← Strength of Materials · RSMSSB JE Civil (Diploma)

Chapter 8 of 9

Cylinders and Springs

In the RSMSSB JE Civil (Diploma) syllabus under Strength of Materials · 2 parts

Thin & Thick Cylinders · Springs

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📑 Contents (16 sections)

Part 1 of 2

Thin & Thick Cylinders

Last reviewed 16 Sept 2026 · 6 min read

Thin and thick cylinders

Boilers, water mains, penstocks, gas cylinders and storage tanks are pressure vessels. A cylinder is treated as thin when its wall thickness $t$ is small compared with its diameter — commonly $t < d /20$ (some books use $d / t > 20$ ). The stresses are then taken as uniform through the thickness and radial stress is neglected. Otherwise it is a thick cylinder, analysed by Lamé's theory.

Stresses in a thin cylinder

Internal pressure $p$ , internal diameter $d$ , thickness $t$ .

Hoop (circumferential) stress

Cut the cylinder along its length. Bursting force on a length $L$ $= p d L$ ; resisting force of the two walls $= 2 σ_{c} t L$ .

σ_{c} = \frac{p d}{2 t}

Longitudinal stress

Cut across the cylinder. Bursting force on the end $= p π d^{2} /4$ ; resisting area $= π d t$ .

σ_{l} = \frac{p d}{4 t}

FormulaThin cylinder

σ_{c} = \frac{p d}{2 t}, σ_{l} = \frac{p d}{4 t} = \frac{σ _{c}}{2}, τ_{ma x (in - pl an e)} = \frac{σ _{c} - σ _{l}}{2} = \frac{p d}{8 t}

Hoop stress is twice the longitudinal stress, so a thin cylinder bursts along a longitudinal seam. Considering the third (radial) principal stress ≈ 0, the absolute maximum shear stress is $σ_{c} /2 = p d /4 t$ .

Joint efficiency

Riveted or welded seams are weaker than the plate. With longitudinal joint efficiency $η_{l}$ and circumferential joint efficiency $η_{c}$ :

σ_{c} = \frac{p d}{2 t η _{l}}, σ_{l} = \frac{p d}{4 t η _{c}}

Note the pairing: the longitudinal joint resists the hoop stress; the circumferential joint resists the longitudinal stress.

Thin sphere

By symmetry the stress is the same in every direction in the wall:

σ = \frac{p d}{4 t}

For the same diameter, pressure and allowable stress, a sphere needs half the thickness of a cylinder — spheres are used for high-pressure gas storage.

Changes in dimensions of a thin cylinder

With $E$ and Poisson's ratio $μ$ :

ε_{c} = \frac{σ _{c} - μ σ _{l}}{E} = \frac{p d}{4 tE} (2 - μ), ε_{l} = \frac{σ _{l} - μ σ _{c}}{E} = \frac{p d}{4 tE} (1 - 2 μ)

FormulaChanges in size

Change in diameter: $δ d = ε_{c} d = \frac{p d ^{2}}{4 tE} (2 - μ)$
Change in length: $δ L = ε_{l} L = \frac{p d L}{4 tE} (1 - 2 μ)$
Volumetric strain (cylinder): $ε_{v} = 2 ε_{c} + ε_{l} = \frac{p d}{4 tE} (5 - 4 μ)$
Volumetric strain (sphere): $ε_{v} = 3 ε = \frac{3 p d}{4 tE} (1 - μ)$

The extra volume of fluid that can be pumped in at pressure $p$ = increase in vessel volume + compression of the fluid ( $p V / K$ ).

Part 2 of 2

Springs

Last reviewed 16 Sept 2026 · 6 min read

Springs and their uses

A spring is an elastic member that deflects under load and recovers when the load is removed. Springs are used to absorb shock (vehicle suspension, railway buffers), store energy (clocks, toys), apply force (valve springs, brakes, clutches) and measure force (spring balances).

Type	Main loading in the material	Typical use
Close-coiled helical (tension or compression)	Torsion of the wire	Valve springs, buffers, balances
Open-coiled helical	Torsion + bending	Special mechanisms
Torsion (spiral, helical torsion)	Bending	Clocks, clothes pegs, door hinges
Leaf (laminated, carriage)	Bending	Truck and railway wagon suspension
Disc (Belleville)	Bending	Short-stroke, high-load applications

Stiffness (spring constant, spring rate) $k$ = load per unit deflection: $k = W / δ$ (N/mm). Flexibility = $1/ k$ . Proof load = the maximum load a spring can carry without permanent set; the corresponding deflection is the proof deflection; energy stored up to it is the proof resilience.

Close-coiled helical spring under axial load

A spring is close-coiled when the helix angle is so small (under about 10°) that each coil is nearly in a plane perpendicular to the axis. An axial load $W$ then produces an almost pure twisting moment $T = W R$ in the wire, where $R$ is the mean coil radius.

Notation: wire diameter $d$ , mean coil radius $R$ (mean diameter $D = 2 R$ ), number of active coils $n$ , modulus of rigidity $G$ . Length of wire $l = 2 π R n$ .

Shear stress

From the torsion formula for the wire:

τ = \frac{16 T}{π d ^{3}} = \frac{16 W R}{π d ^{3}} = \frac{8 W D}{π d ^{3}}

Deflection

The twist of the wire $θ = \frac{T l}{G J}$ ; the axial deflection $δ = R θ$ :

δ = \frac{W R \cdot 2 π R n \cdot R}{G \cdot π d ^{4} /32} = \frac{64 W R ^{3} n}{G d ^{4}}

FormulaClose-coiled helical spring

τ = \frac{16 W R}{π d ^{3}} = \frac{8 W D}{π d ^{3}}, δ = \frac{64 W R ^{3} n}{G d ^{4}} = \frac{8 W D ^{3} n}{G d ^{4}}

k = \frac{W}{δ} = \frac{G d ^{4}}{64 R ^{3} n} = \frac{G d ^{4}}{8 D ^{3} n}

Energy stored: $U = \frac{1}{2} W δ = \frac{32 W ^{2} R ^{3} n}{G d ^{4}}$ .

Exam TipWhat changes stiffness

$k \propto d^{4} / (D^{3} n)$ . Doubling the wire diameter makes the spring 16 times stiffer; doubling the coil diameter makes it 8 times more flexible; doubling the number of coils halves the stiffness. Cutting a spring into two equal halves doubles the stiffness of each half.

Wahl's correction factor

The simple formula ignores direct shear $W / (π d^{2} /4)$ and the curvature of the wire (the inner side of the coil is more highly stressed, as in a curved beam). With spring index $C = D / d$ :

τ_{ma x} = K_{w} \frac{8 W D}{π d ^{3}}, K_{w} = \frac{4 C - 1}{4 C - 4} + \frac{0.615}{C}

For $C$ = 6, $K_{w} = 23/20 + 0.1025 = 1.2525$ . Spring index is usually kept between about 4 and 12.

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