Question 9.4: The beam shown has uniform elastic modulus E and second mome......
The beam shown has uniform elastic modulus E and second moment of area I. Determine
1. The reactions at the left wall.
2. The beam’s deflection w as a function of x.
3. The maximum allowable value of load intensity q_0 if the beam has a square cross-section with sides of 4 in and length L = 96 in, and is made from a material with E = 15 × 10^6 psi and maximum allowable normal stress 110 ksi.
Given: Loading conditions; properties of beam.
Find: Reactions, deflection, maximum allowable intensity q_0.
Assume: Hooke’s law applies; long, slender beam.
Since there are no applied axial loads we know that the supports exert equal and opposite axial forces, which may occur due to the sagging of the beam, but we do not have a method to find them and they are not considered in this problem. An FBD of the system can thus be constructed:
And, summing forces and moments, we have
\begin{aligned} \sum F_z=0 & =-q_0 L+R_A+R_B ,\\ \curvearrowleft \sum M_A & =0=M_A-M_B+R_B L-\frac{q_0 L^2}{2}. \end{aligned}By symmetry, we can reasonably assume that R_A = R_B and M_A = M_B; however, this assumption will not help us solve the equations of statics for the reaction moments. We will need more than just statics to find all four reactions. As in Example 9.3, we will proceed with the solution for deflection w(x) and hope that the boundary conditions will help us identify our unknowns. At the two fixed supports, we know that both deflection and slope must equal zero, that is, w(0) = w(L) = 0 and \mathrm{d} w /\left.\mathrm{d} x\right|_{x=0}=\mathrm{d} w /\left.\mathrm{d} x\right|_{x=L}=0 .
We use the method of sections and make a cut at a distance x to find the internal bending moment M(x):
Balancing moments on this x-long segment, we have
M(x)=M_A-R_A x+\frac{q_0 x^2}{2} .Next, we integrate for the deflection w(x):
\begin{aligned} & E I \frac{\mathrm{d}^2 w}{\mathrm{~d} x^2}=-M(x)=-M_A+R_A x-\frac{q_0 x^2}{2}, \\ & E I \frac{\mathrm{d} w}{\mathrm{~d} x}=-M_A x+\frac{R_A x^2}{2}-\frac{q_0 x^3}{6}+C_1, \\ & E I w(x)=-\frac{M_A x^2}{2}+\frac{R_A x^3}{6}-\frac{q_0 x^4}{24}+C_1 x+C_2. \end{aligned}Applying our BCs we have
\begin{aligned} w(0) & =0 \rightarrow C_2=0, \\ \left.\frac{\mathrm{d} w}{\mathrm{~d} x}\right|_{x=0} & =0 \rightarrow C_1=0, \\ w(L) & =0 \rightarrow-\frac{M_A L^2}{2}+\frac{R_A L^3}{6}-\frac{q_0 L^4}{24}=0, \\ \left.\frac{\mathrm{d} w}{\mathrm{~d} x}\right|_{x=L} & =0 \rightarrow-M_A L+\frac{R_A L^2}{2}-\frac{q_0 L^3}{6}=0 . \end{aligned}We solve these last two equations together with the two equilibrium equations for our four unknowns, and find
\begin{aligned} R_A & =R_B=\frac{q_0 L}{2} ,\\ M_A & =M_B=\frac{q_0 L^2}{12}. \end{aligned}We can now substitute these values into the expression for w(x) above:
\begin{gathered} E I w(x)=-\frac{M_A x^2}{2}+\frac{R_A x^3}{6}-\frac{q_0 x^4}{24}+C_1 x+C_2, \\ w(x)=\frac{1}{E I}\left[-\frac{q_0 L^2 x^2}{24}+\frac{q_0 L x^3}{12}-\frac{q_0 x^4}{24}\right], \end{gathered}or
w(x)=\frac{q_0 x^2}{24 E I}\left[-L^2+2 L x-x^2\right] .To find the maximum allowable load intensity q_0 based on the given normal stress limitation, we must calculate the maximum normal stress induced in the beam in terms of q_0. Because normal stress is linearly proportional to bending moment, we will do this by finding the maximum internal bending moment in the beam. We return to our general equation for M(x) and now put in the known values for the reactions.
M(x)=M_A-R_A x+\frac{q_0 x^2}{2}=\frac{q_0 L^2}{12}-\frac{q_0 L x}{2}+\frac{q_0 x^2}{2} .The maximum M(x) will occur where \mathrm{d} M / \mathrm{d} x=0: \mathrm{d} M / \mathrm{d} x=-q_0 \frac{L}{2}+q_0 x=0 \text { at } x= \frac{L}{2} . We must consider the end points of the beam, which also have zero slope, as well. The bending moment at the center of the beam is
M\left(\frac{L}{2}\right)=-\frac{q_0 L^2}{24}This has a lower magnitude than the bending moment q_0L^2/12 at the ends of the beam, so the ends are the critical points.
Below, we sketch the form of M(x).
The second moment of area of the given cross-section is I = bh³/12 = (4 in)^4/12 = 21.3 in^4. The maximum normal stress is given by
\sigma_{\max }=\frac{M_{\max } c}{I} \leq \sigma_{\text {allow }} .
So working with the magnitude of M_{max} only since there are equal and opposite tensile and compressive stresses at each x location in the beam,
\begin{gathered} \left|M_{\max }\right|=\frac{q_0 L^2}{12} \leq \frac{\sigma_{\max } I}{c}, \\ q_0 \leq \frac{(110 \mathrm{ksi}) \cdot\left(21.3 \mathrm{in}^4\right)}{2 \mathrm{in}} \frac{12}{(96 \mathrm{in})^2}=1.53 \mathrm{kips} / \mathrm{in} . \end{gathered}Note that this result is independent of the Young’s modulus of the beam, E.
