MAGAZINE OF SCIENCE AND ART. 250; 
feet, area 6 “ X4,” and depth “6,” 
Example 1.—Required the weight a 
beam of iron bark 1 foot square could sus¬ 
tain in the middle, its length being 20 feet. 
The tabular value of S being 2’28fc, the 
depth 12 inches, the area 144 inches, 
the length 240 inches, consequently 
8288 X 4 x 44 H,s, and the beam 
240 . 65644 
may be loaded in practice with 4 — 
16411 lbs, without injuring its texture. 
From the equation S 4 where 
f=length, A 
o—breadth, >in inches, and IV, the break- 
d=depth, J 
ing weight in lbs, any three oi the lour 
quantities l, a, d, to, being given, the other 
may be determined. 
Example 2. Let it be required to deter¬ 
mine the size of a girder of iron bark for a 
warehouse where the distance between the 
points of support is 20 feet, . and the 
greatest possible stress in the middle, in¬ 
cluding the weight of the floor, is oO 
tons=67200. Let us assume that the 
greatest depth that we can obtain this 
timber is 20 inches, then in practice a 
4 lw _ 4 X 240 X 67200_— 1700 
must -— 2 4 X 2288 X 20 X 20 ' 
inches. « 
Thus, two girders of 20 inches depth and 
nine inches breadth resting on the same 
sustaining piece would he sufficient to 
support the weight of 30 tons. 
If the 30 tons are to be uniformly dis¬ 
tributed over the surface of the floor one 
girder of 20 “ X9” would be sufficient. 
Problem II. To determine the strength 
of a rectangular piece of timber fixed at 
one end and loaded at the other. Multiply 
the value of S by the area and depth of 
• the section in inches, and divide that pro¬ 
duct by the leverage or length without 
support in inches, the quotient will be 
the weight required. In this case the 
formula is— 
l to 
S=- 
ad 2 
In practice the load ought not to exceed 
one-fourth the weight found by this rule. 
If the weight be distributed uniformly 
over the piece the distance of the centie of 
, gravity from the point of support must be 
taken* for the length or leverage of the 
piece. 
Example 1. For iron bark, the breaking 
weight of a piece having a leverage of 5 
*> 
would be ^*1*^ = 5491.2 and 
5i£hr = i373 ip s the load it would hear 
4 
without injury. 
Example 2. A balcony to bear one ton 
is to be supported by two cantilevers, the 
projection of the balcony to be four feet. 
Assuming the cantilevers to be iron 
bark, that the weight will be uniformly 
distributed over the whole length, and that 
the depth of the cantilevers shall be four 
inches. 
The distance of the centre of gravity 
from the wall will be two feet and 
S=2288 as before, and it will be found by 
the formula that the breadth of each canti¬ 
lever must be 2.35 inches to bear the 
weight required without injury. 
The value of E in the table is essential 
for the determination of the dimensions of 
a beam capable of supporting a given 
weight, with as given degree of defection 
when fixed at one end, or when supported 
at both ends. 
When the beam is supported at one end 
the rule is as follows Divide the weight 
in lbs by and by the breadth and de¬ 
flection, both in inches, the cube root of 
the quotient, multiplied by the length in 
feet will be the depth required in inches. 
The formula being— 
l * w 
E=-— 
a d 3 S 
l, a, and d representing the same as before, 
W the greatest weight, S the deflection 
while elasticity remained perfect. Any 
one of the quantities, the others being 
given, may be found by the reversion of 
the formula. If the beam be loaded uni¬ 
formly throughout— 
l 3 to 
When the beam is supported at both 
ends multiply the weight to be supported 
in lbs. cube of the length in feet, divide 
this product by 16 times ^ multiplied 
into the given deflection in inches, and the 
quotient is the breadth multiplied by the 
cube of the depth in inches. The formula 
for this is— 
l 1 tv 
E=- 
16 a d * S 
« 
