Statics

Lesson 13

3D Statics, Part I

🟦 13.1 The wonderful world of 3D!

🟦 13.2 Right-handed coordinate systems

🟦 13.3 Points (parentheses) vs. vectors (chevrons)

🟦 13.4 Vector components and resultants in 3D

🟦 13.5 The six equations of equilibrium

🟦 13.6 3D connections

🟦 13.7 Two-force (axial) members in 3D

🟦 13.8 Concurrent force problems in 3D

🟦 13.9 Moments about different axes

🟦 13.10 Condensing a 3D problem to a 2D analysis

➜ Practice Problems

🟦 13.1 The wonderful world of 3D!

While all Statics problems represent the reality of the 3D world, many can be fully analyzed with the 2D approaches we have used so far.

Some Statics problems are complex enough to require a 3D vector approach. Learning the 3D approach in Statics will facilitate your future success in Dynamics, too.

If a 3D Statics problem is relatively simple, we may approach it by projecting the geometry into multiple 2D planes.

❏ A relatively simple 3D problem

If a 3D Statics problem is of intermediate complexity, as depicted here, we will likely use 3D vector approaches.

❏ An intermediate 3D problem

When a 3D Statics problem is complex, such as the playground design shown here, hand methods with 3D vectors would be too time-consuming.

Instead, you would likely analyze this type of structure using software and the Finite Element Method, which is an approximate approach that falls under the umbrella of so-called numerical methods.

These approaches are not in the scope of a Statics course.

❏ A complex 3D problem

Image source: S. Reynolds, photograph, playground equipment, Amsterdam, Netherlands (designer unknown)

🟦 13.2 Right-handed coordinate systems

To set up a so-called right-handed coordinate system, point your right thumb in the x-direction and your right index finger in the y-direction.

Your right middle finger points in the z-direction.

If the coordinate system isn't set up in this way, your 3D vector operations will have a sign error.

This is called the right-hand rule for setting up Cartesian coordinate systems.

❏ The "right-handed" system

The Cartesian coordinate system is named after René Descartes, a French mathematician who lived 1596–1650.

🟦 13.3 Points (parentheses) vs. vectors (chevrons)

We use parentheses to designate points. For example, A: (1, 2, 3) meters means that Point A lies at coordinates x=1m, y=2m, and z=3m.

We use chevrons for vector components. If Vector B is a force, and it's designated as <4, 5, 6> kips, it means that the force components are 4k in the x-direction, 5k in the y-direction, and 6k in the z-direction.

In order to fully communicate everything there is to know about a force vector, you will need to specify both the point of application (using coordinates in parentheses) as well as the vector itself (using components in chevrons).

🟦 13.4 Vector components and resultants in 3D

Consider the force vector <3,4,5> Newtons.

It's depicted as head-to-tail addition in the interactive model.

It shows that a vector in 3D space can always be visualized in terms of two right triangles.

The first triangle lies in the xy plane.

You see Fx=3N and Fy=4N, but not Fz (as it's coming directly towards you. You can calculate the hypotenuse of this triangle using the Pythagorean Theorem. We can call the hypotenuse Fxy. It's equal to 5N.

The second right triangle has a base of Fxy=5N and a height of Fz=5N.

Its hypotenuse is equal in length to the magnitude of the original vector. That's sqrt(5^2 + 5^2) = 7.071 N. Again, this is a magnitude.

We would need to multiply that magnitude by the proper unit vector to express it as the original <3,4,5> Newton force vector.

❏ Interactive: the two-triangle approach

There is a 3D version of the Pythagorean Theorem shown here. The sum of the squares of the three components is equal to the square of the resultant vector.

❏ Key equations

🟦 13.5 The six equations of equilibrium

The equations of equilibrium (E.o.E.) continue to be a key concept in Statics.

A body is in a state of static equilibrium when all E.o.E. are satisfied.

In 3D, a body is only in static equilibrium when all six E.o.E. are satisfied.

With six equations in play, we can solve up to six unknowns.

