enter image description here

I am trying to calculate the force sustained on the three labelled members of this frame (Members 1, 2 & 3). The frame will experience a uniform pressure of 1000 kPa in the area enclosed by the members. It will be screwed into walls on all 4 sides. You can imagine it as being installed in a 4m by 4m shaft. And the goal is to brace a door in the shaft, which might experience wind pressures of a 1000 kPa. Once I have found the force I will calculate the deflection in order to select the correct steel members for the frame.

For those curious, I just made up the numbers and the scenario but I am working on a similar situation and it’s been many years since I studied statics. Any help or guidance would be appreciated.


2 Answers 2


The force on the perimeter will depend on the boundary conditions at the edges of each plate. Determining the precise distribution of load at each point of the supported edges of the panels is going to be a headache if done by hand. I would determine the maximum force per unit length of the edges and use that as a uniform linear force on the supporting members.

The way I would approach this is to use formulations from Roark's Formulas for Stress and Strain, 7th Edition, Table 11.4:

Edges Simply-Supported

simply supported

Edges Fixed


You will want to solve for $R_{max}$. For the Edges Fixed case, I believe $R_{max}$ is the same as for the simply-supported case. Also, in the Fixed case, you will have to include the edge moments into your design.

Otherwise, if you're just looking for an approximate method to determine the force on the supporting members, first determine the total force on a panel:

$$F = (1000 kPa)A_{panel}$$

Then, take that force and divide it by the perimeter of the panel ($P_{panel}$) to determine your uniform load per unit length:

$$w = F/P_{panel}$$

This will give you a load analogous to the $R_{max}$ determined if you follow Roark's above, though this will be an average instead of the peak lineal load. It will under-estimate the loads on the long edges and overestimate the forces on the short edges.


Adding to @grfrazee's answer, there is an intermediate method, less rigorous than Rourke but more precise than simply dividing by the perimeter.

That is the trapezoidal method, where one divides the load being applied to each supporting member. The only relevant information is the boundary conditions of the plate: whether it is free, pinned or fixed. Usually, slabs are said to fix each other, so if you have a beam supporting two slabs, one to either side, then both slabs are considered fixed on that face (the only exception is if one of the slabs is a cantilever, in which case the cantilevered slab is considered fixed but the other is pinned, since a cantilevered slab won't resist rotations due to loading on the other slab). On the other hand, if a beam is supporting only one slab, then that slab-face is considered pinned. If the edge of a slab is not supported by a beam, then it is considered free, of course. This method doesn't work for slabs supported directly by columns. It also assumes the slab is under uniform loads which cover the entire slab and is not strictly valid for other loading conditions (concentrated loads or variable loading within a slab).

Now, if two adjacent and perpendicular faces of a slab are equally supported (both are pinned or both are fixed), then one draws a 45° line starting from their shared vertex. If, however, one face is pinned and the other is fixed, then the line must be 60° (from the fixed face). These lines describe the load division to each supporting member. (After a quick googling, I've discovered some people simply consider 45° in all cases. Do as you wish, but I think giving more load to the stiffer member makes sense. That being said, I can't prove it, nor can I demonstrate why 60° is a good angle, I simply learned this in college).

Extending these lines until they fully divide the slab area, you now know the fraction of load that will go to each face.

This can be easily understood with an example such as this one:

enter image description here

Here we have three slabs A, B and C. Slab A is supported by beams B2.1, B3.1 and B4.1 and has a free end (the dotted line). Slab B is supported by beams B2.2, B3.2, B4.1 and B5.1. Slab C is supported by beams B1, B2.2, B4.2 and B5.2. In this example, lets just look at the load distribution of Slabs A and B.

Slab A is free on one face, pinned on faces B2.1 and B3.1 and fixed on B4.1. This is because the stiffness of slab B on the other side of B4.1 impedes (or at least limits) the rotation around face B4.1, while the torsional rigidity of beams B2.1 and B3.1 is considered insufficient to influence the slab's rotation. Given the rules described above, lines are drawn from the vertices between the faces. In this case, since B4.1 is fixed and B2.1 and B3.1 are pinned, the lines must be at 60° from B4.1. The remaining load is distributed between the pinned faces. The load distribution is therefore:

enter image description here

For the same reasons as described above, Slab B is fixed on faces B2.2 and B4.1 (due to the slabs on the other side of the beams) and pinned on faces B3.2 and B5.1. Since both B2.2 and B4.1 are fixed, the line between them is at 45°. Since both B3.2 and B5.1 are pinned, the line between them is also 45°. Since B4.1 and B2.2 are fixed and B3.2 and B5.1 are pinned, the lines between them are at 60° from the fixed faces. The load distribution is therefore:

enter image description here

Now one can calculate how much load goes to each beam by multiplying the "tributary area" of the beam by the applied load. In the case of beam B3, for example, the applied load is:

$$ q_{B_3} = \begin{cases} \dfrac{A_3 \cdot q_A}{L_{B_{3.1}}} \text{ along B3.1} \\ \dfrac{B_3 \cdot q_B}{L_{B_{3.2}}} \text{ along B3.2} \end{cases}$$ where $q_{B_3}$ is the uniform load on beam B3, $A_3$ and $B_3$ are the tributary areas found in the figures and $q_A$ and $q_B$ are the loads applied in each slab (in force-per-area units). The loading of beam B3 is therefore

enter image description here

In the case of beam B4, the loads would be $$ q_{B_4} = \begin{cases} \dfrac{A_4 \cdot q_A + B_4 \cdot q_B}{L_{B_{4.1}}} \text{ along B4.1} \\ \dfrac{C_4 \cdot q_C}{L_{B_{4.1}}} \text{ along B4.2} \end{cases}$$ (C4 is the tributary area of beam B4 in slab C, which hasn't been drawn). Along B4.1 one must obviously add the loads applied by both slabs supported by the beam.

These equations are somewhat simplified, since they consider the loading along the beam to be uniform, which isn't really the case. The loading should actually be trapezoidal according to the tributary area. To calculate this more precise form, the loading on the beam is found by $$q(x) = p \cdot h(x)$$ where $q(x)$ is the value of the distributed load at point $x$, $p$ is the area-load on the slab and $h(x)$ is the height of the tributary area perpendicular to the beam's axis (displayed as $h_{3.1}(x)$ in the second figure above). The loading therefore becomes:

enter image description here

  • $\begingroup$ Thanks for your help Wasabi. I have a question about this formula qB3={A3⋅qA along B3.1, B3⋅qB along B3.2. In this case is q, the loading, pressure or force? and is the parenthesis above indicating that the all the terms are going to be summed up? for example qB4=(A3.qA + B3.qB) ? $\endgroup$
    – user3233
    Oct 6, 2015 at 15:13
  • $\begingroup$ for the last line qB3=(A3.qA + B3.qB)* $\endgroup$
    – user3233
    Oct 6, 2015 at 15:20
  • $\begingroup$ also I am unable to find a reference to this method any where else. Do you have a source for the above? $\endgroup$
    – user3233
    Oct 6, 2015 at 15:30
  • $\begingroup$ @user3233, I have edited my answer with more information. I believe it should now be clearer. I can't find an absolute source, but googling "trapezoidal slab load distribution" comes up with a few articles and pages and whatnot. Looking at image results is also interesting. That being said, doing this search I've noticed that some people simply adopt 45° in all cases, even when one face is pinned and the other fixed. I think giving more load to the stiffer face makes sense, but I lack the arguments to really push that point. $\endgroup$
    – Wasabi
    Oct 6, 2015 at 16:46

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