Strap Foundation

DESIGN OF STRAP BEAM

EXT COL

1. 0 LOADS INPUT

Pw1 Mwy1 Pu1 Muy1

EXT COLUMN LOAD = 190.96 KN = 42.947 KN-m = =

IN Pw2 Mwy2

242.51 KN 59.542 KN-m

2.0 CONSIDER WEIGHT OF FOOTING

Pu2 Muy2

=

10

%

EXTERIOR COLUMN Serv!e C"#um$ L"&

=

190.96

Serv!e C

'e()* "+ ,""*$(

=

19.096

'e()* "+

=

210.056

 T"*#  T"*# C"$

 T"*#  T"*# C"$!e$*r*e& L"&

3.0 FOOTING AREA  T"*#  T"*# C"m$e& er*!# er*!# L"& Pw1 = 210.056 KN Pw2 = 316.602 KN

/e SC r&

Pw*

=

526.66

KN

Leng! "n# A$e" %& E'e$(%$ F%%(ng 1 = 2.10 .m L = 1.45 m L 1

= =

1.60 2.56

m .m

Leng! 2 L L 2

/emr

COMBINED AREA 1  2

=

5.)

.m

!*  r&8

*.0 CENTROID OF FOOTING +WOR,ING LOAD-

:(

CG OF LOADS = P1:1  P2:2  Mwy1  Mwy2

526.65;

:(

:(

=

5.;

CG OF AREA = 1:1  2:2

:(

=

2.56

m +r"m re+ere$!e #$

=

-0.27

m

&
2.)3

m +r"m re+ere$!e #$

/e+ #$e

Leng! %& &%%(ng 1

1.6

Ce"$ /#/

Ce$*r"& "+ !"#um$ L"& X =

P1 =

2.;3

210.06

L1 2 = 0.)

R1

X1 0.)

P* =

 %  %& %4n 

5.0 SOLING FOR STRAP BEAM FORCES +ULTIMATE FORCESEXTERIOR COLUMN 242.5

KN

0.;

M >KN-m? = 59.5 

 >S)er?

M 

= =

0.;

/1 M1 1

/1 =

210.91

= = =

210.91 34.26 31.60

KN

KN KN-m KN

@NT COL

TERIOR COLUMN LOAD = 2;7.;2 KN = 2;.6;3 KN-m = =

371.99 40

KN KN-m

INTERIOR COLUMN #um$ L"&

=

2;7.;2

,""*$(

=

2;.7;2

e$*r*e& L"&

=

316.602

ure& T"*# re "+ +""*$( = 100 KA =

5.27

.m

 "n# A$e" %& Ine$(%$ F%%(ng = 3.17 .m = 1.7; m =

1.;0 3.24

m .m



e

e

e. C%  1.*5

Leng! %& &%%(ng 2 

1.)

P2  316.60

L2 2 = 0.

526.65) R2

3.15

4

0.

*.)5

INTERIOR COLUMN

372.0

KN

0.9

M >KN-m? = 40.0

 >S)er? 0.9

/2 M2 2

34.26 31.60

KN-m KN

/2 =

403.59

= = =

403.59 11.56 31.60

KN

KN KN-m KN