Swinburne University CVE80008 Seismic Design Report: Building Analysis

This report presents a preliminary seismic design analysis for a 5-storey reinforced concrete (RC) building located in Melbourne, considering a 1000-year return period event and an importance level of 3. The analysis follows the guidelines of AS1170.4 (2007) and includes the determination of the Earthquake Design Category (EDC). The report details the calculation of the base shear force, considering the building's seismic weight and structural performance factor. It further outlines the lateral force distribution across the building's floors, accounting for the Ks factors for each level. The bending moments at each floor level are computed based on the shear forces and floor heights. Additionally, torsion moments are determined considering the eccentricity between the center of mass and center of stiffness. Finally, the report calculates the acceleration of the floors and the lateral force on the building components, ensuring compliance with the minimum requirements.

SWINBURNE UNIVERSITY OF TECHNOLOGY
BUILDING DESIGN REPORT
CVE 80008
DEPARTMENT OF CIVIL AND CONSTRUCTION ENGINEERING
CASE SCENARIO
The report entails preliminary design of a 5-storey RC building (non-storage) located on a rock
site (class B) in Melbourne for a 1000 year return period event. The importance level is 3. The
building measurements are: 20 x 20m plan, height- 15m (each story has equal height). Centre of
mass is offset from the centre of stiffness by 2m in plan in each direction. The dead and
superimposed loads are 8kPa and imposed loads for the typical floors are 4kPa and roof is 1kPa.
DESIGN
1. The earthquake design category (EDC) is found based on the given information.
We know that the building seats on a rocky place (class B), IL=3, 1000 year return
period.
From AS 1170.4 (2007) and using the parameters: IL=3, class B, structure height- 15m,
table 2.1 provides the EDC for this case as EDC II
2. Base shear is determined from the formula :
V= [KpZCh(T1)Sp/u]Wt ….(1) (obtained from section 6 of the standard)
Now, Wt is the seismic weight given by the sum of level weights and this is obtained
from clause 6.2.2 as shown below:
Wi= Gi+ϕQi where ϕ= 0.3 since this is a nonstorage
Substituting in equation: = 8+0.3x 4+1 = 10.2kPa per level
Hence total Wt= 5x 10.2= 51kPa
Sp/u= Structural performance factor per structural ductility factor
This is obtained from clause 6.5, since structure is RC with ordinary moment resisting
frames, we select the 3rd entry hence , Sp/U= 0.38

Ch(T1) is the spectral shape factor and this is obtained from clause 6.4) using the
following parameters: 1000 year return period, hence P (secs) = 1.0, subsoil classification
is Be (from table 2.1) hence Ch(T1) = 0.88
KpZ is obtained from table 3.5, since we know that: return period is 1000 years hence
probability of exceedance is 1/1000, KpZ= 0.10
Therefore, the base shear is obtained by substituting these parameters in the formula (eqn
1 above)
V= [0.1 x 0.88x 0.38]x51= 1.70544kPa
2(b) Lateral force distribution
For this case, we use clause 5.4.2.3 by first checking if criteria is met: remember L= 15m
hence minimum requirements to invoke this clause is met
We therefore use the formula: Fi= Ks [KpZSp/u] Wi at each level
For individual Ks we deduce the following information:
Floor type Ks
5th 3.1
4th 2.5
3rd 1.8
2nd 1.2
1st 0.6
For 1st floor with Ks=0.6,
F1= 0.6[0.1x0.38] x51= 1.1628kPa
And other floors same:
F2= 1.2(0.1x0.38) x51= 2.3256kPa
F3= 1.8(0.1x0.38) x51= 3.4884kPa
F4= 2.5(0.1x0.38) x51= 4.845kPa
F5= 3.1 (0.1x0.38)x 51= 6.0078kPa
3. Distribution of Bending Moment
This is given by the formula:
BM= Vx where V is the shear force and X is the height between floor levels.
Therefore BM for each floor is determined:
1st level: Mx= 1.70544 x 3 = 5.116kN-m

2nd floor Mx= 6x 1.79544= 10.232kN-m
3rd floor Mx= 9x1.70544= 15.348kN
4th floor = 12 x 1.70544= 20.464kN-m
5th floor Mx= 15 x 1.70544 = 25.58kN-m
4. Torsion moments determination
Eccentricity e:
e= 0.1b, 0.1x 20/2= 1.0
Φ= dstxWjx(HsiUxFj) (this is obtained from section 5, 6 and 7)
= 3.117 x 10.2x 1.1628x2.0x1.5= 110.9kN-m
5. Acceleration of the floors:
Acceleration for floors (obtained from section 6):
Hence:
a1 = force x Rc/Icxac= 1.1628 x2.5= 2.907m/s2
a2= 2.5 x 2.3256= 5.814m/s2
a3= 2.5 x 3.484= 8.72m/s2
a4= 4.845 x 2.5 = 12.1125m/s2
a5= 6.0078 x 2.5 = 15.0195m/s2
6. Lateral force
F= afloor [Icx ac/Rc]Wc<0.5Wc
Wc= 3kN, the minimum criteria to be met: 0.5x3= 1.5kN ( answer obtained must not be
more than this)
From section 8.2, we obtaine the following:
Ic= 1.0 (for all other components), ac= 1.0 and Rc= 2.5 (since it is noncritical ductile)
and af= KpZCh= 0.38 x 0.88 = 0.3344
Hence F= 0.3344 (1x1/2.5)x 3= 0.40128kN

REFERENCE
Australian Standard. (2007). Structural Design Actions: Earthquake Actions in Australia.
AS 1170.4. Council of Standards Australia, Sydney. Government publication.