## 🟦 15.1 3D connections

In a 2D, planar problem, there are three main types of connections: the pin, the pin-roller, and the fixed connection. If you need a review or refresher on this, refer back to Lesson 04.

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

To review:

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. Be sure you 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 in the x-direction means there's an x-direction force reaction

a constraint in the y-direction means there's an y-direction force reaction

a constraint in the z-direction 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 is a wheel (or roller) that only develops an Fy reaction. We would not put a reaction in the z-direction, because if you want to transfer a z-direction force, friction is a dumb way to do it.

The second can be called a 3D pin connection or a ball-and-socket joint. Translation in x, y, and z is constrained, but the member is not constrained from rotation. Therefore three force reactions can develop: Fx, Fy, and Fz.

The third is a 3D fixed connection. They are powerful: you can support the weight of any body with one 3D fixed connection! They develop all 6 potential reaction forces and reaction moments: Fx, Fy, Fz, Mx, My, Mz.

### ❏ 3D connection assemblies and reactions

There are infinite combinations of 3D connection assemblies. A few are shown below. The design engineer is the one that chooses how to properly account for each connection assembly in the loading diagram.

### ❏ a 3D connection

### ❏ another one

### ❏ yet another one

### ❏ four is probably enough

## 🟦 15.2 Axial (or "two-force") members in 3D

In 2D, we learned that the geometry of an axial member can help us solve for unknown forces or force components. This is because the angles (alpha and beta) in the geometry diagram are congruent to the angles in the component force diagram.

The same logic holds true for 3D.

An axial member in 3D space 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#

## 🟦 15.3 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, you may find it helpful to draw the point as a cube instead of a circle or sphere. This helps you visualize the x-, y-, and z-component forces in their proper orientation, as illustrated below.

### ❏ FBD (3 unknown axial forces)

### ❏ Nodal FBD using a cube for the point

## 🟦 15.4 Computing the center of weight of mass in 3D

This sculpture is comprised of three identical cuboids.

Each cuboid measures 1" x 1" x 10" and weighs three pounds.

We want to determine the sculpture's center of weight. This is because the sculptor wants to suspend the sculpture from a single cable, and the center of mass isn't computed correctly, the sculpture will tilt.

Thinking through the E.o.E., we quickly deduce that the upward force in the cable must equal 3# x 3 cuboids = 9 pounds.

Next, we project the three planar views of the sculpture, with the 9# force located at (xbar, ybar, zbar) from the origin.

You'll complete this problem on your own; the projections are given to you to help you think through it.

### ❏ Center of a 3D sculpture

### ❏ Three 2D FBD projections of the sculpture

## 🟦 15.5 Moment summations about an axis in 3D space

Let's consider a force, F, acting on a yellow-colored geode.

For some reason, you want to compute the tendency to rotate (the moment) about line CE.

The first step is to compute the moment at C using the cross-product of the force and position vector.

For the second step, we need to do a vector projection operation.

### ❏ Moment about an axis

We have already learned to use the dot product to project a vector to an axis. This produces a scalar. A positive scalar means it points double arrow from O to A. A negative scalar points from A to O.

To solve for that moment in vector notation, simply multiply the dot product back by the unit vector.

Remember:

If you just do the dot product, the result is a scalar moment.

If you multiply the dot product back by the unit vector, then you can write the projected moment in vector form.

### ❏ Projecting a moment to an axis

The cool thing about that equation is that line OA doesn't have to be parallel to x, y, or z. It can be any random line in 3D space. That fact will be key to our problem-solving process in 3D space. We sum moments strategically about different axes to solve for unknowns.

Don't waste time projecting to the x-, y-, or z- axes, though. Those moments are the three components of the moment vector itself!

Are you a math geek? You can also project a moment to an axis in one big equation, called the scalar triple product! Dig in if you're interested.

## 🟦 15.6 Example problem

In this example problem, we have a heavy plate (800 pounds) that supports two loads (PD and MC). It's supported by a cable at E, a 3D pin connection at O, and a linear cylinder that goes from A to B.

