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ARCH5023/ARCH4333 - ARCHITECTURAL STRUCTURES I

][ STEEL STRUCTURES ][
THE UNIVERSITY OF OKLAHOMA - COLLEGE OF ARCHITECTURE
____________________________________________________

STEEL COLUMN

BASE PLATES

Purpose:
=======
		connect column to its support
		ample bearing area
		seat at proper elevation
		level surface
		uniform distribution of load

Application:
=========
		transfer bearing to concrete footing
		steel plate welded to column
		light load - small plates
		high-strength steel - heavy plates

Plate Design:
=========
Design according to AISC Manual
and AISC Code
Based on allowable bearing pressure
(typically allowable bearing stress on concrete)


Determination of Plate:
=================
A 1 	= P / Fp
A 1	= Plan Area of bearing plate 
		= B x N
		N = thickness of the wall
		B = length
			 for excessive length use shallow 
			 I-beam instead of flat plate
P 	= compression load of the column
Fp 	= allowable bearing stress on the
	   supporting material typically concrete 
fc	= allowable bearing stress on the concrete
	   f'c  = 0.3 fc 
		when plate covers entire support
	   for larger areas stress may be increased
	   by a factor
	    	   ----------	 
		V A2/A1
	 maximum 2

due to the limitations of the word processor
the following symbol
  
   ----------

v

stands for the square root 
which includes the nominator and denominator

  


Thickness of Plate Formula:
====================
P&A 336

          -------------------	 
t = V  3 fp x n x n
           ----------------
    	Fb


t	= thickness of bearing plate (inches)

fp 	= actual bearing pressure  P/A
Fp 	= allowable bending stress in the plate
		0.75 Fy    (PA 267) 
m 	= edge distance x
n	= edged distance y

COLUMN BASE PLATE EXAMPLE PROBLEM

Column Base Plate:
==============
Example Problem:
Design a base plate
of A36 steel
for a W 12 x 58 column
with a load of 250 kips.
The column bears on a large
3,000 psi reinforced concrete footing
with fc' = 3 ksi

A. GIVEN

base plate of A36 steel
for a W 12 x 58 column
with a load of 250 kips.
The column bears on a large
3,000 psi reinforced concrete footing
with fc' = 3 ksi

B. ASKED

Design the column base plate N = length B = width t = thickness m = edge distance x m = edge distance y

C. GRAPH

D. SOLUTION

1. We assume the footing are is considerably
	larger than the plate area
	thus
	Fp = 0.3 fc' for plate = concrete
	2 = maximum factor to increase
	Fp = 0.3 fc' x 2 
	Fp = 0.6 fc' 
     	Fp = 0.6 x 3 ksi
	Fp = 1.8 ksi
	 
2. Area A1
	P = 250 kips
	Fp = 1.8 ksi
	A 1 = P / Fp	
	A 1 = 250 kips / 1.8 ksi
	A 1 = 138.9 sq in

3. We assume the plate is square
	B = N
	m = n
	         _________	
	B = V A 1
	          __________	
	B = V 138.9 sq in
	B = 11.8 in

4. Determine n:
	according to layout rules
	given W 12 x 58
	Table 4.1 PA 171
	d = depth 	=	12.19 	inch
	b = width	=	10.014 	inch

5.	0.8 b
	B - 2n = 0.8 b
	B - 2n = 0.8 x 10.014 inch 
	B - 2n = 8.01 inch

6.  	edged distance both sides
	2n 	=  B - 0.8 b 	
	2n = 12 - 8.01
	 	= 3.99 inch
		= 4 inch

7. 	edge distance side n
	n 	= 	2n /2
	n 	=	4 inch / 2
	n	=	2 inch

8. 	d	= 12.19"
	N - 2m = 0.95 x d 
	N - 2m = 0.95 x 12.19	
	N - 2m = 11.58"

9. 	m assumed based on 
		welding needs
		anchorbolt layout
	m = 1.25 inch
	m = 1.21 inch actual

10. N = 0.95 d + 2m
	N = 11.58 + 2 x 1.21
	N = 11.58 + 2.42 inch
	N = 14 inch  

11.
A 1	= Plan Area of bearing plate 
		= B x N
		N = width 
		B = length
P 	= compression load of the column
fp 	= actual bearing pressure

	fp = P / A1
	fp = 250 / 12 x 14
	fp = 250 kips / 168 inch
	fp = 1.49 ksi

12. 
	Fp = allowable bending stress in the plate
		0.75 Fy    (PA 267) 
	Fp = 0.75 x 36
	Fp = 27 ksi
	
13.

        -----------------	 
t = V  3 fp x n x n
         ---------------
	Fb

as it is a square column m = n 

        -------------------	 
t = V  3 fp x m x m
         -----------------
	Fb

t	= thickness of bearing plate (inches)
fp 	= actual bearing pressure  P/A
Fp 	= allowable bending stress in the plate
		0.75 Fy    (PA 267) 
n 	= m = edge distance x = y

        -------------------	 
t = V  3 fp x n x n
         ---------------
	Fb

        -----------------------	 
t = V  3 x 1.49 x 2 x 2
         ---------------------
	0.75 x 36

        -----------------------	 
t = V  3 x 1.49 x 4
         ---------------------
	27


        ---------------	 
t = V  4.47 x 4
         ------------
	27

        -----------	 
t = V  17.88
         ---------
	27

        -----------	 
t = V  0.662

t = 0.814 inch
plates are usually specified in thickness
increments of 1/8 inch
so the minimum thickness would be 7/8 inch.
0.875 inch