GRAIN PRESSURES IN STORAGE BINS. 13 
From Table 3a: A wall 6 inches thick with a depth to steel of 5 
inches has a resisting moment of 32,300 inch-pounds, requiring 0.462 
square inch of steel per foot in height, which 1s slightly greater than 
actually required. 
2 
Using the formula — 
300 x 10 x 10 : 
we have, a= 70000 = 0:43 square inch. 
iven by 2’’ rounds 84’’ on center. 
8 2 
If it is desirable to have the bars spaced in even inches, we can 
increase the spacing to 9 inches and investigate to determine the 
corresponding stresses in steel and concrete. 
The area of 3’’ rounds 9’’ on center equals 0.41 square inch. 
p=~-2" = 0.0068 
for p=0.0068 and n=15, k=0.361, and 7=0.879. 
M+ 180000 
Dad 0410.879x5 10040 Pounds. 
2*Se 2xf.Xp_ 2X 16640 x 0.0068 
Jo= 0.361 
The stress in the ee is less than the allowable assumed and 
the stress in the steel is not excessive. 
To find the height at which the spacing may be increased to 10 
inches: The area of $’’ rounds 10’’ on center equals 0.37 square 
inches. 
= 630 pounds. 
Using the formula L =“ ae 
_ 0.37 X 70000 _ a 
we have L= 10X10 = 0.37 X 700 = 259 
LL 259 
D= 10 72>: .9, for which ee) (see Table 1). 
H, then, is equal to 20, and ata distance of 20 feet from the top we 
can increase the spacing of the bars to 10 inches. 
40 
The area of 2 rounds 12’’ on center equals 0.31 square inch. 
L=0. 31 X 700 = 217 
JE GAZ H 
Dorn oe 7, for which oo Be 
At a distance of 13 feet from the top we can increase the spacing to 
12 inches. 
a 
The area of 5 rounds 12’’ on center = 0.20 square inch. 
L=0.20 x 700 = 140. 
Te WA( HT 
ie 10 for which Daan 6. 
