Question 7.SP.2: Draw the shear and bending-moment diagrams for the beam and ......
Draw the shear and bending-moment diagrams for the beam and loading shown.
STRATEGY: Treat the entire beam as a free body to determine the reactions, then cut the beam just before and just after each external concentrated force (Fig. 1) to see how the shear and bending moment change along the length of the beam.
Beam and loading shown
External concentrated forces on the beam
Need to draw shear and bending-moment diagrams for the beam
Free-Body, Entire Beam We start by drawing the free-body diagram of the entire beam and finding the reactions at B and D. From the diagram, we determine that the reaction at B is 46 kN upward and the reaction at D is 14 kN upward.
In summary, we analyzed the beam by considering it as a series of free bodies and applying the equilibrium equations. This allowed us to determine the reactions at certain points and create the shear and bending moment diagrams.
Final Answer
MODELING and ANALYSIS:
Free-Body, Entire Beam. From the free-body diagram of the entire beam, find the reactions at B and D:
\mathbf{R}_B=46 ~\mathrm{kN} \uparrow \quad \mathbf{R}_D=14 ~\mathrm{kN} \uparrow
Shear and Bending Moment. First, determine the internal forces just to the right of the 20-kN load at A. Consider the stub of beam to the left of point 1 as a free body, and assume V and M are positive (according to the standard convention). Then you have
+\uparrow \Sigma F_y=0: \quad-20 ~\mathrm{kN}-V_1=0 \quad V_1=-20 ~\mathrm{kN}
+↺ \Sigma M_1=0: \quad(20 ~\mathrm{kN})(0 \mathrm{~m})+M_1=0 \quad M_1=0
Next, consider the portion of the beam to the left of point 2 as a free body:
+\uparrow \Sigma F_y=0: \quad-20 ~\mathrm{kN}-V_2=0 \quad V_2=-20 ~\mathrm{kN}
+↺ \Sigma M_2=0: \quad(20 ~\mathrm{kN})(2.5 \mathrm{~m})+M_2=0 \quad M_2=-50 ~\mathrm{kN} \cdot \mathrm{m}
Determine the shear and bending moment at sections 3, 4, 5, and 6 in a similar way from the free-body diagrams. The results are
\begin{array}{ll}V_3=+26 ~\mathrm{kN} & M_3=-50 ~\mathrm{kN} \cdot \mathrm{m} \\V_4=+26 ~\mathrm{kN} & M_4=+28 ~\mathrm{kN} \cdot \mathrm{m} \\V_5=-14 ~\mathrm{kN} & M_5=+28 ~\mathrm{kN} \cdot \mathrm{m} \\V_6=-14 ~\mathrm{kN} & M_6=0\end{array}
For several of the later cuts, the results are easier to obtain by considering as a free body the portion of the beam to the right of the cut. For example, consider the portion of the beam to the right of point 4. You have
+\uparrow \Sigma F_y=0: \quad V_4-40 ~\mathrm{kN}+14 ~\mathrm{kN}=0 \quad V_4=+26 ~\mathrm{kN}
+↺ \Sigma M_4=0: \quad-M_4+(14 ~\mathrm{kN})(2 \mathrm{~m})=0 \quad M_4=+28 ~\mathrm{kN} \cdot \mathrm{m}
Shear and Bending-Moment Diagrams. Now plot the six points shown on the shear and bending-moment diagrams. As indicated in Sec. 7.2C, the shear is of constant value between concentrated loads, and the bending moment varies linearly. You therefore obtain the shear and bending-moment diagrams shown in Fig. 1.
REFLECT and THINK: The calculations are pretty similar for each new choice of free body. However, moving along the beam, the shear changes magnitude whenever you pass a transverse force and the graph of the bending moment changes slope at these points.
