CAPACITY OF LAG SCREWS
In Fig. 6, the cottonwood members are connected wtih three^5 Xs-In (15.88-mm) lag screws
8 in (203.2 mm) long. Determine the load P that may be applied to this connection.
Calculation Procedure:
- Determine the member group
The National Design Specification shows that cottonwood is classified in group IV. - Find the allowable screw/ loads
The National Design Specification gives the following values for each screw: allowable
load parallel to grain = 550 Ib (2446.4 N); allowable load normal to grain = 330 Ib
(1467.8 N). - Compute the allowable load on the connection
Use the Scholten nonibgram, or N= PQI(P sin^2 B + Q cos^2 O), with 0 = 50°, and solve as
given earlier. Either solution gives P = 3(395) = 1185 Ib (5270.9 N).
DESIGN OFA BOLTED SPLICE
A 6 x 12 in (152.4 x 304.8 mm) southern pine member carrying a tensile force of 56 kips
(249.1 kN) parallel to the grain is to be spliced with steel side plates. Design the splice.
Calculation Procedure:
- Determine the number of bolts, and bolt size, required
Find the bolt capacity from the National Design Specification. The Specification allows a
25 percent increase in capacity of the parallel-to-grain loading when steel plates are used
as side members.
Determine the number of bolts from n = /Vcapacity per bolt, Ib (N), where P = load,
Ib (N). By assuming^5 / 8 -in (22.2-mm) diameter bolts, n = 56,000/[3940(1.25)] = 11.4;
use 12 bolts. The value 1.25 in the denominator is the increase in bolt load mentioned
above.
As a trial, use three rows of four bolts each, as shown in Fig. 7.
FIGURE 6