FIGURE 1
sion and compression flange. Then A = 3(8.93) = 26.79 in^2 (170.268 cm^2 ); I 0 = 3(22.5) =
67.5 in^4 (2809.56 cm^4 ).
- Design the webs
Use the approximation tw = \25Vldvn = 1.25(9900)/[40(192)] = 1.61 in (40.894 mm). Try
two^7 /s-in (22.2-mm) thick plywood webs. A catalog of plywood properties reveals that
the^7 /8-in (22.2mm) member consists of seven plies and that the parallel plies have an ag-
gregate thickness of 0.5 in (12.7 mm). Draw the trial section as shown in Fig. 1.
5. Check the bending stress in the member
For simplicity, disregard the webs in evaluating the moment of inertia. Thus, the moment
of inertia of the flanges 7^= 2(67.5 + 26.79 x 17.25^2 ) = 16,080 in^4 (669,299.448 cm^4 );
then the stress/= McII= 1,070,000(20)716,080 = 1330 < 1350 lb/in^2 (9308.25 kPa). This
is acceptable.
6. Check the shearing stress at the contact surface of the flange
and web
Use the relation Qf=Ad = 26.79(17.25) = 462 in^3 (7572.2 cm^2 ). The q per surface =
VQf f(2If) = 9900(462)/[2(16,080)] = 142 Ib/lin in (24.8 kN/m). Assume that the shearing
stress is uniform across the surface, and apply 36 lb/in^2 (248.2 kPa), as noted earlier, as
the allowable stress. Then, v = 142/5.5 = 26 lb/in^2 (179.3 kPa) < 36 lb/in^2 (248.2 kPa).
This is acceptable.
7. Check the shearing stress in the webs
For this purpose, include the webs in evaluating the moment of inertia but apply solely
the area of the parallel plies. At the neutral axis Q = Qf+ Qw = 462 + 2(0.5)(20)(10) =
662 in^3 (10,850.2 cm^3 ); / - If+ Iw = 16,080 + 2(1/12)(0.5)(40)^3 = 21,410 in^4 (89.115
dm^4 ). Then v = VQI(If) = 9900(662)/[21,410(2)(0.875)] = 175 lb/in^2 (1206.6 kPa) < 192
lb/in^2 (1323.8 kPa). This is acceptable.
8. Check the deflection^ applying the moment of inertia of only
the flanges
Thus, A = (7.5/384)wL^4 /(£7/) = 7.5(550)(36)^4 (1728)/[384(1,760,00O)(16,08O)] = 1.10 in
(27.94 mm); ML = 1.10/[36(12)] < 1/360. This is acceptable, and the trial section is there-
fore satisfactory in all respects.
