Statics

Lesson 9

Trusses: Explorations in Triangulation

🟦  9.1  Introduction to trusses

Trusses are structures that are composed of members connected at their ends with pin connections.Β 

For stability, each truss panel is comprised of three members at the perimeter. As a result, trusses tend to take the shape of a sequence of connected triangles. This phenomenon is called triangulation.

True trusses are only loaded at the nodes (at the pins).

As an introduction to trusses, let's consider the structure in the images below. While few people would describe this particular structure as a truss, it behaves like one, and for that reason will serve as an excellent introduction to a new and tricky topic.

Let's assume that the three rubber bands are identical in every way, and stretched equal amounts. That means that each carries the same magnitude of tension force.

Let's also assume that the discs are heavy enough that the system depicted is in static equilibrium.Β 

Consider the FBD of global equilibrium below. The structure has removed from the hexagonal table and the effects of the table are depicted as the three normal forces (N). Note that in this FBD, the forces exchanged between the nodes and members aren't shown.

Then, inspect the exploded FBDs. Each node and each member has been detached and drawn in equilibrium.

❏ Global equilibrium of the truss

❏ Exploded FBDs of the nodes and members

Did you notice the Newton's Third Law (N3L) pairs in the exploded FBD?Β 

Remember that Newton's Third Law pairs are equal in magnitude but opposite in direction. They represent interactions between adjacent elements.

Did you also notice that the rubber bands were cut to expose the tension? That's important too.

If you make a cut through a member, you can accurately depict the force as one that induces internal tension in the rubber band.

This is a hard concept to understand, because the rubber band is in contact with the nail shaft. That means that the transfer of force between the bodies is compressive (a push).Β 

As we're far more interested in what the connection does to the rubber band, we would always depict the force on the rubber band as tension (arrows away from the body).

❏ Internal normal or axial force

Takeaways so far:

(1) Global equilibrium is still important. We can use the three planar (2D) equations of equilibrium to solve for unknown reactions.

(2) We can also investigate nodal equilibrium. Since the forces in each nodal FBD are concurrent, we can solve for unknowns using two equations of equilibrium: force equilibrium in the x-direction and in the y-direction.

❏ Equations of equilibrium

🟦  9.2  Trusses in the wild

Trusses can be of any size or scale, but we often associate them with bridge structures, such as the Okuma bridge depicted at left, located in Yokohama, Japan.

Trusses act a little like beams -- except that most of the material is cut away.Β 

A truss that spans a certain distance is significantly more efficient (cheaper, uses less material) than a beam with the same span and load.


Image source: Public domainΒ 

❏ A large-scale truss structure in Yokohama, Japan

🟦  9.3  "Two-force" or axial members

True trusses are comprised solely of two-force members or axial members. The former terminology is common in lower-level courses (like Statics) and the latter is common in higher-level courses (like Structural Analysis).

❏ What are 2FMs? 

A two-force member (2FM) is defined as one that:

❏ Why are 2FMs important? 

When you identify two-force members early in your problem-solving process, you will save yourself lots of time. This is because each pin has only one unknown (the axial tension or compression).

The key is that the x-direction and y-direction components of that force are coupled. The force components are similar to the x-direction and y-direction components of the member's geometry. We use similar triangles to solve for unknowns.

Let's say that you know Fx and want to solve Fy. Divide Fx by Lx, and then multiply by Ly. That's Fy.

And if you need to know the axial force (the tension or compression along the longitudinal axis), you can solve it with ratios or with the Pythagorean Theorem.

❏ 2FM experience tension or compression:

❏ Which members are 2FM?

(Answer: only B is a 2FM)

❏ How to use the ratio of lengths to solve for force components for a 2FM

❏ 2FM example problem

🟦  9.4  Introduction to Method of Joints

In a truss, we are usually interested in solving for unknown reactions, as well as the forces transferred between various members at the pin connections. The Method of Joints is a systematic way to solve for all of the unknowns in a truss.

The flipbook below shows the procedure. Start by using global equilibrium to solve unknown reactions. Then, move systematically from joint to joint (or node to node) throughout the structure. At each node, use two equations:

You have to work strategically: start at a node that is solvable (no more than two unknowns). After each node is solved, apply Newton's Third Law to transfer equal and opposite forces to the next node.

❏ Method of Joints introductory example

🟦  9.5  Zero-force members

Sometimes, under a given loading, a truss member carries no load. When the internal force in a member (under a given loading) is zero, we call that member a zero-force member (ZFM).

Consider Node A. Members 1 and 2 are pin-connected at A. Node A is not subjected to a point load (it is not the location of an applied load or an external reaction force). Member 1 can only exert a horizontal force on A. Since member 1 can't transfer a y-direction force, then the y-direction force in Member 2 must also equal zero. Both members have zero internal force. They are both zero-force members (ZFM).

How about node B? It's unloaded. Members 3 and 4 are collinear (they share the same longitudinal axis. This means that F3 = F4. If you solve for either force, you know that 100% of the force travels through pin B and is picked up by the other member.Β 

Finally, inspect node C. It's unlloaded as well. Three members frame into C. If we only consider the ij coordinate system, we can't identify F7 as a zero-force member. But if we use the xy coordinate system, we can make some important deductions. Members 5 and 6 only have x-direction forces. Since there isn't a force to balance a potential y-direction force in Member 7, the force in Member 7 has to be zero. By summing forces in the x-direction we see that the force in Member 5 must be equal to that in Member 6.

