Question 3.19: The cantilever beam in Figure 3–29 is a steel American Stand...

Eccentricity of the load = e = 6.0 in from the neutral axis of the beam to the line of action of the horizontal component of the applied load.

Analysis: The analysis takes the following steps:
1. Resolve the applied force into its vertical and horizontal components.
2. Transfer the horizontal component to an equivalent loading at the neutral axis having a direct tensile force and a moment due to the eccentric placement of the force.
3. Prepare the free-body diagram using the techniques from Section 3–16.
4. Draw the shearing force and bending moment diagrams and determine where the maximum bending moment occurs.
5. Complete the stress analysis at that section, computing both the maximum tensile and maximum compressive stresses.
Results: The components of the applied force are:

\begin{aligned}F_{x}=F \cos \left(30^{\circ}\right) &=(10000 \mathrm{lb})\left[\cos \left(30^{\circ}\right)\right]=8660\mathrm{lb} \text { acting to the right } \\F_{y}=F \sin \left(30^{\circ}\right) &=(10000 \mathrm{lb})\left[\sin \left(30^{\circ}\right)\right]=5000 \mathrm{lb} \text { acting downward }\end{aligned}
The horizontal force produces a counterclockwise concentrated moment at the right end of the beam with a magnitude of:
M_{2}=F_{x}(6.0 \mathrm{in})=(8660 \mathrm{lb})(6.0 \mathrm{in})=51960 \mathrm{lb} \cdot \mathrm{in}

The free-body diagram of the beam is shown in Figure 3–29(b).
Figure 3–29(c) shows the shearing force and bending moment diagrams.
The maximum bending moment, 68 040 lb in, occurs at the left end of the beam where it is attached
firmly to a column.
The bending moment, taken alone, produces a tensile stress (+) on the top surface at point B and a compressive stress (-) on the bottom surface at C. The magnitudes of these stresses are:

\sigma_{1}=\pm M_{1} / S=\pm(68040 \mathrm{lb} \mathrm{in}) /\left(7.37 \mathrm{in}^{3}\right)=\pm 9232 \mathrm{psi}
Figure 3-29(d) shows the stress distribution due only to the bending stress.
Now we compute the tensile stress due to the axial force of 8660 lb.
\sigma_{2}=F_{\lambda} / A=(8660 \mathrm{lb}) /\left(3.67 \mathrm{in}^{2}\right)=2360 \mathrm{psi}
Figure 3-29(e) shows this stress distribution, uniform across the entire section.
Next, let’s compute the combined stress at B on the top of the beam.
\sigma_{B}=+\sigma_{1}+\sigma_{2}=9232 \mathrm{psi}+2360 \mathrm{psi}=11592 \mathrm{psi}-\text { Tensile }
At C on the bottom of the beam, the stress is:
\sigma_{C}=-\sigma_{1}+\sigma_{2}=-9232 \mathrm{psi}+2360 \mathrm{psi}=-6872 \mathrm{psi}-\text { Compressive }
Figure 3-29(f) shows the combined stress condition that exists on the cross section of the beam at its left end at the support. It is a superposition of the component stresses shown in Figure 3-29(d) and (e).