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CVE20003, Design of Concrete Structures Project 1 – Analysis of a multistorey concrete building Semester 2, 2019 Due date: 10th September 2019 Group Project: No more than 4 members per group....

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CVE20003, Design of Concrete Structures
Project 1 – Analysis of a multistorey concrete building
Semester 2, 2019
Due date: 10th September 2019
Group Project: No more than 4 members per group.
Weighting: 15%
A five story building is to be designed and constructed in Melbourne’s eastern suburbs. A preliminary assessment by
the engineer has nominated that a concrete framed building would be most suitable. The engineer has nominated the
geometry of the structure as per figures 1-4 below. It is proposed that each floor be comprised of a suspended beam and
slab system consisting of reinforced concrete (RC) beams (1200mm wide x 600mm deep along grid C and 1200mm
wide x 450mm deep along grids 1 & 2) spanning between 550mm x 550mm RC columns. The beams will be supporting
a one way spanning RC slab. It is proposed that a 180 mm thick RC slab is required to span between the beams (refer
figure 1).
The beams and columns will be rigidly connected using steel reinforcement and should be designed as a moment
resisting frame to resist the horizontal forces which will be imposed on the structure due to wind actions (Wu) in the
east-west direction (parallel with grid lines 1-3). The wind loads will be applied to the external faces of 250mm thick
precast concrete shear walls in the easterly direction as shown in figures 1 and 2. The precast concrete 250mm thick
shear walls will be design to take the horizontal loads from wind actions in the north-south directions. As a result, the
beams, slabs and columns are not required to be designed for horizontal loads in the north-south directions (parallel to
grid lines A-D). The 250mm thick shear wall will be connected to the concrete beams and slabs using steel dowel bars
at each level (refer detail A). There is a 100mm gap between the walls and the face of the edge columns, therefore the
columns will only experience horizontal loads from the beams connected to them and the walls will not transfer loads
to the columns directly along their height. The walls will not support the beams or slabs vertically and it can be assumed
that all the vertical loads will be transferred from the beams to the columns. However, the walls are capable of supporting
their own self-weight since they are tied into footings at the base and can be assumed to have no out of plane flexural
stiffness to resist horizontal loading in the out of plane east-west direction. I.e the wall can only transfer horizontal wind
loads in the east-west direction to the beams at each level and do not contribute to providing any resistance the horizontal
loads in the east-west direction. The columns are fully fixed into footings at their base and footing to column connection
can be assumed as rigid for design purposes.
Answered 116 days After May 15, 2022

Solution

Banasree answered on Sep 08 2022
63 Votes
1.Ans.
DL = 1kpa + self weight
Self wt = BDY
Imposed load = 1kpa span = 5500mm
Let assume concrete density = Y = 24kn/m
Assume effective cover = 25mm
Slab thk = 180mm
Span/d = 26 x M.F [let, M.F = 1]
5500/d = 26*1
d = 211.53
total depth D = 211.53 + 25 = 236.53 = 234mm
effective depth = d = 234-25 = 209mm
Self wt = 0.234*1*24 = 5.616
DL = 1kn/m
Floor finish = Live load = 1kn
Imposed = 1kn
Total DL = 5.616+1+1 =7.616KN/m
2. Ans
Exterior wind load to be applied to the 5 storey frames comprising beams and col =
Same as immediately above plus for the ASD load combination
D+W
D+0.7E+0.5L^2+0.2S^2
Same as immediately above plus for the LRFD load combination
1.2D+1.5W
1.2D+1.0E+0.5L^2+0.2S^4
D= DL = 1kn
W= Design wind load
E =Design earth wake load
L =Design live load = kn
S = Design roof snow load.
3. Ans
Tributary area of the floor grades = 5.5*3 = 16.5mm
FL = 7.2
Tributary area of the col = (1.3+1.3)*(0.55*0.55) =1.992mm
4. Ans.
Consider as 550 mm x 550mm square in crossed.
Given data –
Superimposed deal load including finishes to be taken as 1.0 kpa.
Refe
ing data of table 3.1 AS1170.1 classification E - height of 2m.
Loading analysis – a constant positive and wind load of (Wu) of 1.5 at each level.
DL (G) = 1 kpa Q = 4.0kpa, Wu = 1.5kpa
Ed 1 = Factored load FL = 1.2G + Wu + 0.4Q
        = 1.2*1 +1.5+0.4*4
        = 4.3kpa
Ed 2 = 1.2 G + 1.5Q = 1.2*1 + 1.5*4
        = 7.2kpa
Critical FL = 7.2kpa
Longer dimension=on of span (Ly) = 0.71m
Small dimension of span (Lx) = 0.387m
Load on beam = (( FL X Lx)/ 6)[3-(Lx/Ly)^2]
        = ((7.2*0.387)/6)[3-(0.387/0.71)^2]
        = 0.464*2.702
        = 1.253 KN/m
Self wt of beam grid line C = 0.387*0.6*25 = 5.8kn/m
Self wt of beam grid line1- 2= 0.387*0.45*25 = 4.35kn/m
Total load = 1.253+5.8+4.35 = 11.40kn/m
Load on column – (11.40*0.387)/2 = 2.205kn
5. Ans
For portal method, we will use some assumptions in order to approximate analysis.
Assumption
1. POC is assumed to be located at the mid poibt of each beam and column.
2. Horizontal shear (H) is assumed to be divided among all columns such...
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