micro and macro economics in building indusrtyFull description
Full description
ANALYSIS & DESIGN OF MULTI STOREYED BUILDINGS V.M.RAJAN,M.E.(Struct),FIE, Former CE/Civil/TANGEDCO ******
Example: SINGLE BAY SINGLE STOREY FRAME WITH HINGED END : Data: Centre to centre of frame Span of frame
: 3.50m : 10.50m
Number of bay
: One
Number of storey
: One
Storey Height
: 4.85 m
Intended Use of structure
: Control Room Building
Loads to be considered
: IS 875- 2007
Member Sizes
: Column 300 x 500 mm : Main Beam 300 x 750 mm : End Beam 230 x 300 mm : Slab Thickness 120 mm
Concrete Grade M20 Steel Grade Fe 415 Support condition
: Bottom Hinged. Hinged.
Step 1: Design of Slab Same as in the Previous problem.
Step 2: Preliminary size of beams and columns Same as in previous problem. Beam size : 300 x 750mm Column size: 500 x 300 mm End Beam : 230 x 300 mm Step 3: Frame Analysis: 3.1 Calculate Load/m run on intermediate frame 3.2 calculation of moment of Inertia 3.3 Distribution factor 3.4 Fixed End Moment
3.5 . Moment Distribution by HARDY CROSS Method (Full Frame) Joint Member DF FEM
A AB -
B BA 0.24
BC 0.76 -252.66 192.02
CB 0.76 +252.66 -192.02
-96.01
96.01
72.97
-72.97
-36.49
36.49
8.76
27.73
-27.73
-8.76
92.44
-92.44
+92.44
-92.44
60.64
Distribution Carry over Distribution
23.04
Carry over Distribution Total
0
C CD 0.24
D DC -
-60.64
-23.04
0
Beam Bending moment: At Support MBC= -92.44 KNm At Mid span: Simply supported bending moment at centre= wL 2 /8 = 27.50 x 10.502 /8 = 378.98 KN.m Maximum +ve Bending moment at mid span BC = 378.98 – 92.44 =286.54 KN.m Bending moment for Column: M AB= 41.84 KNm. (At Bottom) MBA = 92.44 KNm (At Top) Factored B.M at bottom=1.5 x 41.84=62.76 KNm Factored B.M. at top
= 1.5 x 92.44 =138.66 KNm
Bending Moment Diagram:
Shear Force: Max. Shear Force = wL/2 = 27.50 x 10.50/2= 144.38 KN Factored Shear Force =1.5 x 144.38 =216.57 KN Working Shear face at A =[ M A + MB]/ H=[41.84 +92.44] /4.85=27.69 KN Factored shear Force at A =V uA = 1.5 x 27.69=41.54 KN Design of Beam section : At Mid Span Section ( TEE BEAM): The Mid span section of this beam is designed as a T-beam. Working BM
= 286.54 KNm
Design Moment = 1.5 x 286.54 =429.81 KNm Flange Width bf = L0 /6 + bw + 6 Df = 0.7 x 10.50/6 +0.30 + 6 x 0.12=2.245m say 2.25m. OR Centre to centre of Beam=3.50m whichever is less. Take b f =2.25m D=750 mm d= 750-25-8-20/2=707mm Moment capacity of flange alone =Muf =0.36 f ck bf Df (d- 0.42 Df ) = 0.36 x 20 x 2250 x 120 x (707- 0.42 x 120) = 1276.43 x 106 Nm =1276.43 KNm
Mu i.e. 429.81 KNm Hence the NA lies within the flange. OR
b /b f w = 2.25/0.3=7.5
D /d = 0.12/0.707=0.17 f
Referring to Table 58 of SP16, the moment resistance factor is given by K T=0.589, Mulim=K T bwd2 f ck = 0.589 x 300 x 7072 x 20/1x106 = 1766.47 kN-m > Mu i.e 429.81 KNm . Hence Safe. The reinforcement is computed using Table 2 of SP16. Mu /b d2 = 429.81 x 106 /2250 x 7072 =0.38 Refering Table 2 of SP 16 p t=0.108 Ast= 0.108/100 x 2250 x 707= 1718 mm2 No of 20 mm dia bar = 1718/( ∏x202 /4) = 5.47 say 6 bars Hence (3 +3) Nos. of #20 at bottom and 3 Nos. #20 at top in the mid span
At Support ( Rectangular section) : Design moment MuB=1.5 x 92.44 = 138.66 kN-m MuB /bd2= 138.66 x106 /300x7072=0.92 From table 2 of SP16 p t=0.270 Ast=0.270 x 300 x 707 / 100 = 573 mm2 No of 20 mm dia bar = 573 /(∏ x202 /4) =1.82 say 3 Hence 3 Nos. of #20 at the top near the ends for a distance of o.25 L = 2 .50 m from face of the column as shown in Fig 6.6 Check for Shear: Factored SF =1.5 x 144.38= 216.57 KN Nominal shear stress Ʈv= 216.57 x 103 /300 x 707=1.02 N/mm2 < 2.80 N/mm2
For pt=100x 942/(300x707)=0.44 Permissible stress for pt=0.44 from table 19 Ʈc=0.4512 < Ʈv Shear carrying capacity of concrete Vuc =Ʈc b d = 0.4512 x 300 x 707/1000 = 95.70 KN Balance Shear to be taken by stirrups Vus=216.57- 95.70 = 120.97 kN Spacing 2 legged 8 mm dia stirrup s v= 0.87 f y A sv d/ Vus = 0.87 x 415 x 2 x50 x 707 /120.97 x 10 3 =211 mm Provide #8 Two legged stirrups at 200 mm c/c (equal to maximum spacing) Web Reinforcement: As the depth is equal to 750mm, web reinforcement is necessary. As per clause 25.5.1.3 of IS 456-2000 Total area of side reinforcement =0.1/100 x 300 x 750=225 mm 2 Provide 2 Nos #8 RTS on either side at mid depth (Ast=452 mm2 > 225 mm2)
Check for deflection: For pt = 100 x 6 x 314/2250 x 707 =0.12% Modification Factor from Fig 3 of IS 4562000 = α1= 1.90 For flanged section bw /bf = 300/2250=0.133 < 0.3
Modification factor α3 Required effective depth = L/26 x α 1 xα3 =10500 /26 x 1.9 x 0.8= 265.68mm <707 mm Ok.
Design of Column Section : (i) At Top Section: Axial Load at Top: Load from Main Beam
27.50 x 10.50/2
= 144.38 KN
Load from end beam 0.23 x (0.30-0.12) x 25x3.50
=
3.62 KN
Load due to parapet wall 0.23 x 0.90x 3.50x 19.2
=
13.91KN
Total Axial Load
=161.91 KN
Factored Axial Load
1.5 x 161.91=242.87 KN
Ultimate Bending moment at top
1.5 x 92.44 =138.66 KNm
Height of column
=4.85 m
Unsupported length of column =4.85 – 0.75/2 = 4.475 m Bending about X –axis bisecting depth of section: Emini= L/500 + D/30 = 4475/500 + 400/30=22.28 mm Mmini = Pu emini= 242.87 x 22.28/1000=5.41 KNm < 138.66 KNm Mu1=138.66 KNm Since the column is subjected to vertical load only and its top end is held in position hence Leff is assumed to 0.8 x unsupported length of column. Leff = 0.8 x 4.475 =3.58 m Leff /D= 3.58/0.40= 8.95 <12 Column is Short. Assuming 8 mm lateral ties and 20 mm Dia bars Effective cover =40+8+20/2 =58 mm d/D =58/500 =0.116 =0.12 Pu /f ck b D= 242.87 x 103 / 20 x 300 x 500= 0.08 Mu /f ck b D2 = 138.66 x 106 /20 x 300 x 500 2 = 0.09 say 0.10
Referring to charts 44 & 45
of SP 16
d/D 0.10 0.15 0.12
Pu /f ck b D 0.08 0.08 0.08
Mu /f ck b D2 0.10 0.10 0.10
p/f ck 0.058 0.062 0.060
p= 0.060 x 20 = 1.20 A st = 1.20 /100 x 300 x 500 = 1800 mm2 Mini. A st = 0.80/100 x 300 x 500= 1200 mm2 Provide (4 Nos. 20 mm Dia + 4 Nos. 16mm Dia) RTS Distributed equally on each face. Total A st Provided = 2060 mm2 > 1800 mm2 Bending about Y axis bisecting width of column: The Column is braced at lintel level by beam of size 230 mm x 300mm at height of 2.50m above plinth level. Unsupported length of top section of column, = 4.85 – (2.5 +0.6+0.3) =1.45 m Eminiy= L/500 +b/30 = 1450 /500 + 300/30 = 12.90 mm Mininy = Pu eminiy = 242.87 x 12.90/1000=3.13 KNm Ieff / b= 14500/300 =4.83 < 12.0
Column is short.
d/ b = 58/300 = 0.19 say 0.20 Pu /f ck b D= 242.87 x 103 / 20 x 300 x 500= 0.08 Mu /f ck b D2 = 3.13 x 106 /20 x 300 x 500 2 = 0.0021 Referring to chart 46 reinforcement.
of SP 16 p/f ck < Mini. Chart value. Hence adopt same
Column at Bottom Section: Assuming depth of Footing = 400mm Ultimate load at top of Footing: Load at top of Column
= 161.91 KN.
