Truss Analysis

Now that you have a basic understanding of trusses, let's explore how to analyse them! 
AssumptionsSome assumptions that are considered for Truss Analysis in an ideal environment: 
Members are subject to axial forces only (shear force and bending moment are neglected, which exist but are nonetheless small) 
Self weight of members may be neglected 
Members assumed to be linear/straight
Members are connected at joints by frictionless pins
Applied loads act at joints only 
One end support is idealised as a pin, the other end support is idealised as a roller. This is a useful assumption that simplifies the calculations. As the vertical load is applied, the truss or beam deflects and if one end is a roller this deflection can occur without a change in the total length of the truss. Therefore no extra tension is induced in the truss. 
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💡 General sign conventions:
Compression - negative axial force 
Tension - positive axial force 
Members in tension - force acting away from the joint 
Members in compression - forces acting toward the joint
Determinacy Determining the determinacy of your truss will help you choose which analysis method to use.
Generally a perfect truss satisfies:
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m=2j−3m=2j-3m=2j−3
Where:
m = number of members 
j = number of joints 
If:
m < 2j-3, it is a deficient truss and would collapse
m > 2j-3, it is classified as a redundant truss
Statically Determinate Truss A structure is said to be statically determinate when the number of unknown forces (reactions and internal forces) can be determined from the equilibrium equations alone, which means it is easy to calculate by hand!
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Statically Determinant Truss
Notice here that there are:
5 members 
4 joints 
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m=2j−3m=2(4)−3m=5m=2j-3\\m=2(4)-3\\m=5m=2j−3m=2(4)−3m=5
We will look at analysis of statically determinate trusses further down the page, specially at Method of Joints and Method of Sections. You will need the three equilibrium equations:
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∑ Fx=0∑ Fy=0∑ M=0\sum\ F_x=0\\\sum\ F_y=0\\\sum\ M =0∑ Fx​=0∑ Fy​=0∑ M=0
Statically Indeterminate Truss On the other hand, statically indeterminate trusses present a challenge since they have unknown forces that can't be resolved solely through the equilibrium equations, and so it is harder to calculate by hand!
How do you analyse them? 
Force method - redundant forces are treated as unknowns
Displacement method - displacements are treated as unknowns  
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Statically Indeterminant Structure
Notice here that there are:
6 members 
4 joints  
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m=2j−3m=2(4)−3m=5≠6m=2j-3\\m=2(4)-3\\m=5\neq6m=2j−3m=2(4)−3m=5=6
We will not look at analysis of statically indeterminate trusses in this Design Guide. 
Zero Force Members Some members actually don't carry any load. You may be thinking, what is their role in the structure? They create a "robust" structure which means the truss won't fail by unexpected loads. 
We want to learn how to identify them so we can analyse the rest of the truss by hand quicker!
You can find zero force members at a pin joint not subjected to any external force: 
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Case 1
Case 1: If there are three members and two of them are co-linear then the force in the third member is zero. 
Here, member BC is a zero force member.
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Case 2
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Case 2: If there are only two members that are not co-linear, then there is no reaction at the joint and the force in both members is zero.  
Here, both members CB and CD are zero force members.
A special case where the pin joint is subjected to an external force: 
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Case 3
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Case 3: If there are two members and an applied force at the pin joint is parallel to one of the members and perpendicular to the other then the force in the member perpendicular to the applied force is zero.  
Here, member AF is a zero force member.
Method of JointsThis method finds the unknown forces acting on members through the equilibrium of a joint. 
General steps: 
Draw a Free Body Diagram (FBD) of the whole truss and solve for the reaction forces at supports using the equilibrium equations
Draw a FBD of a joint and solve for the internal forces
Repeat step 2 for all joints 
✏️ Worked Example
Method of SectionsThis method finds the unknown forces acting on members through the equilibrium of a 'section cut'. The best places to make your cut is through a section with three members (or less), otherwise there will be more unknown forces than there are equilibrium equations to solve. The section could be horizontal, inclined, or vertical.
General steps: 
Draw a Free Body Diagram (FBD) of the whole truss and solve for the reaction forces at supports using the equilibrium equations
Draw a FBD of a 'section cut' and solve for the internal member forces
Repeat step 2 until all members are solved
✏️ Worked Example
Let's look at the same Pratt truss from the previous example.
Step 1: Draw a FBD of the entire problem and calculate the reactions at the supports using the three equilibrium equations.
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Support reactions
Identify the section(s) where the member forces are required. 
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Step 2: Create a FBD of the 'section cut'. Make sure each sections does not contain more than three unknown member forces. 
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Section 1 
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Taking the moments about F: 
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Σ MF=0↺(+ve)30⋅8−NCD⋅6=0     NCD=40kN (T)\Sigma\,M_F=0\quad\circlearrowleft(+ve)\\30\cdot8-N_{CD}\cdot6=0\\\qquad\qquad\;\;\,N_{CD}=40kN\,(T)ΣMF​=0↺(+ve)30⋅8−NCD​⋅6=0NCD​=40kN(T)
Taking the moments about C: 
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Σ MF=0↺(+ve)30⋅16−20⋅8−NGF⋅6=0 NGF=40kN (C)\Sigma\,M_F=0\quad\circlearrowleft(+ve)\\30\cdot16-20\cdot8-N_{GF}\cdot6=0\\\qquad\qquad\quad\,N_{GF}=40kN\,(C)ΣMF​=0↺(+ve)30⋅16−20⋅8−NGF​⋅6=0NGF​=40kN(C)
Equilibrium of forces in the y-direction:
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Σ Fy=0↑(+ve)30−20−0.6⋅ NCF=0 NCF=16.67kN (C)\Sigma\,F_y=0\quad\uparrow(+ve)\\30-20-0.6\cdot\,N_{CF}=0\\\qquad\qquad\quad\,N_{CF}=16.67kN\,(C)ΣFy​=0↑(+ve)30−20−0.6⋅NCF​=0NCF​=16.67kN(C)
Step 3: And so on!
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Assumptions

Determinacy

Statically Determinate Truss

Statically Indeterminate Truss

Zero Force Members

Method of Joints

✏️ Worked Example

Method of Sections

✏️ Worked Example

CalcTree

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