Date post: | 13-Sep-2015 |
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HIGH RISE BUILDINGS
UNDER THE GUIDANCE OF
PROF. B.V. SARMA
ABSTRACT
As the population is increasing, the available land for housing is decreasing to have individual dwelling units. So the High Rise Buildings is the best option to overcome this situation.
OBJECTIVE
As the increasing population results in the requirement of more land at higher cost which restricts horizontal growth of construction industry. High rise buildings allows accommodating more number of people in limited space to vertical growth.
Requirements
Building plan
Knowledge on types of loads to be considered in designing and their distribution.
Knowledge on analyzing methods.
Knowledge on designing
Knowledge on code books to be used.
Column layout
Loads to be considered
Gravity loads
Wind forces or earth quake forces
Load distribution pattern on slabs
Gravity loads analysis
Converting triangular load into equivalent U.D.L
Total load: x Lx x Lx/2 x w = (w lx2) / 4 kN.m [ triangular loading]
EQUIVALENT SHEAR FORCE :
Shear force = load/2 = (w lx2) / 8 kN [ triangular loading]
shear force = (wq lx)/ 2 kN.m [ U.D.L]
now equating the shear force of triangular loading to uniformly distributed loading to get triangular equivalent UDL
(wq lx)/ 2 = (w lx2 ) / 8
wq = ( wlx ) / 4
EQUIVALENT BENDING MOMENT: Bending Moment
[triangular loading ] Bending Moment [Uniform loading ] Equating both to get equivalent load :
Converting simply supported trapezium loading into Equivalent U.D.L
Total load since EQUIVALENT SHEAR FORCE : [Trapezium loading ] Shear force Shear force for Equivalent UDL Equating both to get equivalent load:
EQUIVALENT BENDING MOMENT:
Bending moment
UDL bending moment
Equating both to get equivalent load
CALCULATION OF EQUIVALENT SHEAR FORCE AND BENDING MOMENT OF FIXED BEAM
FOR TRIANGULAR LOADING : MAB =MBA =5/96 wl
2
BMBM @ center =wl2 /12
@ ends:
@ centre:
SHEAR FORCE:
FOR TRAPEZOIDAL:
@Ends: a=
SUNKEN SLAB
Sunken slabs are used in the toilets, bathrooms and washing place where we have our washing machines. The purpose of having a sunken slab is to conceal all the pipes below the floor. Since the pipes that carry water are concealed below the floor, care has to be taken to avoid leakage problems.
Assuming 120 mm slab thickness Dead load = 0.12 x 1 x 1x 25
= 3 kN / m2
Floor finish = 1 kN / m2 = 4 kN / m2
live load = 3 kN / m2
PRELIMINARY COLUMN DESIGN FOR DEAD LOAD
FOR BEAM AB :
Total Load = area of loaded portion x load intensity
= 2( X 5.41 X 2.705 X 4 )
= 58.536 kN
Load on column = 58.536 / 2
= 29.268 kN
Similarly
Load on column
from beam BC = 42.904 kN
from beam BD = 67.127 kN
from beam BE = 50.026 kN
= 189.325 kN
Similarly for live load
Load on column = 141.994 kN
As per IS 875 (part 2) Clause 3.2 Reduction in Imposed Loads on Floors
Number of Floors ( Including the Roof) Reduction in Total to be carried Distributed Imposed by Member under Load on all Floors to Consideration be Carried by the Member under Consideration ( Percent ) 1 0 2 10 3 20 4 30 5 to 10 40 Over 10 50
Load on bottom most floor column excel.xlsx
= DL + LL
= 6626.38 +4942.64
= 11569.02 kN
Assuming fck = 30 N/mm2
fy = 415 N/mm2 and Ast = 1%
Pu = 0.4 fck Ag + 0.67 fy Ast 1.5 x 11569.02 x 103 = 0.4 x 30 x Ag + 0.67 x 415 x 0.01Ag
Ag = 1183733 mm2
Assuming 750 mm width
Depth of the column = 1183733 / 750
= 1578 mm
Provide 750 x 1800 mm column
Yet to design
Have to analyze and design structure for wind and earthquake forces
Wind pressure calculations by using IS 875 (part 3)
Wind analysis by using portal or cantilever method.
Earthquake resistant analysis and design by using IS 1893 and IS 13920
CONCLUSION
We learnt distribution of loads on slab to beam and then to column.
From preliminary analysis and design, we got the approximate size of column as 750 x 1000mm.
Thank you