1. Find Reaction of each Support.

If the loads on a girder are uniformly or symmetrically distributed, each support carries or reacts with a force equal to one-half of the total load. If the weights are unevenly distributed, each support carries, or the reaction of each support is equal to, the sum of the products of each load into its distance from the other support, divided by the whole length of span. See Formulae (14), (15), (16), and (17).

2. Find Point of Greatest Bending Moment. The greatest bending moment of a uniformly or symmetrically distributed load is always at the centre. To find the point of greatest bending moment, when the loads are unevenly distributed, begin at either support and pass over load after load until an amount of loads has been passed equal to the amount of reaction at the support from which the start was made, and this is the desired point. In a cantilever the point of greatest bending moment is always at the wall.

3. Find the Amount of the Greatest Bending Moment. In a beam (supported at both ends) the greatest bending moment is at the centre, of the beam, provided the load is uniform, and this moment is equal to the product of the whole load into one-eighth of the length of span, or m =u.l/8 (21)

Where m = the greatest bending moment (at centre), in lbs. inch, of a uniformly-loaded beam supported at both ends.

Where u = the total amount of uniform load in pounds.

Where I = the length of span in inches.

If the above beam carried a central load, in place of a uniform load, the greatest bending moment would still be at the centre, but would be equal to the product of the load into one-quarter of the length of span, or m =w.l/4 (22)

Where m = the greatest bending moment (at centre), in lbs. inch, of a beam with concentrated load at centre, and supported at both ends.

Where w = the amount of load in pounds.

Where I = the length of span in inches.

To find the greatest bending moment of a beam, supported at both ends, with loads unevenly distributed, imagine the girder cut at the point (previously found) where the greatest bending moment is known to exist; then the amount of the bending moment at that point will be equal to the product of the reaction (of either support) into its distance from said point, less the sum of the products of all the loads on the same side into their respective distances from said point, i. e., the point where the beam is supposed to be cut. To check the whole calculation, try the reaction and loads of the discarded side of the beam, and the same result should be obtained.

To put the above in a formula, we should have: mA =p.x-Σ(wl x1 + w11x11 +w111 x111, etc.) (23) And as a check to above: mA - q.(l-x)-Σ (w111 x111 + uv, xv + wVl xy1) etc. 24)

Where A == is the point of greatest bending moment. Where mA = is the amount of bending moment, in lbs. inches, at A.

Where p = is the left-hand reaction, in pounds.

Where q = is the right-hand reaction, in pounds.

Where x and (l-x) = the respective distances in inches, of the left and right reactions from A.

Where x1, x11, x111, etc., = the respective distances, in inches, from A, of the loads w1, w11, w111 ,etc.

Where wt, w11, w111, etc., = the loads, in pounds. Where Σ = the sign of summation.

The same formulae, of course, would hold good for any point of beam. In a cantilever (supported and built in at one end only), the greatest bending moment is always at the point of support.

For a uniform load, it is equal to the product of the whole load into one-half of the length of the free end of cantilever, or m =u.l/2 (25)

Where m = the amount, in lbs. inch, of the greatest bending moment (at point of support).

Where u = the amount of the whole uniform load, in pounds.

Where l = the length, in inches, of the free end of cantilever.

For a load concentrated at the free end of a cantilever, the greatest bending moment is at the point of support, and is equal to the product of the load into the length of the free end of cantilever, or m = w. l (26)

Where m = the amount, in lbs. inch, of the greatest bending moment (at point of support).

Amount of greatest bending moment.

Where w = the load, in pounds, concentrated at free end.

Where l = the length, in inches, of free end of cantilever.

For a load concentrated at any point of a cantilever, the greatest bending moment is at the point of support, and is equal to the product of the load into its distance from the point of support, or m = w. x (27)

Where m = the amount, in lbs. inch, of the greatest bending moment (at point of support).

Where w = the load, in pounds, at any point.

Where x = the distance, in inches, from load to point of support of cantilever.

Note, that in all cases, when measuring the distance of a load, we must take the shortest distance (at right angles) of the vertical neutral axis of the load, (that is, of a vertical line through the centre of gravity of the load.)

4. Find the Required Cross-section.

To do this it is necessary first to find what will be the required moment of resistance.

