Lesson 14

3D Statics, Part II

🟦 14.1  The six equations of equilibrium

The equations of equilibrium (E.o.E.) continue to be the most important concept in our course. A body is in a state of static equilibrium when all E.o.E. are satisfied.

In 3D, we have six equations of equilibrium to satisfied. A body is only in static equilibrium when all six equations are satisfied. 

With six equations, we can also solve up to six unknowns in a 3D problem.

🟦 14.2  An introductory example problem

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 supports a pulley, and someone seems to be hoisting a 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.

You remember how cantilever beams work right away. 

The support (in this case, the brick wall) provides a vertical reaction and a reaction moment.

Just as you did in Lesson 8 (Beams), you draw a side view of the structure in the xy plane. We would say that you drew an xy elevation or projected the geometry with respect to z.

You compute the moment at the support as 160 N•m (or 160 Nm ↻) as shown.

What we didn't focus on before is that this moment is about the z-axis that is coincident with B. This fact is now critically important as we move into 3D space.

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 Nm

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 Nm.

We could therefore express the moment reaction vector as <14.4, 0, 160> Nm.

🟦 14.3  Illustrating moment as a double-headed vector

The curly arrow symbol for moment is useful in 2D projections, when the moment is about the axis that is perpendicular to the screen.

For other views -- and in 3D illustrations, it's best to use a double-arrow vector instead.

Moment has always been a vector, because it has both a magnitude and direction. The "direction" of a moment vector is more abstract than the direction of a force vector. It's a reference to the axis about which there is a tendency to rotate.

We use the right-hand rule to help us convert double-arrow notation to curly-arrow notation and vice versa. The steps are laid out in the graphic: read them carefully.

Components of moment vectors

Double-arrow notation

🟦 14.4  Right-handed Cartesian coordinate systems

In engineering and science, there's not just one right-hand rule. There are actually several different ones. Here is another one. 

We use a right-hand rule to set up Cartesian (named after René Descartes) coordinate systems. (If the coordinate system isn't set up properly, your vector operations will have a sign error.)

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

Using this logic, given any two axes, you can use this right-hand-rule to set up (or determine) the direction of the third axis.

The "right-handed" coordinate system

🟦 14.5  Sign conventions for double-arrow moment vectors

This image depicts forces (single vectors) and moments in 3D space.

When the double-arrow points in the +x direction, we say it's a positive moment about x.

When the double-arrow points in the -y direction, we say it's a negative moment about y.

🟦 14.6  Moment of a force in 3D space: equation and example

In this flipbook you will learn how to calculate a moment of a force in 3D space.

The equation for moment of force in vector notation is:

This equation can be pronounced "MooooooRoooaaaaaxxxxxxFFFFFFF."

It's also common to simply say "M equals r cross F."

🟦  14.7  Moment of a force: more advanced 3D problem

In this example problem, a 520 N force has been applied to the end of bar OA.

Point O can be considered to be fixed, so that the bar is in static equilibrium.

Our goal is to compute the moment at O caused by the force F.

First, explore the interactive. Then, work it independently. Then, check your work with the written solution.

Attribution: I believe that this problem originated from the Hibbeler Statics textbook, although I'm not 100% certain.

Alternate solution with vector projection:

➜ Practice Problems

Summer 2024:

Please work problems 3.4, 3.5, 3.7, and 3.8 at