GATE GUIDE

GATE ME Strength of Materials Formula Sheet 2027

By MD ANISH AHAMADUpdated 4 Oct 20267 min read
GATE ME Strength of Materials Formula Sheet 2027

Strength of materials is the subject GATE ME's syllabus calls Mechanics of Materials. Its questions are short once the right formula is chosen, and slow when it is guessed. This sheet gives the formulas the subject runs on, a worked one-line example for each group, and the trap that costs marks.

In this guide
  1. Key takeaways
  2. The terms this sheet uses
  3. Elastic constants and Hooke's law
  4. Stress transformation and Mohr's circle
  5. Thin-walled pressure vessels
  6. Shear force and bending moment
  7. Bending, shear stress and shear centre
  8. Deflection and energy methods
  9. Torsion of circular shafts
  10. Columns and thermal stress
  11. Strain gauges, rosettes and mechanical properties
  12. Using this sheet in the exam
  13. Quick revision

Key takeaways

The terms this sheet uses

Stress σ\sigma is internal force per unit area, and strain ε\varepsilon is change in length per unit length. A principal plane carries no shear, and the normal stress on it is a principal stress. EE, GG and KK are the Young's, shear and bulk moduli, and ν\nu is Poisson's ratio.

II is the second moment of area about the bending axis; JJ is the polar moment for torsion. Tension is positive throughout. Keep MPa with millimetres, since 1 MPa=1 N/mm21\ \text{MPa} = 1\ \text{N/mm}^2.

Elastic constants and Hooke's law

Formula Watch out for
E=2G(1+ν)=3K(1−2ν)E = 2G(1 + \nu) = 3K(1 - 2\nu); E=9KG3K+G\displaystyle E = \frac{9KG}{3K + G} 0≤ν≤0.50 \le \nu \le 0.5; new 2027 wording names the relationships
εx=σx−ν(σy+σz)E\displaystyle \varepsilon_x = \frac{\sigma_x - \nu(\sigma_y + \sigma_z)}{E}; γ=τG\displaystyle \gamma = \frac{\tau}{G} Lateral stresses reduce the strain
εv=(1−2ν)(σx+σy+σz)E\displaystyle \varepsilon_v = \frac{(1 - 2\nu)(\sigma_x + \sigma_y + \sigma_z)}{E} Zero at ν=0.5\nu = 0.5
δ=PLAE\displaystyle \delta = \frac{PL}{AE} Stepped bar: add the pieces

Example: E=210E = 210 GPa and ν=0.25\nu = 0.25 give G=2102(1.25)=84\displaystyle G = \frac{210}{2(1.25)} = 84 GPa and K=2103(0.5)=140\displaystyle K = \frac{210}{3(0.5)} = 140 GPa. With σx=100\sigma_x = 100 MPa, σy=50\sigma_y = 50 MPa and E=200E = 200 GPa, ν=0.3\nu = 0.3: εx=100−15200 000=4.25×10−4\displaystyle \varepsilon_x = \frac{100 - 15}{200\,000} = 4.25 \times 10^{-4}.

Stress transformation and Mohr's circle

The normal stress on a plane at angle θ\theta is the mean stress plus a part that turns with 2θ2\theta:

σx′=σx+σy2+σx−σy2cos⁡2θ+τxysin⁡2θ\sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta
σ1,2=σx+σy2±(σx−σy2)2+τxy2,tan⁡2θp=2τxyσx−σy\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}, \qquad \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}

Mohr's circle has its centre at the mean stress and its radius equal to the maximum in-plane shear. Example: σx=80\sigma_x = 80, σy=20\sigma_y = 20, τxy=40\tau_{xy} = 40 MPa give centre 50 and radius 302+402=50\sqrt{30^2 + 40^2} = 50, so σ1=100\sigma_1 = 100 MPa, σ2=0\sigma_2 = 0 and θp=26.57∘\theta_p = 26.57^\circ.

Trap: in plane stress with σ3=0\sigma_3 = 0, the absolute maximum shear is half the largest of ∣σ1−σ2∣\lvert\sigma_1 - \sigma_2\rvert, ∣σ1∣\lvert\sigma_1\rvert and ∣σ2∣\lvert\sigma_2\rvert. When both principal stresses share a sign, it exceeds the circle's radius.

