Base Plate+wind Load.xlsx

Load Calculation for Truss Wind load F= (Cpe-Cpi)xAxPd Where Cpe = External pressure coefficient, Cpi = Internal Pressure coefficient, A= Surface area of structural element or cladding unit, pd= Design wind pressure = 0.6Vz2 Where, Vz= Design wind velocity (m/s) at any height z above ground = Vbxk1xk2xk3 Basic Wind Speed ( Vb ) Vb

47 m/s

Risk Cofficient ( k1 Factor ) Class of structure 1 = General Building 3 = Isolated tower, farm buildings

4 2= Temporary sheds , formwork , etc. 4= Important like hospitals, communication Tower, power plant

k1= 1.07 Terrrain, Height, Structure Size Factor ( k2 factor ) 2 B 4.3 m 0.98

Terrain Category Structure Class Eaves Height k2

Design Wind Speed Basic wind speed (Vb) 47 m/sec k1 1.07 k2 0.98 k3 1 Design Wind Speed( Vz )=Vb* k1 * k2 * k3 = 49.2842 m/sec Pressure Coefficient Design Wind Pressure ( pz ) = 0.6 Vz2 =

1457.35942 N/m2

Pressure coefficient Slope of Truss Eaves Height ( h ) Width of the Truss ( w ) h/w Wind Pressure Calculation A) Wind Normal To The Ridge ( 0o )

18 4.3 m 13.4 m 0.32

Direction

Cpe

Wind ward(EF)

-0.56

Lee ward(GH)

-0.4

Cpi 0.2 -0.2 0.2 -0.2

Pressure (KN/m2) -524.6493918422 -1107.5931605558 -291.4718843568 -874.4156530704

B) Wind Parallel To The Ridge ( 90o) Direction

Cpe

Wind ward(EF)

-0.72

Lee ward(GH)

-0.6

Hence

Wind Pressure

Cpi 0.2 -0.2 0.2 -0.2

Pressure (KN/m2) -757.8268993277 -1340.7706680413 -582.9437687136 -1165.8875374272

-1.3407707 KN/m2 -0.2914719 KN/m2

LOADING PARAMETERS Live Load Reduced Live Load Dead Load LOADING ON PURLINS Spacing between purlins Dead load Live Load Wind Load

0.75 0.59 0.2

KN/m2 KN/m2 KN/m2

0.88 m 0.176 KN/m 0.5192 KN/m -1.1798782 KN/m -0.2564953 KN/m

Upwind Upwind

mwork , etc. ls, communication Tower,

Pressure (KN/m2) -524.6493918422 -1107.5931605558 -291.4718843568 -874.4156530704

Pressure (KN/m2) -757.8268993277 -1340.7706680413 -582.9437687136 -1165.8875374272

roof ang alpha wind angle theta 20 -0.4 18 -0.48 30 0

Height

k2 0 4.3 10

0.98 0.98 0.98

roof ang a wind angle theta 10 -0.8 18 -0.72 20 -0.7

Design of Bolted Gusseted Base with Axial Load and Moment Grade of Concrete Bearing Strength of concrete = 0.45* f ck

M 25

Grade pf Steel fy fu fub Member from which the load is tranferred Thickness of flange ( tf )

Fe

ISNB

11.25 N/mm2 250 250 N/mm2 410 N/mm2 400 N/mm2 165.1 4.8 mm

Partial Safety Factor for Materials

Resistance,governed by yeilding( gm0 )

1.1

Resistance of member to buckling ( gm0 )

1.1

Resistance, governed by ultimate stress ( gm0 )

1.25

Resistance of connection:

Shop Fabrication

Field Fabrication

i) Bolts-Friction Type ( gmf )

1.25

1.25

ii) Bolts-Bearing Type ( gmb )

1.25

1.25

iii) Rivets ( gmr )

1.25

1.25

iv) Welds ( gmw )

1.25

1.5

Design Compressive Load (P) Design Bending Moment (M) Eccentricty, ( e ) =

95 KN 0 KN-m M P

Provide Two Cleat Angles of specification Height = 0 mm Length = 0 mm Thickness = 0 mm Area of Base Plate Required = P 0.45*fck Length of base plate Provide Length of Base Plate ( L ) =

0 mm

ISA 100x65x8 mm

8444.444 mm2

91.89366 mm 400 mm

Now, e L Now , We know

when

0 (e/L) < (1/6)

(e/L) < (1/6) P(1+((6*e)/L)) LB

Pressure distribution =

This pressure distribution should not be greater than bearing strength of concrete Equating Bearing Strength with the pressure distribution formula We get, Breadth of Base Plate = Provide Base Plate breadth ( B ) = Provided size of Base Plate (L x B )= Area of base plate ( A ) = Elastic Section Modulous (Ze)

400

x

21.11111 mm 400 mm 400 mm 160000 mm2 10666667 mm3

Projection from end of angle to end of base plate ( Lp)

117.45 mm

Now, Maximum Pressure = (P/A) + (M/Ze)

0.59375 < 11.25N/mm2 OK

Minimum Pressure = (P/A) - (M/Ze)

0.59375 N/mm2

Critical Distance where the moment is acting ( c )

117.45 mm

Pressure at the critical section ( Px)

0.59375 N/mm2 bb*Zp*fy

Moment at critical section ( Mc ) =

gm0 where,

bb = Zp =

1 b*t2 6

Moment carrying capacity ( Md ) =

1.2*fy*Ze gm0

where, Ze =

b*t2 6

Equating Mc and Md,

=

4095.243

N-mm

Required thickness of base plate Provide Thickness of Base Plate ( tb ) Dimension of Base Plate Number of plates L= 400 mm B= 400 mm tb = 10 mm

9.491857 mm 10 >tf= 4.8 mm OK

2

BOLTED CONNECTION ( Check For Shear Force ) Shear Force ( F ) Assume Dia of Bolts ( d ) = Dia of Holes ( d0 ) =

375 KN 24 mm 26 mm

Anb*(fub/√3)

Design Shear Strength Bolt (Vdsb) =

65.1923 KN

gmb

Minimum Edge Distance = 1.5*dia of hole (d 0) Provide Edge distance ( e ) =

39 mm 40.8 mm

Minimum Pitch = 2.5 * dia of bolt ( d ) Provide pitch ( p ) =

60 mm 100 mm

Kb = min of (

e 3*do

or

p 3*do

-0.25

Kb =

or

fub

or 1 )

fu 0.523077

Now, Design Bearing Strength of Bolt ( Vdpb ) =

2.5 *Kb * d * t * fu gmb

102.9415 KN

Strength of Bolt = min ( Vdsb or Vdpb ) 65.1923 KN Assuming complete bearing, so 50% of axial load is transferred to base plate and another 50% will be transferred through angle connection Number of Bolts ( n ) =

P Bolt value

Provide number of bolts ( n ) Providing configuration such that there are Two bolts in 1 row.

1.457227 nos.

4 nos

ANCHOR BOLT FOR UPLIFT ( From N. Subramaniam ) Uplift ( Nu )

180 KN

Nu where k= k heff = heff heff

k * sqrt(fck) * (heff)^1.5

=

ACI 318-2005

13.5 for post installed anchors and 15.5 for cast insitu headed studs and headed anchor bolts 15.5 embedment length = =

1.753801895 m 1753.801895 mm

This length is required if 1 bolt is provided We have provided So, heff for So, provide

4

4

nos

nos of bolt 4

nos.

440 mm 24

mm dia

450 mm

long anchor bolts

nd headed anchor bolts

long anchor bolts