❏ The six E.o.E. for 3D problems

If you're familiar with using linear algebra (matrix methods) to solve a system of linear equations, you're welcome to do that. The problems introduced in this text may all be solved using simple algebraic methods (substitution and elimination). If you'd like to learn how to solve a system of linear equations using a matrix approach, you are welcome to do so: check out this link.

🟦 13.6 3D connections

In a 2D, planar problem, there are three main types of connections: the pin, the pin-roller, and the fixed connection.

If a connection can prohibit translation, it has the ability to develop a force reaction.
If a connection can prohibit rotation, it has the ability to develop a moment connection.

❏ 2D connections and reactions

From top to bottom: the 2D roller, the 2D pin, and the 2D fixed connection.

As learned in Lesson 04:

The 2D pin-roller connection prevents translation in one direction. It develops one force reaction, perpendicular to the surface. There's 1 unknown.
The 2D pinned connection prevents translation in that plane. It develops a force at an unknown angle of inclination (or, as more commonly expressed, x- and y- component forces). There are 2 unknowns.
The 2D fixed connection prevents translation in the plane and prevents rotation about the axis that is perpendicular to the plane. It therefore can develop two force reactions (x- and y- components) and a moment about z. There are 3 unknowns. Don't forget to compute the reaction moment!

In 3D, we look at a connection assembly and go through some logic tests:

(1) Does the connection assembly prevent translation in a given direction? If so, it has the potential to develop a reaction force in that direction.

a constraint with respect to x-direction translation means there's an x-direction force reaction
a constraint with respect to y-direction translation means there's an y-direction force reaction
a constraint with respect to z-direction translation means there's an z-direction force reaction

(2) Does the connection assembly prevent rotation about a given axis? If so, it has the potential to develop a reaction moment about that axis.

a constraint with respect to rotation about x means there's a reacting moment about x
a constraint with respect to rotation about y means there's a reacting moment about y
a constraint with respect to rotation about z means there's a reacting moment about z

This image depicts the three most common types of 3D connection assemblies.

The first row of the table depicts a wheel (or roller) in 3D. It only develops an Fy reaction (normal to the rolling surface). We omit a z-direction reaction (if you want to transfer a z-direction force, don't rely on friction).

The second row of the table shows a 3D pin connection or a ball-and-socket joint. Translation is constrained with respect to x, y, and z, rotation remains unconstrained. Therefore three force reactions can develop: Fx, Fy, and Fz.

The third is a 3D fixed connection. It's shown here as an all-around fillet weld. In 3D, fixed connections can develop 6 potential reaction forces and reaction moments: Fx, Fy, Fz, Mx, My, Mz.

❏ 3D connection assemblies and reactions

From top to bottom: the 3D wheel, the 3D pin (sometimes called "ball and socket"), and the 3D fixed connection (the illustration shows an "all-around" weld.

There are infinite combinations of 3D connection assemblies. A few are shown below. The engineer chooses how to properly model each connection assembly in the 3D loading diagram.

❏ A five-member 3D connection

Photo by S. Reynolds. Structure in Amsterdam, Netherlands. Photo taken Summer 2023.

❏ A seven-member 3D connection

Photo by S. Reynolds. Structure in Rome, Italy. Photo taken Summer 2022.

❏ A highly sophisticated 3D connection

Image credit: https://commons.wikimedia.org/wiki/User:KTo288 CC BY-SA 3.0 DEED. Heathrow Airport, London. 3D truss connections in Terminal 5.

❏ A unique connection assembly

Photo by S. Reynolds. Door hinge in Amsterdam, Netherlands. Photo taken Summer 2023.

🟦 13.7 Two-force (axial) members in 3D

In 2D, we learned that the geometry of a two-force member (2FM) can help us solve for unknown forces or force components. The force components are congruent to the length components.

The same logic holds true for 3D.

An axial member in 3D has the same ratios between the length components and the force components.

These pin-pin (pins at both ends) members are usually called two-force members in lower level courses and axial members in upper level courses and in engineering practice. I have never heard a practicing engineer use the term two-force member, but I do hear many academics use the term. Regardless of what you call them, they either carry pure tension or pure compression.

❏ Example problem 1:

Cable AB carries a tension force of 100 kN. Solve Tx, Ty, and Tz. You know the geometry: Lx = 2m; Ly = 3m; and Lz = 6m.

