Question 5.14: Draw the shear, bending moment, and axial force diagrams and...
Draw the shear, bending moment, and axial force diagrams and the qualitative deflected shape for the frame shown in Fig. 5.22(a).
Static Determinacy m = 2, j = 3, r = 3, and e_c = 0. Because 3m + r = 3j + e_c and the frame is geometrically stable, it is statically determinate.
Reactions (See Fig. 5.22(b).)
+\longrightarrow \sum{F_x}=0
-A_x + 110 = 0 A_x = 110 kN\longleftarrow
+\uparrow \sum{F_y}=0
A_y – 25(5) = 0 A_y = 125 kN\uparrow
+\circlearrowleft \sum{M_A}=0
M_A – 110(3.5) – 25(5)(2.5) = 0 M_A = 697.5 kN-m \circlearrowleft
Member End Forces (See Fig. 5.22(c).)
Joint A By applying the equilibrium equations \sum{F_X}=0 , \sum{F_Y}=0, and \sum{M_A}=0, we obtain
A^{AB}_X = -110 kN A^{AB}_Y =125 kN M^{AB}_A =697.5 kN-m
Member AB Next, considering the equilibrium of member AB, we write
+\longrightarrow \sum{F_x}=0 -110 + 110 + B^{AB}_X =0 B^{AB}_X =0
+\uparrow \sum{F_y}=0 125 +B^{AB}_Y =0 B^{AB}_Y = -125 kN
+\circlearrowleft \sum{M_B}=0 697.5 – 110(3.5) + M^{AB}_B=0 M^{AB}_B = -312.5 kN-m
Joint B Applying the three equations of equilibrium, we obtain
B^{BC}_X = 0 B^{BC}_Y = 125 kN M^{BC}_B = 312.5 kN-m
Member BC (Checking computations.)
+\longrightarrow \sum{F_x}=0 Checks
+\uparrow \sum{F_y}=0 125 – 25(5) = 0 Checks
+\circlearrowleft \sum{M_B}=0 312.5 – 25(5)(2.5) = 0 Checks
The member end forces are shown in Fig. 5.22(d).
Shear Diagrams See Fig. 5.22(e).
Bending Moment Diagrams See Fig. 5.22(f ).
Axial Force Diagrams See Fig. 5.22(g).
Qualitative Deflected Shape See Fig. 5.22(h).
