With (f> = 30°, sin 30° = 0.500; sin 15° = 0.259; cos 15° = 0.966; tan 15° = 0.268.
- Find the lengths AC and BC
Express these two lengths as functions of AB. Or, AB = b/sin </>); AC = [b sin (<|>/2)]/
sin <£); BC = [b cos (</>)/2)]/sin <£); AC = 3.625(0.259/0.500) = 1.9 in (48.26 mm);
BC = 3.625(0.966/0.500) = 7.0 in (177.8 mm). The projection into Ml is therefore not ex-
cessive. - Evaluate the stresses f 1 and f 2
Resolve F into components parallel to AC and BC. Thus,/! = (F sin <f))/(A tan </>/2);f 2 = (F
sin <£)[tan (<£/2)]A4, where A = crossectional area of Ml. Substituting gives/j = 783 lb/in^2
(5399 kPa);/ 2 = 56 lb/in^2 (386.1 kPa). - Calculate the allowable stresses
Compute the allowable stresses NI and N 2 on A C and BC, respectively, and compare these
with the actual stresses. Thus, by using Hankinson's equation from the previous calcula-
tion procedure, N 1 = 1200(390)/(1200 x 0.259^2 + 390 x 0.966^2 ) = 1053 lb/in^2 (7260.4
kPa). This is acceptable because it is greater than the actual stress. Also, N 2 =
1200(390)/(1200 x 0.966^2 + 390 x 0.259^2 ) = 408 lb/in^2 (2813.2 kPa). This is also accept-
able, and the joint is therefore satisfactory.
ALLOWABLE LATERAL LOAD ON NAILS
In Fig. 5, the Western hemlock members are connected with six 5OJ common nails. Cal-
culate the lateral load P that may be applied to this connection.
Calculation Procedure:
- Determine the member group
The capacity of this connection is calculated
in conformity with Part VIII of the National
Design Specification. Refer to the Specifica-
tion to ascertain the classification of the
species. Western hemlock is in group III. - Determine the properties of
the nail
Refer to the Specification to determine the
properties of the nail. Calculate the penetra-
tion-diameter ratio, and compare this value
with that stipulated in the Specification. Thus, FIGURE 5
length = 5.5 in (139.7 mm); diameter = 0.244
in (6. 1976 mm); penetration/diameter ratio =
(5.5 - 1.63)70.244 = 15.9 > 13. This is accept-
able. - Find the capacity of the connection
Using the Specification, find the capacity of the nail. Then the capacity of the connection
= P = 6(165) = 990 Ib (4403.5 N).