For plane strain, use the same equations with ε\varepsilon for σ\sigma and γ/2\gamma/2 for τ\tau. Plotting γ\gamma instead of γ/2\gamma/2 doubles the radius.

Thin-walled pressure vessels

Formula Watch out for
Cylinder: hoop σh=pd2t\displaystyle \sigma_h = \frac{pd}{2t}, longitudinal σl=pd4t\displaystyle \sigma_l = \frac{pd}{4t} Thin only, tt below about d/20d/20
Sphere: pd4t\displaystyle \frac{pd}{4t} in every direction Half the cylinder's hoop stress
Hoop strain σh−νσlE\displaystyle \frac{\sigma_h - \nu\sigma_l}{E}; εv=2εh+εl=pd4tE(5−4ν)\displaystyle \varepsilon_v = 2\varepsilon_h + \varepsilon_l = \frac{pd}{4tE}(5 - 4\nu) Volumetric strain counts hoop twice
In-plane maximum shear pd8t\displaystyle \frac{pd}{8t}; absolute pd4t\displaystyle \frac{pd}{4t} The radial direction makes the absolute value larger

Example: p=1.5p = 1.5 MPa, d=600d = 600 mm, t=6t = 6 mm give σh=75\sigma_h = 75 MPa and σl=37.5\sigma_l = 37.5 MPa. With E=200E = 200 GPa and ν=0.3\nu = 0.3, the hoop strain is 75−11.25200 000=3.19×10−4\displaystyle \frac{75 - 11.25}{200\,000} = 3.19 \times 10^{-4} and the volumetric strain 7.13×10−47.13 \times 10^{-4}. A helical seam is a transformation from these two directions.

Shear force and bending moment

Formula Watch out for
dVdx=−w\displaystyle \frac{dV}{dx} = -w; dMdx=V\displaystyle \frac{dM}{dx} = V MM is extreme where V=0V = 0 or changes sign
Simply supported: central PP gives PL4\displaystyle \frac{PL}{4}; UDL gives wL28\displaystyle \frac{wL^2}{8} Reactions wL2\displaystyle \frac{wL}{2} each
Cantilever: end PP gives PLPL; UDL gives wL22\displaystyle \frac{wL^2}{2} Maximum at the fixed end
Triangular load, simply supported: Mmax⁡=wL293\displaystyle M_{\max} = \frac{wL^2}{9\sqrt{3}} at L3\displaystyle \frac{L}{\sqrt{3}} from the zero end ww is the peak intensity

Example: a 6 m simply supported beam under w=10w = 10 kN/m has reactions of 30 kN and Mmax⁡=10×368=45\displaystyle M_{\max} = \frac{10 \times 36}{8} = 45 kN m. This group appeared in 2023, 2024, 2025 and 2026, more steadily than any other in the subject.

Remember: an applied couple makes the bending moment diagram jump by its value and leaves the shear force unchanged. Contraflexure is where MM changes sign, not where VV is zero.

The book's last-minute sheet lists every formula on this page with its condition, and the Mechanics of Materials chapter traces each topic through the six counted papers. Both are in the GATE ME 2027 book, with 942 questions with worked solutions.

Bending, shear stress and shear centre

Formula Watch out for
MI=σy=ER\displaystyle \frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}; Z=Iymax⁡\displaystyle Z = \frac{I}{y_{\max}} yy from the centroidal axis
Rectangle I=bd312\displaystyle I = \frac{bd^3}{12}, Z=bd26\displaystyle Z = \frac{bd^2}{6}; circle I=πd464\displaystyle I = \frac{\pi d^4}{64}, Z=πd332\displaystyle Z = \frac{\pi d^3}{32} Triangle about its centroid bh336\displaystyle \frac{bh^3}{36}
τ=VQIb\displaystyle \tau = \frac{VQ}{Ib} Rectangle τmax⁡=1.5VA\displaystyle \tau_{\max} = 1.5\frac{V}{A}; circle 43VA\displaystyle \frac{4}{3}\frac{V}{A}
Channel shear centre e=3b2h+6b\displaystyle e = \frac{3b^2}{h + 6b} from the web Lies outside the section, away from the flanges

Example: b=100b = 100 mm, d=200d = 200 mm and M=20M = 20 kN m give Z=666 667 mm3Z = 666\,667\ \text{mm}^3 and σ=30\sigma = 30 MPa. A shear force of 40 kN on the same section gives a mean of 2 MPa and a maximum of 3 MPa.

