# How do I calculate the forces on a desk and its legs?

I have a design for a desk, and I'd like to not just guess at how strong it'll be, but I can't find an explanation on how to figure out all the forces involved that doesn't assume I already know a lot about engineering already.

So, if I were to apply 300lbf (1334 newtons) straight down on the front corner of the desk, how could I calculate the stress from the desktop to the upright beams, to the diagonal braces, to the ground?

Assume A500 steel, 1x2x16ga.

### Diagrams

• You seem confused between "force" and "strength". Calculating the forces that will be in a desk doesn't tell you how strong it'll be, but it'll tell you how strong it needs to be. Please clarify your question as to which you are actually looking for. Jan 30 '15 at 13:35
• This is more complicated than it seems and it will probably be pretty hard (though maybe not impossible) to get a satisfactory answer to your question in this venue. The field of study that will go a long way in teaching you how to do this is called Statics. There are lots of free Statics/Mechanics of Solids courses available online now. Here's a good one. Jan 30 '15 at 18:25
• To reiterate and expand on what's been said, you can't find info that doesn't assume engineering knowledge because you do need engineering knowledge. This isn't a hard statics problem, but it definitely is a statics problem, and it's more than just linear forces, there are bending moments involved as well. You can probably eliminate most of them by calculating the linear forces in the correct manner, but you learn to do that in a Statics course. When I have some more time, I can go more in depth with an answer, but know that it's not as simple as you might think. Jan 31 '15 at 5:35
• Sorry, just pound-force, equal to 1334 newtons. Feb 6 '15 at 18:49
• This question is still unclear. It mixes up "force", "strength" and "stress". They are different terms! "stress from the desktop to the upright beams" makes no grammatical sense. Additionally, this is supposed to be a site for experts to ask questions of experts; I'm afraid the actual analysis (if the question was clarified) is basic statics. Feb 16 '15 at 12:41

## 4 Answers

To begin with I am assuming each of your horizontal surfaces: desktop and the three selves are each made the same material. To use a ridiculous and exaggerated case, the left half if the desktop is not heavy marble and the right side is not light weight balsa wood. The desktop is composed of one uniform material and each self is made of its own uniform material: wood, glass, metal, particle board and laminex, plywood, whatever.

As shown, each of the shelves and the desktop are independently attached to the vertical supports that acts as the legs. Hence, the weight of each horizontal surface is directly transferred to the vertical supports. All horizontal surfaces are of uniform materials that have a uniform weight distribution. Consequently, each leg is carrying half of the combined weight of all the horizontal surfaces.

The section of each leg that experiences the full weight of what is above it is the short section between the two triangular braces: the one for the desktop and the other for the stand/foot of the desk.

The stress in each of these short sections of leg will be the weight carried by each leg divided the cross sectional area of the leg in the z-plane (breadth by width of the leg)

The sloping part of the desktop brace will carry some of the weight of the desktop. Whereas, the short vertical part of the desktop brace carries all the weight of the vertical support above the desktop, the three shelves and some of the weight of the desktop. What the proportion of the desktop weight carried will be will depend on the perpendicular (normal) distance from the vertical support.

Likewise at the base of the leg, the triangular brace will redistribute the load in the foot according to your triangular configuration.

This is just a general overview of how to think about things relating to your design. As @Rick Teachey states, you really need to do a course in statics, get numbers for weights and cross-sectional dimensions of supports and plug it all into some formulae.

Since you want to know what happens with a load applied to the corner of the desk, I'm going to simplify this question into two dimensions, assuming that the leg on that corner resists the load alone. Considering the fact that the rigidity of the steel members is orders of magnitude larger than that of the wooden desktop, this probably isn't too far from the truth.

I'm also going to assume that the desk is made of magical materials that don't have self-weight and that the desk is otherwise empty of other loads, just to keep things simple. Also, as others have mentioned, this is effectively impossible to do without some knowledge of statics. I can't give an entire lesson here, but I'm going to explain things as best I can.

The structure you effectively have is the following (removing the tail end of the desk after the foot, which is irrelevant, and the diagonal at the base of the foot, which just complicates matters and doesn't actually change the relevant internal stresses):

This particular case can actually be resolved by hand, so here goes: The load at the very edge of the table is $300\text{lb}$ and is $12\text{in} = 1\text{ft}$ from the diagonal. That means that the beam has to withstand a bending moment of $M=300\cdot1=300\text{ft-lb}$ and a shear force equal to the applied load of $Q=-300\text{lb}$ (negative because it's pointing down).

Now we are at the point where the diagonal starts to help the horizontal beam, so we need to figure out how much force goes to each of them. For this, we have to look a bit ahead and notice that the horizontal beam meets the column at another pinned joint (those "balls" in the figure). These joints allow parts to rotate relative to each other, which (and this is something you learn in statics) means that the bending moment at that point is zero. Since there are no other external loads applied along those $20\text{in}$ (between the horizontal bar's connection with the diagonal and with the column), the shear force must be constant along that stretch. And since the shear force is the derivative of the bending moment, the moment must vary linearly. And since the diagonal is pinned ("ball" connection) to the horizontal, it didn't steal any of the moment. That means that the horizontal beam goes from a bending moment of 300 at the beginning of the diagonal to zero at the column. The constant shear force along that stretch is therefore equal to the tangent of that linear variation, which is

$$Q = \dfrac{300\text{ft-lb}}{20\text{in} = \frac{5}{3}ft} = 180\text{lb}$$.

