of rupture. For beams more than 16 in (406.4 mm) deep, it is necessary to allow for this
reduction in strength by introducing a depth factor F.
Thus, F= 0.81 (d
2
+ 143)/(</
2
+ 88), where d = dressed depth of beam, in. Substituting
gives F= 0.81(19.5^2 + 143)/(19.5^2 + 88) = 0.905.- Apply the result of step 1 to obtain the moment capacity
Use the relation M = FfS 9 where the symbols are as given earlier. Thus, M = 0.905 x
(1.5)(728.8)/12 = 82.4 ft-kips (111.73 kN-m).
DESIGN OFA WOOD-PLYWOOD BEAM
A girder having a 36-ft (11.0-m) span is to carry a uniform load of 550 Ib/lin ft (8026.6
N/m), which includes its estimated weight. Design a box-type member of glued construc-
tion, using the allowable stresses given in the table. The modulus of elasticity of both ma-
terials is 1,760,000 lb/in^2 (12,135.2 MPa), and the ratio of deflection to span cannot ex-
ceed 1/360. Architectural details limit the member depth to 40 in (101.6 cm).
Lumber Plywood
Tension, lb/in^2 (kPa) 1,500 (10,342.5) 2,000 (13,790.0)
Compression parallel to grain, lb/in^2 (kPa) 1,350 (9,308.3) 1,460 (10,066.7)
Compression normal to grain, lb/in^2 (kPa) 390 (2,689.1) 405 (2,792.5)
Shear parallel to plane of plies, lb/in^2 (kPa) 72* (496.4)
Shear normal to plane of plies, lb/in^2 (kPa) 192 (1,323.8)*Use 36 lb/in^2 (248.2 kPa) at contact surface of flange and web to allow for stress concentration.Calculation Procedure:- Compute the maximum shear and bending moment
Thus, V =^1 / 2 (550)(36) = 9900 Ib (44,035.2 N); M =^1 A(WL^2 )^ =^1 / 8 (550)(36)^212 =
1,070,000 in-lb (120,888.6 N-m). To preclude the possibility of field error, make the ten-
sion and compression flanges alike. - Calculate the beam depth for a balanced condition
Assume that the member precisely satisfies the requirements for flexure and deflection,
and calculate the depth associated with this balanced condition. To allow for the deflec-
tion caused by shear, which is substantial when a thin web is used, increase the deflection
as computed in the conventional manner by one-half. Thus, M =fllc = 2fl/d = 27007/c/,
Eq. 0.A = (1.5/4S)L^2 Mf(EI) = £/360, Eq. b.
Substitute in Eq. b the value of M given by Eq. a; solve for d to obtain d = 37.3 in
(947.42 mm). Use the permissible depth of 40 in (1016.0 mm). As a result of this increase
in depth, a section that satisfies the requirement for flexure will satisfy the requirement
for deflection as well. - Design the flanges
Approximate the required area of the compression flange; design the flanges. For this pur-
pose, assume that the flanges will be 51 A in (139.7 mm) deep. The lever arm of the result-
ant forces in the flanges will be 34.8 in (883.92 mm), and the average fiber stress will be
1165 lb/in^2 (8032.7 kPa). Then A = l,070,000/[ 1165(34.8)] = 26.4 in^2 (170.33 cm^2 ). Use
three 2 x 6 in (50.8 x 152.4 mm) sections with glued vertical laminations for both the ten-