This post shows the internal force analysis of our rigid gable frame example for every member. It shows how to determine the axial, shear, and moment diagrams.
Continuing from this example, let's learn how to model a rigid gable frame's axial, shear, and moment using equations and diagrams.
The solution presented is in SI. The author will update the post soon to reflect English units.
Local Coordinate System
Part of the analysis is to establish a well-defined coordinate system per member. We recall the local coordinate system of our frame example as follows:
In member \(AB\), let the origin be joint \(A\), the local coordinates of the member are:
\(A(0.0m, 0.0m), A(0.0ft, 0.0ft)\).
\(B(5.0m, 0.0m), B(16.40ft, 0.0ft)\).
In member \(BC\), let the origin be joint \(B\), the local coordinates of the member are:
\(B(0.0m, 0.0m), B(0.0ft, 0.0ft)\).
\(C(5.0m, 0.0m), C(16.40ft, 0.0ft)\).
In member \(CD\), let the origin be joint \(C\), the local coordinates of the member are:
\(C(0.0m, 0.0m), C(0.0ft, 0.0ft)\).
\(F(2.236m, 0.0m), F(7.34ft, 0.0ft)\).
\(D(6.71m, 0.0m), D(22.0ft, 0.0ft)\).
In member \(DE\), let the origin be joint \(D\), the local coordinates of the member are:
\(D(0.0m, 0.0m), D(0.0ft, 0.0ft)\).
\(G(2.5m, 0.0m), G(8.20ft, 0.0ft)\).
\(E(5.0m, 0.0m), E(16.40ft, 0.0ft)\).
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The following solution shows how to model the axial \(N\), shear \(V\), and moment \(M\) of the structural frame per member.
Joint A
We begin by analyzing joint \(A\). The forces acting on it are the following:
Hinge reaction at \(A\)
The end forces of adjacent member \(AB\), which are its axial \(N_{AB}\), shear \(V_{AB}\), and moment \(M_{AB}\) components.
Our goal here is to find these unknowns. We already have the hinge support forces at \(A\), thanks to reaction analysis - all we need are the member end forces of \(AB\).
We assume the directions of these forces at first. We let:
\(N_{AB}\) act downward
\(V_{AB}\) act to the right
\(M_{AB}\) act clockwise
Using equilibrium equations at joint \(A\), we solve for it:
\(\sum{F_h}=0]\space{\rightarrow_+}\)
\(V_{AB}-15=0\)
\(V_{AB}=15kN\)
\(\sum{F_v}=0]\space{\uparrow_+}\)
\(47.25-N_{AB}=0\)
\(N_{AB}=47.25kN\)
\(\sum{M_A}=0]\space{\circlearrowright_+}\)
\(M_{AB}=0\)
Member AB
After analyzing joint \(A\), we analyze the adjacent component, member \(AB\).
End Forces
We start by finding the end forces of the member, which are:
\(N_{AB}\), \(V_{AB}\), and \(M_{AB}\)
\(N_{BA}\), \(V_{BA}\), and \(M_{BA}\)
From Newton's Third Law, we transfer the values of \(N_{AB}\), \(V_{AB}\), and \(M_{AB}\) from our analysis of joint \(A\). All that is left are the end forces of the other end.
To solve for these forces, we apply equilibrium equations to member \(AB\). We'll assume the directions of these components first:
\(N_{BA}\) act to the left
\(V_{BA}\) act downward
\(M_{BA}\) act counterclockwise
If our answers are negative after the calculations, the correct directional sense should be the opposite.
\(\left.\sum F_x=0\right] \rightarrow_{+}\)
\(47.25-N_{B A}=0\)
\(N_{B A}=47.25kN\)
\(\left.\sum F_y=0\right] \uparrow_{+}\)
\(15-V_{B A}=0\)
\(V_{B A}=15kN\)
\(\left.\sum M_B=0\right] \circlearrowright_{+}\)
\(15(5)-M_{B A}=0\)
\(M_{B A}=75k N \bullet m\)
Modeling Axial, Shear, and Moment
After identifying the end forces, we can model the axial, shear, and moment of member \(AB\). Creating these models is the same as analyzing a beam. We can use axial, shear, or moment equations and diagrams. We highly recommend reading our posts on these topics if we need a refresher on making these models.
Below are the axial, shear, and moment diagrams of member \(AB\).
Axial Diagram
Shear Diagram
Moment Diagram
In the axial diagram, the tensile forces are positive, while compressive forces are negative-sense.
In the shear and moment diagrams, we adopt the sign convention in which these are positive if it causes concaving of the member (not convex).
Joint B
After investigating member \(AB\), we move to adjacent joint \(B\).
The forces acting on it are the following:
The end forces of adjacent member \(AB\), which are its axial \(N_{BA}\), shear \(V_{BA}\), and moment \(M_{BA}\) components.
The end forces of adjacent member \(BC\), which are its axial \(N_{BC}\), shear \(V_{BC}\), and moment \(M_{BC}\) components.