So, going back to the connection between the horizontal and the diagonal, we now know that the horizontal beam went from a shear force of $-300\text{lb}$ to $+180\text{lb}$. That means that the diagonal must have applied a vertical force equal to $+480\text{lb}$ onto the horizontal. However, since the diagonal is pinned on both ends and has no external loads applied on it, it can only contain axial loads. That means that those $480\text{lb}$ are actually just a component of the force actually applied by the diagonal. The horizontal component can be easily found by the tangent and is equal to $480\cdot\frac{20}{5} = 1920\text{lb}$. The total axial force on the diagonal can be found by Pythogoras: $\sqrt{480^2+1920^2} = 1979\text{lb}$, and is of compression. Meanwhile, the horizontal component of this force has to be restrained by the horizontal beam, which therefore suffers a tension of $1920\text{lb}$.

All that's left now is the column. Since the horizontal beam is suffering a tension of $1920\text{lb}$, that needs to be absorbed by the column, which transforms that tension into a shear of $1920\text{lb}$. That shear, however, is cancelled out by the connection with the diagonal, which applies the same force (but onto a different side, therefore with a different sign... statics). Between those points, however, the shear is alive and well. And where there's shear, there's bending moment. A constant shear of $1920\text{lb}$ over $5\text{in}$ generates a bending moment of $1920\cdot\frac{5}{12}=800\text{ft-lb}$. Between the base of the column and the connection of the diagonal, there no longer is any shear, so the moment is constant.

Also, the horizontal beam had a shear of $+180\text{lb}$ which gets transmitted to the column as an axial tension of equal value (that part of the column is being stretched, not squished!). However, after the connection with the diagonal, which also dumps its horizontal component of $-480\text{lb}$ (it was positive at the top because it pointed up. Here it points down, so it is negative). Therefore, between the base and the diagonal, the column suffers a compression of $300\text{lb}$, which makes sense since that part of the column would have to withstand the entire external load which was applied at the edge of the table. If its compression weren't equal to that applied load, something would be wrong.

At the end of the day, you end up with a structure undergoing the following (click to expand):

However, knowing the internal forces is not sufficient to know whether your desk will support it. That, however, is highly dependent on where you live and which codes apply (and I'm sure desks don't have to follow structural codes, but I'm sure there's some relevant code) and can't adequately be answered here.

That being said, for tension and shear there's usually little mystery to it. For tension, divide the tensile force by the cross-section area and compare that stress to the steel's strength (the weakest A500 is 45ksi), with some safety factor (allowable stress design often uses 60% of the steel's strength). For shear, divide the shear force by the "shear area", which in your case is equal to the area of the "vertical" sides of the cross-sections. This gives you the shear stress, which should be compared to the steel's strength (allowable stress design uses 40% of the tensile strength).

Bending and compression, however, are more complicated due to the risk of buckling and need to be done by the relevant codes. If one ignores buckling (one really shouldn't), then it's just a matter of getting the relevant stress and comparing it to the strength again. For compression, that's the same as for tension. For bending, divide the bending moment by the elastic modulus to get the maximum tension/compression stress (see below) and compare against the allowable stress as well:

$$\sigma = \dfrac{6Mh_1}{b_1h_1^3-b_2h_2^3}$$

And, for what it's worth, the diagonal at the base of the foot might be relevant for buckling analysis, though if I had to guess I'd say the upper diagonal aiding the horizontal beam would be the controlling member (for buckling).

What you are asking for is a statics analysis or really something that an engineer would learn in a "Mechanics of Materials" course. You need to know how much stress is placed on the desk members as a result of the 300 lb force and whether it can hold the load.

I have solved this problem for the cross beam support on the desk. However, the highest load will be seen on the support member when the load is directly above it and not when the load is at the end.

The analysis can be carried out for the remaining members but to do a thorough analysis you need to look at the connection points as those would be the likely choke points.

The document linked above was done on a platform I am developing called CADWOLF. You can alter the load to see the resulting forces.

The result of the load you described is a load of 74.49 lbf on the cross member that supports the desk and a reactionary force of 274.5 lbf at the point where the desk connects to the legs.

The document describes the process of summing the forces and moments to get these results. This same process can be used with the loads on the cross member and the reactionary force to calculate the load on the cross member connecting the vertical legs to the lower horizontal legs.

I would get a free 3-year student version of Autodesk Inventor (because I am familiar with it, SolidWorks, CATIA, work as well). Then model the desk and perform static analysis. The days of force diagrams on A0 sheets are long gone.

• How does this help the OP understand the principles involved with performing the analysis they're interested in? Having a tool doesn't convey knowledge of how to use the tool.
– user16
Feb 15 '15 at 13:30
• @GlenH7, of course OP will need to figure out how to use the tool. From then on, he will be able to do simulation-correction-simulation cycles. Furthermore, knowledge of statics will for sure aid greatly in the analysis of the results. Feb 16 '15 at 10:44