Let Fig. 249 represent the section through a roof and ceiling, showing the shape of the truss and the position of the purlins, and Fig. 250 a plan of the same roof showing the location of the trusses. Then the roof area supported by joint 2 of truss No. 3 will be that portion enclosed within the rectangle h i m l, Fig. 250. As the area of this rectangle is supposed to be measured on the slant of the roof, the true length of the lines l m and h i, is the distance a, Fig. 249, and the area of the rectangle will be the product of h I by the distance a. The point h should be taken half way between truss 2 and truss 3, and the point / half way between truss 3 and the front wall. The distance a should be one half of the distance from 1 to 2, Fig. 249, plus one half of the distance 10'4" + 12'4" /2 = from 2 to 3. In this particular example, a= 11'4" and h I = 5' + 6' = 11 ft., hence the roof area supported at joint 2 of truss 3 ='11' x 11 1/3'= 124 2/3 sq. ft.

Example I 300252

Fig. 249.

Example I 300253

Fig. 250.

The roof area supported by joint 3 of truss 3 is the portion enclosed within the rectangle i k m n, whose area = i mx2b (Fig. 249), which in this example = 11'x 12' 4" = 135 2/3 sq. ft.

The lengths a and b apply to all trusses of the same roof, but the horizontal distances may vary as in Fig. 250. Unless there is some particular reason for spacing the trusses unevenly, they are generally spaced uniformly, so that the trusses will all have the same loads.

Where the roof is hipped or has valleys, the valleys and hips are apt to change the loads slightly from what they would be if the roof were straight, but as a rule the difference is so slight that it is customary to estimate the loads as though the roof were straight. Thus the roof area supported at joint 3 of truss 2 may be taken as the area efki=(4' 5" + 6') x 12' 4" = 128 1/2 sq. ft.

The roof area at joint 3 of truss 1 may be taken as bcfe = bc xb c. In this case b c = a = 11' 4" and b c = half the horizontal distance between joints 2 and 4 = 8'10". [Note. - Truss I is shown by the dotted lines in Fig. 249.] The roof loads on the right hand side of the truss are the same as those on the left hand side.

112. Exact Method Of Figuring Roof Area Supported At Joints When The Roof Is Hipped

When the spacing between trusses and purlins is 16 ft. or more, and the roof rather flat, the exact loads transmitted by the hips should be calculated after the following manner: -

Let Fig. 251 represent the plan of a portion of a hip roof, the dot and dash line passing through joints 2 and 4, being half way between side walls, and the line passing through k I s, being half way to the next truss.

The dimensions between arrow heads are the horizontal dimensions. The pitch of the roof is such that the length of the lines b d, e d, i h, n m, etc., measured on the slant of the roof, is exactly 6 ft. Assuming that the rafters are well spiked to the hips, then the latter will act as beams transferring the roof load to the joints in the same manner as a horizontal beam. On this assumption the roof areas supported by the hips are shown by the shaded portions.

Fig. 231.   Partial Plan of High Roof.

Fig. 231. - Partial Plan of High Roof.

The joints of the trusses are supposed to be at the points 1, 2, 3 and 4. The roof area supported at joint 1 is made up of the reaction of the two hips; of the rectangles b d h i and c f g d, and also of a portion of each of the triangles d n m and d t v. The area supported by the hip 0-1 equals the sum of the triangles o b d and b d x L 6x9

0 e d. The area of the first is equal to = = 27 sq. ft.

As the triangles have the same area, the total area supported by the hip = 54 sq. ft. To find what portion of this load is transmitted to 1 and what to o, we must find the centre of gravity of the triangles. This can readily be done by drawing a line from two angles to the centre of the opposite sides, as b a and 0 x. The point where they intersect will be the centre of gravity of the triangle. Joining the two points c and c' by a line, the point C where the line crosses the hip will be the centre of gravity of the entire shaded portion. By the principle of the lever, the load at entire area x 0 C

1 is to the entire load as 0 C is 01, or load at 1 = ------------------ .

01.

For the lengths of 0 C and 01 we can take the horizontal measurement, as all that we require is the ratio. 0 C measures 89" and 54 x 89 01, 153". Then the area supported at 1 = = 31.4 sq. ft.

As the roof area supported by the hip 1-4 is the same as that supported by 0-1, the reaction at 4 from one hip will also be 31.4, and the reaction at 1 from hip 1-4 must be 54 - 31.4= 18.6 sq. ft. Consequently joint 1 supports 54 sq. ft. of roof area transmitted by the two hips, which is just one half of what it would have to support if the roof were straight.

The area of each of the rectangles b dih and d e f g = 6' x 4 1/2' = 27 sq. ft. The purlin 1-3 must support the roof area represented by the triangle d n m, and the top chord of truss 1 the triangle d v t. The centre of gravity of triangle d n m is at p', and the proportion of the area transmitted to joint 3 is equal to dp/dn, and as in a right triangle the point p is at 1/3 the distance d n from 3,

---- = 2/3. The area of d n m = ----- = 27, and 1/3 of this is transmitted to joint 1. By the same analysis, 1/3 of dvt is transmitted to joint 1. Then total roof area supported at joint 1 =

Reaction of two hips...................

=

54

Areas of rectangles b d h i and d e f g ....................

=

54

1/3 of each of triangles d n m and d v t ..................

=

18

Total area......................

126 sq. ft.

Roof Area Supported at Joint 3. - The roof area supported at this joint will be § of the area of the triangle dn m plus the area of the. L-shaped figure hiklmn. The area of dmn we found before to be 27 sq. ft., and § of this is 18 sq. ft.

The area of the L-shaped figure equals three times the area of b d h 1 = 3x27 = 81 sq. ft. Therefore the entire roof area supported at joint 3 = 81 + 18 = 99 sq. ft.

Roof Area Supported at Joint 2. - As but one half of the roof area supported at this joint is shown in Fig. 251, we will find half the area and multiply by two. From an inspection of the figure we find that the area at one side of the centre line is made up of the rectangle f w vg - 27 sq. ft. + 2/3 of the triangle dvt = 18 sq. ft. Hence the half area = 45 sq. ft., and the whole area 90 sq. ft.

Roof Area Supported at Joint 4. - The roof area contributory to joint 4, on each side of the centre line, is its proportional part of the shaded area supported by the hip, which we found before to be 31.4 sq. ft., and the area of the rectangle Imrs, which is 27 sq. ft. Consequently, the entire roof area supported at joint 4 = 116.8 sq. ft.

Comparison of Actual Roof Areas With What They Would Be For a Straight Gable Roof. - From the above calculations we find the entire roof area which must be supported by the six joints of trusses 1 and 2 to be -

Area at joint 2...................

=

90.0

sq. ft.

Area at joint 4..................

=

116.8

"

Areas at joints 1 and 1'.............

=

252.0

"

Areas at joints 3 and 3'....

=

198.0

"

656.8

sq. ft.

[Note. - Joints 1' and 3' are the joints corresponding to joints 1 and 3 for the other half of the roof, not shown.]

If the roof were straight without hips the area supported at each of the six joints would be 9' x 12' = 108 sq. ft., or a total of 648 sq. ft. Therefore the actual area is 8.8 sq. ft. in excess of what it would be if the roof were straight.

Taking the trusses separately, truss 1 has to support an actual roof area of 342 sq. ft., or 18 sq. ft. more; and truss 2, 314.8 sq. ft., or 9.4 less than if the roof were straight. These differences, however, amount to only 5 1/2% for truss 1 and less than 3% for truss 2, so that except where the trusses are spaced 16 ft. or more apart, and the roof is of heavy construction, the error in figuring the roof areas as in Example 1, is not sufficient to effect the actual construction of the trusses.

With steep roofs, moreover, it is probable that the hips would not take their full share of the load, so that the joint loads obtained by assuming the areas to be the same as for a straight roof would be as close to the actual conditions as can be estimated.

The method above explained for figuring the reaction from hips also applies to valleys, only that with valleys the bottom of the rafters are supported and the full estimated load is sure to come upon them, and possibly more.