# Fixed-Free End Column Design

I am currently working on a problem where I have to design a column. The structure I am currently designing currently takes the following form,

When designing the column, I know that I have to calculate what the maximum axial and bending stresses acting on the column are due to the horizontal beam. However, one thing I am unsure of is how do I calculate the bucking load/stress for the column? I'm not sure if I can treat the column as axially loaded due to the bending moment that is applied to it by the horizontal beam. Should I treat the load due to the beam as an eccentric load to account for the moment it applies on the column?

Also, when designing the column, other than than calculating the axial stress, bending stress and the buckling stress, are there any other parameters that I should calculate?

Thank you!

If your sketch is up to scale you need not worry about buckling under compression.

This column will fail under the bending moment long before it reaches the critical buckling load.

$$P_{cr} = \pi^2\,E\,I\,/\,L^2$$

And this is the Secant formula for an eccentrically loaded column. Where e is the length of your beam, c is half the depth of the column, I is the column moment of inertia and A is its area.

$$\sigma_{max} = P \, \left[ {1 \over A} + {e \, c \over I} \sec \left( {\pi \over 2} \sqrt{P \over P_{cr}} \, \right) \right]$$

We use the $$\sigma \text{ alloowable}, \$$ which is after applying load and safety factors.

To answer this question I need to assume that

1. the material is linear
2. the column is fixed at the bottom.
3. the beam to the column joint is fixed.
4. the load is applied at the tip of the beam

It is noted that buckling of frames is complicated by nature, the code provisions do not provide a simple formula for buckling of frames. For more advanced information, please look at the following paper which exactly solved the problem.

https://doi.org/10.1016/j.tws.2018.10.006

However, for simplicity, I assume that we are dealing with a beam-column case:

It is noted that when we are dealing with beam-columns, we need to consider the second-order deformation of the members. the differential equation of this beam-column is: $$EI y^{\prime\prime\prime\prime} + P_0y^{\prime\prime} =0$$ Where $$E$$ is Young's module, $$P_0$$ axial force, and $$I$$, the moment of inertia in minor direction. This equation can be simplified

$$y^{\prime\prime\prime\prime} + k^{2}y^{\prime\prime} =0$$

where $$k = \sqrt{\frac{P_0}{IE}}$$. The solution of this Differential equation is: $$y = A + B + C cos(kz) + D sin(kz)$$ For finding the costant, you need to apply the boundary conditons. For fixded- free boundary conditions we have $$y(0) =0$$ , $$y^{\prime}(L) = 0$$, $$y^{\prime\prime}(0) = \frac{M_0}{EI}$$ and $$y^{\prime\prime}(L) = \frac{M_0}{EI}$$. Where $$L$$ is the length of the memeber.

Applying the boundary conditions gives you the constants. If you substitute the constants, the second-order deformation of the beam-column can be calculated. the moment distribution can be calculated by the formula: $$M=-EI y^{\prime\prime}$$

Substitute $$y^{\prime\prime}$$ into the above equation to find the moment distribution: $$M = M_0 \frac{cos(kz)sin(kL)-sin(kz)cos(kL)+sin(kz)}{sin(kL)}$$ Before considering the second-order effects, the moment was constant along the laength of the column, $$M_0$$,. But now it is not constant. By taking the derivative $$M$$ with respect ot $$z$$, and setting it to zero, you can find the maximum point. The maximum moment can be found as $$M_{max} = M_0 \frac{\sqrt{2}}{\sqrt{cos(kL)+1}}$$ or $$M_{max} = M_0 \frac{\sqrt{2}}{\sqrt{cos(\pi \sqrt{\frac{P_0}{P_E}})+1}}$$ Where $$P_E = \frac{\pi^2 E I}{L^2}$$. This equation shows that maximum moment depends on both $$P_0$$ and $$M_0$$. You may define a moment amplification factor: $$M_{max} = \phi M_0$$ where $$\phi = \frac{\sqrt{2}}{\sqrt{cos(\pi \sqrt{\frac{P_0}{P_E}})+1}}$$. This amplification can be only used for your case. In different codes, you might find approximate formulas for $$\phi$$, which is cool. For design, the maximum normal stress should be smaller then yielding stress, i.e., $$\frac{P}{A}+\frac{M_{max}}{S}=f_y$$ $$A$$ is the area, $$S$$ the elastic section module, and $$f_y$$ yielding stress. In your case, the quantity of $$M_0$$ depends on beam loading.

Manipulation on the following equation will give the interaction relationship, but I digress. in