In this lecture, we are going to discuss the analysis of statically determinate trusses,
using The Method of Joints.
The lecture addresses two main questions:
What is a statically determinate truss, and how do we analyze one?
A truss structure is an interconnected network of slender members,
each capable of carrying an axial force only.
The force is either compressive,
or tensile.
Theoretically speaking, no shear force or bending moment is present in truss members.
Here are a few truss configurations, usually seen in building roofs and bridges.
Howe truss,
Warren truss,
Pratt truss,
and K-truss.
Generally, we assume truss members are connected to each other using frictionless pins.
This means if we have two truss members that are joined together,
one is free to rotate relative to the other.
In other words, truss joints cannot resist any bending moment,
since neither the joints, nor the members of a truss are designed to carry any bending moment,
applied loads need to be placed at the joints only.
No loads should be applied directly to a truss member.
Here is a simple truss structure with five members:
The frictionless joints of the truss are labeled: A, B, C and D.
The truss is subjected to a horizontal force of 5 N at joint C.
The entire structure rests on a pin support at B, and a roller support at A.
The truss is said to be statically determinate,
if its member forces can be calculated solely using the equilibrium equations.
Algebraically speaking, if the number of unknown forces equals to the number of equilibrium equations,
then the truss is said to be statically determinate.
So, how many equilibrium equations can be written for a truss structure?
The total number of equilibrium equations equals two times the number of joints in the structure.
If the truss has four joints, we can write eight equilibrium equations for it.
Why two times the number of joints? I'm going to answer that question in a minute.
For now, let's just accept it as a fact.
But what about the number of unknown forces?
How many of them do we have?
The total number of unknown forces in a truss equals to the number of its members,
plus the number of its support reactions.
Our truss has five members.
Each member carries an axial force, so there are five unknown member forces.
Further, since the structure rests on a pin and a roller, there are three unknown reactions.
The pin support provides two reaction forces, and the roller support has one reaction force.
Therefore, the total number of unknowns is five plus three, or eight.
In this case, since the number of unknowns equals to the number of equations,
the truss is said to be statically determinate.
How do we analyze such a truss?
How do we calculate the member forces?
There are several truss analysis techniques.
Here we are going to explain and illustrate a technique called The Method of Joints.
This technique is a direct application of the principle of static equilibrium.
According to this principle, if a structure is in equilibrium,
then not only the equilibrium equations must be satisfied for the structure as a whole,
but they also must be satisfied for any of its parts.
Suppose we have a simply supported beam, with three support reactions.
then the beam would be in equilibrium, if the following equilibrium equations can be satisfied:
Further, if indeed the beam is in equilibrium, then if cut into three segments like this:
The equilibrium equations must also be satisfied for each segment:
We can apply this basic principle to analyze statically determinate trusses.
Here is our simple truss:
Let's draw the free-body diagram of the entire structure,
keep in mind that each truss member carries an axial force only.
Let's refer to the axial force in member AC as Fac.
Similarly, let's call the force in member BC, Fbc.
We label the force in each of the remaining members in a similar manner.
Now let's separate the members from the joints,
that is, we divide the entire structure into nine segments.
There are five member segments, and four joint segments.
Let's look at the segment containing member AC.
We know that the member carries an axial force,
but we don't know the magnitude of the force.
We also don't know if the force is compressive or tensile.
Since I want to draw the free-body diagram for the segment,
I need to assume a direction for the force.
I am going to assume the member carries a tensile force.
So, the free-body diagram for the segment looks like this:
The only force present in this diagram is the unknown tensile force which we have labeled Fac.
I am going to draw the free-body diagram for the other members in a similar manner,
where each member is assumed to carry an unknown tensile force.
Now, let's draw the free-body diagram for the joint segments.
We examined joint A.
It connects members AC and AD.
Since member AC is attached to the joint, the axial force and the member also acts on the joint.
We show this by placing Fac on the joint.
Note how this joint force acts in the opposite direction to the same force acting on member AC.
This is the case because the two forces must cancel each other out.
That is, if we reconnect member AC to joint A, the sum of the internal forces must vanish.
This could happen only if the two forces act in the opposite directions, so that they would add up to zero.
There is one more member force acting on the joint, the force in member AD.
Again, we place this force at A, in the opposite direction to the force shown on member AD.
This completes the free-body diagram of joint A.
At joint B,
there are two member forces: the force in member BC, and the force in member BD.
