As 2870-2011 Residential Slabs & FootingsFull description
Naval EngineeringFull description
Descrição completa
Design of Machine Elements. Virgil Moring Faires. 0023359501 "A new machine is born because there is a real or imagined need for it. It evolves from someone's conception of a device wit…Full description
Design of Machine Elements. Virgil Moring Faires. 0023359501 "A new machine is born because there is a real or imagined need for it. It evolves from someone's conception of a device with which...
Design of Trickling FiltersFull description
Design of Machine Elements. Virgil Moring Faires. 0023359501 "A new machine is born because there is a real or imagined need for it. It evolves from someone's conception of a device wit…Descrição completa
Design of ColumnFull description
Full description
Full description
Design of Glass BalustradeFull description
Design of Absorber
SCAD - Foundation Studies
procedure for designing steel structures
Descripción completa
DESIGN OF FOOTINGS Size of column= 500x500mm Pu= 363.60 T P= 242.40 T Mu= 3.42 T-m Hu= 1.85 T Assume depth of the footing as 500mm Total M= (3.42/1.5) + (1.85 x 0.5/1.5) M= 3.23 T-m The SBC of the soil is given as 30 T/m2. Check for upward bearing pressure
/ = ± ()--------------------------- (1) Let us assume size of the footing as 2.9x2.9m 2. 9x2.9m A=2.9x2.9=8.41m2 Z= bd2/6= (2.9x2.92/6)= 4.06m3 Pmax= (242.40/8.41) + (3.23/4.06) Pmax= 29.61 T/m2 Similarly
We can assume pedestal of size 900x900mm and 500mm depth. Considering bending moment at the face of the pedestal we get Mt= (29.06x12/2) + ((0.55x1/2) x (1x2/3)) Mt= 14.71 T-m Mu=1.5x14.71= 22.07 T-m (Mu/bd2)= (22.07x107/2900x4402) = 0.393 Referring SP-16 handbook we get
pt = 0.093 % min Ast= 0.15%bD = (0.15x2900x500/100) = 2175 mm2 I) Check for one way shear Area of footing at critical section (Ac) = 2900x440 =1.276x106mm2 From IS-456 2000 table 19
τc= 0.29 N/mm2 Vc= Ac x τc Vc = 1.276x106 x 0.29 Vc = 37 T Shear subjected Vu= Wu x b x (1-d) = 29.66x 2.9 x (1-0.44) Vu =47.64 T > Vc Let us increase pt to 0.27%
τc= 0.38 N/mm2 Vc = 1.276x106 x 0.38 Vc = 48.48 T > Vu Hence safe. II) Check for two way shear We take perimeter at (d/2) from face of the pedestal P= (900+440) x 4= 5360mm Area of footing at critical section (Ac) = 5360x440 =2.358x106mm2 Vc= Ac x τc2 = 2.358x106 x 1.5x √30 x0.25
(where τc2= 1.5x 0.25 x √f ck )
= 481.1 T Shear subjected Vu= Wu x {L x B - (D+d) (b+d)} = 29.33 x {2.92 – (0.9+0.44)2}
= 242.06 T < Vc Hence OK
Therefore pt =0.27% Ast = 3445 mm2 Provide 19#16mm dia bars in both directions. Top r/f or temperature r/f = 50% of min. A st = 0.5 x 0.15x2900x500/100 =1087.5 mm2 Provide 12mm dia @ 250mm c/c in both directions.
DESIGN OF BEAMS MAINBEAM
Secondary Beam 1 Dimensions of beam: 300x800 mm Span: 9.7m Loads Dead Load = 0.3x0.8x2.4=0.576 T/m Live Load = 1.5 T/m2 =1.5x2.5 =3.75 T/m Floor finishes = 0.2 T/m2 = 0.2x2.5 =0.5 T/m Moments due DL and LL (Mt) = (w x l 2/8) = (0.576x9.32/8) + (3.75x9.32/8) + (0.5x9.32/8) Mt = 52.17 T-m Moment obtained from Etabs 2013: Mt =56.71 T-m Ast= (0.5 fck / f y) [1-
Mu/ f ck b d²)]
= (0.5 30/500) [1-
bd
52.17 x 1.5 10^7/ 30 300 740²)]
300 740
=3124mm2 Check for shear Shear due DL and LL (Vc) = (w x l/2) = (0.576x9.3/2) + (3.75x9.3/2) + (0.5x9.3/2) Vc = 22.44 T Vu = 33.66 T
τv=(33.66x104/300x740) = 1.51N/mm2 pt= 1.41%
τv=0.742 N/mm2 Using 10mm two legged stirrups Asv=157mm2
Vus = 33.66 x104 – (0.742x300x740) Vus = 17.18 T Sv = (0.85x500x157x740/17.18x104) = 287.40mm Provide two legged 10mm stirrups @ 250mm c/c.
