Let A B C D be a floor plan of a building, A B and C D are the walls, E and F the columns, with a girder between, the other lines being floor beams, all 12" between centres; on the left side a well-hole is framed 2' x 2'. Let the load assumed be 100 pounds per square foot of floor, which includes the weight of construction. Each of the right-hand beams, also the three left-hand beams E L, K P and F Q will each carry, of course, ten square feet of floor, or

10.100= 1000 pounds each uniform load. Each will transfer one-half of this load to the girder and the other half to the wall. The tail beam S N will carry 8 square feet of floor, or

8.100 = 800 pounds uniform load. One-half of this load will be transferred to the wall, the other half to the header R T, which will therefore carry a load of 400 pounds at its centre, one-half of which will be transferred to each trimmer.

The trimmer beam G M carries a uniform load, one-half foot wide, its entire length, or fifty pounds a foot (on the off-side from well-hole), or

50.10 = 500 pounds uniform load, one-half of which is transferred to the girder and the other half to the wall. The trimmer also carries a similar load of fifty pounds a foot on the well-hole side, but only between M and R, which is eight feet long, or

50.8 = 400 pounds, the centre of this load is located, of course, half way between M and R, or four feet from support M, and six feet from support G, therefore M will carry (react)

6.400/10 = 240 pounds and G will carry

4.400/10= 160 pounds.

See Formulae (14) and (15).

We also have a load of 200 pounds at R, transferred from the header on to the trimmer; as B is two feet from G, and eight feet from M, we will find by the same formulae, that G carries

8.200/10= 160 pounds and M carries

2.200/10 = 40 pounds.

Beams.

So that we find the loads which the trimmer transfers to G and M, as follows:

 At M = 250 + 240 + 40 = 530 pounds. ,, G = 250 + 160 + 160 = 570 pounds.

The loads which trimmer O I transfers to wall and girder will, of course, be similar. We therefore find the total loading, as follows:

Walls.

 On the wall A B: At L= 500 pounds. ,, M= 530 pounds. ,, N = 400 pounds. ,, O= 530 pounds. ,, P = 500 pounds. ,, Q = 500 pounds. Total on wall AB = 2960 pounds.

On the wall C D we have six equal loads of 500 pounds each, a total of 3000 pounds.

Fig. 36.

 On the girder E F, we have: At E from the left side 500 pounds, from the right 500 pounds. Total 1000 pounds. At G from the left side 570 pounds, from the right 500 pounds. Total 1070 pounds. At H from the left side nothing, from the right 500 pounds. Total 500 pounds. At I from the left side 570 pounds, from the right 500 pounds. Total 1070 pounds. At K from the left side 500 pounds, from the right 500 pounds. Total 1000 pounds.

At F from the left side 500 pounds, from the right 500 pounds.

Total 1000 pounds.

Total on girder 5640 pounds.

As the girder is neither uniformly nor symmetrically loaded, we must calculate by Formulas (16) and (17), the amount of each reaction, which will, of course, give the load coming on the columns E and F. (These columns will, of course carry additional loads, from the girders on opposite side, further, the weight of the column should be added, also whatever load comes on the column at floor above.)

Girder E F then transfers to columns,

At E = 1000+ (4/5. 1070) + (3/5. 500) + (2/5.1070) + (1/5. 1000) + (0. 1000)== 2784 pounds.

At F = 1000 + (4/5. 1000) + (3/5. 1070) + (2/5. 500) + (1/5. 1070) + (0. 1000) = 2856 pounds.

As a check the loads at E and F must equal the whole load on the girder, and we have, in effect,

2784 + 2856 = 5640.

Now as a check on the whole calculation the load on the two columns and two walls should equal the whole load. The whole load being 20'x6'x 100 pounds minus the well-hole 2'x2'x 100 pounds, or 12000 - 400=11600 pounds. And we have in effect,

 Load on A B = 2960 pounds. " CD = 3000 pounds. " two columns 5640 pounds. Total loads = 11600 pounds.

We therefore can calculate the strength of all the beams, headers and trimmers and girders, with loads on, as above given.

For the columns and walls, we must however add, the weight of walls and columns above, including all the loads coming on walls and columns above the point we are calculating for, also whatever load comes on the columns from the other sides. If there are openings in a wall, one-half the load over each opening goes to the pier each side of the opening, including, of course, all loads on the wall above the opening.

Thus in Figure 37, the weight of walls would be distributed, as