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

When a load is placed upon a beam supported at each end, the beam bends more or less; the distance that the beam descends under the operation of the load, measured at the middle of .its length, is termed its deflection. In an investigation of the laws of deflection it has been demonstrated, and experiments have confirmed it, that while the elasticity of the material remains uninjured by the pressure, or is injured in but a small degree, the amount of deflection is directly in proportion to the weight producing it; for example, if 1000 pounds laid upon a beam is found to cause it to deflect or descend at the middle a quarter of an inch, then 2000 pounds will cause it to deflect half an inch, 3000 pounds will deflect it three fourths of an inch, and so on.

134. - Deflection: Relation to Dimensions - In Table III. are recorded the results of experiments made to test the resistance of the materials named to deflection. The figures in the third column designated by the letter F (for flexure) show the number of pounds required to deflect a unit of material one inch. This is an extreme state of the case, for in most kinds of material this amount of depression would exceed the limits of elasticity; and hence the rule would here fail to give the correct relation as between the dimensions and pressure. For the law of deflection as above stated (the deflections being in proportion to the weights) is true only while the depressions are small in comparison with the length. Nothing useful is, therefore, derived from this position of the question, except to give an idea of the nature of the quantity represented by the constant F; it being in reality an index of the stiffness of the kind of material used in comparing one material with another. Whatever be the dimensions of the beam, F will always be the same quantity for the same material; but among various materials F will vary according to the flexibility or stiffness of each particular material. For example, F will be much greater for iron than for wood; and again, among the various kinds of wood, it will be larger for the stiff woods than for those that are flexible. The value of F, therefore, is the weight which would deflect the unit of material one inch, upon the supposition that the deflections, from zero to the depth of one inch, continue regularly in proportion to the increments of weight producing the deflections, or, for each deflection -

F: I:: W: δ,

from which we have -

W=F δ; or F= W/δ,

in which δ represents the deflection in inches corresponding to W, the weight producing it. This is for the unit of material. For beams of larger dimensions, investigations have shown (Transverse Strains, Chapters XIII. and XIV.) that the power of a beam to resist deflection by a weight at middle is in proportion to its breadth and the cube of its depth, and it is inversely in proportion to the cube of the length;

or, when the resistance of the unit of material is measured, as above, by W/δ, we have the relation between it and a larger beam of -

W/δ:bd3/l3

Putting this ratio in a proportion with that of the unit of material, we have -

F:I:: W/δ:bd3/l3

which gives -

W/δ=Fbd3/l3

from which we have -

W = Fbd2δ/l3 (31.)

135. - Deflection: Weight when at Middle. - In equation (31.) we have a rule by which to ascertain what weight is required to deflect a given beam to a given depth of deflection; this, in words at length, is -

Rule XXVI. - Multiply the breadth of the beam by the cube of its depth, and by the given deflection, 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 is required at the middle of a 4x12 inch Georgia-pine beam 20 feet long to deflect it three quarters of an inch? The value of F for Georgia pine, in Table III., is 5900; therefore, by the rule, we have 4 x 123 x 0.75 x 5900 = 30585600, which divided by (20x20 x 20 =) 8000 gives a quotient of 3823.2, the required weight in pounds.

136. - Deflection: Breadth or Depth, Weight at middle.

- By a transposition of equation (31.), we obtain -

bd3= Wl3/Fδ, (32.)

a rule by which may be found the breadth or depth of a beam, with a given load at middle and with a given deflection; this, in words at length, is -

Rule XXVII. - Multiply the given load by the cube of the length in feet, and divide the product by the product of the deflection into the value of F for the material of the beam, in Table III.; then the quotient will be equal to the breadth of the beam multiplied by the cube of its depth, both in inches.

Now, to obtain the breadth, divide the said quotient by the cube of the depth, and this quotient will be the required breadth. But if, instead of the breadth, the depth be required, then divide the said quotient by the breadth, and the cube root of this quotient will be the required depth. But if neither breadth nor depth be previously fixed, but it be required that they bear a certain proportion to each other; such that d: b:: 1: r, r being a decimal, then b = r d, and b d3 - rd4; then, to find the depth, divide the aforesaid quotient by the decimal r, and the fourth root (or the square root of the square root) will be the required depth, and this multiplied by the decimal r will give the breadth.

Example. - What should be the size of a spruce beam 20 feet long between bearings, sustaining 2000 pounds at the middle, with a deflection of one inch? By the rule, the weight into the cube of the length is 2000 x 8000 = 16000000. The value of F for spruce, in Table III., is 3500; this by the deflection == 1 gives 3500, which used as a divisor in dividing the above 16000000 gives a quotient of 4571.43. Now, if the breadth be required, the depth being fixed, say, at 10, then 4571.43 divided by (10 x 10 x 10 =) 1000 gives 4. 57, the required breadth. The beam should be, say, 4 5/8 by 10 inches. But if the depth be required, the breadth being fixed, say, at 4, then 4571.43 divided by 4 gives 1142.86, the cube root of which is 10.46; so in this case, therefore, the beam is required to be 4 x 10 1/2 inches. Again, if the breadth is to bear a certain proportion to the depth, or that the ratio between them is to be, say, 0.6 to 1, then let r = 0.6, and then 4571.43 = 0.6d, and dividing by 0.6, we have 7619.05 = d4 This equals d2xd2; therefore the square root of 7619 is 87.29, and the square root of this is 9.343, the required depth in inches. Now 9.343x0.6 equals the breadth, or 9.343x0.6=5.6; therefore the beam is required to be 5.6 x 9.34 inches, or, say, 5 5/8 x 9 1/3 inches.

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