(from IS-456 2000 table 19)
Secondary Beam 2 Dimensions of beam: 300x800 mm Span: 10m Loads Dead Load = 0.3x0.8x2.4=0.576 T/m Live Load = 1.5 T/m2 =1.5x2.5 =3.75 T/m Floor finishes = 0.2 T/m2 = 0.2x2.5 =0.5 T/m Moments due DL and LL (Mt) = (w x l 2/8) = (0.576x9.62/8) + (3.75x9.62/8) + (0.5x9.62/8) Mt = 55.59 T-m Moment obtained from Etabs 2013: Mt = 60.52 T-m
Ast= (0.5 fck / f y) [1-
Mu/ f ck b d²)]
= (0.5 30/500) [1-
bd
60.52x1.5 10^7/ 30 300 740²)]
300 740
=3314mm2 Check for shear Shear due DL and LL (Vc) = (w x l/2) = (0.576x9.3/2) + (3.75x9.3/2) + (0.5x9.3/2) Vc = 22.44 T Vu = 33.66 T
τv=(33.66x104/300x740) = 1.51N/mm2 pt= 1.41%
τv=0.742 N/mm2 Using 10mm two legged stirrups Asv=157mm2 Vus = 33.66 x104 – (0.742x300x740) Vus = 17.18 T Sv = (0.85x500x157x740/17.18x104) = 287.40mm Provide two legged 10mm stirrups @ 250mm c/c.
(from IS-456 2000 table 19)
DESIGN OF SLAB This will be a continuous one way slab, therefore IS-456 2000 Clause 22.5.1 Loads considered DL= 0.375 T/m2 LL= 1.50 T/m2 FF= 0.20 T/m2 Effective Span (l) = 2.5-0.18=2.32m End Span Maximum +ve BM= (0.575x2.322/12) + (1.50x2.322/10) = 1.07 T-m (Mu/bd2)= (1.07x107x1.5/1000x1152) = 1.213 Referring SP-16 handbook we get pt = 0.303 % Ast= 348.45mm2 min Ast= 0.15%bD = (0.15x1000x150/100) = 225 mm2 Provide 10mm @ 150mm c/c. 1st Internal support Maximum -ve BM= (0.575x2.322/10) + (1.50x2.322/9) = 1.21 T-m (Mu/bd2)= (1.21x107x1.5/1000x1152) = 1.37 Referring SP-16 handbook we get pt = 0.322 % Ast= 370.30mm2 Provide 10mm @ 150mm c/c.
Check for shear At end span Max shear = 0.575x 0.4x2.32 + 1.5x0.45x2.32 = 2.1 T At support next to end support Outer side = 0.6 x0.575x2.32 + 0.6 x 1.5 x2.32 = 2.9 T Inner side = 0.55 x0.575x2.32 + 0.6 x 1.5 x2.32 = 2.82 T At all other interior supports = 0.50 x0.575x2.32 + 0.6 x 1.5 x2.32 = 2.8 T Check for shear stress
τc= 0.37 + (0.5-0.37/0.5-0.25) x (0.334-0.25) τc= 0.413 N/mm2 τcmax= 3.5 N/mm2 For slabs k = 1.3
Therefore τ c=1.3x0.413 = 0.537 N/mm2 Since τv < τc < τcmax Hence OK
DESIGN OF TRUSS Calculation of wind pressure In accordance to IS 875-part 3 Basic Wind Pressure = 39 m/s Design Wind speed (Vz) = V b x k 1xk 2xk 3 Class B, Terrain Category B At height h=10m, k 2= 0.88 At height h=10m, k 3= 1.00 Vz=39x1x0.88x1 Vz= 34.32m/s Say Vz=35m/s Case (1) -----WL1 F= (C pe - C pi) Ax Pz C pe-External wind pressure coefficient C pi- Internal wind pressure coefficient A1= 1.26x1.5 = 6.3m2 A2= 1.80x5.0= 9.0m2 Case (2) -----WL2
Loads considered for design in SAP2000 Services = 15kg/m2 False ceiling = 25kg/m2 Sheets =12.5kg/m2 Dead load = 7.5kg/m2 Total Dead load= 60kg/m 2 = 60x1.25= 75kg/m Live Load = 50kg/m 2 = 50x1.25= 62.5kg/m
Wind load as above Load combinations
1.5(DL+LL) 1.5(DL+LL+WL1) 1.5(DL+LL+WL2) (DL+0.5LL) All the forces in the members are obtained from SAP2000 TYPE 1 TOP CHORD Force =29 T (compression) Let us assume a stress of 1.5T/cm2 Area required= (29/1.5) = 19.33cm2 Let us assume 91.5x91.5x4.5mm [SHS] Area provided = 17.85cm2 leff = 126x0.9 = 113.4cm r = 3.48cm (leff /r) = (113.4/3.48) =32.58 Referring the values of compressive stresses we get Allowable stress = 175.67MPa Capacity of the member (compression) = 1.757x 17.85 = 31.36 T (OK) For reversal of stresses = 3.3 T BOTTOM CHORD MEMBERS Force =17 T (tension) Allowable stress = 1.86T/cm2 Area required= (17/1.86) = 9.31cm2 Let us assume 72x72x4mm [SHS]
Area provided = 10.47cm2 Capacity of the member (tension) = 1.86x 10.47 = 19.47 T (OK) leff = 126x0.9 = 113.4cm r = 2.75cm (leff /r) = (113.4/2.75) =41.23 Referring the values of compressive stresses we get Allowable stress = 1.69T/cm2 Capacity of the member (compression) = 1.69x 10.47 = 17.69 T (OK)
