Preliminary Design of 5-Storey RC Building on Rock Site in Melbourne

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