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A new engineer, I was getting used to the “excitement” of cubicle life when something (finally) happened. We’d stiffened part of our flow meters a month prior, and one of our customers started complaining about breakage. The graybeards upstairs were defending the change, while my engineering partner was struggling to explain why the minor tweak was causing the breakage.
And that’s where I came in. Fresh out of school, I had just spent four years honing my engineering analysis skills. A few minutes at the whiteboard later, we had the foundation of our argument: shear-moment diagrams simplifying the tubes as cantilever beams, clearly demonstrating why a little more strength was causing breakage at the base.
We made the change and were back to shipping product within the week. I felt like a hero.
This dry little anecdote isn’t going to hook any adrenaline junkies on a career in engineering, but it has been a reminder to me of the basic foundations of engineering. After a decade in the field, you sometimes forget what the fresh-faced graduates just learned: We don’t need a design to fail to know that it will. From free body diagrams (freshman physics, anyone?) to more complex static force analysis, even to FMEA on a supercomputer, engineers have been working for years to give us tools to understand the world around us and, more importantly, to predict real-world outcomes, so we can improve our designs before going to prototype.
While specific types of analysis are needed for unique situations (waterproof housing design, snap-fit design for plastics, or tolerance stackup, to mention a few), most engineering analysis examines how forces interact between parts of a design. Though studied as freshman, no tool rivals the free body diagram (FBD) for understanding forces on a design element in mechanical or structural designs. Or even a toboggan race down the stairs.
What? A toboggan race down the stairs? I’m glad you asked. Let’s hypothetically assume that young engineers in a large corporation got bored of cubicles and decided to create toboggans from packing cardboard to see who could get the best time down the staircase in the center of the building.
(Of course, please keep in mind this didn’t happen. And I will swear that to Eileen in HR all day long—there’s no video. On a completely unrelated note, I’d like to give a shout-out to my friend Frank in security.) If we wanted to know how fast these young engineers would be moving, how could we figure out the net acceleration?
Here’s where the FBD comes into its own. With an FBD, we look at all the forces acting on a body (element) and the direction of those forces to create a net force and net moment, and then predict what will happen.
Because the cardboard was stiff enough to contact multiple stairs, we can simplify this into a classic ramp and block FBD, where we look at the engineer and the cardboard as a block and the stairs as a ramp. What are the forces, and what direction are they pulling?
Gravity force (g times mass) pulls vertically down on the engineer. For our use, we need to break this down into the normal force (perpendicular to the stairs, m*g*cos θ) and the parallel force (heading down the stairs, m*g*sin θ).
The stairs support (push) on the cardboard and engineer in a perpendicular direction to the angle of the stairs. Because the engineer neither falls through the stairs nor flies off them, we know this normal force is equal in magnitude to the perpendicular (normal) force of gravity (m*g*cos θ).
Friction (not enough of it, as it turns out, to keep the engineer from running into the plate glass window across from the stairs) pushes up the stairs parallel to the angle of the stairs, (friction = fk*m*g*cos θ)
Though it’s difficult to find a real coefficient of friction between gravity and carpet, when we look up common friction coefficients, we substitute the coefficient for waxed wood on snow, fk=0.15. We can also assume a stair rise-run angle of 30° (θ).
When we sum the force vectors, we get a single sum of a force pointing directly down the stairs equal to:
Fresult = m*g*sin θ – 0.15*m*g*cos θ = (190 lbsf)*sin 30° – 0.15*(190 lbs-f)*cos 30° = 70 lbsf
(about ⅓ the speed of total freefall)
and find that this engineer is going down the stairs really, really fast, which is both intuitively true and also true from experience (had it actually happened, which of course it didn’t).
Too simple, you say? This is a building block for more complex analyses, and we can look at the interaction between elements by combining free body diagrams to analyze forces in machines, which brings us to the next level: statics.
Free body diagrams are great, but when do we have the simplicity of a one-piece machine? Almost never—but we can often reduce machines to a series of elements in mechanical equilibrium, where all the forces cancel each other out, and we end up with an understanding of entire systems using static analysis.
For example, let’s assume that on another long day, the young engineers in this company decided to use the crane in the loading dock area to take a ride and check out the view over the valley. (Again, hypothetically. Geez, Eileen; relax.)
We want to know whether or not the hydraulic cylinder on the crane is capable of lifting two engineers and a crate of beer (which becomes our weight, W). This can be complicated, but we simplify it by looking at a similar question: How much force will the cylinder experience when there is no motion?
And that’s where statics allows us to figure this out. Because there is no motion, then we know that there is no rotation around the pin A, meaning the forces around A are zero, so we get the very important:
∑MA = 0
Looking at each force causing a moment (remembering that a moment is the force at a point, multiplied by the distance from that point, multiplied by the sine of the angle between the force and the lever arm), and we end up with:
Fcd * Lac * sin θ – W * (Lac + Lcb) * sin ɑ = 0
A little algebra yields:
Fcd * Lac * sin θ = W * (Lac + Lcb) * sin ɑ => Fcd = W * (Lac + Lcb) * sin ɑ / Lac * sin θ
We can plug in some numbers, using:
W = 350 lbsf, ɑ = 70°, θ = 10°, Lac = 5 ft, Lcb = 15 ft
Fcd = (350 lbsf) * (5 ft+15 ft) * (sin 70°) / (5 ft)*(sin 10°) = (350 lbsf)*(20 ft)*(0.94)/(5 ft)*(0.17) = 7576 lbsf of force on the cylinder.