Joint C has three member forces acting on it.
The force in member AC, the force in member BC, and the force in member CD.
At joint D,
there are three member forces: Fad, Fbd and Fcd.
Now we have the complete free-body diagram for each member and joint.
As I mentioned before, for the entire truss to be in equilibrium,
each truss member must be in equilibrium, and each joint must be in equilibrium.
We can easily see that the equilibrium equations are satisfied for each member.
The two force vectors shown in each member cancel each other out,
that is, their sum equals zero.
And since there is no bending moment, or shear force acting on the member,
the moment equilibrium equation is automatically satisfied.
Now let's turn our attention to the joints.
Each joint has to be in equilibrium. This means given the free-body diagram of the joint,
sum of the forces in X direction must be zero.
Sum of the forces in Y direction must be zero,
and sum of the moments about the joint must be zero.
Note that the moment equation is automatically satisfied,
since all the forces passed through the joint,
therefore we only need to examine the first two equations.
Let's see how this works for joint A.
Using the geometry of the truss, we can calculate angle alpha.
It is 45 degrees.
Now we can write: Sum of the forces in the X direction equals:
Ax plus Fad plus Fac cosine 45 equals zero.
And, sum of the forces in the Y direction equals:
Ay plus Fac sine 45 equals zero.
Note that we have four unknown forces,
but only two equations here,
therefore we cannot solve for the unknowns directly, using these equations.
Let's number them: one and two.
We need to apply the equilibrium equations to the remaining joints.
Here is joint B.
Angle gamma can easily be calculated from the truss geometry.
The angle is 45 degrees.
Here, the equilibrium equations are:
Sum of the forces in the X direction equals:
negative Fbd minus Fbc cosine 45 equals zero.
Sum of the forces in the Y direction equals:
By plus Fbc sine 45 equals zero.
Let's number these equations: three and four.
For joint C, we get:
Sum of the forces in the X direction equals:
And, sum of the forces in the Y direction equals:
These equations we label: five and six.
Finally, for joint D we have:
Sum of the forces in the X direction equals:
Sum of the forces in the Y direction equals:
These equations are numbered: 7 and 8.
So, we have eight equations and eight unknowns.
We formulated two equations per truss joint,
therefore the number of equations equals two times the number of joints.
And since the truss has five unknown member forces, and three unknown support reactions,
we get a total of 8 unknowns
Since the number of equations equals to the number of unknowns,
we can easily solve the equations for the unknowns.
We can either solve the equations simultaneously, using a standard technique,
such as the Gaussian elimination method,
or we can use shortcuts to speed up the calculations.
Let's solve the problem rather quickly, without using a standard technique.
I'm going to scan the joint free-body diagrams looking for two types of joints:
Either a joint with only two unknown forces,
because if such a joint exists,
then the two forces can be calculated directly using the joint equilibrium equations.
Or, a joint having exactly one unknown force, in either X or Y direction.
Found one! Joint D has only one unknown force in the Y direction,
therefore, using equation eight, I can solve for Fcd.
Fcd equals zero.
Now, if we replace Fcd with zero in the joint free-body diagrams, we get:
Scanning the diagrams again, I notice joint C.
It now has only two unknowns,
So, I should be able to solve for the unknowns,
using equations five and six.
Here they are in simplified form, where Fcd is replaced with zero.
0.707 Fbc minus 0.707 Fac plus five equals 0.
And, negative 0.707 Fbc minus 0.707 Fac equals zero.
Solving these equations for Fac and Fbc, we get:
Fac equals 3.54 N
And Fbc equals negative 3.54 N.
Let's simplify the free-body diagrams again.
Note that point B now has only two unknown forces: Fbd and By.
I'm going to use equations 3 and 4 to find these unknowns.
The equations are:
Solving them for the unknowns, we get:
By equals 2.5 N.
Fbd equals 2.5 N.
Now, if we examine joint D, we can clearly see that Fad is also equal to 2.5 N.
We only have one more joint remaining, joint A.
The free-body diagram for the joint looks like this:
There are only two unknown forces remaining, Ax and Ay.
They can be determined using Equations 1 and 2.
Ax equals -5 N.
Ay equals -2.5 N.
The negative sign here means that the assumed direction for the force is incorrect.
We present the results of the truss analysis by writing the force magnitude on each member,
and by marking compression members with capital C, and tension members with capital T, like this: