For trusses unsymmetrically loaded, the simplest method of finding the amount of each supporting force is by the method of moments, which is as follows:

First draw a diagram of the truss, representing the center lines of the members.

Trusses Unsymmetrically Loaded 300264

Fig. 261.

2d - Indicate the loads in their correct position, by arrows, with the amount of the load in figures above, as in Figs. 260 and 261, and then scale the horizontal distances from the loads to the right-hand support (corresponding to the distances a, b, c, etc.) as accurately as possible.

3d - Multiply each load by its distance from the right support, add together the products, and divide their sum by the span; the quotient will be the amount of the left support, or P1.*

4th - From the sum of the loads subtract the value of Pl ,and the remainder will be the amount of P2. Thus in either of Figs. 260 or 261.

W1 x a + W2 x b + W3 x c + W4 x d

P1= ----------------------------------s

P2 = (W1 + W2 + W3 + W4) - P1.

It is important to take the measurements a, b, c, etc., from the truss diagram, or from the intersections of the center lines, because it is these lines which are used in drawing the stress diagram, and that the latter shall work out correctly, the supporting forces must be determined as accurately as possible. The method of drawing the truss diagram is more fully explained in Section 129.

120. Example 9

Find the supporting forces for the truss shown by Fig. 262, the loads being in pounds.

Ans. - Multiplying each load by its distance from the right support, and adding together the products, we have:.

3,800

X

28.......................................

=

100,400

4,300

X

20.......................................

=

86,000

11,500

X

20.......................................

=

230,000

4,800

X

10......................................

=

48,000

24,400

470,400

Dividing the sum of the products (470,400) by the span (36), we obtain for the quotient 13,066 lbs., which is the value of P1. Subtracting this amount from the sum of the loads, we have 11,334 lbs. as the value of P2.

Example 10

A good example of an unsymmetrical truss, unsymmetrically loaded, is shown in Fig. 263, which represents one of the half diagonal trusses, shown on the plan, Fig. 162. [Although these trusses have been called half trusses, each of them is a full and complete truss, the term half truss being used because they only extend from the supporting corner to the center of the through truss.] Computing the roof and ceiling areas contributory to the joints, in the manner illustrated by Example 4, and multiplying by 42 lbs. per sq. ft. for the roof, and 15 lbs. for the ceiling, we obtain the joint loads indicated in the figure. The horizontal distances are measured to the intersection of center lines. Taking moments about P., we have for moments

*It is immaterial whether the moments are taken about the right or the left support, but if taken about the left support, the quotient will be the amount of the right support.

(2.360 + 450)

x

7 1/2'..............................

=

21,075

(4.800 + 5,000)

x

15'........................

=

147,000

Sum of loads

=

12,610 lbs.;

sum of moments. .

=

168.075 ft lbs.

Dividing the sum of the moments by the distance between supports, 233 ft., we obtain 7,100 lbs. as the reaction at P2, and P1 must be the difference between the sum of the loads and P2, or 5,510 lbs. As there are two of the half trusses, the through truss must be figured for a load at the center of 14,200 lbs., in addition to the weight of the roof and ceiling which it supports directly.

Example 10 300265

Fig. 262.

Example 10 300266

Fig. 263.

It should be remembered that in computing the moments of a force, the arms must always be measured perpendicular to the direction of the force, and as dead loads always act vertically, the arm must always be measured on a horizontal line.

Other examples of finding the supporting forces are given in Chapter VIII (Architectural Terra Cotta).