Question 1.4: The plane frame in Fig. 1-16 is a modified version of that s...
The plane frame in Fig. 1-16 is a modified version of that shown in Fig. 1-1. Initially, member DF has been replaced with a roller support at D. Moment M_{A} is applied at pin-supported joint A, and load F_{B} is applied at joint B. A uniform load with intensity q_{1} acts on member BC, and a linearly distrib-uted load with peak intensity q_{0} is applied downward on member ED. Find the support reactions at joints A and D, then solve for internal forces at the top of member BC. Use numerical properties given. As a final step, remove the roller at D, insert member DF (as shown in Fig. 1-1) and reanalyze the structure to find the reaction forces at A and F.
Numerical data (Newtons and meters):
\begin{gathered}a=3 m \quad b=2 m \quad c=6 m \quad d=2.5 m \\M_{A}=380 N \cdot m \quad F_{B}=200 N \quad q_{0}=80 N / m \quad q_{1}=160 N / m\end{gathered}
(1) Draw the FBD of the overall frame. The solution for reaction forces at A and D must begin with a proper drawing of the FBD of the overall frame (Fig. 1-17). The FBD shows all applied and reactive forces .
(2) Determine the statically equivalent concentrated forces. Distributed forces are replaced by their statical equivalents \left(F_{q 0} \text { and } F_{q 1}\right). The compo-nents of the inclined concentrated force at B also may be computed :
\begin{aligned}F_{q 0} &=\frac{1}{2} q_{0} c=240 N & F_{q 1}=q_{1} b=320 N \\F_{B x} &=\frac{4}{5} F_{B}=160 N & F_{B y}=\frac{3}{5} F_{B}=120 N\end{aligned}(3) Sum the moments about A to find reaction force D_{y}. This structure is stat-ically determinate because there are three available equations from statics \left(\Sigma F_{x}=0, \Sigma F_{y}=0, \text { and } \Sigma M=0\right) and three reaction unknowns \left(A_{x^{\prime}}, A_{y^{\prime}}, D_{y}\right). It is convenient to start the static analysis using \Sigma M_{A} = 0, because we can isolate one equation with one unknown and then easily find reaction D_{y} .
D_{y}=\frac{1}{d}\left[-M_{A}+F_{B x} a-F_{q 1}\left(a+\frac{b}{2}\right)+F_{q 0}\left(d+\frac{2}{3} c\right)\right]=152 N(4) Sum the forces in the x and y directions to find the reaction forces at A. Now that D_{y} is known, we can find A_{x} and A_{y} using ΣF_{x} = 0 and ΣF_{y} =0, and then find the resultant reaction force at A using components A_{x} and A_{y}.
Sum forces in x direction: \begin{aligned}&A_{x}-F_{B x}+F_{q 1}=0 \quad A_{x}=F_{B x}-F_{q 1} \\&A_{x}=-160 N\end{aligned}
Sum forces in y direction: \begin{aligned}&A_{y}-F_{B y}+D_{y}-F_{q 0}=0 \quad A_{y}=F_{B y}-D_{y}+F_{q 0} \\&A_{y}=208 N\end{aligned}
Resultant force at A: A = \sqrt{A_{x}^{2}+A_{y}^{2}} \quad A=262 N
(5) Find the internal forces and moment at the top of member BC. Now that reaction forces at A and D are known, we can cut a section through the frame just below joint C, creating upper and lower FBDs (Fig. 1-18) .
Section forces N_{c} (axial) and V_{c} (shear) as well as section moment (M_{c}) are exposed and may be computed using statics. Either FBD may be used to find N_{c} , V_{c} , and M_{c} ; the computed stress resultants N_{c} , V_{c} , and M_{c} will be the same.
Calculations based on upper FBD :
\begin{array}{ll}\Sigma F_{x}=0 & V_{c}=0 \\\Sigma F_{y}=0 & N_{c}=D_{y}-F_{q 0}=-88 N \\\Sigma M_{c}=0 &\end{array} \\ M_{c}=-D_{y} d+F_{q 0}\left(d+\frac{2}{3} c\right)=1180 N \cdot mCalculations based on lower FBD:
\begin{array}{ll}\Sigma F_{x}=0 & V_{c}=-F_{q 1}+F_{B x}-A_{x}=0 \\\Sigma F_{y}=0 & N_{c}=F_{B y}-A_{y}=-88 N \\\Sigma M_{c}=0 & \\M_{c}=-F_{q 1} \frac{b}{2}+F_{B x} b-A_{x}(a+b)+M_{A}=1180 N \cdot m\end{array}(6) Remove the roller at D, insert member DF (as shown in Fig. 1-1) and reanalyze the structure to find reaction forces at A and F. Member DF is pin-connected to EDC at D, has a pin support at F, and carries uniform load q2 in the -y direction. See Figs. 1-3a and 1-3b for the FBDs required in the solution. Note that there are now four unknown reaction forces (A_{x}, A_{y}, F_{x} and F_y ) but only three equilibrium equations available (ΣF_{x} = 0, ΣF_{y} = 0, ΣM = 0) for use with the overall FBD in Fig. 1-3a. To find another equation, we will have to separate the structure at pin connec-tion D to take advantage of the fact that the moment at D is known to be zero (friction effects are assumed to be negligible); we can then use ΣM_{D} = 0 for either the upper FBD or the lower FBD in Fig. 1-3b to develop one more independent equation of statics. Recall that a statics sign convention is used here for all equilibrium equations. Dimensions and loads for new member DF:
\begin{gathered}e=5 m & e_{x}=\frac{3}{5} e=3 m & e_{y}=\frac{4}{5}& e=4 m \\q_{2}=180 N / m \quad F_{q 2}=q_{2} e=900 N\end{gathered}First, write equilibrium equations for the entire structure FBD (see Fig. 1-3a).
(a) Sum forces in x direction for entire FBD:
A_{x}+F_{x}-F_{B x}+F_{q 1}=0 (a)
(b) Sum forces in y direction for entire FBD:
A_{y}+F_{y}-F_{q 0}-F_{q 2}-F_{B y}=0 (b)
(c) Sum moments about A for entire FBD:
\begin{aligned}-M_{A}-F_{q 1}\left(a+\frac{b}{2}\right)-F_{x}\left(a+b+e_{y}\right) &+F_{y}\left(e_{x}-d\right)+F_{B x} a \\&+F_{q 0}\left(d+\frac{2}{3} c\right)+F_{q 2}\left(d-\frac{e_{x}}{2}\right)=0\end{aligned}so \begin{aligned}-F_{x}\left(a+b+e_{y}\right) &+F_{y}\left(e_{x}-d\right)=M_{A}+F_{q 1}\left(a+\frac{b}{2}\right) \\-& {\left[F_{B x} a+F_{q 0}\left(d+\frac{2}{3} c\right)+F_{q 2}\left(d-\frac{e_{x}}{2}\right)\right] }\end{aligned} (c)
Next, write another equilibrium equation for the upper FBD in Fig. 1-3b.
(d) Sum moments about D on upper FBD :
-F_{x} e_{y}+F_{y} e_{x}-F_{q 2} \frac{e_{x}}{2}=0 \quad \text { so } \quad-F_{x} e_{y}+F_{y} e_{x}=F_{q 2} \frac{e_{x}}{2} (d)
Solving Eqs. (c) and (d) for F_{x} and F_{y}, we have
\left(\begin{array}{l}F_{x} \\F_{y}\end{array}\right)=\left[\begin{array}{cc}-\left(a+b+e_{y}\right) & e_{x}-d \\-e_{y} & e_{x}\end{array}\right]^{-1}\\ \left[\begin{array}{c}M_{A}+F_{q 1}\left(a+\frac{b}{2}\right)-\left[F_{B x} a+F_{q 0}\left(d+\frac{2}{3} c\right)+F_{q 2}\left(d-\frac{e_{x}}{2}\right)\right] \\F_{q 2} \frac{e_{x}}{2}\end{array}=\left(\begin{array}{l}180.6 \\690.8\end{array}\right) N \right .Now, substitute solutions for F_{x} and F_{y} into Eqs. (a) and (b) to find reac-tions A_{x} and A_{y} :
\begin{array}{ll}A_{x}=-\left(F_{x}-F_{B x}+F_{q 1}\right) & A_{x}=-340.6 N \\A_{y}=-F_{y}+F_{q 0}+F_{q 2}+F_{B y} & A_{y}=569.2 N\end{array}The resultant force at A is A=\sqrt{A_{x}^{2}+A_{y}^{2}} \quad A=663 N
Sum the moments about D on lower FBD as a check; the lower FBD is in equilibrium as required:
F_{q 0}\left(\frac{2}{3} c\right)+F_{q 1} \frac{b}{2}-F_{B x} b-F_{B y} d-M_{A}+A_{x}(a+b)+A_{y} d=0(7) Finally, compute the resultant force in the pin connection at D. Use the summation of forces in the upper FBD to find component forces D_{x} and D_{y}, then find the resultant D (see Fig. 1-3b)
\begin{array}{ll}\Sigma F_{x}=0 & D_{x}=-F_{x}=-180.6 N \\\Sigma F_{y}=0 & D_{y}=-F_{y}+F_{q 2}=209.2 N\end{array}The resultant force at D is D=\sqrt{D_{x}^{2}+D_{y}^{2}}=276 N .
