This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

1 1 150 ÷ 3 130 = 4 shop rivets. Each End of L1 U2 and U2 L3: Rivets ⅝-inch. Plate ¼-inch.

6 240 ÷ 3 070 = 2 shop rivets. Each End of Ul L1 and U3 L3: Rivets ⅝-inch. Plate ¼-inch.

5 580 ÷ 3 070 = 2 shop rivets.

Where U1Ll and U3 L3 join the top chord, two rivets will he required in the top chord.

Since the components of the two diagonals meeting at U2 are parallel and equal, and opposite to the stress in U2 L2, no rivets will be required, theoretically, to hold the plate to the top chord. A sufficient number, however, must be put in to take up the vertical reaction of the purlin. This number is 5 550 ÷ 3 130 = 2 shop rivets. In practice a greater number are usually put in to prevent vibration and to fill out the plate.

At L3 a sufficient number of rivets must be placed in L2 U4 to take up the difference in stress between L3 U4 and L2 L3. The number required is (18 725 - 12 475) ÷ 3 130 = 3 shop rivets.

At the end L0 of the member L0 Lv there is a horizontal stress of 43 700 pounds, and a vertical force equal to the reaction, which is 49 930 ÷ 2= 24 965 pounds (see Fig. 76). The force acting on the rivets in this member is the resultant of these two forces, and is:

= 50 300 pounds.

Since the rivets are ¾-inch and the plates ⅜-inch, the number of rivets required is 50 300 ÷ 5 630 = 9 shop rivets. This number should be placed symmetrically with respect to the intersection of the two chords. In case the point of application of the reaction had not coincided with the intersection of the chords, the number of rivets must be computed according to the formula on page 36.

For the joint at L1, a sufficient number of rivets must be put in, in order to take up the difference in stress between the members L0 L1 and L1 L2. The number required is (43 700 - 37 450) ÷ 3 750 = 2 shop rivets.

The purlins have a horizontal shear at each end, of H ÷ 2 = 2 490 ÷ 2=1 245 pounds. This requires 1 245 ÷ 4 420 = 1 shop rivet or 1 field rivet, to keep them from sliding down on the top chord. Clip angles 5 by 3½ by ⅜-inch will be used as shown in Plate I-

These help in the erecting of the purlins, since they are shop-riveted to the truss and therefore hold the purlins in place while they are being field-riveted to the truss and to the clip angles (see Fig. 77).

Rivets in Lateral Bracing. The plates of the lateral bracing should be made ¼ inch thick. The 3-inch leg of the angle will be placed against the plate. Rivets | inch in diameter can then be used, and the strength of the joint will be governed by bearing in the ¼-inch plate. The stress for which the rivets are to be determined is given on p. 55. It is 27 000 pounds. The number of field rivets in bearing in ¼-inch plate, required to withstand the stress, is (27 000 ÷ 4 420) X 4/3 = 9. The size and shape of the plate can be determined only while making the detailed drawing (see Plate III, p. 01).

Design of the Splice. The general details of the splice will be as shown in Fig. 68. The plate underneath will be made ¼ inch thick, the same thickness as the vertical connection plate at this point. Note that the member on the left-hand side of the splice must have ¾-inch shop rivets, and the member on the right-hand side must have ⅝-inch field rivets. The total number of rivets on either side of the splice must be sufficient to take up the entire stress of the member through which they are driven. If eight ⅝-inch field rivets are driven through the horizontal legs and the bottom splice plate, and five ⅝-inch field rivets are driven through the vertical plate and legs of the angles (see Fig. 78), the total strength of the joint, remembering that the rivets are ⅝-inch, will be:

Fig. 77. Detail Showing Clip Angle Connection.

Fig. 78. Detail of Lower Chord Splice.

8 | x | ⅔ | X | 3 070 | = | 16 370 pounds. |

5 | X | ⅔ | X | 3 130 | = | 10 430 pounds. |

Total | 26 800 pounds. |

Note that the rivets through the bottom splice plates are governed by single shear; and those through the vertical plate, by bearing in the plate. Since 16 370 pounds is the value of the rivets through the bottom splice plate, this amount will be transmitted to the other side, where it must be taken up by shop rivets. Bearing in the plate governs the number of ¾-inch shop rivets required. This number is 16 370 ÷ 3 750 = 5. Since 16 370 pounds of the stress in the member L1 L2 is taken up by these 5 shop rivets, the remainder, 37 150 - 16 370 = 21 0S0 pounds, must be taken up by the rivets through the vertical connection plate. This requires 21 080 ÷ 3 750 = 6 shop rivets.

Since 16 370 pounds is transmitted from one side of the splice to the other by means of the bottom splice plate, this plate should be 16 370 ÷ 15 000 = 1.09 square inches in net section. The net width, the plate being ¼ inch thick, is 1.09 ÷ 0.25 = 4.36 inches. If two ¾-inch rivet-holes are taken out of the section, the entire width of the plate must be 4.36 + 2 (¾ + ⅛ ) = 6.11, say 7 inches wide. The length of the plate must be sufficient to get in the number of rivets, and this length is determined in detailing.

Design of the Masonry Plate. If this truss rested upon a masonry wall, it would require a bearing of (49 930 ÷ 2) ÷ 250 = 100 square inches. The width of the plate cannot be less than twice the width of the legs of the bottom chord angle, nor should it extend outside the legs of the chord angle more than 3 inches on each side. The masonry plate will be assumed as 12 inches wide, in which case it must be 100 ÷12 = 8.34, say 8½ inches long. The thickness should be ½ inch. Temperature Allowance. Slotted holes must be put in one end of the truss, to allow for a variation of 150 degrees in temperature. A common rule is to allow ⅛-inch expansion for every ten feet of span. The total allowance for expansion is 64 X ⅛ = say, 1 inch. Since the bolts which go through this hole are f inch in diameter, the hole must be long enough to allow for ½ the expansion on each side. The width of the hole should be ¼ inch greater than the diameter of the bolt (see Fig. 79).

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