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.

Load on

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:

Load on. .

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.

Load on Clrder.

Load on Clrder

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

Load overwall openings.

indicated by etched lines; where, however, the opening in the wall is very small compared to the mass of wall-spare over, it would, of course, be absurd to consider all tins load as on the arch, and practically, after the mortar has set, it would not be, but only an amount about equal to the part enclosed by dotted lines in Figure 38, the inclined lines being at an angle of 60° with the horizon. Where only part of the wall is calculated to be carried on the opening, the wooden centre should be left in until the mortar of the entire wall has set. In case of beams or lintels the wall should be built up until the intended amount of load is on them, leaving them free underneath; after the intended load is on them, they should be shored up, until the rest of wall is built and thoroughly set. Wind-pressure on a roof is generally assumed at a certain load per square foot superficial measurement of roof, and added to the actual (dead) weight of roof; except in large roofs, or where one foot of truss rests on rollers, when it is important to assume the wind as a separate force, acting at right angles to incline of rafter.

To Ascertain Amount Of Loads 100106

Fig. 3 7.

The load of snow on roofs is generally omitted, when wind is allowed for, as, if the roof is very steep snow will not remain on it, while the wind pressure will be very severe; while, if the roof is flat there will be no wind pressure, the allowance for which will, of course, offset the load of snow.

If the roof should not be steep enough for snow to slide off, a heavy wind would probably blow the snow off.

In case of "continuous girders," that is, beams or girders supported at three or more points and passing over the intermediate supports without being broken, it is usual to allow more load on the central supports, than the formulae (14) to (17) would give. This subject will be more fully dealt with in the chapter on beams and girders.