# Tag Info

2

The graph below shows a fixed-end frame subjected to a concentrated load in the mid-span of the horizontal member and the resulting deflected shape. Let's draw tangent lines on the deflected shapes at joints B, we note that both the curved segments of BF & BE are deflected away from the respective tangent lines with the angle of rotation $\theta = 0$, ...

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Increasing the size of the gusset plate is a bit deceiving first thought. But a disproportionally large gusset can cause stress concentration and even section failure right at the point of termination of the gusset. There are strong joint connections that are designed to be tough and ductile at the same time, this is one example. The stiffeners make the ...

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This is what I might do, with or without the gusset plate. If you use a gusset plate, chop off the corner to allow for continuous welding.

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The reactions at support "C" are as shown below. From here, you can calculate the normal stresses (due to bending) and shear stress (due to shear force and torsion).

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Let's start from the horizontal beam on top. if the joint at the ends of this beam where a pin joint you would get a maximum moment at the center $$M_{c}= \frac{PL}{4}$$ But because they are fixed after the beam deflects under the load the corner joint rotates a little and eventually stops at equilibrium. Both beam and the column bend a bit with the joint ...

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If you are only interested in approximation, the inner tube can be considered as a clamped cantilever after getting in contact with the walls of the outer tube, thus the applied load is resisted by the forces at the points in contact. If you want to go one step further, the clamped inner tube is resembling a beam simply supported on two hinges with an ...

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The best way is to add another HSS c channel back to back because this way you are quadrupling the I in the weak axis. Just make sure the bolts are tight. If the bolts are not tight fusing the two channels as one, you get at least something more than 2*I.

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