A uniform 0.01 N shear force acts on the top of a 1 cm-sided tissue cube. The deformed shear angle is θ=5°. Treat the tissue as linear, isotropic and homogeneous. Find shear modulus in kPa.
The top area is (0.01 m)²=10⁻⁴ m², so shear stress is 0.01/10⁻⁴=100 Pa. Engineering shear strain is tan5°≈0.08749. Thus G=stress/strain≈1143 Pa=1.14 kPa, A. Using the small-angle value 5π/180 gives essentially the same result.
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