Uniform And Concentrated Loads

Fig. 157.

Select pole x distant from load line at random (for the sake of illustration, though it would be better to make xy = (k/f) = 12000 pounds.) We find xy scales 6500 pounds. We now draw x b, xl, xh, xf, etc. And construct figure C E G. Draw xo parallel C G and we find a o (or reaction A) scales = 6333 pounds, and o b (or reaction E) scales = 9667 pounds. The longest vertical is DE = 161" (inch scale) therefore greatest bending-moment is at w1 and from Formula (93) mwl = 161.6500 = 1046500 For the required moment of resistance we have from Formula (18) 1046500 12000 The cheapest or most economical nearest section we find - to this required r (87,2) is the 20" - 200 pounds beam of which the moment of resistance is r= 123,8.

Had we combined the formulae for uniform and concentrated loads and worked out the problem arithmetically it would have been tedious, but we should have had similar results.

We can safely overlook shearing, but note that the real shearing figure would not be the shaded figure, but dotted figure O1 HIJ O For finding the deflection we now draw lower load line g c = the sum of the verticals through C E G, beginning at top with length of wvII, then wv11„ wv, wlv, w111, and w11 in their order. We take no notice of vertical w1 as it does not fall in one of the even divisions of C G or A B into lengths l1 We select pole z distant zj=288" from load line, draw zg, zc, etc., and then figure c1,f1,g1. We now draw z o parallel c1 g1 it divides g c, so that go = 295" and oc = 245", we divide c1 gl in same proportion at f, and carry this up to F at beam, which is the point of greatest deflection of beam, and is distant 163" from B, and 197" from A. We scale ff1, = 106" (inch scale) and have from Formula (97)

δ = = 0,357 27000000.1238 1238 being = i, the moment of inertia of beam as found in Table XX. The beam is therefore amply stiff even to carry plastering. The graphical method lends itself very readily to finding centres of gravity and neutral axes, as explained in the chapter on arches, and also for finding the moments of inertia of difficult cross-sections.

Irregular Cross-sections.

If we have an irregular figure A B C D E (Figure 158) we divide it into simple parts I, II, III and IV. We find the centres of gravity g1, g11, g111 and g1v of each part and draw their respective horizontal neutral axes through these. Anywhcre's make a line ae=area of whole figure and divide it, so that: a b = area of I bc = area of II c d = area of III and de = area of IV. Select pole x at random, draw r a, x b, x c, x d, and x e.

From any point of horizontal g1 draw fh parallel b x till it insects horizontal g11; then draw hj parallel c x to horizontal g111; then j k parallel dx to last horizontal, and finally ko parallel xe; and fo parallel ax till they intersect at o. A horizontal through o is the main neutral axis of the whole.1 If we multiply the area of the figure f o k j h by the area of the figure ABODE (both in square inches) we have the value of moment of inertia i of ABODE in inches, around its horizontal neutral axis o.

Uniform And Concentrated Loads 100262

Fig. 158.

Uniform And Concentrated Loads 100263

Fig. 159.

1The point of intersection of this line with a main neutral axis, found similarlv, in any other direction, would he the centre of gravity of the whole figure.

A simple way of obtaining the area of the figure fok would be to draw horizontal lines through it at equal distances beginning with half distances at top and bottom, and to multiply the sum of these horizontals in length by the distance apart of any two horizontals, all measurements in inches. This will approximate quite closely both the area and moment of inertia. Of course the more parts we take in all of the processes, the closer will be our result.

A practical example will more fully illustrate the above.

To find area.

Example VII.

Find horizontal neutral axis and the corresponding moment of inertia of a 7" - 55 pounds per yard deck beam, resting on its flat flange (Figure 159).

We will take the roll as one part, divide the web into four equal parts, the flange into two parts, one the base which will be practically rectangular, and its upper part which will be practically triangular. The whole area we know is for wrought-iron: a = 55/10 =5, 5 square inches.

The bottom rectangular part of flange will be av11 = 4 1/2. 3/8= 1,7 square inches next triangular part av1 = 4 1/2.3/8/2 = 0,9

The web parts a11 = a111 = alV= av = 5/16.5 1/2/4=0,4 square inches each.

