🟦 13.1 The wonderful world of 3D!
While all Statics problems represent the reality of the 3D world, many can be analyzed in 2D. Others require a 3D approach.
When the geometry requires a 3D approach, we have three analysis approaches to consider.
Option 1:
If the problem is relatively simple, we may decide to conduct three planar two-dimensional analyses (the xy plane, xz plane, and yz plane). For simple problems, this approach is recommended.
Option 2:
For more complex problems, we may decide to apply the mathematical rigor of vector notation. For that reason, this lesson is principally a review of useful 3D vector operations and vector concepts.
Option 3:
When a problem is extremely complex (e.g. the playground in the photograph), you would use software (finite element analysis) for the analysis. This approach is based on an approximate method under the umbrella of so-called numerical methods.
It would be too time-consuming to do these types of calculations by hand. Engineers must be able to solve challenges quickly and efficiently. We don't always have to be accurate; we do always have to be safe.
❏ A complex playground design
A note about linear algebra / matrix methods:
The problems we solve in this class do not require linear algebra or matrix solution methods. Instead, we will use simple algebra (substitution and elimination methods to solve systems of equations). If you'd like to learn how to solve a system of linear equations using a matrix approach, check out this link.
🟦 13.2 Roadmap of the Statics Toolbox for 3D problems
This image provides an overview for our transition from 2D problem-solving to 3D problem-solving.
The concepts are the same. The tools are a bit different.
🟦 13.3 Review: position vectors
For position vectors, we typically use the symbol r.
In this image, rectangles C, D, and E all measure 3 feet by 5 feet.
The position vector from A to B is <-5,-3,-3> feet.
In other words, to get from A to B, you have to move in the negative x-direction, negative y-direction, and negative z-direction.
Practice the position vector from G to A on your own. The answer is <5,-2,-3> feet.
Also practice the position vector from B to G. This one is <0,5,6> feet.
🟦 13.4 Review: magnitude of a vector
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.
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 Review: unit vectors
Unit vectors have a length of one (or unity). They are used to communicate the direction of a vector. They do not have units.
When we want to use a unit vector in the x-direction, y-direction, or z-direction, we use i, j, and k.
These are pronounced as "i-hat," "j-hat," and "k-hat."
When typed, they are bold and italicized.
When handwritten, they wear pointed hats.
Sometimes we want a unit vector that defines some other direction in space. Use u for this type of unit vector. Also, use two subscripts to indicate the tail-to-head direction of the unit vector.
Important: note that the order of the subscripts matter.
Example problem 1:
Points A and B lie at an inclination of 28 degrees, as shown.
You need to write the unit vector that defines the inclination of the line of action that is parallel to line AB. You want to go from A to B.
If an angle is given, the components of the unit vector can be expressed in terms of sine and cosine.
Remember that you can use the unit circle and Pythagorean Theorem to check your work. (cos^2 + sin^2 = 1^2).
Example problem 2:
Points A and B lie 5m apart, as shown.
As before, you want to the unit vector that defines the line of action from A to B.
Since we know the dimensions, use ratios. You don't need to calculate the angle.
In this example, you'd take sqrt(0.8^2 + 0.6^2) = 1 to verify that you have written a unit vector.
Example problem 3:
In 3D space, point A lies at <2, 0, 3> inches and point B lies at <0, 6, 0> inches.
We compute the length AB using the 3D Pythagorean Theorem.
Then, the unit vector is written by dividing the position vector by the magnitude of the length.
Remember: since all unit vectors equal unity by definition, you can always check your unit vector by using the 3D version of the Pythagorean Theorem. Make sure it has a length of 1.
🟦 13.6 The bounding cuboid for 3D vectors
This flipbook provides a review of:
position vectors
vector components and resultants
unit vectors
In 2D, we visualize a bounding box for vector components in a plane.
In 3D, we can do the same kind of thing. We simply need to use a bounding cuboid instead of a bounding box.
We use the bounding cuboid to visualize the three vector components (x-direction, y-direction, and z-direction).
🟦 13.7 Summary of vector symbols and notation
In our 2D problems, we had the luxury of being a little lax in our notation. For instance, we wrote many force vectors, but we didn't go to the trouble of putting a little arrow on top of the signal, because we all knew that force was a vector.
In 3D problems, it's a good idea for us to be a little more formal with our symbols and notation. We will try to remember to put little arrows on top of our vectors. When we use a magnitude, we can use single or double vertical bars (either is OK):
| F | = the magnitude of the vector F
||F|| = the magnitude of the vector F
Subscripts:
Many vectors have subscripts, such as rAB (a position vector). This can be read as the position "from A to B." Note that rAB is equal to (-1)rBA. You reverse the signs of all three components when you reverse the direction.
Unit vectors:
Use i, j, and k for unit vectors in the x-, y-, and z-directions
Use u for unit vectors in other directions
The magnitude, by definition, is 1.
Other vectors:
use r for position (with 2 subscripts)
use F for force (with 2 subscripts)
use M for moment (with 2 subscripts)
Remember: when we handwrite vectors, we (make an effort to remember to) draw arrows on top. When we type vectors, we make them bold instead. And the unit vectors i, j, and k are both bold and italicized.
🟦 13.8 Review: how to compute the dot product of two vectors
Skim this page to review the basics of the dot product. We use the dot product to determine how much of a given vector points in the direction of another vector. This is why when you dot two perpendicular vectors, you receive an answer of zero.
Key take-aways:
For dot-products, the end result is a scalar.
For dot-products, the order of the vectors does NOT matter. (Aٜ·B = B·A). That is, the dot product of two vectors is commutative.
Practice a few problems until you have the ability to compute dot products quickly, either by hand, or by programming your calculator to do it for you.
🟦 13.9 Review: how to project vectors with the dot product
Peruse this flipbook to see how we can use the dot product to project a vector component onto a line (or an axis).
Vector projections give us a way to determine the vector components that are parallel to the ray defined by the unit vector.
In this example problem, a component of force vector F is projected to the line (axis) OA, through the dot product operation.
Here is a model for how to write out these calculations. (It's the same problem.)
🟦 13.10 Review: how to compute the cross-product of two vectors
Skim this overview of the cross-product.
Most people are taught to perform the cross-product with either Method 1 or Method 2 in the image.
Stick with whatever method you were first taught. There's no need to learn the other method.
Key takeaways:
For cross-products, the end result is a vector. It's perpendicular to the plane created by the source vectors.
For cross-products, the order of the vectors is important. AxB ≠ BxA.
If you reverse the order of the operation, then you are reversing the sign of the answer.
Practice until you have the ability to compute cross products quickly. Ideally, use your calculator for this operation. If your calculator doesn't have this functionality pre-built in, program it in yourself.
Stay tuned: we will discover applications for the cross-product in Lesson 14.
🟦 13.11 Condensing to two dimensions
A three-dimensional structure that is symmetric (in terms of both geometry and loading) can be condensed to a 2D problem.
For instance, you might try to tip a sibling's chair by exerting force at B and C on the chair with your two hands.
At first glance, a Statics student might try to use vector notation for this type of problem.
But more experienced students would quickly notice that System I is equivalent to System II.
Since the geometry and loads are symmetric, we can condense to a 2D model (System III). We project (or elevate) the xy plane of the chair and sum moments about E to determine whether the chair tips.
❏ Some 3D problems can be modeled in 2D
➜ Practice Problems
Problem 0.
(a) Watch at least the first half of this video.
(b) Figure out how to do a cross-product with your calculator. Either FIND the program that does a cross-product, or WRITE the program that does a cross-product. Optional: do the same for the dot product.
Problem 1.
Work through Example 2.7.7. Dot Products on https://engineeringstatics.org/dot_products_2D.html.
Problem 2.
Work through Example 2.8.8. 3D Cross Product at https://engineeringstatics.org/cross-product-math.html.
Problem 3.
Write the unit vector that describes the line of action from A to B.
Rectangles C, D, and E all measure 2m by 4m.
C and D lie in the xy plane.
E is parallel to the xz plane.
Problem 4.
Vector B has been created by rotating vector A 30 degrees about z.
Write out the components of Vector B. Express your answers as fractions and radicals.
Bx = ?
By = ?
Bz = ?
Problem 5.
This is a continuation of the previous problem.
Vector C has been created by rotating Vector B (which lies on b-b) 45 degrees about line d-d (which is perpendicular to line b-b).
What are the components of Vector C? Express your answer in fractions and radicals.
Cx = ?
Cy = ?
Cz = ?
Problem 6.
This Vector C is the same one from the previous problem. What are the vector components in the x'y'z coordinate system? (The x'y' axes are rotated 30 degrees about z compared to the xy axes.)
Cx' =
Cy' =
Cz =
Problem 7.
A force vector (F) has components of 1 kN, 3 kN, and 2 kN, as shown.
Project F into the xy plane.
Fxy = ?
Verify that Fxy + Fz = F
Then, project F into the zy plane.
Fzy = ?
Verify that Fzy + Fx = F
Finally, project F into the xz plane.
Fxz = ?
Verify that Fxz + Fy = F
Problem 8.
This is a 3D FBD of Node E (or particle E or point E).
Your friend has already solved the problem and given you the following magnitudes for the forces. The units are kips.
The force in A is 32/(sqrt 3)
The force in the two forces labeled B is 16/(sqrt 3)
The force in C is 16
Your job is to draw all three 2D projections of the node:
an xy view (in which you do not see any z-direction vectors)
a zy view (in which you do not see any x-direction vectors)
a xz view (in which you do not see any y-direction vectors)
After that, use the equations of equilibrium to determine whether or not your friend's answers are correct. If each 2D projection is in static equilibrium, then the forces given to you must be correct.
This will feel like doing three successive concurrent force problems. Refer back to Lesson 03 if you need a refresher on how to solve concurrent force problems.
Fall 2024 students: everything is done now! The TAs and I are now working on solutions and will get solutions / answers posted to the website as soon as we can. I will also bring solutions to class today (9/25/2024). -S