As a civil engineer, it’s always important to learn the key points needed for designing any member. The article on design strength describes easy steps to compute the compressive design strength, $P_d$ of the column as per IS 800:2007. In this article, we will see simple steps on how to design a column.

Note: You can apply this concept to the design of I-section, Channel section, and Double-angle sections columns but not to single-angle sections.

## What are the basic informations needed to start the design?

**To design a axially loaded compression member, the following things are required:**

- Length of the member
- Boundary conditions
- Load it has to support $($also called design load, $P)$

Now a compression member should be designed such that it’s design compressive strength, $P_d$ should be greater than design factored load, $P$ i.e., $P_d>P$.

## Design: an AB process → Assumption and Backing

### Assumption:

It is obvious that the designer has to select a cross-section such that $P_d>P$. Now, to compute $P_d$, the key ingredient is slenderness ratio $\lambda = KL/r$. From the given information, we know the value of $KL$ which thus sums up to finding appropriate radius of gyration, $r$.

**Thus, a designer has to adopt following two steps:**

**Step A:**Assumption → Assume $\lambda$ to find a trial cross-section for which $P_d>P$.**Step B:**Backing → Use design equations of IS 800:2007 to compute $f_{cd}$ and confirm $P_d>P$.

Let us look at some thumb rule to design a with an example for better understanding the design process.

## Design Example with step-by-step procedure for design of a compression member

**Design a 4.0 meters long column to carry a factored design load of 900 kN. The column is restrained in both rotation and translation. Use Fe410.**

**Step 1: Assume** $\lambda$

As a thumb rule for columns of height 3.5 m to 5 m, a value between 40 to 60 you can assume. If the factored load is more (>1000 kN) smaller value of slenderness (~40) you have to assume. Whereas for longer lengths of columns (> 5, m), a higher value of slenderness should be assumed (60 < $\lambda$ <90).

For the given problem, as the column height is 4.0 meters and factored load is 900 kN, let us assume $KL/r=50$.

**Step 2: Choose a buckling class for the member**

For the given problem, determine the buckling class for the member. Using $\lambda$, $f_y$, and Table 9, IS 800:2007 to determine compressive stress, $f_{cd}$.

**Tips: For I-Section, unless specified, assume member to fail minor axis buckling as $(r_y<r_x)$.**

For the given problem, the cross-section type is I-section which will undergo minor axis flexural buckling as $(r_y<r_x)$. The buckling class as per Table 10, IS 800:2007 is “* buckling class b*” provided $h/b_f>1.2$ and $t_f<40\; mm$.

We assumed $KL/r=50$ and $f_y=250\; MPa$ and for buckling class b, $f_{cd}$ from Table 10, IS 800:2007 is $194 \; MPa$.

## Remaining steps for the design a compression member

**Step 3: Determine required cross-sectional area** $A$

The required cross-sectional $A$ you can calculate as

$$ A=\frac{800\times 10^3}{194}=4639 \; mm^2 $$

**Step 4: Choose a suitable cross-section from IS Handbook SP 6:(1)**

The required area as per assumption is $4639 \; mm^2$. Choosing section ISHB 200 @ 37.3 kg/m as $A=4754 \; mm^2>4639\; mm^2$.

Step 1 to Step 4 concludes Step A: Assumption.

Now,

**Step 5: Computing** $f_{cd}$ for chosen section.

Follow blog to find out elaborate steps on computing $f_{cd}$.

In the current example,

For section ISHB 200 @ 37.3 kg/m;

$$ A=4754 \; mm^2\; I_{xx}=3608.4\times 10^{4} mm^4;\; I_{yy}=967.1\times 10^4 \; mm^4 $$

$$ r_{x}=87.1\;mm ; \; r_{y}=45.1 \; mm $$

You know that the column is having to restrain in rotation and translations. Hence, from Table 11, IS 800:2007,

$$ KL=0.65 \times 4000=2600\; mm $$

Now,

$$ \lambda_x=\frac{2600}{87.1}=29.85\quad \quad \lambda_y=\frac{2600}{45.1}=56.65 $$

Since, $\lambda_y<\lambda_x$, Minor axis buckling. From Table 10, IS 800:2007, buckling class is “* b*”.

Now, using Table 9(b) for $f_{cd}$ for $KL/r=56.65$ and $f_y=250\; MPa$; we have

$$ f_{cd}= 185.35\; MPa\;\implies P_d=A\times f_{cd}=4754\times 185.35 = 881 \; kN < 900 \; kN $$

The current section ISHB 200 @ 37.3 kg/m is not sufficient to resist the factored load of 900 kN. Thus, we have to adopt a different cross-section.

## Is design always an iterative process? → No

In the current design example we chose area of cross-section just above the required cross-section. It is advisable to increase the the required area of cross-section by 20% while choosing a cross-section in order to not make the design an iterative process.

So now, lets check for ISHB 225 @ 43.1 kg. Now we have

$$ A=5494\; mm^2\\ I_{xx}=5279.5\times 10^{4} mm^4;\;I_{yy}=1353.8\times 10^4\; mm^4 $$

$$ r_{x}=98\;mm; \; r_{y}=49.6\;mm $$

Now,

$$ KL=2600\;mm \quad \therefore \lambda_x=\frac{2600}{98}=26.53\quad \quad \lambda_y=\frac{2600}{49.6}=52.42 $$

Since, $\lambda_y<\lambda_x$, Minor axis buckling. From Table 10, IS 800:2007, buckling class is “* b*”.

Now, using Table 9(b) for $f_{cd}$ for $KL/r=52.42$ and $f_y=250\; MPa$; we have

$$ f_{cd}= 190.85\; MPa\;\implies P_d=A\times f_{cd}=5494\times 190.85= 1049\; kN > 900 \; kN $$

Thus, adopt ISHB 225 @ 43.1 kg/m as desired cross-section to resist the factored load of 900 kN

## Conclusions

While the design of compression members might be tricky, it requires just experience and practice to do the the simple process **AB i.e., A**ssumption and **B**acking. In this article we covered the design example and have seen simple steps to design a compression member. In the upcoming articles, we will encounter problems where single section column will not alone be able to resist the heavier factored load.

**From this article, you have learned the following key points:**

**Design:**You have learned the simple AB process to design a column.**Assumptions:**We have also learned how to take initial assumptions while we are designing the column.

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