Shear Modulus Calculator

The modulus of rigidity G from shear stress and shear strain, step by step
Find the shear modulus if a 10 kPa shear stress produces a shear strain of 0.020
Find the shear modulus of steel with E = 200 GPa and Poisson's ratio 0.30
A 40 mm solid shaft 1.5 m long carries 500 N·m with G = 79 GPa. Find the angle of twist.
Convert a shear modulus of 79 GPa to psi

What Shear Modulus Is

The shear modulus GG, also called the modulus of rigidity, measures a material's resistance to being skewed — one face sliding parallel to the opposite face, without any change in volume.

G=τγ=F/AΔx/LG = \frac{\tau}{\gamma} = \frac{F/A}{\Delta x / L}

Symbols and SI units:

  • GG — shear modulus, pascals (Pa), normally quoted in GPa
  • τ\tau — shear stress, Pa; the force FF in newtons divided by the area AA in m² parallel to the force
  • γ\gamma — shear strain, dimensionless (radians); the sideways displacement Δx\Delta x in m divided by the separation LL in m
  • Δx\Delta x — lateral displacement of the sheared face, m

Typical values: structural steel 79\approx 79 GPa, aluminium 26\approx 26 GPa, copper 45\approx 45 GPa, rubber 0.5\approx 0.5 MPa.

When it applies: small elastic strains, below the proportional limit, in an isotropic material.

The assumption people forget: AA is the area parallel to the applied force, not the cross-section perpendicular to it as in tension. That single difference separates GG from Young's modulus EE.

Relating G to E, and Using It in Torsion

For an isotropic material the three elastic constants are not independent:

G=E2(1+ν)G = \frac{E}{2(1 + \nu)}

with EE Young's modulus in Pa and ν\nu Poisson's ratio (dimensionless, about 0.300.30 for metals). Since ν\nu is near 0.30.3, GG lands near 0.385E0.385E — steel's 200200 GPa gives roughly 7777 GPa. It follows that G<EG < E for every ordinary material.

GG is the constant that governs torsion of a circular shaft:

θ=TLGJ,J=πd432 (solid shaft)\theta = \frac{TL}{GJ}, \qquad J = \frac{\pi d^4}{32}\ \text{(solid shaft)}

  • θ\theta — angle of twist, radians
  • TT — applied torque, N·m
  • LL — shaft length, m
  • JJ — polar second moment of area, m⁴

A stiffer GG, or a fatter shaft, means less twist — and because JJ goes as d4d^4, diameter dominates.

The assumption people forget: J=πd4/32J = \pi d^4/32 is for a solid circular shaft. Hollow and non-circular sections need their own JJ.

Common Mistakes to Avoid

  • Using the perpendicular cross-section for AA — shear stress divides by the area the force slides along.
  • Confusing GG with EE — they answer different questions, and GG is roughly 0.4E0.4E for metals.
  • Treating shear strain as a lengthγ=Δx/L\gamma = \Delta x/L is dimensionless, an angle in radians for small distortions.
  • Quoting GG in newtons — it is a stress-like quantity, so Pa or GPa.
  • Using the radius in J=πd4/32J = \pi d^4/32 — that formula takes the diameter; with the radius the constant becomes πr4/2\pi r^4/2.
  • Leaving θ\theta in radians when degrees were asked for — multiply by 180/π180/\pi.
  • Applying G=E/[2(1+ν)]G = E/[2(1+\nu)] to an anisotropic material — wood and composites do not obey it.

Examples

Step 1: τ=F/A=(200 N)÷(0.020 m2)=1.0×104 Pa\tau = F/A = (200\ \text{N}) \div (0.020\ \text{m}^2) = 1.0 \times 10^4\ \text{Pa}
Step 2: Convert the displacement: Δx=2.0 mm=2.0×103 m\Delta x = 2.0\ \text{mm} = 2.0 \times 10^{-3}\ \text{m}
Step 3: γ=Δx/L=(2.0×103 m)÷(0.10 m)=0.020\gamma = \Delta x/L = (2.0 \times 10^{-3}\ \text{m}) \div (0.10\ \text{m}) = 0.020 (dimensionless)
Step 4: G=τ/γ=(1.0×104 Pa)÷0.020=5.0×105 PaG = \tau/\gamma = (1.0 \times 10^4\ \text{Pa}) \div 0.020 = 5.0 \times 10^5\ \text{Pa}
Answer: G=5.0×105G = 5.0 \times 10^5 Pa =0.50= 0.50 MPa

Step 1: G=E/[2(1+ν)]G = E/\left[2(1+\nu)\right]
Step 2: 1+ν=1.301 + \nu = 1.30 (dimensionless)
Step 3: 2(1.30)=2.602(1.30) = 2.60
Step 4: G=(200 GPa)÷2.60=76.9 GPaG = (200\ \text{GPa}) \div 2.60 = 76.9\ \text{GPa}, matching the handbook value of about 7979 GPa
Answer: G76.9G \approx 76.9 GPa

Step 1: J=πd4/32=π(0.040 m)4/32=π(2.56×106 m4)/32=2.513×107 m4J = \pi d^4/32 = \pi (0.040\ \text{m})^4/32 = \pi(2.56 \times 10^{-6}\ \text{m}^4)/32 = 2.513 \times 10^{-7}\ \text{m}^4
Step 2: GJ=(79×109 Pa)(2.513×107 m4)=1.986×104 N\cdotpm2GJ = (79 \times 10^9\ \text{Pa})(2.513 \times 10^{-7}\ \text{m}^4) = 1.986 \times 10^{4}\ \text{N·m}^2
Step 3: θ=TL/(GJ)=(500 N\cdotpm)(1.5 m)÷(1.986×104 N\cdotpm2)\theta = TL/(GJ) = (500\ \text{N·m})(1.5\ \text{m}) \div (1.986 \times 10^{4}\ \text{N·m}^2)
Step 4: θ=0.0378 rad=0.0378×180/π=2.16°\theta = 0.0378\ \text{rad} = 0.0378 \times 180/\pi = 2.16°
Answer: θ0.0378\theta \approx 0.0378 rad =2.16°= 2.16°

Frequently Asked Questions

It is the ratio of shear stress to shear strain, G = τ/γ, and it measures how strongly a material resists a change in shape at constant volume. It is also called the modulus of rigidity and is quoted in pascals, usually GPa.

About 79 GPa (roughly 11.5 × 10⁶ psi) for structural steel. It follows from G = E/[2(1+ν)] with E = 200 GPa and ν = 0.30, which gives 76.9 GPa — close to the handbook figure, which varies slightly by alloy.

Pascals, since shear strain is dimensionless and shear stress is a pressure. Engineering tables use GPa in SI and Msi (10⁶ psi) in US units; 1 GPa ≈ 145,038 psi.

Young's modulus E relates a pulling stress to a stretch along the same axis; shear modulus G relates a sliding stress to an angular distortion. For an isotropic material G = E/[2(1+ν)], so G is roughly 0.4E for typical metals.

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