Question 5.15: A gable frame is subjected to a snow loading, as shown in Fi...
A gable frame is subjected to a snow loading, as shown in Fig. 5.23(a). Draw the shear, bending moment, and axial force diagrams and the qualitative deflected shape for the frame.
Static Determinacy m = 4, j = 5, r = 4, and e_c = 1. Because 3m + r = 3j + e_c and the frame is geometrically stable, it is statically determinate.
Reactions (See Fig. 5.23(b).)
+\circlearrowleft \sum{M_E}=0
-A_Y (8) + 12(8)(4) = 0 A_Y = 48 kN \uparrow
+\uparrow \sum{F_Y}=0
48 – 12(8) + E_Y = 0 E_Y = 48 kN \uparrow
+\circlearrowleft \sum{M^{AC}_C}=0
A_X (8) – 48(4) + 12(4)(2) = 0 A_X = 12 kN \longrightarrow
+\longrightarrow \sum{F_X}=0
12 + E_X = 0
E_X = -12 kN E_X = 12 kN\longleftarrow
Member End Forces (See Fig. 5.23(c).)
Joint A By applying the equations of equilibrium \sum{F_X}=0 and \sum{F_Y}=0, we obtain
A^{AB}_X = 12 kN A^{AB}_Y = 48 kN
Member AB Considering the equilibrium of member AB, we obtain
B^{AB}_X = -12 kN B^{AB}_Y = -48 kN M^{AB}_B = -60 kN-m
Joint B Applying the three equilibrium equations, we obtain
B^{BC}_X = 12 kN B^{BC}_Y = 48 kN M^{BC}_B = 60 kN-m
Member BC
+\longrightarrow \sum{F_X}=0 C^{BC}_X = -12 kN
+\uparrow \sum{F_Y}=0
48 – 12(4) + C^{BC}_Y =0 C^{BC}_Y = 0
+\circlearrowleft \sum{M_B}=0
60 – 12(4)(2) + 12(3) = 0 Checks
Joint C Considering the equilibrium of joint C, we determine
C^{CD}_X =12 kN C^{CD}_Y = 0
Member CD
+\longrightarrow \sum{F_X}=0 D^{CD}_X = -12 kN
+\uparrow \sum{F_Y}=0
-12(4) + D^{CD}_Y =0 D^{CD}_Y = 48 kN
+\circlearrowleft \sum{M_D} = 0
-12(3) + 12(4)(2) + M^{CD}_D = 0 M^{CD}_D =-60 kN-m
Joint D Applying the three equilibrium equations, we obtain
D^{DE}_X = 12 kN D^{DE}_Y =-48 kN M^{DE}_D = 60 kN-m
Member DE
+\longrightarrow \sum{F_X}=0 E^{DE}_X = -12 kN
+\uparrow \sum{F_Y}=0 E^{DE}_Y = 48 kN
+\circlearrowleft \sum{M_E} = 0
60 – 12(5) = 0 Checks
Joint E
+\longrightarrow \sum{F_X}=0 -12 + 12 = 0 Checks
+\uparrow \sum{F_Y}=0 48 – 48 = 0 Checks
Distributed Loads on Inclined Members BC and CD As the 12-kN/m snow loading is specified per horizontal meter, it is necessary to resolve it into components parallel and perpendicular to the directions of members BC and CD. Consider, for example, member BC, as shown in Fig. 5.23(d). The total vertical load acting on this member is (12 kN/m)(4 m) = 48 kN. Dividing this total vertical load by the length of the member, we obtain the intensity of the vertical distributed load per meter along the inclined length of the member as 48/5 = 9.6 kN/m. The components of this vertical distributed load in the directions parallel and perpendicular to the axis of the member are (3/5)(9.6) = 5.76 kN/m and (4/5)(9.6) = 7.68 kN/m, respectively, as shown in Fig. 5.23(d). The distributed loading for member CD is computed similarly and is shown in Fig. 5.23(e).
Shear and Bending Moment Diagrams See Fig. 5.23(f ) and (g).
Axial Force Diagrams The equations for axial force in the members of the frame are:
Member AB Q = -48
Member BC Q = -38.4 + 5.76x
Member CD Q = -9.6 – 5.76x
Member DE Q = -48
The axial force diagrams are shown in Fig. 5.23(h).
Qualitative Deflected Shape See Fig. 5.23(i).