The \(20kN\) horizontal joint force acts to the right.
Our goal here is to find these unknowns. Thanks to the previous section, we already have the end components of member \(AB\). We transfer these forces using Newton's Third Law. All that is left are the member end forces of \(BC\).
We assume the directions of these forces at first. We let:
\(N_{BC}\) act towards the joint (compression)
\(V_{BC}\) act downward (to bottom-right)
\(M_{BC}\) act counterclockwise
An interesting take on this joint \(B\) is that we must break down the end forces of member \(BC\) to its components before applying the equilibrium conditions. Using such equations, we solve the unknowns:
We can solve these two equations simultaneously to get the following:
\(N_{B C}=32.35kN\)
\(V_{B C}=34.8kN\)
\(\left.\sum M_B=0\right] \circlearrowright_{+}\)
\(75-M_{B C}=0\)
\(M_{BC}=75k N \bullet m\)
Member BC
After analyzing joint \(B\), we analyze the adjacent component, member \(BC\).
An interesting note is that the \(10kN/m\) uniform load acts on the \(4m\) projection. Before we solve for the internal \(N\), \(V\), and \(M\), we need to find a way to translate the effect of this load to the member.
We do that by finding the resultant force of the uniform load and solving for its components oriented to member \(BC\):
\(R=10(4)=40\)
\(\frac{R_x}{40}=\frac{3}{5}\)
\(R_x=\frac{3}{5}(40)=24kN\)
\(\frac{R_y}{40^{k N}}=\frac{4}{5}\)
\(R_y=\frac{4}{5}(40)=32k N\)
Next, we divide the resultant components by the length of the member to get the equivalent uniform load acting on \(BC\):
\(w_x=\frac{24}{5}=4.8\frac{k N}{m}\)
\(w_y=\frac{32}{5}=6.4\frac{k N}{m}\)
End Forces
We start by finding the end forces of the member, which are:
\(N_{BC}\), \(V_{BC}\), and \(M_{BC}\)
\(N_{CB}\), \(V_{CB}\), and \(M_{CB}\)
From Newton's Third Law, we transfer the values of \(N_{BC}\), \(V_{BC}\), and \(M_{BC}\) from our analysis of joint \(B\). All left is the end forces of the other end.
To solve for these forces, we apply equilibrium equations to member \(BC\). We'll assume the directions of these components first:
\(N_{CB}\) act to the left
\(V_{CB}\) act downward
\(M_{CB}\) act counterclockwise
If our answers are negative after the calculations, the correct directional sense should be the opposite.
\(\left.\sum F_x=0\right] \rightarrow_{+}\)
\(32.35-4.8(5)-N_{C B}=0\)
\(N_{C B}=8.35k N\)
\(\left.\sum F_y=0\right] \uparrow_{+}\)
\(34.8-6.4(5)-V_{C B}=0\)
\(V_{C B}=2.8k N\)
\(\left.\sum M_C=0\right] \circlearrowright_{+}\)
\(34.8(5)+75-6.4(5)(2.5)-M_{C B}=0\)
\(M_{C B}=169k N \bullet m\)
Modeling Axial, Shear, and Moment
After identifying the end forces, we can model the axial, shear, and moment of member \(BC\). Creating these models is the same as analyzing a beam. We can use axial, shear, or moment equations and diagrams. We highly recommend reading our posts on these topics if we need a refresher on making these models.
Below are the axial, shear, and moment diagrams of member \(BC\).
Axial Diagram
Shear Diagram
Moment Diagram
In the axial diagram, the tensile forces are positive, while compressive forces are negative-sense.
In the shear and moment diagrams, we adopt the sign convention in which these are positive if it causes concaving of the member (not convex).
It's good to note that because of \(w_x\), the axial diagram is a first-degree graph. It means that member \(BC\) experiences a uniform axial load throughout its length.
Joint E
After analyzing member \(BC\), we can continue to adjacent joint \(C\); however, for this example, let's explore the other end of the frame instead and work our way back to \(C\).
There is no hard rule on which joint to start or follow a specific sequence. The main important thing is to model the behavior of each member.
With that, let's proceed. The forces acting on joint \(E\) are the following:
Roller reaction at \(E\)
The end forces of adjacent member \(DE\), which are its axial \(N_{ED}\), shear \(V_{ED}\), and moment \(M_{ED}\) components.
Our goal here is to find these unknowns. We already have the roller support force at \(E\), thanks to reaction analysis - all we need are the member end forces of \(DE\).
We assume the directions of these forces at first. We let:
\(N_{ED}\) act downward
\(V_{ED}\) act to the right
\(M_{ED}\) act clockwise
Using equilibrium equations at joint \(E\), we solve for it:
\(\sum{F_h}=0]\space{\rightarrow_+}\)
\(V_{ED}=0\)
\(\sum{F_v}=0]\space{\uparrow_+}\)
\(52.75-N_{ED}=0\)
\(N_{ED}=52.75kN\)
\(\sum{M_E}=0]\space{\circlearrowright_+}\)
\(M_{ED}=0\)
Member DE
After analyzing joint \(E\), we analyze the adjacent component, member \(DE\).