Self weight of column 0.3 x 0.5 x 4.85 x 25
= 18.19 KN
Outer Wall Load .23 x (4.85-0.75-0.3) x 19.2 x 3.20 = 54.41 KN Weight of Grade beam 0.3 x 0.6 x 10.50/2 x 25
= 23.63 KN
0.23 x 0.3 x 3.50 x 25
=
Total load on Column at Footing level
6.04 KN
= 264.18 KN
Ultimate Load on Column 1.5 x 264.18
= 396.27 KNm
Ultimate Bending Moment at Footing level 1.5 x 41.84
= 62.76 KNm
d/D =58/500 =0.116 =0.12 Pu /f ck b D= 396.27 x 103 / 20 x 300 x 500= 0.13 Mu /f ck b D2 = 62.76 x 106 /20 x 300 x 500 2 = 0.04 Referring to charts 44 & 45 d/D 0.10 0.15 0.12
of SP 16
Pu /f ck b D 0.13 0.13 0.13
Mu /f ck b D2 0.04 0.04 0.04
p/f ck < Chart value < Chart value < Chart value
Mini. A st = 0.80/100 x 300 x 500= 1200 mm2 Provide (4 Nos. 20 Dia +4 Nos. 16 mm Dia) RTS Distributed equally on each face. Total A st Provided = 2060 mm2 > 1200 mm2
Design of End beam (230 x 300 mm): Using one way continuous slab coefficients given in Table 9 of IS 456-2000, the bending moment for End beam may be calculated and designed similar to Main Beam.
Design of Footing: Size of the Footing: Axial Load on footing
=396.27/1.5 =264.18 KN
Self weight of footing 10%
= 26.42 KN
Total load
=290.60 KN
SBC of soil
=150 KN/m2
Working moment at top of footing
M
Eccentricity e =M/P = 41.84/290.60
=41.84 KNm = 0.14m
To avoid tension in the foundation breadth b =6 e =6 x 0.14=0.84 m =290.60/150=1.91m 2
Area of footing required
Provide 1.50 x 1.30m size of footing with equal projection on both sides. =264.18/1.50 x 1.30=135.48 KN/m 2
Actual soil pressure
< 150 KN/m2 (SBC) Factored soil pressure 1.5 x 135.48 =203.22 KN/m2 Bending moment & Steel:
Cantilever projection = 1500 -500 =500mm 2 Mux =Muy = 203.22 x 0.502 /2 x 1.50 =38.10 KNm
Depth required d= √ 38.10 x 106 /2.76 x 1500 =95.93 mm Provide D =350 mm & d=350-50-10-10/2=285 mm
K=Mu /b d2 = 38.10 x 106 /1500 x2852 =0.31 pt from Table 2 of SP 16 =0.0879 , 0.205 i.e Minimum steel Minimum A st= 0.205 x 1500 x 285 = 876.38 mm2 100 Provide 12 Nos. 10 mm RTS bothways. (Ast = 942 mm2 > 876.38mm2) Development length: Development length = 47 # = 47 x 10 =470mm Available anchorage = 500-50 (cover) =450mm < 470mm Hence bent the rod at the end upwards for footing depth. Check for one way shear:
Shear at d=285mm from face of the column = 0.215 x 1.3 x 203.22 =56.80KN 100 A s /b d =100 x 942/1500 x 285= 0.22 % t
Ʈc from Table of IS 456-2000 =0.342 N/mm2 Actual shear stress = 56.80 x 103 /1500 x 285 = 0.13 N/mm 2 < 0.342 N/mm2 Check for two way shear :
Factored shear at d= 285/2 from face of the column Vu ={ 1.50 x 1.3 – (0.785 x 0.585)} x 203.22 =302.96 KN d=285 mm
b= (500+285) + (300 + 285) =1370mm
Ʈv = Vu / Perimeter x d= 302.96 x 103 /1370 x 285 = 0.776 N /mm2 Design shear =k s Ʈc = K s=(0.5 +ẞc) =1.5 K s=1
k s should not greater than 1
Ʈc=0.25 √f ck = 0.25 √20 = 1.12 N/mm2
K s Ʈc = 1 x 1.12 =1.12 N/mm2
Ʈv < k s Ʈc Hence Ok. Check for cracking: Clear distance between bars = (1500 -100-10)/ (12-1) =126.36mm <180mm Ok. Check for bearing stress : Allowable bearing stress = 0.45 f ck √A1 /A 2
√A1 /A 2 should not greater than 2 Allowable bearing stress = 0.45 f ck x 2 =0.45 x 20 x 2= 18 N/mm 2 Actual Shear stress = Column load/Area of top of footing = 1.5 x 264.18 x 10 3 500 x 300 =2.64 N/mm2 < 18 N/mm2 Hence Ok. Also dowels are not required. Check for self weight of footing:
Weight of footing at base =1.5 x 1.3 x 0.35 x 25 =17.06 KN < 26.42KN Ok. Sketch :
Grade Beam Design : the Grade Beam may be Designed for wl 2 /30 as per IS 2911.
