Question

A disk between vertebrae in the spine is subjected to a shearing force of 525 N. Find its shear deformation, taking it to have a shear modulus of 1.30×109 N/m2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter.

Answer 1

To find the shear deformation of a disk between vertebrae subjected to a shearing force, the shear deformation is calculated using the formula (shear deformation = shearing force / (area * shear modulus)). Plugging in the given values, the shear deformation is around 0.315 micrometers.

Step-by-step explanation:

To calculate the shear deformation of a disk between vertebrae under a shearing force, we use the relationship:

Shear deformation (\( \Delta x \)) = \( \frac{F}{A} \) / G

where F is the force applied, A is the area over which the force is applied, and G is the shear modulus of the material. Given:

• Shearing force (F) = 525 N
• Shear modulus (G) = 1.30 \( \times \) 109 N/m2
• Height of the cylinder (h) = 0.700 cm = 0.007 m
• Diameter of the cylinder (d) = 4.00 cm = 0.04 m

The area (A) is the cross-sectional area of the cylinder (the circular top), which can be calculated using \( A = \pi r^2 \), where r is the radius of the cylinder (half of the diameter).

Radius (r) = \( \frac{d}{2} \) = 0.02 m

So, A = \( \pi \times (0.02 m)^2 \) = 1.2566 \( \times \) 10-3 m2

Now, we can find the shear deformation:

Shear deformation (\( \Delta x \)) = \( \frac{525 N}{1.2566 \times 10^{-3} m^2} \) / 1.30 \( \times \) 109 N/m2

Shear deformation (\( \Delta x \)) = \( \frac{525}{1.2566 \times 10^{-3} \times 1.30 \times 10^9} \) m ≈ 3.15 \( \times \) 10-7 m or 0.315 \( \mu \)m