In this session, I am going to discuss the analysis of statically determinate trusses.

This lecture addresses two questions:

What is a statically determinate truss, and how to analyze it?

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 roofs and bridges.

Howe truss,

Warren truss,

Pratt truss,

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 has the freedom 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 the truss member itself.

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 and a roller.

A 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.

So, 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 this question in a minute.

For now, let's just accept it as a fact.

But what about the number of unknown forces?

The total number of unknown forces in a truss equals to the number of its members,

plus the number of its support reactions.

Here, the truss has five members.

Each member carries an axial force, which needs to be determined.

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 provides two reactions, and the roller has one reaction.

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 axial force in each member?

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 the 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.

For example, if conceptually we cut a beam into three segments,

with the correct internal forces shown at the cut points,

then the equilibrium equations must be satisfied for each segment,

if the entire structure is to remain in equilibrium.

We can apply this principle in order 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 F_ac.

Similarly, let's call the force in member BC, F_bc.

We denote the force in each of the remaining members in a similar manner.

Now let's conceptually 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 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 force is tensile.

So, the free-body diagram for the segment looks like this:

The only force present in the diagram is the unknown tensile axial force in the member labeled F_ac.

I am going to draw the free-body diagram for the other member segments in a similar manner,

Each member is assumed to carry an unknown force.

The member is assumed to be in tension.

Now, let's draw the free-body diagram for the joint segments.

Let's examine joint A first.

It connects members AC and AD.

Since member AC is attached to the joint, the axial force in the member also acts on the joint.

We show this by placing F_ac on the joint.

Note that the direction of F_ac at the joint is opposite to the direction of the same force shown on the member.

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.

Obviously, this happens only when the two forces are kept in the opposite directions,

so that they cancel each other out.

There is one more member force acting on the joint, the force in member AD.

Again, we place this force at joint A, in a direction opposite to how the force is acting 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:

F_ad, F_bd and F_cd.

Now we have the complete free-body diagram for each segment.

As I mentioned before, for the entire truss to be in equilibrium,

each segment must be in equilibrium.

We can easily see that the equilibrium equations are satisfied for each member segment.

The two force vectors shown in each segment cancel each other,

that is, their sum equals zero.

And since there is no bending moment,

or shear force acting on a free-body diagram,

the moment equilibrium equation is automatically satisfied.

So now, let's turn our attention to the joint segments.

When we say each joint has to be in equilibrium,

we are saying, at each joint:

sum of the forces in the X direction must be zero.

Sum of the forces in the 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 joint forces pass through the joint,

so we only need to show that the other two equations are satisfied.

Let's see how this works for joint A.

Using the geometry of the truss, we can calculate angle alpha.

It is 45 degrees.

Then, we can write:

Sum of the forces in the X direction equals:

A_x plus F_ad plus F_ac cosine 45 equals zero.

And, sum of the forces in the Y direction equals:

A_y plus F_ac sine 45 equals zero.

Note that we have four unknown forces,

but only two equations here,

it is not possible to find these unknowns yet.

Let's number these equations one and two.

We need to repeat this formulation for the other joints as well.

For joint B,

angle gamma can be easily calculated from the truss geometry.

The angle is 45 degrees.

The two force equilibrium equations for this joint are:

Sum of the forces in the X direction equals:

negative F_bd minus F_bc cosine 45 equals zero.

Sum of the forces in the Y direction equals:

B_y plus F_bc sine 45 equals zero.

Let's number these equations: three and four.

For joint C, we have:

Sum of the forces in the X direction equals:

F_bc sine 45 - F_ac sine 45 + 5 = 0

And, sum of the forces in the Y direction equals:

-F_ac cosine 45 - F_bc cosine 45 - F_cd = 0

These equations are labeled: five and six.

Finally, for joint D we have:

Sum of the forces in the X direction equals:

F_bd - F_ad = 0

Sum of the forces in the Y direction equals:

F_cd = 0

These equations are numbered: 7 and 8.

So, we have eight equations and eight unknowns.

We formulated two equations per truss joint,

that is 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 eight unknowns.

Since the number of equations equal to the number of unknowns,

we can easily solve 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:

One: 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, two, 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 F_cd.

F_cd equals zero.

Now, if I replace F_cd in the joint free- body diagrams with zero, we get:

Scanning the diagrams again, I notice joint C,

now has only two unknowns.

Then, I should be able to solve for the unknown, using equations five and six.

Here they are in simplified form, where F_cd is replaced with zero.

0.707 F_bc minus 0.707 F_ac plus five equals 0.

And, negative 0.707 F_bc minus 0.707 F_ac equals zero.

Solving them for F_ac and F_bc, we get:

F_ac equals 3.55 N.

And F_bc equals negative 3.55 N.

The free-body diagrams are now further simplified.

Note that joint B now has only two unknown forces: F_bd and B_y.

So, I'm going to use equations 3 and 4 to find these unknowns.

The equations are:

Solving them for the unknown, I get:

B_y equals 2.5 N.

F_bd equals 2.5 N.

If we now examine joint D, we can clearly see that F_ad is also equal to 2.5 N.

The remaining joint, joint A, now looks like this:

There are only two unknown forces remaining, A_x and A_y.

They can be determined using Equations 1 and 2.

A_x equals -5 N.

A_y equals -2.5 N.

The negative sign for sum of the calculated values means

that the assumed direction for the force is incorrect.

More specifically, we assumed A_y to be upward,

in reality, it is a downward force.

We generally present the results of the truss analysis by writing the force magnitudes on the members,

and marking the compression member with a capital C,

and tension members with a capital T.