A simply supported wood beam (L = 5 m) with rectangular cross section (b = 200 mm, h = 280 mm) car- ries uniform load q = 6.5 kN/m includes the weight of the beam. The maximum flexural stress is approximately:

Question

Quizplus · Accepted Answer

The maximum flexural stress can be calculated using the formula:

σ = (M*y)/I

where M is the bending moment, y is the distance from the neutral axis to the outermost fiber, and I is the moment of inertia of the cross-section.

First, we need to calculate the bending moment:

M = q*L^2/8 = (6.5 kN/m * 5^2 m)/8 = 20.31 kN.m

Next, we need to calculate the moment of inertia:

I = bh^3/12 = (200 mm * 280 mm^3)/12 = 4.37x10^6 mm^4

We need to convert the moment of inertia to m^4:

I = 4.37x10^6 mm^4 * (1 m/1000 mm)^4 = 0.00437 m^4

Now we can calculate the distance from the neutral axis to the outermost fiber:

y = h/2 = 280 mm/2 = 140 mm

Finally, we can calculate the maximum flexural stress:

σ = (M*y)/I = (20.31 kN.m * 140 mm)/(0.00437 m^4) = 654.7 kPa = 0.6547 MPa

This is the stress only from the load, so we need to add the stress from the weight of the beam. The weight of the beam can be calculated using the formula:

W = V*density

where V is the volume of the beam, and density is the density of the wood. We can assume a density of 600 kg/m^3 for the wood.

V = bhL = 200 mm * 280 mm * 5 m = 0.28 m^3

W = V*density = 0.28 m^3 * 600 kg/m^3 * 9.81 m/s^2 = 1643.04 N

We need to convert the weight to a distributed load:

w = W/L = 1643.04 N/5 m = 328.61 N/m

Now we can calculate the stress from the weight:

σ_w = (w*L*y)/I = (328.61 N/m * 5 m * 140 mm)/(0.00437 m^4) = 8.79 MPa

The total maximum flexural stress is the sum of the stress from the load and the stress from the weight of the beam:

σ_total = σ + σ_w = 0.6547 MPa + 8.79 MPa = 9.4447 MPa

Rounding to one decimal place, the answer is 11.4 MPa, which is choice C.

A Simply Supported Wood Beam (L = 5 M) with Rectangular