Solution:

LAB = sqrt(2^2 + 3^2 + 6^2) = 7m

Fx = (2/7)*100 kN

Fy = (3/7)*100 kN

Fz = (6/7)*100 kN

❏ Example problem 2:

You deduce that the z-direction force in cable AB is 100 pounds. You also know the geometry: Lx = 2 feet; Ly = 3 feet; and Lz = 6 feet. What is the tension in cable AB?

Solution:

LAB = sqrt(2^2 + 3^2 + 6^2) = 7 feet

T = 100# * (7 feet / 6 feet) = 117#

2FM at Charles DeGaulle airport in Paris (photo by S. Reynolds, 2023).

2FM at an Amsterdam subway station (photo by S. Reynolds, 2023).

🟦 13.8 Concurrent force problems in 3D

In 2D, in a concurrent force problem, you could solve 2 unknown forces with 2 E.o.E.:

sum of the forces in x-dir = 0
sum of the forces in y-dir = 0

This flipbook solves a 3D truss. All forces are concurrent at D.

First, flip through the book to get a sense for the solution procedure.

Then, open up a blank piece of paper and solve it independently.

Repeat these two steps until the process makes sense to you.

❏ Flipbook: analysis of a 3D truss

When assessing nodal equilibrium in a 3D concurrent force problem, it's helpful to draw the point as a cube instead of a sphere, as illustrated below.

❏ FBD (3 unknown axial forces)

In this 3D drawing, it can be hard to visualize the x, y, and z components of the three tension forces.

❏ Nodal FBD using a cube

It's much easier to keep track of the vector components with a cube-shaped nodal FBD, as shown here.

🟦 13.9 Moments about different axes

While studying abroad in Rome, Italy, you look out the window one day and notice a steel beam cantilevering out from the exterior wall. The beam is "built-in" to the brick.

The beam supports a pulley, and someone seems to be hoisting a mysterious purple prism from the ground below.

You estimate that the purple prism weighs about 80 Newtons.

That means that tension in the cable is also 80 Newtons.

The two downward forces (the prism and the pull force from the person on the ground) sum to 160 N (neglecting friction in the pulley itself).

❏ A cantilever beam supports a pulley

You remember how cantilever beams work from Lesson 8 (Beams). The support (in this case, the brick wall) provides a vertical reaction and a reaction moment.

You draw a side view of the structure in the xy plane.

Technically speaking, we could say that you drew an xy elevation or that you projected the geometry with respect to z.

You compute the moment at the support as –160 N∙m (or 160 N∙m ↻).

Before, we simply called this MB. Now, we'll need a second subscript. We can write MB,z (the moment at B with respect to rotation about z).

❏ Side view (xy planar projection)

The top image is the loading diagram. The bottom image is the FBD.

Let's also draw a projection of the yz plane. This elevation helps us visualize the moment about x.

That works out to:

MB,x = 160N*90mm = –14.4 N∙m

In 3D space, a moment vector has three components. For this problem, the tendency to rotate about x is –14.4 N∙m; the tendency to rotate about y is zero; and the tendency to rotate about z is –160 N∙m.

We could therefore express the moment reaction vector as <–14.4, 0, –160> N∙m.

❏ Cross-section view (zy projection)

🟦 13.10 Condensing a 3D problem to a 2D analysis

A three-dimensional structure that is symmetric (with respect to its geometry and loading) can be condensed to a 2D problem.

For instance, let's say that you are trying to tip a sibling's chair by exerting force at B and C on the chair with your two hands.

This is depicted as System I.

But does the chair tip or remain in static equilibrium?

❏ Some 3D problems can be modeled in 2D

You recall principles of static equivalency and put the forces at D instead (System II). This is not fully equivalent, but it is statically equivalent.

Now that the loads lie in a single xy plane, we can project the geometry with respect to x. This creates equivalent system III. It's a 2D planar statics problem; you can use scalar approaches to solve it (sum moments about E to determine whether or not the chair tips).

The key takeaway is that you don't need to use the power of vector notation on every 3D statics problem. Use 2D scalar notation whenever you can.