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