A floor has an area of 100 m^{2}. Its consists of two very stiff
plates, separated by coil springs to give it some "bounce". The
equilibrium separation is 5 cm. There are 100 evenly spaced identical
springs per square meter. The spring constant of each spring is k = 10^{3}
N/m.

(a) Find the Young's modulus of this floor.

(b) When 20 people with a combine mass of 1200 kg rest on this floor (in
equilibrium), by
how much does the distance between the plates change?

(c) In (b), what is the elastic (strain) energy stored in the floor?

Solution:

- Concepts:

Young's modulus - Reasoning:

The Young's modulus is defined through F/A = Y ΔL/L,

Y = (tensile stress)/(tensile strain). - Details of the calculation:

(a) The effective spring constant of the system of 10^{4 }springs in parallel is 10^{7}N/m.

Hooke's law: F/ΔL = k = YA/L = 10^{7}N/m.

Y = (10^{7}N/m)*(5*10^{-2}m)/100 m^{2}= 5*10^{3}N/m^{2}.

(b) ΔL = FL/(AY) = (1200 kg)*(9.8 m/s^{2})*(5*10^{-2}m)/(100 m^{2}* 5*10^{3}N/m^{2}) = 1.18 mm.

The distance between the plates change by 1.18 mm.

(c) We model the floor as a compressed spring. The spring constant k = 10^{7}N/m.

The elastic energy stored in the floor is ½k ΔL^{2}= 7 J.

Think about: Why not mgΔL?

A block of gelatin has dimensions 60 mm by 60 mm by 20 mm when unstressed. It rests on one of the 60 mm by 60 mm sides. A force of 0.245 N is applied tangentially to the upper surface causing a 5 mm displacement relative to the lower surface. Find the shear stress, the shear strain, and shear modulus.

Solutions

- Concepts:

Shear stress and strain, shear modulus - Reasoning:

Shear modulus S = (shear stress)/(shear strain) = (F_{||}/A)/(Δx/L_{0})

- Details of the calculation:

Shear stress: 0.245 N/(6*10^{-2}m)^{2}= 68 Pa.

Shear strain = (5 mm)/(20 mm) = ¼.

Shear modulus S = 272 Pa.

Two rigid parallel plates are connected by a composite material consisting of parallel fibers in a plastic matrix as shown in the figure.

The volume fraction of the fibers relative to the total volume of the
composite is f. A stress σ is applied to the plates. The Young's
modulus of the fiber is Y_{f}, and that of the matrix is Y_{m}. Assuming there is no
delamination across the fiber/matrix interface, the strains ε in the components
of the composite are equal (isostrain).

(a) Write Hooke's law in terms of σ, Y and strain, ε.

(b) Write the expression of the condition of isostrain in terms of
σ, Y, ε for the fiber and matrix.

(c) Derive an expression for the Young's modulus, Y_{eff}, of the
composite.

Solution:

- Concepts:

Young's modulus - Reasoning:

The equation F/A = YΔL/L defines the Young's modulus Y of a material.

We need to find the effective Young's modulus, when materials are arranged in parallel. - Details of the calculation:

(a) σ = Yε

(b) σ_{f }= Y_{f}ε_{f}, σ_{m }= Y_{m}ε_{m}, ε_{f}= ε_{m}, σ_{f}/Y_{f}= σ_{m}/Y_{m }= ε.

(c) Let the volume V of the composite be the distance between the parallel plates, L times the cross sectional area A of the plates. V = AL.

The volume fraction of the composite that is fiber is fV = fAL. fV = fAL= A_{f}L, or A_{f}= fA.

The remaining volume is matrix. Because the plates are parallel, L for the matrix is the same as the fiber length, so the cross sectional area of the matrix is A_{m}= (1 - f)A.

The force F exerted on the composite is F = σA and because the plates are rigid this force must be the sum of the forces on each component,

F = σA = Y_{eff}εA = F_{f }+ F_{m}= σ_{f}A_{f}+ σ_{m}A_{m}= [Y_{f}f + Y_{m}(1 - f)]εA.

Y_{eff}= Y_{f}f + Y_{m}(1 - f).