Question 1.4: A plane frame with an overhang is supported at points A and ...
A plane frame with an overhang is supported at points A and D (Fig.1-16). (This is the beam of Example 1-2 to which column BD has been added.) A linearly varying distributed load of peak intensity q_{0} = 160 N/m acts on span AB. Concentrated moment M_{0} = 380 N·m is applied at A, and an inclined concentrated load P = 200 N acts at C. Force P also acts at mid-height of column BD. The lengths of segments AB and BD are L = 4 m, and the length of the overhang BC is 2 m.
Find support reactions at A and D and then calculate the axial force N, shear force V, and bending moment M at the top of column BD.
The four-step problem-solving approach for this plane frame follows the procedures presented for the beam in Example 1-2.
1. Conceptualize : Find the reaction forces A_{y} , D_{x} , and D_{y} using the FBD of the overall structure shown in Fig.1-17. Internal axial force N, shear force V, and bending moment M at the top of column BD (Fig.1-18) are obtained by cutting the column at that location. View column BD with joint D on your left and B on your right to establish the assumed positive directions of N, V, and M on either side of the cut section, as shown in Fig.1-18.
2. Categorize : First find reaction forces A_{y} , D_{x} , and D_{y}. Then use either the upper or lower free-body diagram in Fig.1-18 to find N, V, and M. The free-body diagrams in Fig.1-18 show internal axial force N, shear force V, and bending moment M in their assumed positive directions based on a deformation sign convention.
3. Analyze :
Solution for external reactions: Sum forces in the x direction to find reaction force component D_{x}. Next, sum moments about D to find reaction component A_{y}.
Finally, sum forces in the y direction to find reaction D_{y}. Use a statics sign convention in the solution as follows :
\Sigma F_{x}=0 \quad D_{x}=\frac{3}{5} P+P=320 N
\Sigma M_{D}=0 \quad A_{y}=\frac{1}{L}\left[-M_{0}+\frac{1}{2} q_{0} L\left[\frac{L}{3}\right]+P \frac{1}{2}-\frac{4}{5} P\left[\frac{L}{2}\right]+\frac{3}{5} P L\right]=152 N
\Sigma F_{y}=0 \quad D_{y}=-A_{y}+\frac{1}{2} q_{0} L+\frac{4}{5} P=328 N
All reaction force components are positive, so they act in the directions shown in Figs.1-17 and 1-18. The resultant reaction force at D is
D_{ res }=\sqrt{D_{x}^{2}+D_{y}^{2}}=458 N.
Solution for internal axial force N, shear force V, and moment M at top of column BD: Using the lower free-body diagram in Fig.1-18,
\Sigma F_{y}=0 \quad N=-D_{y}=-328 N \quad \Sigma F_{x}=0 \quad V=-D_{x}+P=-120 N
\Sigma M=0 \quad M=-D_{x} L+P \frac{L}{2}=-880 N \cdot mAlternatively, the upper free-body diagram can be used to compute N, V, and M (Fig.1-18):
\Sigma F_{y}=0 \quad N=A_{y}-\frac{1}{2} q_{0} L-\frac{4}{5} P=-328 N \quad \Sigma F_{x}=0 \quad V=-\frac{3}{5} P=-120 N
\Sigma M=0 \quad M=-M_{0}-A_{y} L-\frac{4}{5} P \frac{L}{2}+\frac{1}{2} q_{0} L\left[\frac{L}{3}\right]=-880 N \cdot mThe minus signs on internal axial force N, shear force V, and moment M indicate that all three quantities act opposite to directions assumed in Fig.1-18.
4. Finalize : Either the lower or upper free-body diagram can be used to find internal forces (N and V) and internal moment (M) at the top of column BD. Section forces and moments at any other location on the frame are found using the same approach. A properly drawn free-body diagram is an important first step in the solution.
