Question 5.9: 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. 5.13(a).
Reactions (See Fig. 5.13(b).)
+\circlearrowleft \sum{M^{BD}_B}=0-20(10)(5) + C_y(10) – 100(15) = 0
C_y = 250 kN C_y = 250 kN \uparrow
+\uparrow\sum{F_y}=0
A_y – 20(10) + 250 – 100 = 0
A_y = 50 kN A_y = 50 kN \uparrow
+\circlearrowleft \sum{M_A}=0
M_A – 20(10)(15) + 250(20) – 100(25) = 0
M_A = 500 kN-m M_A = 500 kN-m\circlearrowleft
Shear Diagram
Point A S_{A,R} = 50 kN
Point B S_B = 50 + 0 = 50 kN
Point C S_{C,L} = 50 – 20(10) = -150 kN
S_{C,R} = -150 + 250 = 100 kN
Point D S_{D,L} = 100 + 0 = 100 kN
S_{D,R} = 100 – 100 = 0 Checks
The shear diagram is shown in Fig. 5.13(c).
Bending Moment Diagram
Point A M_{A,R} = -500 kN-m
Point B M_B = -500 + 500 = 0
Point E M_E = 0 + 62.5 = 62.5 kN-m
Point C M_C = 62.5 – 562.5 = -500 kN-m
Point D M_D = -500 + 500 = 0 Checks
The bending moment diagram is shown in Fig. 5.13(d). The point of inflection F can be located by setting equal to zero the equation for bending moment in segment BC, in terms of the distance x_1 from the right support point C (Fig. 5.13(b)):
M = -100(5 + x_1) + 250x_1 – 20(x_1)(\frac{x_1}{2}) =0or
-10x^2_1 + 150x_1 – 500 = 0from which x_1 = 5 m and x_1 = 10 m from C. Note that the solution x_1 =10 m represents the location of the internal hinge at B, at which the bending moment is zero. Thus, the point of inflection F is located at a distance of 5 m to the left of C, as shown in Fig. 5.13(d).
Qualitative Deflected Shape A qualitative deflected shape of the beam is shown in Fig. 5.13(e). Note that at the fixed support A, both the deflection and the slope of the beam are zero, whereas at the roller support C, only the deflection is zero, but the slope is not. The internal hinge B does not provide any rotational restraint, so the slope at B can be discontinuous.
