There are two possible location to shear the screws. The first location
is the horizontal joints and involves the full top section (label Area A in
first diagram). The left side is part of the left 2x6 and it is assumed the wood
will not shear. The second location is the vertical joints and involves only
part of the top section (label Area B in the second diagram). In both cases,
there are two surfaces that will carry the shear load.
First determine Q for both cases.
Q_{A} = A_{A}y_{A} = (7.0)(1.5)(3.5  0.75)
= 28.875 in^{3}
Q_{B} = A_{B}y_{B} =
(4.0)(1.5)(3.5  0.75) = 16.5 in^{3}
Next, the moment of inertia for the full cross sections needs to be known.
I = 7(7)^{3}/12  4(4)^{3}/12
= 178.75 in^{4}
Using the shear stress equation gives,
This shear stress needs to be carried by the screws. Each screw can carry 800
lb. The minimum spacing (s_{A} and s_{B}) of the screws can be
found using
F_{Screw} = τ A
For each surface, this gives the spacing as
800 lb = (172.3 lb/in^{2}) [(1.5 in)
s_{A}]
s_{A} = 3.095 in
800 lb = (98.46 lb/in^{2}) [(1.5 in) s_{B}]
s_{B} = 5.417 in
Thus, the screws should be every 3 inches on the horizontal joints and every
5 inches on the vertical surfaces.