Students often want to know why engineers would design a truss with members that don't carry any force. Consider a long-span truss that supports moving traffic. Car wheels can be modeled as point loads (forces), and as they drive across the bridge, certain truss members are in the load path, and transfer the weight of the car to the bridge supports. Other truss members don't engage until the car drives to a position that puts them in the load path between the car and the support.

❏ Unloaded nodes

🟦  9.6  Problem-solving approach for Method of Joints

In this flipbook, you'll see a mapping approach that can be used to fully solve a truss problem using the Method of Joints.Β 

Work this problem yourself.

Emulate this approach when you work any problem that requires the Method of Joints.

❏ Flipbook: Method of Joints example problem

🟦  9.7  Method of Sections

All (statically determinate) trusses can be solved with the Method of Joints. It's a systematic way to solve for all unknowns.

Sometimes, though, we aren't asked to solve for the entire truss. Maybe we just need to solve one particular unknown, such as the internal force in just one member.

The Method of Sections is a useful tool for solving a specific unknown. You completely cut through the truss to expose no more than three unknown internal forces, including the force you want to solve for. Then, you can use the equations of equilibrium on the cut piece to solve the unknowns.

Inspect the flipbook to see how this works.Β 

Important: it is only because all truss members are two-force members that we can cut members and represent their internal workings with a single axial force. Later in this course we'll learn about internal shear force and internal bending moment. Both are zero in 2FM.

❏ Flipbook: Method of Sections example problem

🟦  9.8  Triangulation and stability

A truss panel is stable when it consists of 3 members connected by three pins.Β 

This is due to triangulation.

If a truss were to have a pin-connected panel that was bounded with four or more members, it would not be a stable structure.Β 

❏ Stable trusses

This four-member pin-connected panel is called a collapse mechanism if you intended to design a truss.

On the other hand, if you intended to design a machine (an assembly of members that is designed for motion), the four-member panel could be part of your design.

We'll study simple machines a little later in this course. For now, please don't call this type of structure a truss.

❏ A machine in motion

Compare these two trusses. The top truss, with the yellow triangular panels, is stable and statically determinate. This means that such a truss can be solved using the equations of equilibrium.

The bottom truss has four extra members (in green) connecting the nodes.

When a truss has more members than are necessary, the extra members can be called redundant. This type of truss is called statically indeterminate. That means that in addition to the equations of equilibrium, you need a different type of equation to solve such a truss. Civil engineering majors will study these types of trusses in subsequent courses.

❏ Determinate and indeterminate trusses

➜ Practice Problems

Here are scrambled answers for Problems 1, 2, 3, and 4. Use these to check your work.

0k Β  0k 5.79k (T) 7.71k β†’ 15k (T) 80k (C) 100k (T) 157 kN (T) 157 kN (T)

Problem 1.

External force Pc has been applied to Truss ABC.

Use the FBD of Node A to solve for F1 and F2.

Report tension as (T).

Report compression as (C).

Problem 2.

An external force of 140 kN has been applied to Truss ABC.

Use the FBD of Node B to solve for F1 and F2.

Report tension as (T).

Report compression as (C).

Hint: this is a symmetric truss, so your calculations should prove that F1 = F2.

Problem 3.

Three external forces have been applied to this truss.

Use the FBD of Node A to solve for F1 and Ax.

For F1, report tension as (T) or compression as (C). It's an internal force, so we report tension or compression.

For Ax, use a cartoon arrow (β†’ or ←) to report your solution. We report reaction forces with arrows.

Problem 4.

A 60 kip external force has been applied to truss ABC. You need to solve F1, F2, and F3.

To do so, you'll first solve the reactions, using global equilibrium.

Then, use the Method of Joints.

Remember: report internal forces as (T) or (C). Report reactions with cartoon arrows (β†’, ←, etc.).

Here are scrambled answers for Problems 5, 6, and 7. Use these to check your work.

0 0 0 0 0 0 0 13.3 kN (C) 16.7 kN (T)

18.9k (C) 20 kN (C) 26k (C) 30 kN (C)

120 kN (C) 120 kN (C) 170 kN (T) 220 kN (T)

Problem 5.

Three loads have been applied to this symmetric truss. You need to solve F1, F2, F3, F4, and F5.

Take advantage of symmetry.

Starting with this problem, I'm not making the "map" for you (like in the last problem). You will need to do that for yourself for the remainder of the problem set.

Problem 6.

This truss supports two applied loads.Β 

Solve F1, F2, F3, F4, and F5.

Do so using the Method of Joints.

Problem 7.

This truss is perfectly symmetric and supports four loads.

You need to solve F1, F2, F3, F4, F5, F6, and F7.

Do so using the Method of Joints.

Here are scrambled answers for Problems 8, 9, and 10. Use these to check your work.

10k (T) 13.1k (T) 15.6k (C)

Problem 8.

Use the Method of Sections on this problem.

Solve for the force in Member 5.

Problem 9.

Use the Method of Sections on this problem.

Solve for the force in Member 6.

Problem 10.

This is a symmetric truss.

Solve for the force in Member 1.