Deflection and energy methods

Case Deflection Stiffness
Cantilever, end load PL33EI\displaystyle \frac{PL^3}{3EI}, slope PL22EI\displaystyle \frac{PL^2}{2EI} 3EIL3\displaystyle \frac{3EI}{L^3}
Cantilever, UDL wL48EI\displaystyle \frac{wL^4}{8EI} —
Simply supported, central load PL348EI\displaystyle \frac{PL^3}{48EI} 48EIL3\displaystyle \frac{48EI}{L^3}
Simply supported, UDL 5wL4384EI\displaystyle \frac{5wL^4}{384EI} —
Fixed–fixed, central load PL3192EI\displaystyle \frac{PL^3}{192EI} 192EIL3\displaystyle \frac{192EI}{L^3}

Strain energy is P2L2AE\displaystyle \frac{P^2L}{2AE} in tension, ∫M22EI dx\displaystyle \int\frac{M^2}{2EI}\,dx in bending and T2L2GJ\displaystyle \frac{T^2L}{2GJ} in torsion. Castigliano gives the deflection under a load as δ=∂U∂P\displaystyle \delta = \frac{\partial U}{\partial P}. A suddenly applied load doubles the static stress, σ=2PA\displaystyle \sigma = \frac{2P}{A}.

Example: a 2 m cantilever with EI=2×106 N m2EI = 2 \times 10^6\ \text{N m}^2 has stiffness 3×2×1068=750\displaystyle \frac{3 \times 2 \times 10^6}{8} = 750 kN/m, so a 10 kN tip load deflects it 13.3 mm.

Torsion of circular shafts

TJ=τr=GθL,J=πd432,τmax⁡=16Tπd3,P=2πNT60\frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L}, \qquad J = \frac{\pi d^4}{32}, \qquad \tau_{\max} = \frac{16T}{\pi d^3}, \qquad P = \frac{2\pi NT}{60}

Example: T=1T = 1 kN m on a 50 mm shaft gives τmax⁡=16×106π×125 000=40.7\displaystyle \tau_{\max} = \frac{16 \times 10^6}{\pi \times 125\,000} = 40.7 MPa; at 300 rpm it transmits 31.4 kW. Shafts in series add twists; a bonded core and sleeve share one twist and add torques, in proportion to GJGJ.

Columns and thermal stress

Formula Watch out for
Pcr=π2EILe2\displaystyle P_{cr} = \frac{\pi^2 EI}{L_e^2} Le=LL_e = L, 2L2L, L2\displaystyle \frac{L}{2}, L2\displaystyle \frac{L}{\sqrt{2}}; least II
Rankine 1P=1Pc+1PE\displaystyle \frac{1}{P} = \frac{1}{P_c} + \frac{1}{P_E} For intermediate columns
Fully restrained σ=EαΔT\sigma = E\alpha\Delta T No stress if free to expand
With gap δ\delta: σ=E(αΔTL−δ)L\displaystyle \sigma = \frac{E(\alpha\Delta TL - \delta)}{L} Subtract the gap first

Example: a 2 m pinned column with E=200E = 200 GPa and I=8×10−7 m4I = 8 \times 10^{-7}\ \text{m}^4 has Pcr=395P_{cr} = 395 kN. A steel bar with α=12×10−6\alpha = 12 \times 10^{-6} per K heated 50 K is stressed to 120 MPa if fully restrained, and to 60 MPa if a 0.3 mm gap is first taken up over 1 m.

Strain gauges, rosettes and mechanical properties

Formula Watch out for
Gauge factor GF=ΔR/Rε\displaystyle GF = \frac{\Delta R/R}{\varepsilon} About 2 for metal foil (rule of thumb)
45∘45^\circ rosette: γxy=2εb−εa−εc\gamma_{xy} = 2\varepsilon_b - \varepsilon_a - \varepsilon_c εx=εa\varepsilon_x = \varepsilon_a, εy=εc\varepsilon_y = \varepsilon_c
Plane stress σ1=E(ε1+νε2)1−ν2\displaystyle \sigma_1 = \frac{E(\varepsilon_1 + \nu\varepsilon_2)}{1 - \nu^2} Not simply Eε1E\varepsilon_1
Modulus of resilience σy22E\displaystyle \frac{\sigma_y^2}{2E} Toughness is the whole area to fracture
Brinell BHN=2PπD(D−D2−d2)\displaystyle BHN = \frac{2P}{\pi D\left(D - \sqrt{D^2 - d^2}\right)} PP in kgf, DD and dd in mm

