Question 6.4: A wood beam AB of rectangular cross section serving as a roo...
A wood beam AB of rectangular cross section serving as a roof purlin (Figs. 6-16a and b) is simply supported by the top chords of two adjacent roof trusses. The beam supports the weight of the roof sheathing and the roofing material, plus its own weight and any additional loads that affect the roof (such as wind, snow, and earthquake loads).
In this example, we will consider only the effects of a uniformly distributed load of intensity q = 3.0 kN/m acting in the vertical direction through the centroids of the cross sections (Fig. 6-16c). The load acts along the entire length of the beam and includes the weight of the beam. The top chords of the trusses have a slope of 1 on 2 (α = 26.57°), and the beam has width b = 100 mm, height h = 150 mm, and span L = 1.6 m.
Determine the maximum tensile and compressive stresses in the beam and locate the neutral axis.
Loads and bending moments. The uniform load q acting in the vertical direction can be resolved into components in the y and z directions (Fig. 6-17a):
q_{y} = q \cos \alpha q_{z} = q \sin \alpha (6-24a,b)
The maximum bending moments occur at the midpoint of the beam and are found from the general formula M = qL²/8; hence,
M_{y} = \frac{q_{z}L^{2}}{8} = \frac{qL^{2}\sin \alpha}{8} M_{z} = \frac{q_{y}L^{2}}{} = \frac{qL^{2}\cos \alpha}{8} (6-25a,b)
Both of these moments are positive because their vectors are in the positive directions of the y and z axes (Fig. 6-17b).
Moments of inertia. The moments of inertia of the cross-sectional area with respect to the y and z axes are as follows:
I_{y} = \frac{hb^{3}}{12} I_{z} = \frac{bh^{3}}{12} (6-26a,b)
Bending stresses. The stresses at the midsection of the beam are obtained from Eq. (6-18) with the bending moments given by Eqs. (6-25) and the moments of inertia given by Eqs. (6-26):
\sigma_{x} = \frac{M_{y}z}{}-\frac{M_{z}y}{I_{z}}=\frac{ql^{2}\sin \alpha}{8hb^{3}/12}z-\frac{qL^{2}\cos \alpha}{8bh^{3}/12}y
=\frac{3qL^{2}}{2bh}\left(\frac{\sin\alpha}{b^{2}}z-\frac{\cos\alpha}{h^{2}}y\right) (6-27)
The stress at any point in the cross section can be obtained from this equation by substituting the coordinates y and z of the point.
From the orientation of the cross section and the directions of the loads and bending moments (Fig. 6-17), it is apparent that the maximum compressive stress occurs at point D (where y = h/2 and z = -b/2) and the maximum tensile stress occurs at point E (where y = -h/2 and z = b/2). Substituting these coordinates into Eq. (6-27) and then simplifying, we obtain expressions for the maximum and minimum stresses in the beam:
\sigma_{E}=-\sigma_{D}=\frac{3qL^{2}}{4bh}\left(\frac{\sin\alpha}{b}+\frac{\cos\alpha}{h} \right) (6-28)
Numerical values. The maximum tensile and compressive stresses can be calculated from the preceding equation by substituting the given data:
q = 3.0 kN/m L = 1.6 m b = 100 mm h = 150 mm a = 26.57°
The results are
\sigma_{E}=-\sigma_{D}=4.01_\ MPaNeutral axis. In addition to finding the stresses in the beam, it is often useful to locate the neutral axis. The equation of this line is obtained by setting the stress (Eq. 6-27) equal to zero:
\frac{\sin\alpha}{b^{2}}z-\frac{\cos\alpha}{h^{2}}y=0 (6-29)
The neutral axis is shown in Fig. 6-17b as line nn. The angle β from the z axis to the neutral axis is obtained from Eq. (6-29) as follows:
\tan\beta = \frac{y}{z}=\frac{h^{2}}{b^{2}}\tan\alpha (6-30)
Substituting numerical values, we get
\tan\beta = \frac{h^{2}}{b^{2}}\tan\alpha=\frac{\left(150_\ mm\right)^{2}}{\left(100_\ mm\right)^{2}}\left(\tan 26.57°\right)=1.125 β = 48.4°
Since the angle β is not equal to the angle α, the neutral axis is inclined to the plane of loading (which is vertical).
From the orientation of the neutral axis (Fig. 6-17b), we see that points D and E are the farthest from the neutral axis, thus confirming our assumption that the maximum stresses occur at those points. The part of the beam above and to the right of the neutral axis is in compression, and the part to the left and below the neutral axis is in tension.
