Roof > Slabs > Beam > Column > Floor > Foundation
2 types of loads : Imposed Load & Dead Load
Imposed Load (IL):
- Furnitures
- Human
- Any Temporary Loads
Dead Load (DL):
- Beam
- Slab
- Column
- The weight of the building itself
If ratio of Lx : Ly equal or smaller than 2, = one -way slab = Lx / Ly > or = 2
If ratio of Lx : Ly bigger than 2, = two -way slab = Lx / Ly < 2
Standard Factor Of Safety (F.O.S.) of:
- Imposed Load : 1.4 Gk
- Dead Load : 1.6 Qk
- Thus load combination formulae : 1.4 Gk (DL) + 1.6 Qk (IL)
- Assuming that 1.4 and 1.6 is a constant.
- The Dead Load (DL) and Imposed Load (IL) is a variable.
Example :
(A)
1.
List out
the available information (assuming that IL and DL values had been given) :
a.
IL = 2.5 kN/m2
b.
DL = 3.0 kN/ m2
c.
Density of
Concrete = 24 kN/ m2
2.
Determine
the type of load transfer on the slab :
a.
Lx
/ Ly = 9/3 = 3>2 , thus a one-way
slab
3.
Determine
the beam that react to the weight of the slab :
a.
Since it’s
a one-way beam , only half of the slab weight acted on the beam
4.
Calculate
the Combination load using the
formula:
a.
Design
Load, a = 1.4 Gk (DL) + 1.6 Qk
(IL)
b.
Design
Load, a = 1.4 Gk (3.0) + 1.6 Qk
(2.5)
c.
Design
Load, a = 8.2 kN/ m2
d.
*note that
the answer is actually amount of weight acting on 1 m2 area
5.
Calculate
the Effective Area, Aeff :
a.
Effective
Area, Aeff = Lx x ½ (Ly)
b.
Effective
Area, Aeff = 9 x 1.5
c.
Effective
Area, Aeff = 13.5 m2
6.
Calculate
the Concentrated Load, W :
a.
Concentrated
Load, W = 8.2 kN/ m2 x 13.5 m2
b.
Concentrated
Load, W = 110.7 kN
c.
Thus, the
final answer, W = 110.7 kN, which is
the total load acted on the beam
How much is the load acted
each on 1m across the beam?
1.
Just
simply divide the final answer from (A) with the total length of the beam.
a.
Design
Load, a = 110.7 kN / 9 m
b.
Design
Load, a = 0.91 kN/ m
