Question 5.4: Draw the shear and bending moment diagrams for the beam show...
Draw the shear and bending moment diagrams for the beam shown in Fig. 5.6(a).
Reactions See Fig. 5.6(b).
+\longrightarrow \sum{F_x} = 0 B_x = 0
+\circlearrowleft \sum{M_c} = 0
(\frac{1}{2})(9)(27)(\frac{9}{3}) – B_y(6) = 0 B_y = 60.75 kN \uparrow
+\uparrow \sum{F_y} = 0
-(\frac{1}{2})(9)(27) + 60.75 + C_y = 0 C_y = 60.75 kN \uparrow
Shear Diagram To determine the equations for shear in segments AB and BC of the beam, we pass sections aa and bb through the beam, as shown in Fig. 5.6(b). Considering the free bodies to the left of these sections and realizing that the load intensity, w(x), at a point at a distance x from end A is w(x) = (\frac{27}{9})x = 3x kN/m, we obtain the following equations for shear in segments AB and BC, respectively:
S = -(\frac{1}{2})(x)(3x) = -\frac{3x^2}{2} for 0 ≤ x < 3 m
S = -\frac{3x^2}{2} + 60.75 for 3 m < x < 9 m
The values of S computed from these equations are plotted to obtain the shear diagram shown in Fig. 5.6(c). The point D at which the shear is zero is obtained from the equation
S = -\frac{3x^2}{2} + 60.75 = 0from which x = 6.36 m.
Bending Moment Diagram Using the same sections employed previously for computing shear, we determine the following equations for bending moment in segments AB and BC, respectively:
M = -(\frac{1}{2})(x)(3x)(\frac{x}{3}) = -\frac{x^3}{2} for 0 ≤ x ≤ 3 m
M = -(\frac{x^3}{2}) + 60.75(x – 3) for 3 m ≤ x ≤ 9 m
The values of M computed from these equations are plotted to obtain the bending moment diagram shown in Fig. 5.6(d). To locate the point at which the bending moment is maximum, we di¤erentiate the equation for M in segment BC with respect to x and set the derivative dM/dx equal to zero; that is,
\frac{dM}{dx} = (-\frac{3x^2}{2}) + 60.75 = 0from which x = 6.36 m. This indicates that the maximum bending moment occurs at the same point at which the shear is zero. Also, a comparison of the expressions for dM/dx and S in segment BC indicates that the two equations are identical; that is, the slope of the bending moment diagram at a point is equal to the shear at that point. (This relationship, which is generally valid, is discussed in detail in a subsequent section.)
Finally, the magnitude of the maximum moment is determined by substituting x = 6.36 m into the equation for M in segment BC:
M_{max} = -[\frac{(6.36)^3}{2}] + 60.75(6.36 – 3) = 75.5 kN-m
