This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

The deflection of a lever is the same as that of a beam of the same breadth and depth, but of twice the length, and loaded at the middle with a load equal to twice that which is at the end of the lever. Therefore, if Prepresents the weight at the end of a lever, and n the length of the lever in feet, then 2 P= W and 2 n = l, and if these values of W and l be substituted for those in equation (33.), we obtain a result 16 times that in equation (33.), which is the deflection in a beam. Therefore, when a beam and a lever equal in sectional area and in length be loaded by equal weights, the one at the middle, the other at one end, the deflection of the lever will be 16 times that of the beam. This proportion is based upon the condition that neither the beam nor the lever shall be deflected beyond the limits of elasticity.

which reduces to -

δ = 16 Pn3/Fbd3 (37.)

143. - Deflection of a Lever: Load at End. - Equation (37.), in words at length, is -

Rule XXXII. - Multiply 16 times 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 multiplied by the value of F fore material of the lever, in Table III., and the quotient will be the required deflection.

Example. - What would be the deflection of a bar of American wrought iron one inch broad, two inches deep, loaded with 150 pounds at a point 5 feet distant from the wall in which the bar is imbedded? The value of F for American wrought iron, in Table III., is 62000. Therefore, by the rule, 16 x 150 x 53= 300000. This divided by (1 x 23 x 62000 =) 496000 gives 0.048, the required deflection - nearly 5/8 an inch.

144. - Deflection of a Lever: Weight when at End. - By a transposition of the factors in equation (37.), we obtain -

P = Fbd3δ16n3 (38.)

This result is equal to one sixteenth of that shown in equation (31.), a rule for the weight at the middle. Therefore, for -

Rule XXXIII. - Proceed as directed in Rule XXVII.; divide the quotient there obtained by 16, and the resulting quotient will be the required weight in pounds.

Example. - What weight is required at the end of a 4 x 12 inch Georgia-pine lever 20 feet long to deflect it three quarters of an inch? Proceeding by Rule XXVII., we obtain a quotient of 3823.; this divided by 16 gives 238.5, say 239, the required weight in pounds.

145. - Deflection of a Lever: Breadth or Depth, Load at End - A transposition of the factors of equation (38.) gives -

bd3 = 16 Pn3/Fδ (39.)

a rule by which to obtain the sectional area of the lever. By comparison with equation (32.) it is seen that the result in (39.) is 16 times that found by (32.). Therefore, the dimensions for a lever loaded at the end may be found by -

Rule XXXIV. - Multiply by 16 the first quotient found by Rule XXVIT., and then proceed as farther directed in Rule XXVII., using the product of 16 times the quotient, instead of the said quotient.

Example. - What should be the size of a spruce lever 20 feet long, between weight and wall, to sustain 2000 pounds at the end with a deflection of 1 inch? Proceeding by Rule XXVII., we obtain a first quotient of 4571.3. By Rule XXXIV., 4571.3 x 16 = 73144.8. Now, if the depth be fixed, say, at 20, then 73144.8 divided by (20 x 20 x 20 =) 8000 gives 9.143, the required breadth. But to obtain the depth, fixing the breadth, say, at 9, we have for 73144.8 divided by 9 = 8127.1, the cube root of which is 20.055, the required depth. Again, if the breadth and depth are to be in proportion, say, as 0. to 1.0hen 73144.8 divided by 0.7gives 104492.7, the square root of which is 323.254, of which the square root is 17.98, the required depth in inches; and 17.98 x 0.7 = 12.586, the required breadth in inches. The lever, therefore, should be, say, 12 5/8x 18 inches.

146. - Deflection of Levers: Weight Uniformly Distributed. - A comparison of the effects of loads upon levers shows (Transverse Strains, Art. 347) that the deflection by a uniformly distributed load is equal to that which would be produced by three eighths of that load if suspended from the end of the lever. Or, P - 3/8 U. Substituting this value of P, in equation (37.), gives -

δ = 16x3/8 Un3 / Fbd2/Fbd2

which reduces to -

δ = 6 Un3/Fbd3, (40.)

a rule for the deflection of levers loaded with an equally distributed load.

147. - Deflection of Levers with Uniformly Distributed

Load. - The deflection shown in equation (40.) is just six times that shown in equation (33.). The result by (33.) multiplied by 6 will equal the result by (40.); therefore, we have -

Rule XXXV. - Proceed as directed in Rule XXVIII.; the result thereby obtained multiplied by 6 will give the required deflection.

Example. - To what depth will 500 pounds deflect a 3 x 10 inch white-pine lever 10 feet long, the weight uniformly distributed over the lever? Here, by Rule XXVIII., we obtain the result 0.05747; this multiplied by 6 gives 0.3448, the required deflection.

148. - Deflection of Levers: Weight when Uniformly Distributed. - By a transposition of factors in (40.), we obtain -

U = Fbd3δ (41.)

6n3

This is equal to one sixth that of equation (31.); therefore, we have -

Rule XXXVI.-Proceed as directed in Rule XXVI.; the quotient thereby obtained divide by 6, and the quotient thus obtained will be the required weight.

Example. - What weight will be required to deflect a 4x5 inch spruce lever 1 inch, the weight uniformly distributed over its length? Proceeding as directed in Rule XXVI., the result thereby obtained is 1750; this divided by 6 gives 291 2/3, the required weight in pounds.

149. - Deflection of Levers: Breadth or Depth, Load Uniformly Distributed. - A transposition of factors in equation (41.) gives -

bd3 = 6 Un3/Fδ. (42.)

This result is just six times that of equation (32.); we, therefore, have -

Rule XXXVII. - Proceed as directed in Rule XXVII.; multiply the first quotient thereby obtained by 6; then in the subsequent directions use this multiplied quotient instead of the said first quotient, to obtain the required breadth and depth.

Example. - What should be the size of a spruce lever 10 feet long, sustaining 2666 2/3 pounds, uniformly distributed over its length, with a deflection of 1 inch? Proceeding by Rule XXVII., the first quotient obtained is 761.905; this multiplied by 6 gives 4571.43, the multiplied quotient which is to be used in place of the said first quotient. Now, to obtain the breadth, the depth being fixed, say, at 10; 4571 .43 divided by (cube of 10 =) 1000, the quotient, 4.57, is the required breadth. But if the breadth be fixed, say, at 4, then, to obtain the depth, 4571.43 divided by 4 gives 1142.86, the cube root of which is 10.46, the required depth. Again, if the breadth and depth are to be in proportion, say, as 0.6 to 1 .0, then 4571 .43 divided by 0.6 gives 7619.05, the square root of which is 87.27, of which the square root is 9.343, the required depth in inches; and 9.343 x 0.6 equals 5.6, the required breadth in inches; or, the lever may be, say, 5 5/8 x 9 1/3 inches.

Object Clearly Defined. - In the various parts of timber construction, known as floors, partitions, roofs, bridges, etc., each has a specific object, and in all designs for such constructions this object should be kept clearly in view, the various parts being so disposed as to serve the design with the least quantity of material. The simplest form is the best, not only because it is the most economical, but for many other reasons. The great number of joints, in a complex design, render the construction liable to derangement by multiplied compressions, shrinkage, and, in consequence, highly increased oblique strains; by which its stability and durability are greatly lessened.

Continue to: