Page 181 - Handbook of Civil Engineering Calculations, Second Edition
P. 181

1.164           STRUCTURAL STEEL ENGINEERING AND DESIGN



















                               FIGURE 50


                              Corresponding to the applicable limit states are Eqs. (K1-3) and (K1-5), each of which
                            has N, the length of bearing, as a parameter.
                              Solving for N, we obtain

                                                R u     R n    (2.5k + N)F t t w
                                  100 kips   1.0(2.5   1 /8 in. + N)   36 kips/sq.in.   0.40 in. (1.01 cm) (K1-3)
                                                   3
                                                    N   3.5 in. (8.89 cm)

                                                                     1.5
                                                               N
                                                                  t w
                                           R u    R n    68t w 1 + 3                   (K1-5)
                                                        2
                                                                          F y t f
                                                               d
                                                                   t f
                                                                           t w
                                           N   8.6 in. (21.8 cm)
                              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

                                                                    A 2
                                                   c P p     c   0.85f c 
A 1
                                                                    A 1

                            where  A  2 /A  1     2.
                              Substituting in Eq. [11.6], we obtain
                                                                 kips
                                           100 kips   0.60   0.85   3       A 1   2
                                                                 sq.in.
                                                                    2
                            The area of the bearing plate A 1   32.7 sq.in. (210.9 cm ).
                              Because the bearing plate dimensions are
                                                     A 1  32.7 sq.in.
                                       BN   A 1 :        B             3.6 in. (9.14 cm)
                                                      N      9 in.
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