137. - Deflection: when Weight is at middle. - By a transposition of the factors in (32.), we obtain -

δ= wl3/Fbd3, (33.)

a rule by which the deflection of any given beam may be ascertained, and which, in words at length, is -

Rule XXVIII. - Multiply the given weight by the cube of the length in feet; divide the product by the product of the breadth into the cube of the depth in inches, multiplied by the value of .F for the material of the beam, in Table III., and the quotient will be the required deflection in inches;

Example. - To what depth will 1000 pounds deflect a 3x10 inch white-pine beam 20 feet long, the weight being at the middle of the beam? By the rule, we have 1000 x 203 = 8000000; then, since the value of F for white pine, in Table III., is 2900, we have 3 x 103 x 2900 = 8700000; using this product as a divisor and by it dividing the former product, we obtain a quotient of 0.9195, the required deflection in inches.

138. - Deflection: Load. Uniformly Distributed. - In two beams of equal capacity, suppose the one loaded at the middle, and the other with its load uniformly distributed over its length, and so loaded that the deflection in one beam shall equal that in the other; then the weight at the middle of the former beam will be equal to five eighths of that on the latter. This proportion between the two has been demonstrated by writers on the strength of materials. (See p. 484, Mechanics of Eng. and Arch., by Prof. Mosely, Am. ed. by Prof. Mahan, 1856.) Hence, when U is put to represent the uniformly distributed load, we have or, when an equally distributed load deflects a beam to a certain depth, five eighths of that load, if concentrated at the middle, would cause an equal deflection. This value of W may therefore be substituted for it in equation (31.), and give -

5/8 U= Fbd3δ/l3.

from which we obtain -

U= 1.6 Fbd3δ/l3. (34.)

a rule for a uniformly distributed load.

139. - Deflection: Weight when Uniformly Distributed. - In equation (34.) we have a rule by which we may ascertain what weight is required to deflect to a given depth any given beam. This, in words at length, is -

Rule XXIX. - Multiply 1.6 times the deflection by the breadth of the beam, and by the cube of its depth, all in inches, and by the value of F for the material of the beam, in Table III.; divide the product by the cube of the length in feet, and the quotient will be the required weight in pounds.

Example. - What weight, uniformly distributed over the length of a spruce beam, will be required to deflect it to the depth of 0.5 of an inch, the beam being 3 x 10 inches and 10 feet long? The value of F for spruce, in Table III., is 3500. Therefore, by the rule, we have 1.6x0.5 x 3 x 103 x 3500 = 8400000, and this divided by (10x10x10=) 1000 gives 8400, the required weight in pounds.

140.-Deflection: Breadth or Depth, Load Uniformly

Distributed. - By transposition of the factors in equation (34.), we obtain -

bd3 = Ul3/1.6 Fδ, (35.)

a rule for the dimensions, which, in words at length, is -

Rule XXX. - Multiply the given weight by the cube of the length of the beam; divide the product by 1.6 times the given deflection in inches, multiplied by the value of F for the material of the beam, in Table III., and the quotient will equal the breadth into the cube of the depth. Now, to obtain the breadth, divide this quotient by the cube of the depth, and the resulting quotient will be the required breadth in inches. But if, instead of the breadth, the depth be required, then divide the aforesaid quotient by the breadth, and the cube root of the resulting quotient will be the required depth in inches. Again, if neither breadth nor depth be previously determined, but to be in proportion to each other at a given ratio, as r to I, r being a decimal fixed at pleasure, then divide the aforesaid quotient by the value of r, and take the square root of the quotient; then the square root of this square root will be the required depth in inches. The breadth will equal the depth multiplied by the value of the decimal r.

Example. - What should be the size of a locust beam 10 feet long which is to be loaded with 6000 pounds equally distributed over the length, and with which the beam is to be deflected 3/4 of an inch? The value of F for locust, in Table III., is 5050. By the rule, we have 6000 x( 10 x 10 x 10 =) 1000 = 6000000, which is to be divided by (1 .6x0.75 x 5050 =) 6060, giving a quotient of 990.1. Now, if the depth be, say, 6 inches, then 990.1 divided by (6 x 6 x 6 =) 216 gives a quotient of 4.584, the required breadth in inches, say 4 5/8. But if the breadth be assumed at 4 inches, then 990.1 divided by 4.gives a quotient of 247.525, the cube root of which is 6.279, the required depth in inches, or, say, 6 1/4. And, again, if the ratio between the breadth and depth be as 0. 7 to 1, then 990.1 divided by 0.7 gives a quotient of 1414.43, the square root of which is 37.609, of which the square root is 6.1326, the required depth in inches, or, say, 6 1/8; and then 6.1326x0.7 = 4.293, the required breadth in inches; or, the beam should be 4 3/10 x 6 1/8 inches.

141. - Deflection: when Weight is Uniformly Distributed. - By a transposition of the factors of equation (35.), we obtain a result nearly the same as that in equation (33.), which is a rule for deflection by a weight at middle, and which by slight modifications may be used for deflection by an equally distributed load. Thus by Rule XXXI. - Proceed as directed in Rule XXVIII. (Art. 137), using the equally distributed weight instead of a concentrated weight, and then divide the result there obtained for deflection by 1 -6; then the quotient will be the required deflection in inches.

δ= Ul3/1.6F bd3, (36.)

Example. - Taking the example given under Rule XXVIII., in Art. 137, and assuming that the 1000 pounds load with which the beam is loaded be equally distributed, then 0.9195, the result for deflection as there found, divided by 1.6, as by the above rule, gives 0.5747, the required deflection. This result is just five eighths of 0.9195, the deflection by the load at middle.

N.B. - The deflection by a uniformly distributed load is just five eighths of that produced by the same load when concentrated at the middle of the beam; therefore, five eighths of the deflection obtained by Rule XXVIII. will be the deflection of the same beam when the same weight is uniformly distributed.