In transverse strains the (vertical) cross-shearing is generally not equal to the load, but varies at different points of the beam or cantilever. The manner of calculating transverse strains, however, allows for straining only the edges (extreme fibres) up to the maximum; so that the intermediate fibres, not being so severely tested, generally have a sufficient margin of unstrained strength left to more than offset the shearing strain. In solid beams it can, therefore, as a rule, be overlooked, except at the points of support. (In plate-girders it must be calculated at the different points where weights are applied.) The amount of the shearing at each support is equal to the amount of load coming on or carried by the support.

We must, therefore, substitute for w in Formula (7) either/) or q, as the case may be, and have at the left-hand end of beam for the safe resistance to shearing: p = a.(g/f) (9)

And at the right-hand end of beam: q = a.(g/f) (10)

Where p = the amount of load, in pounds, carried on the left-hand support.

Where q = the amount of load, in pounds, carried on the right-hand support.

Where a = the area of cross-section, in inches, at the respective support.

Where (g/f) = the safe resistance, per square inch, to cross-shearing.

Example.

A spruce beam of 5' clear span is 24" deep and 3" wide; how much uniform load will it carry safely to avoid the danger of shearing off at either point of support?

The beam being uniformly loaded, the supports will each carry one-half of the load; if, therefore, we find the safe resistance to shearing at either support, we need only double it to get the safe load (instead of calculating for the other support, too, and adding the results).

Let us take the left-hand support. From Formula (9) we have: p = a (g/f)

Now, we know that a == 24. 3 = 72 square inches.

The ultimate resistance of spruce to cross-shearing is about thirty-six hundred pounds per square inch; using a factor-of-safety of ten, we have for the safe resistance per square inch: (g/f) =3600/10=360 pounds.

We have, now: p = 72.360 = 25920 pounds. Similarly, we should have found for the right-hand support: - q = 25920 pounds. And as: - u =p + q = 51840 pounds, that will, of course, be the safe uniform load, so far as danger of shearing is concerned.

The beam must also be calculated for transverse strength, deflection and lateral flexure, before we can consider it entirely safe. These will be taken up later on.

Should it be desired to find the amount of vertical shearing strain x at any point of a beam, other than at the points of support, use: x== p

or q

Σw

(11)

Where x = the amount of vertical shearing strain, in pounds, at any point of a beam.

Where P or q the reaction, in pounds, (that is, the share of the = total loads carried) at the nearer support to the point.

Where Σ w = the sum of all loads, in pounds, between said nearer support and the point.

When x is found, insert it in place of w, in Formula (7), in order to calculate the strength of beam necessary at that point to resist the shearing.

Example.

A spruce beam, 20' long, and 8" deep, carries a uniform load of one hundred pounds per running foot. What should be the thickness of beam 5' from either support, to resist safely vertical shearing?

Each support will carry one-half the total load; that is, one thousand pounds: so that we have for Formula (11): p or q

= 1000 pounds.

The sum of all loads between the nearer support and a point 5' from support will be: Σw = 5. 100 = 500 pounds.

Therefore, the amount of shearing at the point 5' from support will be: x= 1000 - 500 = 500 pounds. Inserting this in Formula (7) we have: 500 500 = a. (g/f), or, a= (g/f)

We have just found that for spruce, (g/f)=:360 pounds.

Therefore, a =500/360 = 1,39 square inches.

And, as b. d = a. or b = a/d, we have, b = 1,39/8 = 1"/6

This is such a small amount that it can be entirely neglected in an 8" wooden beam.

To find the amount of vertical shearing at any point of a cantilever, other than at the point where it is built in, use: x= Σ w (12)

Where x the amount of vertical shearing strain, in pounds, at any point of canti-lever.

Where Σ to the sum of all loads between the free end and. said point.

To find the strength of beam at said point necessary to resist the shearing, insert x for w in Formula (7).

In transverse strains there is also a horizontal shearing along the entire neutral axis of the piece. This stands to reason, as the fibres above the neutral axis are in compression, while those below are in tension, and, of course, the result along the neutral line is a tendency of the fibres just above and just below it, to slide past each other or to shear off along the grain.

We can calculate the intensity (not amount) of this horizontal shearing at any point of the piece under transverse strain.

If x represents the amount of vertical shearing at the point, then the intensity of horizontal shearing at the point is =3 x

2 a

If this intensity of shearing does not exceed the safe-constant (g1/f) for shearing along the fibres, the piece is safe, or: 3/2 x/a = (g1/f) 13

Where x is found by formulae (11) or (12) for any point of beam or.

Where x = P or q the amount of supporting force, in pounds, for either point of support.

Where a = the area of cross-section in square inches.

Where (g1/f) = the amount of safe resistance, per square inch, to shearing along fibres.

Example. Take the same beam as before. The amount of vertical shearing 5' from support we found to be five hundred pounds, or: - x = 500. The area was 8" multiplied by thickness of beam, or: a = 8b. The ultimate shearing along the fibres of spruce is about four hundred pounds per square inch, and with a factor-of-safety of ten, we should have: (g1/f) = 400/10 = 40.

Inserting this in Formula (13) 3/2 . 500/8b = 40 or b= 1500/16.40 = 2,"34.

The beam should, therefore, be at least 2 1/3" thick, to avoid danger of longitudinal shearing at this point. At either point of support the vertical shearing will be equal to the amount supported there; that is, one-half the load, or one thousand pounds. Substituting this for x in Formula (13), we have: 3 1000 3000 = 4,"68.

- . - - = 40, or b = 2 8b 16.40

The beam would, therefore, have to be 4 2/3" thick at the points of support, to avoid danger of longitudinal shearing. The beam, as it is, is much too shallow for one of such span, a fact we would soon discover, if calculating the transverse strength or deflection of beam, which will be taken up later on. It will also be found that the greater the depth of the beam, the smaller will be the danger from longitudinal shearing, and, consequently, to use thinner beams, it would be necessary to make them deeper.