Question 7.SP.4: 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: The beam supports two concentrated loads and one distributed load. You can use the equations in this section between these loads and under the distributed load, but you should expect certain changes in the diagrams at the load points.
Final Answer
MODELING and ANALYSIS:
Free-Body, Entire Beam. Consider the entire beam as a free body and determine the reactions (Fig. 1):
+↺ \Sigma M_A=0:
D(24 ~\mathrm{ft})-(20 ~\mathrm{kips})(6 ~\mathrm{ft})-(12 ~\mathrm{kips})(14 ~\mathrm{ft})-(12 ~\mathrm{kips})(28 ~\mathrm{ft})=0
D=+26 \text { kips } \quad \text { D }=26 \text { kips } \uparrow
+\uparrow \Sigma F_y=0: \quad A_y-20~ ~\mathrm{kips}-12 ~\mathrm{kips}+26 ~\mathrm{kips}-12 ~\mathrm{kips}=0
A_y=+18 \text { kips } \quad \mathbf{A}_y=18~ \mathrm{kips} \uparrow
\stackrel{+}{\rightarrow} \Sigma F_x=0: \quad A_x=0 \quad \mathbf{A}_x=0
Note that the bending moment is zero at both A and E; thus, you know two points (indicated by small circles) on the bending-moment diagram.
Shear Diagram. Since dV/dx = -w, the slope of the shear diagram is zero (i.e., the shear is constant between concentrated loads and reactions). To find the shear at any point, divide the beam into two parts and consider either part as a free body. For example, using the portion of the beam to the left of point 1 (Fig. 1), you can obtain the shear between B and C:
+\uparrow \Sigma F_y=0: \quad+18 \text { kips }-20 \text { kips }-V=0 \quad V=-2 \text { kips }
You can also find that the shear is +12 kips just to the right of D and zero at end E. Since the slope dV/dx = -w is constant between D and E, the shear diagram between these two points is a straight line.
Bending-Moment Diagram. Recall that the area under the shear curve between two points is equal to the change in bending moment between the same two points. For convenience, compute the area of each portion of the shear diagram and indicate it on the diagram (Fig. 1). Since you know the bending moment M_A at the left end is zero, you have
\begin{array}{ll}M_B-M_A=+108 & M_B=+108 ~\mathrm{kip} \cdot \mathrm{ft} \\M_C-M_B=-16 & M_C=+92 ~\mathrm{kip} \cdot \mathrm{ft} \\M_D-M_C=-140 & M_D=-48 ~\mathrm{kip} \cdot \mathrm{ft} \\M_E-M_D=+48 & M_E=0\end{array}
Since you know M_E is zero, this gives you a check of the calculations.
Between the concentrated loads and reactions, the shear is constant; thus, the slope dM/dx is constant. Therefore, you can draw the bending-moment diagram by connecting the known points with straight lines.
Between D and E, where the shear diagram is an oblique straight line, the bending-moment diagram is a parabola.
From the V and M diagrams, note that V_{\max }=18 kips and M_{\max }=108 kip·ft.
REFLECT and THINK: As expected, the values of shear and slopes of the bending-moment curves show abrupt changes at the points where concentrated loads act. Useful for design, these diagrams make it easier to determine the maximum values of shear and bending moment for a beam and its loading.
