][ STEEL STRUCTURES ][

					Beam Bearing Plates
					===============
Application:
beams on top of masonry or concrete walls & piers

Purpose:
ample bearing area
seat at proper elevation
level surface
uniform distribution of load

Area of Plate:
Based on allowable bearing pressure
A 	= R / Fp
A	= area of plate = B x N
		N = thickness of the wall
		B = length
			 for excessive lenght use shallow 
			 I-beam instead of flat plate
R 	= reaction of the beam
Fp 	= allowable bearing pressure of the
		supporting material


Thickness of Plate
Formula P&A 295

due to the limitations of the word processor
the following symbol
  ----------
V
stands for the square root 
which includes the nominator and denominator

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

Allowable bearing pressure Fp on masonry and concrete P&A 295

Solid brick
      unreinforced, type S mortar			170 psi

Hollow unit masonry
	unreinforced, type S mortar		225 psi

Concrete bearing
	on full area of support
	fc = 2,000 psi				500 psi
	fc = 3,000 psi				750 psi

Concrete bearing
	on 1/3 or less of support
	fc = 2,000 psi				750 psi
	fc = 3,000 psi		  	            1125 psi

Calculation Steps
Example 1 P&A 297
does not require
computation of bending moment
section modulus

PROBLEM:
=========
A W 21 x 57 member
of A36 Steel
transfers an end reaction of 44 kips (196 KN)
to a wall built of solid brick
by means of a bearing plate of A36 steel.
Assume S mortar and a brick with f'm=1500 psi
The N dimension of the plate is 10 in (254 mm)
Design the bearing plate

A. Given:
A W 21 x 57 member
of A36 Steel
transfers an end reaction of 44 kips (196 KN)
to a wall built of solid brick
by means of a bearing plate of A36 steel.
Assume S mortar and a brick with f'm=1500 psi
The N dimension of the plate is 10 in (254 mm)

B. Asked:
Design the bearing plate
N
B
t

C. Diagram:
=========
			| 44 kips
			|
                                          v
                                           I      W 21 x 57          A36  
		=============      	Solid Brick 
 					Type S Mortar
					f'm = 1500 psi

D. SOLUTION:
============

1. End Reaction to be supported
R = 44 kips = 44,000 p

2. Allowable bearing Pressure
see Table 9.3 page 295
given:
Solid brick
unreinforced
type S mortar
f'm = 1500 psi
Fp =    170 psi   from table 9.3    P&A 295 

3. Preselected Area of Plate
A = R / Fp
A = 44,000 p  /  170 p/sqin
A = 259 sq inch

4. Depth N of Plate:
N = 10 inch given
(usually limited by the thickness of the wall)  


5. Length B of Plate:
B = A / L
B = 259 sq in / 10 in
       Step 3       Step 4
B = 25.9 inch
which is rounded off to 26 inch (650 mm)

6. With the true dimensions of the plate
(10 in  x 26 in) we now compute the thickness
of the plate.

7. Actual Area of the Plate
A = N x B
A = 10 in x 26 in
A = 260 in

8. Actual Bearing Pressure:
fp = R / A
fp = 44,000 p /  260 in
fp = 169 psi    (1.19 Mpa)

9. Compare to allowable unit stress Bearing Fp
Fp = 170 psi     Table 9.3     P&A 295

10. Actual Stress to be smaller than allowable 
169 psi  <  170 psi
fp           <  Fp

11. Distance k1 for Wide Flange (W) Shape
Note: Table 4.1 P&A 170 does not show k1
ASD Manual of Steel Construction 1-20
for
W 21 x 57
k1 = 7/8
k1 = 0.875

12. Determine n:
--------------------
n = B/2 - k1
n = 26 / 2  - 0.875
      step 7    step 10
n = 13  - 0.875
n = 12.125 inch

13. Fb = Allowable Bending Stress in the plate
--------------------------------------------------------
Table 8.1 P&A 266-267
Fb = 0.75 x Fy
Fb = 0.75 x 36
Fb = 27 ksi

14. Thickness t of Plate:
-----------------------------
Formula P&A 295
due to the limitations of the word processor
the following symbol
  ----------
v
stands for the square root 
which includes the nominator and denominator

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

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

        ------------------------------------------------	 
t = V  3 x 169 psi  x 12.125 in x 12.125 in
         ----------------------------------------------
		27,000 psi

         -----------------------------------
t =  V  507 psi x 470.01562 sqin
         -----------------------------------
           27,000 psi

          -----------------------	
t =   V 74536.919
          ------------------------
           27,000


        ------------
t = V 2.7606


t = 1.66 in     
========

E. Answer:

15. The bearing plate design is:
-------------------------------------
N = 10 inches
B = 26 inches
t  =  1.66 inches = 1 2/3 inch