Question 6.5: A 12-foot long cantilever beam (Fig. 6-18a) is constructed f...
A 12-foot long cantilever beam (Fig. 6-18a) is constructed from an S 24 × 80 section (see Table E-2 of Appendix E for the dimensions and properties of this beam). A load P = 10 k acts in the vertical direction at the end of the beam.
Because the beam is very narrow compared to its height (Fig. 6-18b), its moment of inertia about the z axis is much larger than its moment of inertia about the y axis.
(a) Determine the maximum bending stresses in the beam if the y axis of the cross section is vertical and therefore aligned with the load P (Fig. 6-18a).
(b) Determine the maximum bending stresses if the beam is inclined at a small angle α = 1° to the load P (Fig. 6-18b). (A small inclination can be caused by imperfections in the fabrication of the beam, misalignment of the beam during construction, or movement of the supporting structure.)
(a) Maximum bending stresses when the load is aligned with the y axis. If the beam and load are in perfect alignment, the z axis is the neutral axis and the maximum stresses in the beam (at the support) are obtained from the flexure formula:
\sigma_{\max}=\frac{My}{I_{z}}=\frac{PL(h/2)}{I_{z}}in which M = PL is the bending moment at the support, h is the height of the beam, and I_{z} is the moment of inertia about the z axis. Substituting numerical values, we obtain
\sigma_{\max}=\frac{(10 k)(12 ft)(12 in./ft)(12.00 in.)}{}=8230 psiThis stress is tensile at the top of the beam and compressive at the bottom of the beam.
(b) Maximum bending stresses when the load is inclined to the y axis. We now assume that the beam has a small inclination (Fig. 6-18b), so that the angle between the y axis and the load is α = 1° .
The components of the load P are P cos α in the negative y direction and P sin α in the positive z direction. Therefore, the bending moments at the support are
M_{y} = -(P sin α)L = -(10 k)(sin 1°)(12 ft)(12 in./ft) = -25.13 k-in.
M_{z} = -(P cos α)L = -(10 k)(cos 1°)(12 ft)(12 in./ft) = -1440 k-in.
The angle β giving the orientation of the neutral axis nn (Fig. 6-18b) is obtained from Eq. (6-20):
\tan\beta=\frac{y}{z}=\frac{M_{y}I_{z}}{M_{z}I_{y}} (6-20)
\tan\beta=\frac{y}{z}=\frac{M_{y}I_{z}}{}=\frac{(-25.13 k-in.)(2100 in.^{4})}{(-1440 k-in.)(42.2 in.^{4})}=0.8684 β = 41°
This calculation shows that the neutral axis is inclined at an angle of 41° from the z axis even though the plane of the load is inclined only 1° from the y axis. The sensitivity of the position of the neutral axis to the angle of the load is a consequence of the large I_{z} /I_{y} ratio.
From the position of the neutral axis (Fig. 6-18b), we see that the maximum stresses in the beam occur at points A and B, which are located at the farthest distances from the neutral axis. The coordinates of point A are
z_{A} = -3.50 in. y_{A} = 12.0 in.
Therefore, the tensile stress at point A (see Eq. 6-18) is
\sigma_{x}=\frac{M_{y}z}{I_{y}}-\frac{M_{z}y}{I_{z}} (6-18)
\sigma_{A}=\frac{M_{y}z_{A}}{}-\frac{M_{z}y_{A}}{I_{z}}
=\frac{(-25.13 k-in.)(-3.50 in.)}{42.2 in.^{4}}-\frac{(-1440 k-in.)(12.0 in.)}{2100 in.^{4}}
= 2080 psi + 8230 psi = 10,310 psi
The stress at B has the same magnitude but is a compressive stress:
\sigma_{B} = -10,310 psi
These stresses are 25% larger than the stress \sigma_{\max} = 8230 psi for the same beam with a perfectly aligned load. Furthermore, the inclined load produces a lateral deflection in the z direction, whereas the perfectly aligned load does not.
This example shows that beams with I_{z} much larger than I_{y} may develop large stresses if the beam or its loads deviate even a small amount from their planned alignment. Therefore, such beams should be used with caution, because they are highly susceptible to overstress and to lateral (that is, sideways) bending and buckling. The remedy is to provide adequate lateral support for the beam, thereby preventing sideways bending. For instance, wood floor joists in buildings are supported laterally by installing bridging or blocking between the joists.