Solids Finale: Master the Formula Card for Standing Structures

JEE/NEET Physics · Mechanical Properties of Solids series · Part 4 of 4 · All parts →

The final card of the Mechanical Properties of Solids series — the whole chapter on one revision sheet, plus why skyscrapers, bridges and aircraft are really applied materials science.

The Master Formula Card

This single table compresses the entire chapter. If you can reproduce every row from memory — with the meaning in the last column, not just the symbols — you are exam-ready on formulas for Mechanical Properties of Solids.

Reading the card smartly: notice that stress is always “force intensity” and strain is always “fractional deformation” — the only thing that changes from row to row is which deformation you measure (length, shape, or volume). Once that pattern clicks, the twelve formulas collapse into one idea with four costumes.

The One-Rule-Per-Part Recap

Part 1: materials respond to intensity per patch (stress) with fractional stretch (strain) — not to total force, but to how concentrated that force is. Part 2: stiffness is a material identity — Young’s modulus — and every object is a spring (k = YA/L), so geometry lets engineers tune stiffness even with a fixed material. Part 3: slide, squeeze, thin, store, and refuse-to-expand: the moduli family and the ever-present ½-factor in elastic energy.

Together, these three rules mean you never memorise a dozen disconnected formulas. You memorise one idea — deformation per unit intensity — and apply it in the direction the problem asks about.

How to Use the Card in Exams

JEE and NEET questions from this chapter rarely test formulas directly; they test unit discipline and limiting cases. Three habits will save you marks:

• Convert first, calculate second. Areas in mm² must become m² (divide by 10⁶) before touching Y = FL/(AΔL). Most lost marks in this chapter are unit slips, not concept slips.
• Sanity-check with the modulus spread. Y spans roughly 10⁶ Pa (rubber) to 10¹¹–10¹² Pa (diamond, steel). If your ΔL comes out as metres for a steel wire under a few hundred newtons, something is wrong.
• Ask “is expansion blocked?” Thermal stress exists only when a rod is clamped between rigid walls. A free-expanding rod carries zero thermal stress, however large the temperature change — a favourite NEET trap.

Materials in the Wide World

Worked Examples — Watch the Method

Example 1: The stretching wire. A steel wire (Y = 2×10¹¹ Pa) of length 2 m and radius 1 mm hangs a 20 kg mass. How much does it stretch?

Step 1: Force F = mg = 20 × 10 = 200 N. Step 2: Area A = πr² = 3.14 × 10⁻⁶ m². Step 3: ΔL = FL/(AY) = (200 × 2)/(3.14×10⁻⁶ × 2×10¹¹) ≈ 6.4×10⁻⁴ m ≈ 0.64 mm. Notice the sanity check: sub-millimetre stretch for steel — exactly what the six-orders-of-magnitude spread predicts.

Example 2: Energy stored. That same wire stores U = ½FΔL = ½ × 200 × 6.4×10⁻⁴ ≈ 0.064 J. Tiny — but scale the wire thicker or the load larger, and stored energy grows as the square of the load (since ΔL ∝ F). Doubling the load quadruples the stored energy. This quadratic growth is why overloaded structures fail suddenly rather than gently.

Example 3: Thermal stress. A steel rail (Y = 2×10¹¹ Pa, α = 1.2×10⁻⁵ /°C) is clamped and heated by 30 °C. Thermal stress = YαΔT = 2×10¹¹ × 1.2×10⁻⁵ × 30 = 7.2×10⁷ Pa — large enough to buckle tracks, which is exactly why railways have expansion gaps.

Practice set (answers hidden — try first)

Frequently Asked Questions

Recap and Key Takeaways

Closing thought: you began this series asking what happens when you pull on a wire, and you end holding the design logic of the built world. Every bridge you cross and every train you board is a solved problem in stress, strain and modulus selection. Keep the master card close, rehearse the units, and this chapter becomes one of the most reliable score sources in JEE Physics and NEET Physics alike.