Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 x 10^3 kg

Q: Why do we divide the total mass by 4?

The problem states that the structure is supported by four identical columns and assumes a 'uniform local distribution'. This means the total weight is shared equally among the four pillars. So, one pillar only feels 1/4th of the total load.

Q: Why is the area formula $\pi(R^2 - r^2)$ and not just $\pi R^2$?

Because the column is hollow! If you look at it from the top, it looks like a donut (an annulus). The load is only supported by the solid steel part, not the empty air in the middle. So, we take the total outer area and subtract the inner hole's area.

Q: Do I really need to convert the radii from cm to meters?

Absolutely! Young's Modulus is given in Pascals (N/m²), and gravity is in m/s². If you leave the area in cm², your units will clash, and your final answer will be off by a massive factor of $10^4$.

Q: What is the physical difference between stress and strain?

Stress is the 'cause'—it's the internal force per unit area that the material feels when a load is applied. Strain is the 'effect'—it's the physical deformation or change in length relative to the original length. Hooke's law connects the two.

Q: Why is the final strain value so incredibly small ($10^{-7}$)?

Strain is a dimensionless ratio of change in length to original length. Mild steel is incredibly stiff, which is reflected by its huge Young's Modulus ($2.0 \times 10^{11}$ Pa). Even under a massive load, it barely compresses at all, hence the tiny strain value.

Question

Four identical hollow cylindrical columns of mild steel support a big structure of mass $50 	imes 10^3$ kg. The inner and outer radii of each column are $50$ cm and $100$ cm, respectively. Assuming, uniform local distribution, calculate the compression strain of each column. [Use, $Y = 2.0 	imes 10^{11}$ Pa, $g = 9.8$ m/s$^2$].

Options:

(A) $3.60 	imes 10^{-8}$

(B) $2.60 	imes 10^{-7}$

(C) $1.87 	imes 10^{-3}$

(D) $7.07 	imes 10^{-4}$

Sigma Percentile · Accepted Answer

### The Grand Setup: Visualizing the Load

Imagine you are standing in front of a massive architectural structure. This behemoth weighs a staggering $50 	imes 10^3 	ext{ kg}$. Now, this entire weight isn't resting on the ground directly; it is elegantly supported by four identical hollow cylindrical columns made of mild steel.

Our mission is to zoom in on just one of these columns and figure out exactly how much it is being squished under this immense pressure. This 'squishing' is what physicists call **compressive strain**.

Since the load is uniformly distributed, we don't need to worry about the entire $50,000 	ext{ kg}$ at once. We can simply divide the total mass by four.

$$m = \frac{50 	imes 10^3}{4} = 12.5 	imes 10^3 	ext{ kg}$$

This means each individual column is bravely holding up $12,500 	ext{ kg}$. The actual force pushing down on the column is the weight, which is mass times the acceleration due to gravity ($g = 9.8 	ext{ m/s}^2$).

### Diving into the Geometry: The Hollow Cylinder

Now, let's look closely at the column itself. It is not a solid block of steel; it is hollow. This is a brilliant engineering choice because hollow tubes are incredibly strong while saving material and weight.

However, this means the area taking the load isn't a simple solid circle. The material only exists in the annular region between the inner radius $r$ and the outer radius $R$.

To find the **effective cross-sectional area**, we must calculate the area of the large outer circle and subtract the area of the empty inner hole.

$$A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$$

Before we plug in the numbers, we must ensure our units are consistent. The radii are given in centimeters, so we convert them to meters: $r = 0.5 	ext{ m}$ and $R = 1.0 	ext{ m}$.

$$A = 3.14 	imes (1.0^2 - 0.5^2) = 3.14 	imes (1.0 - 0.25) = 3.14 	imes 0.75 = 2.355 	ext{ m}^2$$

### The Physics Engine: Hooke's Law

With the force and the area in hand, we are ready to invoke the fundamental law of elasticity: **Hooke's Law**. For solid materials undergoing small deformations, the stress applied is directly proportional to the strain produced. The constant of proportionality is **Young's Modulus ($Y$)**.

$$Y = \frac{	ext{Stress}}{	ext{Strain}}$$

We are looking for the strain, so let's rearrange this equation. We also know that Stress is defined as Force per unit Area ($\frac{F}{A}$).

$$	ext{Strain} = \frac{	ext{Stress}}{Y} = \frac{F}{A \cdot Y}$$

### The Final Crunch: Putting it all together

This is where the magic happens. We have all the pieces of the puzzle; we just need to assemble them carefully. Let's substitute our known values into the master equation.

$$	ext{Strain} = \frac{12.5 	imes 10^3 	imes 9.8}{2.355 	imes 2.0 	imes 10^{11}}$$

Don't let the large numbers intimidate you. Take it step by step. Multiplying the numerator gives us the total force in Newtons: $122.5 	imes 10^3 	ext{ N}$. Multiplying the denominator gives us $4.71 	imes 10^{11}$.

$$	ext{Strain} = \frac{122.5 	imes 10^3}{4.71 	imes 10^{11}} \approx 26.0 	imes 10^{-8}$$

To write this in proper scientific notation, we adjust the decimal point:

$$	ext{Strain} = 2.60 	imes 10^{-7}$$

And there we have it! The compressive strain is incredibly small, which makes perfect sense. Mild steel is an exceptionally stiff material with a massive Young's modulus, meaning it barely deforms even under the weight of thousands of kilograms.

Select Answer:

Visualized Solution

\text{Understanding the Setup}

\text{Load on Each Column}

\text{Cross-sectional Area}

\text{Calculating Area}

\text{Hooke's Law}

\text{Substituting Values}

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

The Grand Setup

Diving into the Geometry

The Physics Engine

The Final Crunch

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