If the cross-section of the beam is uniform above and below the neutral axis, we use Formula (18), viz.: r = m/ (k/f)

If the cross-section is unsymmetrical, that is, not uniform above and below the neutral axis, we use for the fibres above (he neutral axis, formula (19), viz.: - r = m/(c/f) and for the fibres below the neutral axis, Formula (20), viz.: - r = m/(t/f)

In the latter two cases, for economy, the cross-section should be so designed that the respective distances of the upper and lower edges (extreme fibres), from the neutral axis, should be proportioned to their respective stresses or capacities to resist compression and tension. This will be more fully explained under cast-iron lintels.

A simple example will more fully explain all of the above rules.

Example. Three weights of respectively 500 lbs., 1000 lbs., and 1500 lbs., are placed on a beam of 17' 6" (or 210") clear span, 2' 6" (or 30"), 7' 6" (or 90"), and 10' 0" (or 120") from the left-hand support. The modulus of rupture of the material is 2800 lbs. per square inch. The fuc-tor-ofsafety to be used is 4. The beam to be of uniform cross-section. What size of beam should be used?

1. Find Reactions (see formulae 16 and 17).

Reaction p will be in pounds = 500.180/210+1000.120/210+1500.90/210=

1642 6/7 pounds.

Reaction q will be in pounds, = 500.30/210+1000.90/210+1500.120/210=

1357 1/7 pounds.

Check,p+ q must equal whole load, and we have in effect: - p+-q= 1642 6/7+1357 1/7 = 3000, which being equal to the sum of the loads is correct, for: - 500 + 1000 + 1500 - 3000. 2. Find Point of Greatest Bending Moment.

Begin at p, pass over load 500, plus load 1000, and we still need to pass 142 6/7 pounds of load to make up amount of reaction p (1642 6/7 lbs.); therefore, the greatest bending mo-ment must be at load 1500; check, begin at q and we arrive only at the first load (1500) before passing amount of reaction q (1357 1/7 lbs.), therefore, at load 1500 is the point sought.

3. Find Amount of Greatest Bending Moment.

Suppose the beam cut at load 1500, then take the left-hand side of beam, and we have for the bending moment at the point where the beam is cut.

m - 1642 6/7.120 - (500.90 + 1000.30 + 1500.0)' = 197143 - (45000+30000+0) = 197143 - 75000 = 122143 lbs. inch.

Fig. 10.

As a check on the calculation, take the right-hand side of beam and we should have: m = 1357 1/7 90 - 1500.0 = 122143 - 0 = 122143 lbs. inch, which, of course, proves the correctness of former calculation. 4. Find the Required Cross-section of Beam. We must first find the required moment of resistance, and as the cross-section is to be uniform, we use formula (18), viz.: - r = m/(k/f)

Now, m = 122143, and k/f = 2800/4 = 700, therefore, r = 122143/700 = 174,49 or say = 174,5

Consulting Table I, fourth column, for section No. 2, we find r = bd2/6, we have, therefore, bd2

- = 174,5 or bd2 = 1047. 6

If the size of either b or d is fixed by local conditions, we can, of course, find the other size (d or b) very simply; for instance, if for certain reasons of design we did not want the beam to be more than 4" wide, we should have

6 = 4, therefore, 4.d2= 1047, and d2= 1047/4 = 262, therefore, d= (about) 16"', or, if we did not want the beam to be over 12" deep, we should have d = 12, and d2 = 12.12 = 144, therefore, b.144 = 104 7, and b = 1047/144 = 7,2" or say 7 1/4".

One thing is very important and, must be remembered, that the deeper the beam is, the more economical, and the stiffer will it be. If the beam is too shallow, it might deflect so as to be utterly unserviceable, besides using very much more material. As a rule, it will therefore be necessary to calculate the beam for deflection as well as for its transverse strength. The deflection should not exceed 0,"03 that is, three one-hundredths of an inch for each foot of span, or else the plastering would be apt to crack, we have then the formula: δ = L. 0,03 (28)

The deepest beam the most economical.

Safe deflection.

Deflection

Where δ = the greatest allowable total deflection, in inches, at centre of beam, to prevent plaster cracking.

Where L = the length of span, in feet.

In case the beam is so unevenly loaded that the greatest deflection will not be at the centre, but at some other point, use: δ = X. 0,06 (29)

Where δ = the greatest allowable total deflection, in inches, at point of greatest deflection.

Where X = the distance, in feet, to nearer support from point of greatest deflection.

If the beam is not stiffened sideways, it should also be calculated for lateral flexure. These matters will be more fully explained when treating of beams and girders.