Question 5.8: Draw the shear and bending moment diagrams and the qualitati...
Draw the shear and bending moment diagrams and the qualitative deflected shape for the beam shown in Fig. 5.12(a).
Reactions (See Fig. 5.12(b).)
+\longrightarrow \sum{F_x}=0 B_x = 0
+\circlearrowleft \sum{M_C}=0
\frac{1}{2}(45)(4)(7.83 – B_y(65) + 45(65)(3.25)-\frac{1}{2}(45)(2)(0.67) = 0
B_y = 250.03 kN B_y = 250.03 kN\uparrow
+\uparrow \sum{F_y}=0
-\frac{1}{2}(45)(4) + 250.03 – 45(6.5) -\frac{1}{2}(45)(2) + C_y = 0
C_y = 177.47 kN C_y = 177.47 kN \uparrow
Shear Diagram
Point A S_A = 0
Point B S_{B,L} = 0 -\frac{1}{2}(45)(4) = -90 kN
S_{B,R} = -90 + 250.03 = 160.03 kN
Point C S_{C,L} = 160.03 – 45(6.5) = -132.47 kN
S_{C,R} = -132.47 + 177.47 = 45 kN
Point D S_D = 45 -\frac{1}{2}(45)(2) = 0
The shear diagram is shown in Fig. 5.12(c). The shape of the diagram between the ordinates just computed is obtained by applying the condition that the slope of the shear diagram at any point is equal to the load intensity at that point. For example, as the load intensity at A is zero, so is the slope of the shear diagram at A. Between A and B, the load intensity is negative and it decreases linearly from zero at A to -45 kN/m at B. Thus, the slope of the shear diagram is negative in this segment, and it decreases (becomes more steep) continuously from A to just to the left of B. The rest of the shear diagram is constructed by using similar reasoning.
The point of zero shear, E, is located by using the similar triangles forming the shear diagram between B and C.
To facilitate the construction of the bending moment diagram, the areas of the various segments of the shear diagram have been computed and are shown in parentheses on the shear diagram (Fig. 5.12(c)). It should be noted that the areas of the parabolic spandrels, AB and CD, can be obtained by using the formula for the area of this shape given in Appendix A.
Bending Moment Diagram
Point A M_A = 0
Point B M_B = 0 – 120 = -120 kN-m
Point E M_E = -120 + 284.05 = 164.05 kN-m
Point C M_C = 164.05 – 195.39 = -31.34 kN-m
Point D M_D = -31.34 + 31.34 = 0
The shape of the bending moment diagram between these ordinates is obtained by using the condition that the slope of the bending moment diagram at any point is equal to the shear at that point. The bending moment diagram thus constructed is shown in Fig. 5.12(d).
It can be seen from this figure that the maximum negative bending moment occurs at point B, whereas the maximum positive bending moment, which has the largest absolute value over the entire length of the beam, occurs at point E.
To locate the points of inflection, F and G, we set equal to zero the equation for bending moment in segment BC, in terms of the distance x from the left support point B (Fig. 5.12(b)):
M = -(\frac{1}{2})(45)(4)(1.33 + x) + 250.03x – 45(x)(\frac{x}{2}) = 0or
-22.5x^2 + 160.03x – 119.7 = 0from which x = 0.85 m and x = 6.263 m from B.
Qualitative Deflected Shape A qualitative deflected shape of the beam is shown in Fig. 5.12(e). The bending moment is positive in segment FG, so the beam is bent concave upward in this region. Conversely, since the bending moment is negative in segments AF and GD, the beam is bent concave downward in these segments.
