Handbook of Civil Engineering Calculations

FIGURE 50

Corresponding to the applicable limit states are Eqs. (Kl -3) and (Kl -5), each of which
has N 9 the length of bearing, as a parameter.
Solving for N 9 we obtain

Ru< = </>Rn = <l>(2.5k + N)Fttw

100 kips < 1.0(2.5 x Iy 8 in + TV) x 36 kips/in

2

x 0.40 in (1.01 cm) (Kl-3)

N > 3.5 in (8.89 cm)

J / N\/ ^wV-

5

I IFJf

*u^=^[i+3(f)(^) ]^ (Ki-V

#>8.6in(21.8cm)

The minimum length of bearing is N = 8.6 in (21.8 cm). Rounding up to the next full
inch, let N= 9 in (22.9 cm)

2. Compute the area of the bearing plate
   The area of the bearing plate is determined by the bearing strength of the concrete sup-
   port. Using the following equation, the design bearing strength is

iw^c'o.ss/^yj

where V'A 2 I^1 A 1 < 2.
Substituting in Eq. [77.6], we obtain

100 kips = 0.60 x 0.85 x 3 -^- x A 1 x 2

The area of the bearing plate A 1 = 32.7 in
2

. (210.9 cm
2
)
Because the bearing plate dimensions are

A 1 39 7 in^2

BNSzA 1 :

B

~Jf

=

Q = 3.6 in (9.14 cm)

Laterally supported by floor deck