The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

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To tackle this problem, we'll apply principles from mechanics of materials, specifically Hooke's law, which relates stress to strain for linearly elastic materials.

Given:

• Edge length of the aluminium cube (a) = 10 cm = 0.1 m
• Mass (m) = 100 kg
• Shear modulus (G) = 25 GPa = 25 × 10^9 Pa

First, let's calculate the force exerted on the opposite face of the cube due to the mass attached. The force (F) can be calculated using the formula:

F=mgF=mg

where:

• mm is the mass,
• gg is the acceleration due to gravity (approximately 9.81 m/s29.81m/s2).

So, F=100 kg×9.81 m/s2=981 NF=100kg×9.81m/s2=981N.

Now, let's find the shear stress (ττ) applied on the face of the cube. Since the force is applied parallel to the face, the stress is the shear stress, and it can be calculated using:

τ=FAτ=AF

where:

• AA is the cross-sectional area of the face.

The face of the cube has a square cross-section, so its area is a2a2. Therefore, A=(0.1 m)2=0.01 m2A=(0.1m)2=0.01m2.

Thus, τ=981 N0.01 m2=98100 Paτ=0.01m2981N=98100Pa.

Now, let's use Hooke's law to find the strain (γγ) in the material. Hooke's law states:

τ=Gγτ=Gγ

where:

• GG is the shear modulus.

So, γ=τG=98100 Pa25×109 Pa=3.924×10−6γ=Gτ=25×109Pa98100Pa=3.924×10−6.

Now, let's find the vertical deflection (δδ) of the face. Since the cube is fixed to the wall, the deflection will be due to the shear strain. The vertical deflection can be calculated using the formula:

δ=γ×hδ=γ×h

where:

• hh is the height of the cube.

Since the cube is a cube, its height is equal to the edge length (aa). So, h=0.1 mh=0.1m.

Therefore, δ=3.924×10−6×0.1 m=3.924×10−7 mδ=3.924×10−6×0.1m=3.924×10−7m.

So, the vertical deflection of the face is 3.924×10−73.924×10−7 meters.

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