055 79
1347.3 1000 1440