First, we tally our unknowns. The reaction at E is vertical only, so the first unknown is Ez.

The pin support has the ability to generate three reactions (x-, y-, and -z). Since there are no x- or y- loads in the system, those reactions are zero. The vertical reaction at O is our 2nd unknown, Oz.

The cylinder between A and B can only deliver a z-direction reaction. In sum, there are 3 reactions we need to solve.

### ❏ Example problem

Since all three unknown forces point in the z-direction, summing forces with respect to z won't solve anything.

We strategically decide to sum moments about the axis defined by points O and E. (It happens to be the y-axis, but that's not important.) We pick that axis because two of the three unknowns are coincident with that axis. Solving the moment equation, we determine that cylinder AB exerts an upward reaction force of 650#.

You'll get to solve the other two unknowns in one of the practice problems.

### ❏ Partial solution (xz planar projection)

## ➜ Practice Problems

Problem 0:

Answer these two multiple choice questions:

When you dot two vectors together, the result is a ...

(a) scalar

(b) vector

When you cross two vectors together, the result is a ...

(a) scalar

(b) vector

Problem 1:

Work problem 2.7 at: http://mechanicsmap.psu.edu/homework_problems/Chapter2_Problems.pdf. It's a 3D concurrent force problem.

Problem 2:

Work problem 2.8 at: http://mechanicsmap.psu.edu/homework_problems/Chapter2_Problems.pdf. It's also a concurrent force problem.

Problem 3:

Work problem 3.14 at http://mechanicsmap.psu.edu/homework_problems/Chapter3_Problems.pdf.

Problem 4:

Work problem 3.15 at http://mechanicsmap.psu.edu/homework_problems/Chapter3_Problems.pdf.

Problem 5:

A sculpture has been created by gluing three cuboids together. Each measures 1" x 1" x 10" and weighs 3 pounds.

Find the center of weight of the sculpture. Report your answer as (xbar, ybar, zbar) relative to the given origin.

Answer:

(8, 2, 6) in.

Problem 6:

This sculpture looks the same as before, but it's a little bit different.

It still was created by gluing three cuboids together. Each measures 1" x 1" x 10." But each one is a different material.

Component 1 is the lightest. It has a density of 0.1 pounds per cubic inch. Component 2 has a density of 0.2 pounds per cubic inch. And component 3 is the heaviest, with a density of 0.3 pounds per cubic inch.

Find the center of weight of the sculpture. Report your answer as (xbar, ybar, zbar) relative to the given origin.

Answer:

(13.1, 4.13, 11.8) in.

Problem 7:

A heavy plate supports two loads and is supported as shown. Include the self-weight of the plate in your analysis.

Solve for all reactions.

Note: Please assume that the force of the cylinder on the plate acts halfway between A and B.

Problem 8:

A pin-pin flagpole (member 3) is held in place by three cables (commonly called "guy wires" in this type of application).

Cable 1 is tensioned to 100# as shown.

Solve for the force in Cable 2.

Note: there are two members labeled Cable 2. They are identical and symmetric.

Here is a link to the 3D CAD model (in case the image isn't clear).

Answer: 87.2# (T)

Problem 9:

A box (W = 30 kips) is hung by three cables from three pin supports. The three supports are all at the same elevation and form an equilateral triangle (12' by 12' by 12'). Node D lies 10 feet below the ceiling.

How much tension force is carried in each of the three cables? Here is an interactive model that may help you think through the geometry.

Answer: 12.2 kip (T)

Problem 10:

A 3D truss has been subjected to 2 applied forces at D. The applied force lies in the xz plane.

Node A is a 3D pin support. Nodes B and C are 3D rollers (and have the ability to develop either an upward or a downward reaction).

Use the E.o.E. to solve all unknown reactions.

Az = +40 kN

Ay = 0

Ax = -32 kN

Bz = Cz = -24 kN

Future problems (Fall 2024 students, you can ignore everything below this line).

3D fluids problem - opening swinging gates and such. animations with environmentally-friendly designs for fish.

note to self: put this problem on the set of end of term practice problems