Question

Given: A W14X109 column (A992 steel) requires a base plate (A36 steel) design. Assume the base plate will be on the full area of concrete support. The dead load compression in the column is 320 kips and the live load compression in the column is 600 kips. The concrete support under the base plate has a 28-day strength of fc 5 ksi. Required: Design the base plate with the goal of minimizing the required thickness.

Answer 1

AISC specifications and LRFD approach are used in this design.

Material Properties

For column (A992 steel),

F_y = 50 ksi

F_u = 60 ksi

For base plate (A36 steel),

F_y = 36 ksi

F_u = 58 ksi

Geometric Properties

Column W14x109

d = 14.32 in

b_f = 14.605 in

t_f = 0.86 in

t_w = 0.525 in

Load Description

Dead load = 320 kips

Live load = 600 kips

P_u = 1.2 x 320 + 1.6 x 600 = 1344 kips

Base Plate Dimensions

28-day strength of concrete support, f'_c = 5 ksi

$\phi$_c = 0.65

A1_req. = P_u / ($\phi$_c x 0.85 x f'_c) = 1344 / (0.65 x 0.85 x 5) = 486.52 in²

Lets use a square base plate of dimensions N = 23 in, B = 23 in

i.e, dimensions of base plate = 23 in x 23 in

Base plate area, A1 = 23 x 23 = 529 in² > 486.52 in² .... OK

Concrete Bearing Strength

$\phi$_c = 0.65

For full area of concrete support,

$\phi$_c P_p = $\phi$_c x 0.85 x f'_c x A1 = 0.65 x 0.85 x 5 x 529 = 1461.36 kips > 1344 kips .... OK

Required Base Plate Thickness

m = (N - 0.95 d) / 2 = (23 - 0.95 x 14.32) / 2 = 4.698 in

n = (B - 0.8 b_f) / 2 = (23 - 0.8 x 14.605) / 2 = 5.658 in

n' = (d b_f)^0.5 / 4 = (14.32 x 14.605)^0.5 / 4 = 3.615 in

X = ((4db_f)/(d+b_f)²) x (P_u / ($\phi$_c P_p) = 0.92

$\lambda$ = 2 X^0.5 / (1 + (1-X)^0.5) $\leq$ 1

= 2 x 0.92^0.5 / (1 + (1-0.92)^0.5)

= 1.495 > 1

Use $\lambda$ = 1

$\lambda$n' = 1 x 3.615 = 3.615 in

L = max (m, n, $\lambda$n') = 5.658 in

f_pu = P_u / BN = 1344 / (23 x 23) = 2.541 ksi

t_min = L x (2 f_pu / (0.9 F_y))^0.5 = 5.658 x ((2 x 2.541) / (0.9 x 36))^0.5 = 2.241 in

Use a 2.5 in thick base plate.