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.

Question

A gable frame is subjected to a snow loading, as shown in [latex]Fig. 5.23(a)[/latex]. Draw the shear, bending moment, and axial force diagrams and the qualitative deflected shape for the frame.

Accepted Answer

Static Determinacy [latex]m = 4, j = 5, r = 4[/latex], and [latex]e_c = 1[/latex]. Because [latex]3m+r = 3j+e_c[/latex] and the frame is geometrically stable, it is statically determinate.

Reactions (See Fig. 5.23(b).)

[latex]+\circlearrowleft \sum{M_E}=0[/latex]

[latex]-A_Y(8)+12(8)(4)=0[/latex] [latex]A_Y=48 kN \uparrow[/latex]

[latex]+\uparrow \sum{F_Y}=0[/latex]

[latex]48-12(8)+E_Y=0[/latex] [latex]E_Y=48 kN\uparrow[/latex]

[latex]+\circlearrowleft \sum{M^{AC}_C}=0[/latex]

[latex]A_X(8)-48(4)+12(4)(2)=0[/latex] [latex]A_X=12 kN \longrightarrow[/latex]

[latex]+\longrightarrow \sum{F_X}=0[/latex]

[latex]12+E_X=0[/latex]

[latex]E_X=-12 kN[/latex] [latex]E_X=12 kN \longleftarrow[/latex]

Member End Forces (See Fig. 5.23(c).)

Joint A By applying the equations of equilibrium [latex]\sum{F_X}=0[/latex] and [latex]\sum{F_Y}=0[/latex], we obtain

[latex]A^{AB}_X=12 kN[/latex] [latex]A^{AB}_Y=48 kN[/latex]

Member AB Considering the equilibrium of member AB, we obtain

[latex]B^{AB}_X=-12kN[/latex] [latex]B^{AB}_Y= -48kN[/latex] [latex]M^{AB}_B= -60 kN.m[/latex]

Joint B Applying the three equilibrium equations, we obtain

[latex]B^{BC}_X= 12kN[/latex] [latex]B^{BC}_Y=48 kN[/latex] [latex]M^{BC}_B=60 kN.m[/latex]

Member BC

[latex]+\longrightarrow \sum{F_X}=0[/latex] [latex]C^{BC}_X= -12kN[/latex]

[latex]+\uparrow \sum{F_Y}=0[/latex]

[latex]48-12(4)+C^{BC}_Y=0[/latex] [latex]C^{BC}_Y=0[/latex]

[latex]+\circlearrowleft \sum{M_B}=0[/latex]

[latex]60-12(4)(2)+12(3)=0[/latex] Checks

Joint C Considering the equilibrium of joint C, we determine

[latex]C^{CD}_X=12 kN[/latex] [latex]C^{CD}_Y=0[/latex]

Member CD

[latex]+\longrightarrow \sum{F_X}=0[/latex] [latex]D^{CD}_X= -12kN[/latex]

[latex]+\uparrow \sum{F_Y}=0[/latex]

[latex]-12(4) + D^{CD}_Y=0[/latex] [latex]D^{CD}_Y=48 kN[/latex]

[latex]+\circlearrowleft \sum{M_D}=0[/latex]

[latex]-12(3)+12(4)(2) + M^{CD}_D=0[/latex] [latex]M^{CD}_D=-60 kN.m[/latex]

Joint D Applying the three equilibrium equations, we obtain

[latex]D^{DE}_X=12 kN[/latex] [latex]D^{DE}_Y=-48 kN[/latex] [latex]M^{DE}_D=60 kN.m[/latex]

Member DE

[latex]+\longrightarrow \sum{F_X}=0[/latex] [latex]E^{DE}_X= -12kN[/latex]

[latex]+\uparrow \sum{F_Y}=0[/latex] [latex]E^{DE}_Y=48 kN[/latex]

[latex]+\circlearrowleft \sum{M_E}=0[/latex]

[latex]60-12(5)=0[/latex] Checks

Joint E

[latex]+\longrightarrow \sum{F_X}=0[/latex] [latex]-12+12=0[/latex] Checks

[latex]+\uparrow \sum{F_Y}=0[/latex] [latex]48-48=0[/latex] Checks

Distributed Loads on Inclined Members [latex]BC[/latex] and [latex]CD[/latex] As the [latex]12-kN/m[/latex] snow loading is specified per horizontal meter, it is necessary to resolve it into components parallel and perpendicular to the directions of members [latex]BC[/latex] and [latex]CD[/latex]. Consider, for example, member [latex]BC[/latex], as shown in Fig. 5.23(d). The total vertical load acting on this member is [latex](12 kN/m)(4 m) = 48 kN[/latex]. 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 [latex]48/5 = 9.6 kN/m[/latex]. The components of this vertical distributed load in the directions parallel and perpendicular to the axis of the member are [latex](3/5)(9.6) = 5.76 kN/m[/latex] and [latex](4/5)(9.6) = 7.68 kN/m[/latex], respectively, as shown in Fig. 5.23(d). The distributed loading for member [latex]CD[/latex] is computed similarly and is shown in Fig. 5.23(e).

Shear and Bending Moment Diagrams See Fig. 5.23(f ) and (g). Ans.

Axial Force Diagrams The equations for axial force in the members of the frame are:

Member AB [latex]Q = -48[/latex]

Member BC [latex]Q = -38.4 + 5.76x[/latex]

Member CD [latex]Q = -9.6 - 5.76x[/latex]

Member DE [latex]Q = -48[/latex]

The axial force diagrams are shown in Fig. 5.23(h). Ans.

Qualitative Deflected Shape See Fig. 5.23(i). Ans.

Question 5.15: A gable frame is subjected to a snow loading, as shown in Fi...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions