A simply supported wood beam of rectangular cross section and span length 1.2 m carries a concentrated load P at midspan in addition to its own weight (see gure). The cross section has width 140 mm and height 240 mm. The weight density of the wood is 5.4 kN/m3. Calculate the maximum permissible value of the load P if (a) the allowable bending stress is 8.5 MPa and (b) the allowable shear stress is 0.8 MPa.

(a) P = 37.97 kN

(b) P = 35.62 kN

Explanation:

In the given problem, using the parameters and figure provided, we have:

A = b*h = 0.14*0.24 = 0.0336 m^2

S = (b*h^2) / 6 = (0.14*0.24*0.24) / 6 = 0.001344 m^3

q = 5400*0.0336 = 181.44 N/m

(a) The load P when σ (allowable bending stress) = 8.5 MPa = 8.5*10^6 Pa

σ = M/S

Thus: M = σ*S = 0.001344*8.5*10^6 = 11424 Nm

In addition,

M = (P*L/4) + (q*L^2) / 8

11424 = (P*1.2/4) + (181.44*1.2^2) / 8

11424 = 0.3*P + 32.6592

P = (11424-32.6592) / 0.3 = 37.97 kN

(b) The value of P if τ (allowable shear stress) = 0.8 MPa = 0.8*10^6 Pa

(2*A*τ/3) = P/2 + q*L/2

(2*0.0336*0.8*10^6) / 3 = P/2 + 181.44*1.2/2

17920 = P/2 + 108.864

P = (17920 - 108.864) * 2 = 35.62 kN