Example: rosette readings of 400, 300 and −100-100 microstrain give γxy=300\gamma_{xy} = 300 microstrain, centre 150 and radius 2502+1502=291.5\sqrt{250^2 + 150^2} = 291.5, so the principal strains are 441.5 and −141.5-141.5 microstrain.

In one line: a rosette is Mohr's circle for strain with one extra line of arithmetic.

Using this sheet in the exam

Write I=πd464\displaystyle I = \frac{\pi d^4}{64} and J=πd432\displaystyle J = \frac{\pi d^4}{32} on the rough sheet in the first five minutes; the exam-day rules guide says what you may carry. Keep full precision on the GATE virtual calculator and round only the final NAT entry. Most questions here carry two marks, so check the marking scheme, common to every GATE paper, before guessing on an MCQ.

More formula sheets: all of GATE ME · Fluid Mechanics · Heat Transfer · Thermodynamics

Quick revision

  1. E=2G(1+ν)=3K(1−2ν)E = 2G(1 + \nu) = 3K(1 - 2\nu).
  2. Principal stresses: centre ±\pm radius; an angle doubles on Mohr's circle.
  3. Hoop pd2t\displaystyle \frac{pd}{2t}, longitudinal pd4t\displaystyle \frac{pd}{4t}, sphere pd4t\displaystyle \frac{pd}{4t}.
  4. Simply supported UDL gives wL28\displaystyle \frac{wL^2}{8}; an applied couple makes MM jump.
  5. Cantilever PL33EI\displaystyle \frac{PL^3}{3EI}; simply supported PL348EI\displaystyle \frac{PL^3}{48EI} and 5wL4384EI\displaystyle \frac{5wL^4}{384EI}.
  6. τmax⁡=16Tπd3\displaystyle \tau_{\max} = \frac{16T}{\pi d^3}; P=2πNT60\displaystyle P = \frac{2\pi NT}{60}.
  7. Euler LeL_e: LL, 2L2L, L2\displaystyle \frac{L}{2}, L2\displaystyle \frac{L}{\sqrt{2}}.
  8. Thermal stress only from the restrained expansion, EαΔTE\alpha\Delta T at most.

Frequently asked questions

Is strength of materials the same as mechanics of materials in the GATE ME syllabus?

Yes. The official GATE ME syllabus prints the subject as Mechanics of Materials, under Section 2, Applied Mechanics and Design. Many textbooks and students call it strength of materials. The content is the same: stress and strain, transformations, pressure vessels, beams, torsion, columns, energy methods, thermal stresses, strain gauges and mechanical properties.

What is the relation between E, G and K?

Young's modulus EE, shear modulus GG and bulk modulus KK are linked through Poisson's ratio ν\nu by E=2G(1+ν)=3K(1−2ν)E = 2G(1 + \nu) = 3K(1 - 2\nu), or E=9KG3K+G\displaystyle E = \frac{9KG}{3K + G} without ν\nu. For steel with E=200E = 200 GPa and ν=0.3\nu = 0.3, GG is about 76.9 GPa and KK about 166.7 GPa.

How do I find principal stresses quickly in GATE ME?

Take the centre of Mohr's circle as the mean normal stress σx+σy2\displaystyle \frac{\sigma_x + \sigma_y}{2} and the radius as (σx−σy2)2+τxy2\displaystyle \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}. The principal stresses are the centre plus and minus the radius, and the radius is the maximum in-plane shear stress. Remember that an angle on the element doubles on the circle.

What is the effective length of a column for Euler buckling?

Euler load is Pcr=π2EILe2\displaystyle P_{cr} = \frac{\pi^2 EI}{L_e^2}. The effective length LeL_e is LL for both ends pinned, 2L2L for one end fixed and one free, L/2L/2 for both ends fixed, and L/2L/\sqrt{2} for one end fixed and one pinned. Always use the least second moment of area, because the column buckles about that axis.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.

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