Question 5.5: 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. 5.9(a).
Reactions (See Fig. 5.9(b).)
+\longrightarrow \sum{F_x} = 0 A_x = 0
By proportions,
A_y = 55(\frac{6}{9}) + 135(\frac{3}{9})= 81.67 kN A_y = 81.67 kN \uparrow
+\uparrow \sum{F_y} = 0
81.67 – 55 – 135 + D_y = 0
D_y = 108.33 kN D_y = 108.33 kN \uparrow
Shear Diagram
Point A Since a positive (upward) concentrated force of 81.67-kN magnitude acts at point A, the shear diagram increases abruptly from 0 to +81.67 kN at this point.
Point B The shear just to the left of point B is given by
S_{B,L} = S_{A,R} + area under the load diagram between just to the right of A to just to the left of B
in which the subscripts ‘‘,L’’ and ‘‘,R’’ are used to denote ‘‘just to the left’’ and ‘‘just to the right,’’ respectively. As no load is applied to this segment of the beam,
S_{B,L} = 81.67 + 0 = 81.67 kNBecause a negative (downward) concentrated load of 55 kN magnitude acts at point B, the shear just to the right of B is
S_{B,R} = 81.67 – 55 = 26.67 kNPoint C
S_{C,L} = S_{B,R}+ area under the load diagram between just to the right of B to just to the left of C
S_{C,L} = 26.67 + 0 = 26.67 kNS_{C,R} = 26.67 – 135 = -108.33 kN
Point D S_{D,L} = -108.33 + 0 = -108.33 kN
S_{D,R} = -108.33 + 108.33 = 0 Checks
The numerical values of shear computed at points A, B, C, and D are used to construct the shear diagram as shown in Fig. 5.9(c). The shape of the diagram between these ordinates has been established by applying Eq.(5.3) (slope of shear diagram at a point = intensity of distributed load at that point (5.3)), which states that the slope of the shear diagram at a point is equal to the load intensity at that point. Because no load is applied to the beam between these points, the slope of the shear diagram is zero between these points, and the shear diagram consists of a series of horizontal lines, as shown in the figure. Note that the shear diagram closes (i.e., returns to zero) just to the right of the right end D of the beam, indicating that the analysis has been carried out correctly.
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. 5.9(c)).
Bending Moment Diagram
Point A Because no couple is applied at end A, M_A = 0.
Point B M_B = M_A + area under the shear diagram between A and B
M_B = 0 + 245 = 245 kN-mPoint C M_C = 245 + 80 = 325 kN-m
Point D M_D = 325 – 325 = 0 Checks
The numerical values of bending moment computed at points A, B, C, and D are used to construct the bending moment diagram shown in Fig. 5.9(d). The shape of the diagram between these ordinates has been established by applying Eq.(5.8) (slope of bending moment diagram at a point = shear at that point (5.8)), which states that the slope of the bending moment diagram at a point is equal to the shear at that point. As the shear between these points is constant, the slope of the bending moment diagram must be constant between these points. Therefore, the ordinates of the bending moment diagram are connected by straight, sloping lines. In segment AB, the shear is +81.67 kN. Therefore, the slope of the bending moment diagram in this segment is 81.67:1, and it is positive—that is, upward to the right (/). In segment BC, the shear drops to +26.67 kN; therefore, the slope of the bending moment diagram reduces to 26.67:1 but remains positive. In segment CD, the shear becomes -108.33; consequently, the slope of the bending moment diagram becomes negative—that is, downward to the right (\), as shown in Fig. 5.9(d). Note that the maximum bending moment occurs at point C, where the shear changes from positive to the left to negative to the right.
Qualitative Deflected Shape A qualitative deflected shape of the beam is shown in Fig. 5.9(e). As the bending moment is positive over its entire length, the beam bends concave upward, as shown.
