= 5.96” Step 4 – Determine minimum base plate thickness, tmin: t min = L ΦcPp = Design bearing strength of concrete = 0.6Pp = 0.6(0.85f’cA1) Re-arranging to solve for A1: A1 =Ī1 = 457.5 in2 Step 2 – Determine “Optimized” base plate dimensions: Step 1 – Determine required base plate area, A1 to avoid conc. It bears on a steel base plate using A36 steel. 14-6: t min = LĮxample (LRFD) GIVEN: A W14x82 A992 column has a factored axial load Pu = 700 KIPS. Tmin = minimum base plate thickness per AISC p. Where Ωc = 2.50 ASD Ωc where: f’c = specified compressive strength of concrete, KSI A1 = area of steel base plate concentrically loaded on conc, in2 = BN (where B and N use whole inches if possible) B bf The design of a base plate involves the following steps: Pp = Nominal bearing strength of concrete = 0.85f’cA1 Design Bearing strength of concrete: φcPp where φc = 0.60 LRFD Pp The design of steel base plates is based on the following: ĪISC Spec. 14-9 Steel base plate Base Plate thickness An intermediary steel base plate is used to distribute this column load without crushing the concrete.
Lecture 10 – Column Base Plates Columns must transmit vertical loads to the concrete footing.
