We may now resume the construction of the polygons for the joints on the left side. At joint 4, we know the forces in the members kn and kr hence there are only two unknown forces there. The polygon for the joint is KNORK, NO and OR representing the stresses in no and or. The polygon for joint 5 may be drawn next ; it is ONMCDPO, DP and PO representing the stresses in dp and po. The polygon for joint 6 or joint 7 may be drawn next ; for 6 it is PDEQP, EQ and QP representing the stresses in eq and qp. At joint 7 there is but one unknown force, and it must close the polygon for the known forces there. That polygon is ROPQ ; hence QR represents the unknown force. (If the work has been correctly and accurately done, QR will be parallel to qr).

"When the wind blows upon the right side, the values of the reactions, and the stresses in any two corresponding members, are reversed. Thus, when the wind blows upon the left side, the stresses in kl and hx equal 18,300 and 0,100 pounds respectively ; and when it blows upon the right they are respectively 6,100 and 18,300 pounds. It is not necessary, therefore, to construct a stress diagram for the wind pressure on the right. The numbers in the fifth column (see table, Page 72) relate to wind right, and were obtained from those in the fourth.

4I. Combination of Dead, Snow, and Wind=Load Stresses. After having found the stress in any member due to the separate loads (dead, snow, and wind), we can then find the stress in that member due to any combination of loads, by adding algebraically the stresses due to loads separately. Thus, in a given member, suppose:

Dead-load stress

=

+

10,000 pounds,

Snow-load "

=

+

15,000

Wind-load " (right)

=

-

12,000

" " (left)

=

+

4,000

Since the dead load is permanent (and hence the dead-load stress also) the resultant stress in the member when there is a snow load and no wind pressure, is

+ 10,000 + 15,000 = + 25,000 pounds (tension); when there is wind pressure on the right, the resultant stress equals

+ 10,000 - 12,000 = - 2,000 pounds (compression) ; when there is wind pressure on the left, the resultant stress is

+ 10,000 + 4,000 = + 14,000 pounds (tension) ; and when there is a snow load and wind pressure on the left, the resultant stress is

+ 10,000 + 15,000 + 4,000 = + 29,000 pounds (tension).

If all possible combinations of stress for the preceding case be made, it will be seen that the greatest tension which can come upon the member is 29,000 pounds, and the greatest compression is 2,000 pounds.

In roof trusses it is not often that the wind load produces a " reversal of stress " (that is, changes a tension to compression, or vice versa); but in bridge trusses the rolling loads often produce reversals in some of the members. In a record of stresses the reversals of stress should always be noted, and also the value of the greatest tension and compression for each one.

The numbers in the sixth column of the record (Page 72) are the values of the greatest resultant stress for each member. It is sometimes assumed that the greatest snow and wind loads will not come upon the truss at the same time. On this assumption the resultant stresses are those given in the seventh column.

Example For Practice

1. Compile a complete record for the stresses in the truss of Fig. 24, for dead, snow, and wind loads. See Example 1, Article 27, for values of dead-load stresses, and Example 2, Article 29, for values of the wind-load stresses. Assume the snow load to equal 1.2 times the dead load.

After the record is made, compute the greatest possible stress in each member, assuming that the wind load and snow load will not both come upon the truss at the same time.

The greatest resultant stresses are as follows:

Member.

af

fe

bg

fg

gh

hi

hc

ie

id

Resultant ..

_

14,950

+

17,800

_

10,400

-

8,875

+

7,820

_

8,875

_

11,800

+

11,5(30

_

13,850

42. Truss Sustaining a Roof of Changing Slope. Fig 46 represents such a truss. The weight of the truss itself can be estimated by means of the formula of Art. 19. Thus if the distance between trusses equals 12 feet, the weight of the truss equals

W = 12 X 32 (32/25+ 1)= 875 pounds.

The weight of the roofing equals the product of the area of the roofing and the weight per unit area. The area equals 12 times the sum of the lengths of the membersVII Analysis Of Trusses Continued Method Of Sectio 0300236 ,,and that is, 12 X 36 = 438 square feet. If the roofing weighs 10 pounds per square foot, then the weight of roofing sustained by one truss equals 438 X 10 = 4,380 pounds. The total dead load then equals

875 + 4,380 = 5,255 pounds; and the apex dead loads for joints 2, 3, and 4: equal:

1/4 x 5,255 = 1,314 (or approximately 1,300) pounds; while the loads for joints 1 and 5 equal

1/8 X 5,255 = 657 (or approximately 650) pounds.

The snow loads for the joints are found by computing the snow load on each separate slope of the roof. Thus, if the snow

VII Analysis Of Trusses Continued Method Of Sectio 0300240

Fig. 46.

weighs 20 pounds per square foot (horizontal), the load on the portionVII Analysis Of Trusses Continued Method Of Sectio 0300241 equals 20 times the area of the horizontal projection of the portion of the roof represented byThis horizontal projection equals 8 X 12 (= 96) square feet; snow load equals 96 X 20 (= 1,920) pounds. This load is to be equally divided between joints 1 and 2.

In a similar way the snow load borne byVII Analysis Of Trusses Continued Method Of Sectio 0300243 equals 20 times the area of the horizontal projection of the roof represented by; this horizontal projection equals 8 X 12 ( = 96) square feet as before, and the snow load hence equals 1,920 pounds also. This load is to be equally divided between joints 2 and 3.