Question 5.7: 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. 5.11(a).
Reactions (See Fig. 5.11(b).)
+\longrightarrow \sum{F_x}=0A_x – 30 = 0
A_x = 30 kN A_x = 30 kN\longrightarrow
+\circlearrowleft \sum{M_D}=0
-A_y(27) + 10(15)(19.5) – 162 + 40(6) = 0
A_y = 111.22 kN A_y = 111.22 kN \uparrow
+\uparrow \sum{F_y}=0
111.22 – 10(15) – 40 + D_y = 0
D_y = 78.78 kN D_y = 78.78 kN \uparrow
Shear Diagram
Point A S_{A,R} = 111.22 kN
Point B S_{B} = 111.22 – 10(15) = -38.78 kN
Point C S_{C,L} = -38.78 + 0 = -38.78 kN
S_{C,R} = -38.78 – 40 = -78.78 kN
Point D S_{D,L} = -78.78 + 0 = -78.78 kN
S_{D,R} = -78.78 + 78.78 = 0
The shear diagram is shown in Fig. 5.11(c). In segment AB, the beam is subjected to a downward (negative) uniformly distributed load of 10 kN/m. Because the load intensity is constant and negative in segment AB, the shear diagram in this segment is a straight line with negative slope. No distributed load is applied to the beam in segments BC and CD, so the shear diagram in these segments consists of horizontal lines, indicating zero slopes.
The point of zero shear, E, can be located by using the similar triangles forming the shear diagram between A and B. Thus,
\frac{x}{111.22} = \frac{15}{(111.22 + 38.78)}x = 11.12 m
To facilitate the construction of the bending moment diagram, the areas of the various segments of the shear diagram have been computed; they are shown in parentheses on the shear diagram (Fig. 5.11(c)).
Bending Moment Diagram
Point A M_A = 0
Point E M_E = 0 + 618.38 = 618.38 kN-m
Point B M_{B,L} = 618.38 – 75.23 = 543.15 kN-m
M_{B,R} = 543.15 + 162 = 705.15 kN-m
Point C M_C = 705.15 – 232.68 = 472.47 kN-m
Point D M_D = 472.47 – 472.68 = -0.21 ≈ 0
The bending moment diagram is shown in Fig. 5.11(d). The shape of this diagram between the ordinates just computed has been based on the condition that the slope of the bending moment diagram at any point is equal to the shear at that point. Just to the right of A, the shear is positive, and so is the slope of the bending moment diagram at this point. As we move to the right from A, the shear decreases linearly (but remains positive), until it becomes zero at E. Therefore, the slope of the bending moment diagram gradually decreases, or becomes less steep (but remains positive), as we move to the right from A, until it becomes zero at E. Note that the shear diagram in segment AE is linear, but the bending moment diagram in this segment is parabolic, or a second-degree curve, because the bending moment diagram is obtained by integrating the shear diagram (Eq.5.11). Therefore, the bending moment curve will always be one degree higher than the corresponding shear curve.
+\uparrow \sum F_{y}=0S + P – (S + dS) = 0
dS = P (5.11)
We can see from Fig. 5.11(d) that the bending moment becomes locally maximum at point E, where the shear changes from positive to the left to negative to the right. As we move to the right from E, the shear becomes negative, and it decreases linearly between E and B. Accordingly, the slope of the bending moment diagram becomes negative to the right of E, and it decreases continuously (becomes more steep downward to the right) between E and just to the left of B. A positive (clockwise) couple acts at B, so the bending moment increases abruptly at this point by an amount equal to the magnitude of the moment of the couple. The largest value (global maximum) of the bending moment over the entire length of the beam occurs at just to the right of B. (Note that no abrupt change, or discontinuity, occurs in the shear diagram at this point.) Finally, as the shear in segments BC and CD is constant and negative, the bending moment diagram in these segments consists of straight lines with negative slopes.
Qualitative Deflected Shape See Fig. 5.11(e).
