634
MODERN GASWORKS PRACTICE
or (very nearly) =
24 L2 +D2
10,000 N ’
The area A given by this formula is the cross-section of each column necessary to prevent failure of the structure as a whole.
Bending moment on one bay only.
Consider the standard as a beam supported at both ends and uniformly loaded. Then if—
B = the pitch. of the columns in feet,
L = the total Icngth of each column,
The bending moment = —, where w = the load per foot run. Working on
the assumptions of Cripps, it is customary to consider the wind pressure equal to 10 Ib. per square foot, which is assumed as acting upon a length of standard equal to three-quarters of the depth. of the holder. In the above formula, therefore, we have—■
3
w — 10 Ib. per sq. foot, and Z = - L.
Bending moment (B6)
_ 10 X B X (IL)«
8 X 2240
tons-feet
= (approximately)
k g
t-’
O, LO
The bending moment (B&) on a single bay is, therefore, equal to—
BL2 . , x --- inch-tons.
270
where B = the width in feet of one bay, i.e. distance from column to column, and L = height of standard as before.
(This conclusion appertains only to those holders in which the guide-framing is
carried the whole way up.)
Then B& = / ax
z,
or Z (the section modulus) =
tt
Knowing Z, the required area of the section of the column can be foimd from the following approximate formula :—
Z = 2A?/.
Where A = the area of each flange of the column.
ij — half the depth of the cross-section. of the column.
Z
or, A =