Leaving for the roll at top a,= 1,3

We now make the horizontal line a h = 5.5" and divide it. so that



1,3 inches



cd = de = ef= 0,4 inches



0,9 inches and



1,7 inches.

select x at random and draw x a, x o, xc, etc.

Draw the horizontal neutral axes I, II, III, etc., through their respective parts. Begin anywhere on I and draw j k parallel b x to line II; then kl parallel ex to III; then lm parallel dx to IV; then inn parallel ex to V: then np parallel fx to VI; then pg parallel gx to VII; Now draw from q the line qo parallel xh, and from j the line jo parallel ax till they intersect at o. A horizontal through o is the neutral axis of whole beam. We will now make a new drawing of figure j o q for the sake of clearness. Draw horizontals through it every inch in height beginning at both top and bottom with one-half inch. The top one scales nothing, the next 1/3", then 7/8", then 1 1/2", then 2 1/4", then l 1/8", and the bottom one 1/3", the sum of all being 6 5/12' or 6,416". This multiplied by the height of the parts, which is one inch, would give us, of course, 6,416 square inches area. Multiplying this area by the area of the cross-section of deck beam 5,5 square inches, we should have i = 5,5.6,416 = 35,288.

In Table XX it is given as 35, 1 so that we are not very far out.

If we had taken more parts, of course the result would have been more exact.

When constructing plate girders of large size, much material can be saved by making the flanges heaviest at the point of greatest bending-moment, and gradually reducing the flanges towards the supports.

This is accomplished by making each flange at the point of greatest bending-moment of several thicknesses or layers of iron, the outer layer being the shortest, the next a little longer, etc. Of course the angles, which form part of the flange are kept of uniform size the whole length, as it would be awkward to attempt to use different sized angles. Generally (though not necessarily) the inner or first layer of the flange plates, is also run the entire length. Of course; where the flanges are gradually reduced in this way, it becomes necessary to figure the bending-moment and moment of resistance at many points along the plate girder to find where the plates can be reduced. This would be a wearisome job. By using the graphical method, however, it can be easily accomplished. Referring back to Figure 151, we take the point of greatest bending-moment (at w1,) of the beam A B. The required moment of resistance at this point, it will be remembered was the length (inch scale) of vertical E through C D E F G. We now decide what size angles we propose using and settle the necessary thickness of the flanges by Formula (36), inserting for the value of r, the length (inch scale) of v or vertical at E. Further a, will, of course, be the sum of the area of two angles, d the total depth of girder in inches and b the breadth of flange, in inches, less rivet holes. The above is on the assumption that the distance x y of pole x from load line d a was equal to the safe modulus of rupture (k/f) of steel or wrought-iron according to whichever material we were using, or we should have: x=v/d-a1 b


Thickness of


Where x = the thickness, in inches, of each flange of a plate girder at any point of its length.

Where v = the length of vertical, inch scale, through upper or resistance figure, providing we have assumed the distance of pole from load line (pound scale) = (k/f) of the material.

Where d = the total depth, in inches, of the plate girder.

Where b = the width, less rivet holes, in inches, of the flange.

Where a, = the sum of the areas of cross-section, in square inches, of two of the angles used.

We now calculate as above, the thickness x of flange at point of greatest bending-moment and then decide into how many layers or thicknesses we will divide the flanges. Say, in our case we decide to make the flange of four layers of plates, each - or one quarter x in thickness. Then make

ElE11=al.d (99)

Where E1 E11 = the amount to be substracted (inch scale) from moment of resistance or vertical v and representing the work of two angles.

Where a, = the sum of the area of cross-section, in square inches, of the two angles.

Where d = the total depth, in inches, of the girder.

Now draw through E1l a parallel to base of figure C G, divide E11 E into as many parts as we decide to use thicknesses of plates (four in our case) and draw parallel lines to base C G through these parts. Vertically over the points where these lines intersect the curve or outline of figure C D E F G will be the points at which to break off plates, as illustrated in drawing. This method, of course, is approximate, but it will be found sufficiently accurate for all practical purposes. It is not necessary that x or E E11 be divided into equal parts. Had we decided to use plates of varying thicknesses we should simply divide E E11 in proportions to correspond to thicknesses of plates in their proper order, beginning at E11 with plate immediately next to angles and ending at E with extreme central outside plate. An example, more fully illustrating the above, will be given in the chapter on plate girders.