SOLUTION OF PRACTICAL PROBLEMS, FROM THE PRECEDING DATA.

PROB. I.—To find the Strength of Direct Cohesion of a Piece of Timber of any given Dimensions.

Rule. Multiply the area of the transverse section, in inches, by the value of C, in the preceding table of data, and the product will be the strength required.

Note. If the specific gravity be not the same as the mean tabular specific gravity; say, as the latter is to the former, so is the above product to the correct result.

Ex-What weight will it require to tear asunder a piece of teak 3 inches square, the specific gravity being 745 ?—Ans. 139950 lbs.

PROB. II.-To compute the Deflection of Beams fixed at one End and loaded at the other with any given Weight.

Rule 1.-Multiply the tabular value of E by the breadth and cube of the depth of the given beam, both in inches.

2.-Multiply also the cube of the length in inches by the given weight, and that product again by 32.*

3. Divide the latter product by the former, for the deflection sought.

Ex.-An ash batten, 3 inches square, is fixed in a wall, and projects from it 4 feet. If a weight of 200 lbs. be hung on its extremity, how much will it be deflected?—Ans. 13 inches.

Note. The same rule will apply, when the weight is distributed throughout the length, by multiplying the second product by 12 instead of 32.

PROB. III.-To compute the Deflection of Beams supported at each End, and loaded in the Middle with any given Weight.

Rule 1.-Multiply the tabular value of E by the breadth and cube of the depth, both in inches.

1

2.-Multiply also the cube of the length, in inches, by the given weight in lbs.; then divide the latter product by the former for the deflection sought.

Ex.-A square beam of English oak, whose side is 6 inches, is supported on two walls, 20 feet distant, and is to be loaded at its middle point with 1000 lbs., what will it be deflected?—Ans. 1·8 inch.

Note. If the beam be fixed at each end, the deflection will, with equal weights, be two-thirds of that found by the above rule.

PROB. IV. To compute the Deflection of Beams supported at each End, and loaded uniformly throughout their Length with a given Weight.

Rule.-Compute the deflection the same as in the last problem. Multiply that result by 5, and divide the product by 8, and the quotient will be the answer.

Mr Bevan says this should be 16, and Mr Barlow that it is 32 in the usual method of fixing beams in ordinary erections.

Ex.-A uniform bar of Adriatic oak, 2 inches square, is rested upon two props, distant 24 feet, how much will it be deflected by its own weight, its specific gravity being 960, or 60 lbs. to the cubic foot?-Ans. 9 inches.

PROB. V. To compute the ultimate Deflection of Beams or Rods before their Rupture.

Note. The beams are supposed to be supported at each end. Rule. Multiply the tabular value of U, in the preceding table of data, by the depth of the beam in inches, and divide the square of the length, also in inches, by that product, for the ultimate deflection sought.

Ex.-A square inch rod of ash, 6 feet long, is broken by a weight applied to its centre: how much will it be deflected before it breaks? Ans. 13.1 inches.

PROB. VI. To find the ultimate transverse Strength of any rectangular Beam of Timber, fixed at one End and loaded at the other. Rule I.-Multiply the value of S, in the preceding table of data, by the breadth and square of the depth, both in inches, and divide that product by the length, also in inches, and the quotient will be the weight in lbs. This is approximative.

Rule II.-1. Take the ultimate deflection 8 times that of the last problem, and divide the deflection by the length, which will give the sine of the angle; whence by a table find the secant.

2. Multiply the secant by the breadth and square of the depth in inches, and the product again by the value of S' in the table of data. 3. Divide this last product by the length in inches, and the quotient will be the answer in lbs.

Ex. 1.-What weight will it require to break a piece of Mar forest fir, fixed by one end in a wall, and loaded at the other; the breadth being 2 inches, depth 3 inches, and length 4 feet?-Ans. 518 lbs.

PROB. VII.-To compute the ultimate transverse Strength of any rectangular Beam, when supported at both Ends and loaded in the Centre.

Rule I.-Multiply the tabular value of S by 4 times the breadth and square of the depth in inches, and divide that product by the length, also in inches, for the weight.

Rule II.-1. Compute the ultimate deflection by Prob. V.; square that deflection, and divide it by the square of half the length of the beam, and add the quotient to 1, for the square of the secant of deflection; which multiply by the length in inches.

1. Multiply the tabular value of S' by 4 times the breadth, and the square of the depth; and divide that product by the former answer in lbs.

Ex.-What weight will be necessary to break a piece of larch similar to the third specimen, the length being 8 feet 4 inches, the breadth 8 inches, and depth 10 inches; being supported at each end, and loaded in the middle?-Ans. 36676 lbs.

For farther information on this subject, the works of Barlow, Bevan, Tredgold, Farey, Young, Gregory, (Mathematics for Practical Men in particular), Playfair, Leslie, Robison, Hutton, Smeaton, Rennie, &c., may be consulted.

GUNNERY.

I.-Theorems relative to Projectiles on the Horizontal Plane.

g=

a the altitude or impetus, as it is commonly called, due to

V the velocity; then

1. R 2a S=4 asc=

2. V = √2ag = √

SV2 $ c V • _ \ g ¢ T° _ } g T2 _ 4 H

=

=

2 gT

R

=

=

=

2 sc

√g H

=

t

4. Ha s2=&av=tR=

II.-Theorems relative to Projectiles on Oblique Planes.

Let c denote the cosine of the direction above the horizon

the cosine of the inclination of the plane to the horizon the sine of the direction above the plane

the range on the oblique plane

the time of flight

III.-To find the Velocity of any Shot or Shell.

Let v denote the velocity, B the weight of the ball or shot, and C the weight of the charge of powder; then

v = 1600/2260/by Hutton's experiments.

1. v

IV. Given the Range at one Elevation to find the Range at another

--

Rule.

Elevation.

As the sine of twice the first elevation is to the sine of twice the second elevation, so is the first range to the second range.

V.-Given the Range for one Charge, to find the Range for another Charge.

The ranges are directly as their charges, the elevation being the same; or as one range is to another range, so is the charge corresponding to the first range to the charge corresponding to the second

range.

Example 1.—If a ball of 1 lb. be projected with a velocity of 1600 feet per second, when fired with a charge of 8 ounces of powder, with what velocity will each of the several kinds of shells be discharged by the full charges of powder?

Ex. 2.-If a shell range 1000 yards when discharged at an elevation of 45°, what will be its range when the elevation is 30° 16′ with the same charge of powder ?

Ans. 871 yards, or 2612 feet.

Ex. 3.-At an elevation of 45° the range of a shell was 3750 feet, to what angle must the piece be elevated to strike an object at the distance of 2810 feet?

Ans. 24° 16′, or 65° 44′, equally distant from 45°.

Ex. 4.-With what impetus, velocity, and charge of powder must a 13-inch shell be fired, at an elevation of 32° 12', to strike an object at the distance of 3250 feet?

Ans. impetus 1802, velocity 340 feet, charge 4 lb. 7 oz.

Ex. 5.-A shell ranges 3500 feet, when discharged at an elevation of 25° 12′, how far will it range at an elevation of 36° 15' with the same charge of powder? Ans. 4332 feet.

Ex. 6.-If, with a charge of 9 lb. of powder, a shell range 4000 feet, what charge will throw it 3000 feet at an elevation of 45o in both cases? Ans. 6 lb. of powder.

Ex. 7.-What will be the time of flight for the greatest range at the elevation of 45°?

Ans. The time in seconds is one-fourth of the square root of the range in feet.

Ex. 8.-In what time will a shell range 3250 feet at an elevation of 32° ? Ans. 11 seconds nearly.

Ex. 9.-How far will a shot range on a plane which ascends 8° 15', and on another which descends 8° 15', the impetus being 3000 feet?

Ans. 4244 feet on the ascending plane,

and 6745 feet on the descending plane.

Ex. 10.-How much powder will throw a 13-inch shell 4244 feet on an inclined plane which ascends at an angle of 8° 15', the elevation of the mortar being 32° 30′ ?

Ans. 7.38 lb., or 7 lb. 6 oz. nearly.

Ex. 11.-At what elevation must a 13-inch mortar be pointed to range 6745 feet on a plane descending at an angle of 8° 15′ with a charge of 78 lb. of powder? Ans. 32° 41'.

Ex. 12.-In what time will a 13-inch shell strike a plane which rises 8° 30', at an elevation of 45°, and discharged with an impetus of 2304 feet? Ans. 143 seconds.

For more complete information on this subject, see Hutton's Course of Mathematics.