End Forces
We start by finding the end forces of the member, which are:
\(N_{ED}\), \(V_{ED}\), and \(M_{ED}\)
\(N_{DE}\), \(V_{DE}\), and \(M_{DE}\)
From Newton's Third Law, we transfer the values of \(N_{ED}\), \(V_{ED}\), and \(M_{ED}\) from our analysis of joint \(E\). All left is the end forces of the other end.
To solve for these forces, we apply equilibrium equations to member \(CD\). We'll assume the directions of these components first:
\(N_{DE}\) act to the right
\(V_{DE}\) act upward
\(M_{DE}\) act counterclockwise
If our answers are negative after the calculations, the correct directional sense should be the opposite.
\(\left.\sum F_x=0\right] \rightarrow_{+}\)
\(N_{D E}-52.75=0\)
\(N_{DE}=52.75kN\)
\(\left.\sum F_y=0\right] \uparrow_{+}\)
\(V_{D E}-5=0\)
\(V_{DE}=5kN\)
\(\left.\sum M_D=0\right] \circlearrowright_{+}\)
\(5(2.5)-M_{D E}=0\)
\(M_{DE}=12.5kN \bullet m\)
Modeling Axial, Shear, and Moment
After identifying the end forces, we can model the axial, shear, and moment of member \(DE\). Creating these models is the same as analyzing a beam. We can use axial, shear, or moment equations and diagrams. We highly recommend reading our posts on these topics if we need a refresher on making these models.
Below are the axial, shear, and moment diagrams of member \(DE\).
Axial Diagram
Shear Diagram
Moment Diagram
In the axial diagram, the tensile forces are positive, while compressive forces are negative-sense.
In the shear and moment diagrams, we adopt the sign convention in which these are positive if it causes concaving of the member (not convex).
Joint D
After investigating member \(DE\), we move to adjacent joint \(D\).
The forces acting on it are the following:
The end forces of adjacent member \(DE\), which are its axial \(N_{DE}\), shear \(V_{DE}\), and moment \(M_{DE}\) components.
The end forces of adjacent member \(CD\), which are its axial \(N_{DC}\), shear \(V_{DC}\), and moment \(M_{DC}\) components.
Our goal here is to find these unknowns. Thanks to the previous section, we already have the end components of member \(DE\). We transfer these forces using Newton's Third Law. All left is the member end forces of \(CD\).
We assume the directions of these forces at first. We let:
\(N_{DC}\) act towards the joint (compression)
\(V_{DC}\) act downward (to bottom-left)
\(M_{DC}\) act counterclockwise
An interesting take on this joint \(D\) is that we must break down the end forces of member \(CD\) to its components before applying the equilibrium conditions. Using such equations, we solve the unknowns:
We can solve these two equations simultaneously to get the following:
\(N_{D C}=28.06k N\)
\(V_{D C}=44.94k N\)
\(\left.\sum M_D=0\right] \circlearrowright_{+}\)
\(12.5-M_{D C}=0\)
\(M_{D C}=12.5k N \bullet m\)
Member CD
After analyzing joint \(D\), we analyze the adjacent component, member \(CD\).
End Forces
We start by finding the end forces of the member, which are:
\(N_{CD}\), \(V_{CD}\), and \(M_{CD}\)
\(N_{DC}\), \(V_{DC}\), and \(M_{DC}\)
From Newton's Third Law, we transfer the values of \(N_{DC}\), \(V_{DC}\), and \(M_{DC}\) from our analysis of joint \(D\). All left is the end forces of the other end.
To solve for these forces, we apply equilibrium equations to member \(CD\). We'll assume the directions of these components first:
\(N_{DC}\) act to the right
\(V_{DC}\) act upward
\(M_{DC}\) act clockwise
If our answers are negative after the calculations, the correct directional sense should be the opposite.
After identifying the end forces, we can model the axial, shear, and moment of member \(CD\). Creating these models is the same as analyzing a beam. We can use axial, shear, or moment equations and diagrams. We highly recommend reading our posts on these topics if we need a refresher on making these models.
Below are the axial, shear, and moment diagrams of member \(CD\).
Axial Diagram
Shear Diagram
Moment Diagram
In the axial diagram, the tensile forces are positive, while compressive forces are negative-sense.
In the shear and moment diagrams, we adopt the sign convention in which these are positive if it causes concaving of the member (not convex).
Joint C
After investigating member \(CD\), we move to adjacent joint \(C\).
The forces acting on it are the following:
The end forces of adjacent member \(CB\), which are its axial \(N_{CB}\), shear \(V_{CB}\), and moment \(M_{CB}\) components.
The end forces of adjacent member \(CD\), which are its axial \(N_{DC}\), shear \(V_{DC}\), and moment \(M_{DC}\) components.
At this point, we have solved all of these values. We use this last joint \(C\) to check whether our computations are correct.
Using equilibrium equations at joint \(C\), we check our answers:
