Question 5.13: 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. 5.21(a).
Static Determinacy m = 3, j = 4, r = 3, and e_c = 0. Because 3m + r = 3j + e_c and the frame is geometrically stable, it is statically determinate.
Reactions Considering the equilibrium of the entire frame (Fig. 5.21(b)), we observe that in order to satisfy \sum{F_X}=0, the reaction component A_X must act to the left with a magnitude of 18 k to balance the horizontal load of 80 kN to the right. Thus,
A_X = -80 kN A_X = 80 kN \longleftarrow
We compute the remaining two reactions by applying the two equilibrium equations as follows:
+\circlearrowleft \sum{M_A}= 0 -80(6.5) – 30(10)(5) + D_Y (10) = 0 D_Y = 202 kN\uparrow
+\uparrow \sum{F_Y}=0 A_Y – 30(10) + 202 = 0 A_Y = 98 kN \uparrow
Member End Forces The free-body diagrams of all the members and joints of the frame are shown in Fig. 5.21(c).
We can begin the computation of internal forces either at joint A or at joint D, both of which have only three unknowns.
Joint A Beginning with joint A, we can see from its free-body diagram that in order to satisfy \sum{F_X}=0 , A^{AB}_X must
act to the right with a magnitude of 80 kN to balance the horizontal reaction of 80 kN to the left. Thus,
Similarly, by applying \sum{F_Y}=0,we obtain
A^{AB}_Y = 98 kNMember AB With the magnitudes of A^{AB}_X and A^{AB}_Y now known, member AB has three unknowns, B^{AB}_X , B^{AB}_Y, and M^{AB}_B, which can be determined by applying \sum{F_X}=0 , \sum{F_Y} =0 and \sum{M_A}=0.Thus,
B^{AB}_X = 80 kN B^{AB}_Y =-98 kN M^{AB}_B = 520 kN-m
Joint B Proceeding next to joint B and considering its equilibrium, we obtain
B^{BC}_X = 0 B^{BC}_Y = 98 kN M^{BC}_B = -520 kN-m
Member BC Next, considering the equilibrium of member BC, we write
+\longrightarrow \sum{F_X}=0 C^{BC}_X=0
+\uparrow \sum{F_Y}=0 98 – 30(10) + C^{BC}_Y=0 C^{BC}_Y = 202 kN
+\circlearrowleft \sum{M_B}=0 -520 – 30(10)(5) + 202(10) + M^{BC}_C =0 M^{BC}_C =0
Joint C Applying the three equilibrium equations, we obtain
C^{CD}_X=0 C^{CD}_Y= -202 kN M^{CD}_C = 0
Member CD Applying \sum{F_X} = 0 and \sum{F_Y} = 0 in order, we obtain
D^{CD}_X=0 D^{CD}_Y= 202 kN
Since all unknown forces and moments have been determined, we check our computations by using the third equilibrium equations for member CD.
+\circlearrowleft \sum{M_D}=0 Checks
Joint D (Checking computations)
+\longrightarrow \sum{F_X}=0
+\uparrow \sum{F_Y}=0 202 – 202 = 0 Checks
Shear Diagrams The xy coordinate systems selected for the three members of the frame are shown in Fig. 5.21(d), and the shear diagrams for the members constructed by using the procedure described in Section 5.4 are depicted in Fig. 5.21(e).
Bending Moment Diagrams The bending moment diagrams for the three members of the frame are shown in Fig. 5.21(f ).
Axial Force Diagrams From the free-body diagram of member AB in Fig. 5.21(d), we observe that the axial force throughout the length of this member is compressive, with a constant magnitude of 98 kN. Therefore, the axial force diagram for this member is a straight line parallel to the x axis at a value of -98 kN, as shown in Fig. 5.21(g). Similarly, it can be seen from Fig. 5.21(d) that the axial forces in members BC and CD are also constant, with magnitudes of 0 and -202 kN, respectively. The axial force diagrams thus constructed for these members are shown in Fig. 5.21(g).
Qualitative Deflected Shape From the bending moment diagrams of the members of the frame (Fig. 5.21(f )), we observe that the members AB and BC bend concave to the left and concave upward, respectively. As no bending moment develops in member CD, it does not bend but remains straight. A qualitative deflected shape of the frame obtained by connecting the deflected shapes of the three members at the joints is shown in Fig. 5.21(h). As this figure indicates, the deflection of the frame at support A is zero. Due to the horizontal load at B, joint B deflects to the right to B0. Since the axial deformations of members are neglected and bending deformations are assumed to be small, joint B deflects only in the horizontal direction, and joint C deflects by the same amount as joint B; that is, BB^′ = CC^′. Note that the curvatures of the members are consistent with their bending moment diagrams and that the original 90° angles between members at the rigid joints B and C have been maintained.
