Strap Foundation

DESIGN OF STRAP BEAM

EXT COL

INT COL

1. 0 LOADS INPUT EXT COLUMN LOAD

INTERIOR COLUMN LO

Pw1 Mwy1

= =

190.96 KN 42.947 KN-m

Pw2 Mwy2

= =

2 8 7 .8 2 2 8 .6 8 3

Pu1 Muy1

= =

242.51 KN 59.542 KN-m

Pu2 Muy2

= =

3 7 1 .9 9 40

2.0 CONSIDER WEIGHT OF FOOTING

=

10

%

EXTERIOR COLUMN

INTERIOR COL

Service Column Load

=

190.96

Service Column Load

Weight of Footing

=

19.096

Weight of Footing

Total Concentrated Load

=

210.056

Total Concentrated Load

3.0 FOOTING AREA Total Combined Vertical Load Pw1 = 210.06 KN Pw2 = 3 1 6 .6 KN

Required Total Area of fo SBC = 100 Arqd

Pwt

=

526.66

=

5.27

KN

Length and Area of Exterior Footing

Length and Area of Interior

A1 L

= =

2.10 1.45

sq.m m

A2 L

= =

3 .1 7 1 .7 8

L A1

= =

1.60 2.56

m sq.m

L A2

=

1 .8 0 3 .2 4

Remark

COMBINED AREA

A1 + A2

=

5.8

Aact > Arqd, ok!!

sq.m

4.0 CENTROID OF FOOTING (WORKING LOAD) CG OF LOADS

xg

526.66

=

xg

=

P1x1 + P2x2 + Mwy1 + Mwy2 m from reference reference line

2.83

CG OF AREA

xg

5 .8

=

xg

=

2.56

m from reference reference line

=

-0.27

m

difference

A1x1 + A2x2

Ref line

Length of footing 1 =

1.6

Centr entroi oid d of col column umn Load Load X =

P1 =

Clear "d" bet. Col = 1.45

2.83 2.83

210.06

L1 /2 = 0.8

R1

X1 0.8

Pt = 526.658

c to c of column =

3.15

4.85

m

5.0 SOLVING FOR STRAP BEAM FORCES (ULTIMATE FORCES) EXTERIOR COLUMN

2 4 2 .5

KN

0.8

M (KN-m) (KN-m) = 59.5 59.5 V

V (Shear)

V (Shear) 0.8

R1 M1 V1

R1 =

210.91

= = =

210.91 3 4 .2 6 3 1 .6 0

KN

KN KN-m KN

M V

= =

34.26 31.60

KN-m KN

AD

KN KN-m KN KN-m

UMN

 

=

287.82

=

28.782

=

316.602

oting Kpa sq.m

 

Footing

sq.m m m sq.m

1.8

Length of footing 2 =

P2 =

316.60

L2 /2 = 0.9

R2

0.9

INTERIOR COLUMN

372.0

KN

0.9

M (KN-m) = 40.0

0.9

R2 M2 V2

R2 =

4 0 3 .5 9

= = =

4 0 3 .5 9 11.56 31.60

KN

KN KN-m KN