Base Plate Calculation

Seat-Fitting Design: Base Plate: Axial Load = 82.689 kips; Shear = 11.7598 kips(please refer to page 28) M = 254.64 in.kips.; e = 3.08 in.; (F’c = 4.0.0 ksi) For allowable Bearing capacity of Concrete, Fa=0.35F’c = 0.35(4) = 1.40 ksi Minimum base plate area required = 82.689/1.4 = 59.06 in.2 Assume 450mm x 450mm Base Plate, b

= 17.72 in. A = 313.88 in.2

I

= 8209.83 in.4; c = 8.86 in.

For Combined Flexural stress, f f

= - P/A ± Pec/I = -82.689/313.88 ± 82.689 x 3.08x 8.86 /8209.83 = -0.263 ± 0.275

f

= -0.538; 0.012 ksi< 1.40 ksi ∴ O.K.

∴ USE 450mm x 450mm Base Plate Moment @ 1 = 0 = (0.5x 0.364 x 5.62 x 5.62 x 0.333) + (0.5 x 0.538 x 5.62 x 5.62 x 0.667) = 7.58 kip in. ksi ; Fb =0.75Fy = 27.00 ksi

fb =

M/S

S =

M/fb =7.58/27 = 0.281 in.3

175

36.00

50

Fy =

8 00 175 450 17 5 5 01 7 5 1 7 5 0

=

1.298in. (32.97 mm)

∴

USE 450mm x 450mm x 35mm thick PLATE

175 175

H

800 450

175

Ø6.50"

50

= 6s/b = 6(0.281)/1 = 1.684 in.

2

3 .0 8 " 8 0 0 m m x 8 0 0 m m C o n c re te C o lu m n ( B Y O T H E R S ) V = 8 2 . 6 8 9 k ip s

550

H = 1 1 .7 6 k ip s

LO A D D IA G R A M 5 .6 2 "

S T R E S S D IA G R A M 0.538

0.364

0.012

H

2