Question 13.9: Determine the reactions and draw the shear and bending momen...
Determine the reactions and draw the shear and bending moment diagrams for the beam shown in Fig. 13.14(a) by the method of consistent deformations.
Degree of Indeterminacy i = 2.
Primary Beam The vertical reactions C_y and E_y at the roller supports C and E, respectively, are selected as the redundants. These supports are then removed to obtain the cantilever primary beam shown in Fig. 13.14(b). Next, the primary beam is subjected separately to the external loading and the unit values of the redundants C_y and E_y, as shown in Fig. 13.14(b), (c), and (d), respectively.
Compatibility Equations See Fig. 13.14(a) through (d).
Δ_{CO} + f_{CC} C_y + f_{CE} E_y = 0 (1)
D_{EO} + f_{EC} C_y + f_{EE} E_y = 0 (2)
Deflections of Primary Beam By using the deflection formulas, we obtain
Δ_{CO} = -\frac{82,500 kN-m^3}{EI} D_{EO} = -\frac{230,000 kN-m^3}{EI}
f_{CC} = \frac{333.333 m^3}{EI} f_{EC} = \frac{833.333 m^3}{EI}
f_{EE} = \frac{2,666.667 m^3}{EI}
By applying Maxwell’s law,
f_{CE} = \frac{833.333 m^3}{EI}Magnitudes of the Redundants By substituting the deflections of the primary beam into the compatibility equations, we obtain
– 82,500 + 333.333C_y + 833.333E_y = 0-230,000 + 833.333C_y + 2,666.667E_y = 0
or
333.333C_y + 833.333E_y = 82,500 (1a)
833.333C_y + 2,666.667E_y = 230,000 (2a)
Solving Eqs. (1a) and (2a) simultaneously for C_y and E_y, we obtain
C_y = 145.714 kN ↑ E_y = 40.714 kN ↑
Reactions The remaining reactions can now be determined by applying the three equations of equilibrium to the free body of the indeterminate beam (Fig. 13.14(e)):
+→∑F_x = 0 A_x = 0
+→∑F_y = 0 A_y – 120 + 145.714 – 120 + 40.714 = 0
A_y = 53.572 kN ↑
+\circlearrowleft ∑M_A = 0 M_A – 120(5) + 145.714(10) – 120(15) + 40.714(20) = 0
M_A = 128.58 kN-m\circlearrowleft
Shear and Bending Moment Diagrams See Fig. 13.14(f ).
