W= 186000 Iδ/lmn= 186000 x 292.0.5 x 1.5/25x10x15=21728.52;

or, the required weight is, say, 21,730 pounds.

173. - Rolled-Iron Beams: Dimensions; Weight at any Point. - By transposition of factors in equation (72.), we obtain -

I=Wlmn/186000 δ (73.)

This may be expressed in words as follows:

Rule LII. - Multiply the weight by the length, and by the rectangle of the two parts into which the point where the weight rests divides the length; divide the product by 186000 times the deflection, and the quotient will be the value of I, which (or its next nearest number) may be found in Table IV., opposite to which will be found the required beam.

Example. - What beam 10 feet long will be required to carry 5000 pounds at 3 feet from one end with a deflection of 0.4 inch? Here we have m equal 3, and n equal 7; therefore -

I=Wlmn/186000δ = 5000x10x3x7/186000x0.4=14.113.

The value of I is 14.113, the nearest number to which in the table, is 14.317, the moment of inertia of the Phoenix 5-inch, 36-pound beam; this, therefore, is the beam required.

174. - Rolled-Iron Beams: Dimensions; Weight Uniformly Distributed. - Since 5/8 U - W(Art. 138), equation (69.) may be modified by the substitution of this value of W, when we obtain -

I=5/8 Ul3/744000 δ' which reduces to -

I = Ul3/1190400 δ (74.)

a rule for the dimensions of a beam for a uniformly distributed load, which, in words, is as follows:

Rule LIII. - Multiply the uniformly distributed load by the cube of the length; divide the product by 1190400 times the deflection, and the quotient will be the value of I, corresponding to which, or to its next nearest number will be found in Table IV. the required beam.

Example. - What beam 10 feet long is required to sustain an equally distributed load of 14,000 pounds with a deflection of half an inch? For this we have -

I=14000 x 103/1190400x0.5= 0.5

This is the moment of inertia of the required beam; nearly, the same as 23.761, in Table IV., the value of I for a Trenton 6-inch, 40-pound beam, which will serve as the required beam.

175. - Rolled-Iron Beams: Deflection; Weight Uniformly Distributed. - A transposition of the factors in equation (74.)

gives -

δ=Ul3/1190400 I' (75.)

a rule for the deflection of a uniformly loaded beam, and which may be put in these words, namely:

Rule LIV. - Multiply the uniformly distributed load by the cube of the length; divide the product by 1190400 times the value of I, Table IV., and the quotient will be the required deflection.

Example. - To what depth will 14,000 pounds, uniformly distributed, deflect a Buffalo 10 1/2-inch, 90-pound beam 20 feet long? The value of I for this beam, as per the table, is 151.436; therefore -

δ =14000x203/1190400x151.436 = 0.6213;

or, the required deflection is, say, 5/8 of an inch.

176. - Rolled -Iron Beams: Weight when Uniformly

Distributed. - Equation (75.), by a transposition of factors, gives -

U=1190400 Iδ/13 (76.)

a rule for the weight uniformly distributed, and which may be worded thus:

Rule LV. - Multiply 1190400 times the value of I, Table IV., by the deflection; divide the product by the cube of the length, and the quotient will be the required weight.

Example. - What weight uniformly distributed upon a Buffalo 10 1/2-inch, 105-pound beam 25 feet long between bearings will deflect it 3/4 of an inch?

The value of I for this beam, as per Table IV., is 175.645; therefore -

U= 1190400 x 175.645 x 3/4/253=10036.21;

or, the required weight is, say, 10,036 pounds.

177. - Rolled-Iron Beams: Floors of Dwellings or Assembly-Rooms. - From Transverse Strains, Art. 500, we have -

c=255 I/l3 - y/420, (77.)

a rule for the distance from centres of rolled-iron beams in floors of dwellings, assembly-rooms, or offices, where the spaces between the beams are filled in with brick arches and concrete. In the equation, c is the distance apart from centres in feet, and y is the weight per yard of the beam. This, in words, is thus expressed:

Rule LVI. - Divide 255 times the value of I by the cube of the length; from the quotient deduct one 420th part of the weight of the beam per yard, and the remainder will be the required distance apart from centres.

Example. - What should be the distance apart from centres of Buffalo 12 1/4-inch, 125-pound beams 25 feet long between bearings, in the floor of an assembly-room? For these beams, in Table IV., I equals 286.019, and y = 125; therefore -

c= 255 x 286.019/253-125/420;

c= 72934.8/15625-125/420=4.668-.298=4.37;

or, the required distance from centres is, say, 4 feet 4 1/2 inches.

178. - Rolled-Iron Beams: Floors of First-Class Stores.- From Transverse Strains, Art. 504, we have -

c=148-81/l3 - y/960, (78.)

a rule for the distance from centres of rolled-iron beams in the floor of a first-class store; the spaces between the beams being filled with brick arches and concrete. This rule may be put in words as follows:

Ride LVII. - Divide 148.8 times the value of I by the cube of the length; from the quotient deduct one 960th part of the weight of the beam per yard, and the remainder will be the distance apart of the beams from centres in feet.

Example. - What should be the distance apart from centres of Buffalo 12 1/4-inch, 180-pound beams 20 feet long between bearings, in the floor of a first-class store? For these beams the value of I, Table IV., is 418.945, and the value of y is 180; therefore -

c =148.8 x 418.945/203-180/960=7.60;

or, the required distance from centres is, say, 7 feet 7 1/4 inches.

179. - Floor-Arches: Gneral Considerations. - In filling the spaces between the iron beams of a floor, the arches should be constructed with hard whole brick of good shape, laid upon the supporting centre in contact with each other, and the joints thoroughly filled with cement grout, and keyed with slate. Made in this manner, the arches need not be over four inches thick at the crown for spans extending to 7 or 8 feet, and 8 inches thick at the springing, where they should be started upon a proper skew-back. The rise of the arch should not be less than 1 1/2es for each foot of the span.