Knowing that the hydraulic cylinder was rated for 5 tons, we assume the young engineers had a fantastic evening and made good memories watching the sunset.
Of course, engineers don’t just care about any single part of a machine: We’re responsible for the cylinder, the cable, the attachment and the beam that ties all this together. And because we can often simplify mechanical elements into beams with various attachments, beams are especially worth looking at.
For the example above, we know that (when nothing is moving) the moments about point A are zero, but we also know that the vertical and horizontal forces in the whole machine each must be zero, so we can calculate the reactions at the attachment point A:
Ay – Fcd * sin(90- θ- ɑ ) – W = 0 =>
=>Ay = Fcd * sin(90- θ- ɑ ) + W = 7576 lbs * 0.17 + 350 lbs = 1666 lbs
Ax – Fcd * cos(90- θ- ɑ ) = 0 => Ax = Fcd * cos(90- θ- ɑ ) = 7576 lbs * 0.985 = 7461 lbs
Looking more closely, something interesting is happening here. Point A is known as a pinned end, which means in mechanical equilibrium, it has a reaction force in both the x and y direction, but has no moment reaction (in other words, moment is zero). Knowing about this unique property helps us analyze different beams.
In considering beams, several types of support need consideration and are used in creating mental models for beam elements. Fixed supports (think cantilever beams, like diving boards) have both x and y reaction forces, and a moment reaction, too. A pin support has x (horizontal) and y (vertical) reaction forces, but no moment (it’s like a hinge), while a roller support is like a wheel, where the load is supported vertically, but is free to move horizontally—t has a y reaction force, but neither an x reaction force nor a moment.
A beam can have nearly any combination of these supports, but more importantly, we can often simplify machines into several beams with different types of supports.
For the crane, we simplify the arm as a horizontal beam with a pinned support at one end (point A), a roller support at point C, and a load at point B. So, in the crane example, we know what’s happening at the ends of the beam. But what’s happening in the beam itself?
For failure risk analysis, we move beyond simple statics and look at forces inside the beam itself. Though FEA and other high-tech methods are great for final analysis, an approximation from moment and shear diagrams using beam analysis will suffice early in the design—and train your engineering intuition.
Looking at our simplified crane beam again, we know there is zero moment at point A, and that we’ll have a reaction force at point C, as well as the load at point B. We can also see that the moment at B will be zero. For shear and moment diagrams, we typically ignore any axial force along the beam, and so we have perpendicular forces at B, C, and A of:
B = – W * sin ɑ = -(350 lbsf) * sin 70° = – 329 lbsf
[negative – the force is pointing down]
C = – B * (Lac + Lcb) / Lcb = – (- W * sin ɑ * 20 ft / 5 ft) = 1316 lbsf
[ because ∑MA = 0 ]
A = – B – C = – 987 lbsf
[no movement, so the sum of the y forces is zero]
If we want to know how much bending force the beam will experience or how much force is trying to fracture it, we need to set up moment and shear diagrams. At first, we have a downward force of 987 lbsf, which continues from A to C, and then we have a reaction force of 1316 lbsf upward, with a resulting upward force of 329 lbsf to the end, where the force comes down to zero. The resulting shear diagram would look like this:
Without point moment loads, the moment along the beam will be continuous: the shear force multiplied by the distance, starting at zero (because ∑MA = 0). The moment will reach its greatest magnitude at C, then gradually reach zero again at B:
Intuition check: Where is the beam most likely to bend? At the hydraulic attachment. Intuition is trained and reinforced by sketching these diagrams. (By the way, not to worry: The crane beam held fine with no bending and none in HR the wiser.)
Infinite variations of loads and beam attachments exist, but a few basic beams will train your intuition, giving insight into your mechanical designs. Moving beyond shear and moment, how do we predict when these forces will result in damage? I recommend staying tuned for an upcoming article in this series on stiffness and strength (you can sign up below) and checking out my article How to Design for Stiffness Using Material Properties.
You can also find some great free courses online. For analysis methods like these, I’d recommend looking into MIT’s Structural Mechanics (beam analysis, deflection, and stresses) or the more in-depth Analysis and Design of Feedback Control Systems (time and frequency analysis, types of control loops, and mechanical stability analysis). There are literally more tools for analysis available than any one person could ever learn, much less use.
While I’ll always remember Doc O’s lecture at the end of sophomore year on how most of these tools were for training intuition more than for daily use, after ten years in the field, I can’t help but to also be grateful he spent the time drilling these engineering analysis equations into our heads. Maybe I don’t use them every day, but it’s a rare design project that doesn’t prompt me to at least quickly sketch a free body diagram or use force analysis to see what’s happening in the heart of the design.
Engineering analysis has saved me thousands in broken prototypes—and the embarrassment of explaining those broken pieces to clients. Want more tools like these? Sign up below for great tips sent right to your inbox. And if you’re finished with your analysis, you can check out our prototype machining and printing options to see if you were right!