Let's consider the example of how to analyze pin-connected frames. We'll explore its stability, reaction analysis of its supports, and internal force analysis of every member.
This example shows units in SI and English. The English system is in parentheses.
In terms of computations, however, it will primarily be in the SI system. The author converted the answer from the SI solution to English to reflect the latter system. It only applies to definite values, not equations.
We will update this post to reflect solutions in the English system soon.
Structural Model
Let's consider a pin-connected frame example. An exciting feature of this problem is that we have an internal pin connecting two frames. All members are assumed to have a uniform section of the same material.
Type of Analysis: Classical Approach, Linear Analysis, Static Loads, Plane Frame, Determinate Frame
Preparation
Before analyzing a structure, we'll need to make some preparations first. That includes setting up our references and finding their determinacy.
Set-Up References
An excellent structural analysis must have a uniform mathematical understanding of the structure. It ensures that other people can easily understand your results.
Global References
We first place a Cartesian grid with its origin defined by our preference. In this case, let's assign the origin \(A(0, 0)\) at the bottom-left point of the frame. Consequently, the x-axis will run horizontally with the y-axis perpendicular to it.
We also need to identify the location of all points of interest: the location of supports, change in loads, and differences in the frame's cross-sectional properties. Starting from \(A\) and going around the structure, we identify the following points of interest:
\(A(0.0m, 0.0m), A(0.0ft, 0.0ft)\). The hinge support and the start of the varying distributed load of \(30kN/m (2.06kip/ft)\) →.
\(B(0.0m, 6.0m), B(0.0ft, 19.69ft)\). Frame joint and the end of the varying distributed load.
\(D(10.0m, 0.0m), D(32.81ft, 0.0ft)\). The hinge support,
\(E(4.0m, 6.0m), E(13.12ft, 19.69ft)\). The \(90kN•m (66.38kip•ft)\) clockwise moment, and
\(F(8.0m, 3.0m), F(26.25ft, 9.84ft)\). The \(15kN (3.37kip)\) concentrated load acting to the left.
You can label each joint according to your preference. The most important thing is that its coordinates must be defined appropriately.
Local References
When dealing with structures composed of elements with varying properties and internal forces along their length, we need to create a localized reference system for each. It is a common step, especially when dealing with structural frames.
The first step would be to break the frame into its members.
Then, we analyze each member separately and assign a location of the local origin. Usually, we place it at the left-most part of the member when looking inside the frame. You can put it on the right-most side - there is no hard rule. What's important is consistency, and it must be understandable. The x-axis would go along the length of the member while the y-axis would run perpendicular to it.
Finally, from the selected origin, we label each point of interest.
Let's illustrate how we can create our local system for each member. In our frame example, we have three members: \(AB\), \(BC\), and \(CD\).
In member \(AB\), let the origin be joint \(A\), the local coordinates of the member are:
\(A(0.0m, 0.0m), A(0.0ft, 0.0ft)\).
\(B(6.0m, 0.0m), B(19.69ft, 0.0ft)\).
In member \(BC\), let the origin be joint \(B\), the local coordinates of the member are:
\(B(0.0m, 0.0m), B(0.0ft, 0.0ft)\).
\(E(4.0m, 0.0m), E(13.12ft, 0.0ft)\).
\(C(6.0m, 0.0m), C(19.69ft, 0.0ft)\).
In member \(CD\), let the origin be joint \(C\), the local coordinates of the member are:
\(C(0.0m, 0.0m), C(0.0ft, 0.0ft)\).
\(F(3.61m, 0.0m), F(11.84ft, 0.0ft)\).
\(D(7.22m, 0.0m), D(23.69ft, 0.0ft)\).
Determinacy
We need to find the structure's determinacy \(D\) to know our approach.
For a 2D frame, it is:
\(D=(3m+r)-(3j+c)\)
For this frame example, there are three members: \(AB\), \(BC\), \(CD\), four joints: \(A\), \(B\), \(C\), \(D\), four reaction components: \(A_h\), \(A_v\), \(D_h\), \(D_v\), and one internal connection at joint \(C\) \(e_c = 1\); hence \(D=0\)
\(D=(3(3)+4)-(3(4)+1)=0\)
A determinacy of zero indicates that the structure can be analyzed using only the equilibrium equations.
The reaction components \(A_h, A_v, D_h, D_v\) are not collinear, parallel, or concurrent with each other.
The determinacy is equal to zero.
From these observations, we can conclude the frame is externally stable. In terms of its internal stability, the frame arrangement doesn't pose any risks of excessive deformation or immediate collapse; hence, it's internally stable.
Since the structure is externally and internally stable, we can proceed with the analysis. If it is unstable, we may have to adjust its model before proceeding.
Reactions
The second requirement for a complete analysis is to compute the support loads of the structure. Solving for the components enables us to understand the transfer of loads.
For a determinate structure, solving for its reactions is straightforward. To solve it, always remember that the whole model must obey the laws of equilibrium.
As a demonstration, we first break the support loads into their components (not their resultant) along the axes. In our example, we have three components:
We assume \(A_h\) and \(D_h\) to be acting towards the left,
At this point, we have solved the vertical reactions of the hinge supports. All that is left is to solve for the horizontal components.
To solve these reactions, we need to "disconnect" the pin-connected frame at points of internal connections. In our example, it would be the hinge at point \(C\).
As a result, we will have the free-body diagrams (FBD) of each frame element - FBD ABC and FBD CD.
For each diagram, we have the joint \(C\). At this point, the hinge will provide two forces: \(C_h\) and \(C_v\), which are unknown.
At this hinge, it will provide an additional condition: the summation of moments at that point is zero. We can use this to help us solve the reactions.
In terms of its directions, \(C_h\) and \(C_v\) are also unknown. Normally, one assumes the direction. Said directions must be opposite for each FBD to comply with Newton's Third Law.
In FBD ABC, \(C_h\) acts to the left while \(C_v\) acts upward.
In FBD CD, \(C_h\) acts to the right while \(C_v\) acts downward.
To solve for the reactions, as well as the hinge, all we have to do is to apply the equilibrium principle to these individual FBDs. Let's demonstrate below:
Free Body Diagram CD
Applying the equilibrium equations:
\(\sum{M_C}=0]\space{\circlearrowright_+}\)
\(15(3)-22.5(2+2)+D_h(6)=0\)
\(D_h=7.5kN (1.69kip)\)
\(\sum{F_h}=0]\space{\rightarrow_+}\)
\(C_h-15-7.5=0\)
\(C_h=22.5kN (5.06kip)\)
\(\sum{F_v}=0]\space{\uparrow_+}\)
\(22.5-C_v=0\)
\(C_v=22.5kN (5.06kip)\)
From our calculations, the reaction components and the hinge forces are:
\(D_h=7.5kN (1.69kip)\)
\(C_h=22.5kN (5.06kip)\)
\(C_v=22.5kN (5.06kip)\)
If our answer is negative, the assumed direction is wrong, and the correct one is the opposite.
Free Body Diagram ABC
Applying the equilibrium equations:
\(\sum{F_h}=0]\space{\leftarrow_+}\)
\(22.5+A_h-\frac{1}{2}(30)(6)=0\)
\(A_h=67.5kN (15.17kip)\)
From our calculations, the reaction component is:
\(A_h=67.5kN (15.17kip)\)
If our answer is negative, the assumed direction is wrong, and the correct one is the opposite.
At this point, we have now solved the reaction components.
\(A_h=67.5kN (15.17kip)\)
\(A_v=22.5kN (5.06kip)\)
\(D_h=7.5kN (1.69kip)\)
\(D_v=22.5kN (5.06kip)\)
We can verify the results by applying the equilibrium principle to the whole structure.
Force Analysis
The third requirement for a complete analysis is understanding the internal force and stress developed on the structure due to the applied loads.
We have four types of stresses to analyze: axial, shear, moment, and torsion. Typically, we do these by modeling the behavior using functions and diagrams.
In 2D frames, we limit to axial, shear, and moment only.
Modeling Axial, Shear, and Moment Behavior
We can use axial, shear, and moment equations and graphs to describe the internal behavior of each frame member.
Creating these equations and diagrams deserves a separate section. Learn more about how to construct these using these individual posts:
The following section shows each frame member's axial, shear, and moment behavior in the local system. You can view the complete solution in this link.
Member AB
Axial Diagram
Shear Diagram
Moment Diagram
Member BC
Axial Diagram
Shear Diagram
Moment Diagram
Member CD
Axial Diagram
Shear Diagram
Moment Diagram
Global System
The following section shows all frame members' combined axial, shear, and moment diagrams.
Deflections
The final requirement is to analyze the structure's deflection. In this part, we analyze the translations and rotations of the object from their original position.
We have two types to analyze: rotation and translation. As in the previous part, we can describe their behavior using functions and diagrams.
The topic of deflection deserves a separate section. There are many ways how to explain a structure's movement, such as the following:
