Question

with usual notation derive the relationship between se=wG​

Answer 1

Answer:

The relation between water content,degree of saturation, specific gravity and void ratio is given bySe=WG or (s)=WG/e.

Answer 2

Answer:

We can write the relationship between strain energy (u) and work done (W) as:

u = (1/2)wmδ = (1/2)Wδ = (1/2)Fd

where F is the applied force and d is the displacement.

Explanation:

The relationship between strain energy (U) and work done (W) in a solid is given by the following equation:

U = (1/2)σεV

where σ is the stress, ε is the strain, and V is the volume of the solid.

The strain (ε) is related to the displacement (δ) and the original length of the solid (L) by the following equation:

ε = δ/L

Therefore, we can rewrite the equation for strain energy as:

U = (1/2)σ(δ/L)V

The work done (W) on the solid can be expressed as the product of the force (F) and the displacement (δ):

W = Fδ

Using Hooke's Law, we can express the stress (σ) as the product of the elastic modulus (E) and the strain (ε):

σ = Eε

Substituting this expression for σ into the equation for strain energy, we get:

U = (1/2)Eε²V

Substituting the expression for ε in terms of δ and L, we get:

U = (1/2)E(L^(-2))(δ²)V

Dividing both sides by V and multiplying by the density (ρ), we get:

(u/V)ρ = (1/2)E(L^(-2))(δ²)ρ

The quantity (u/V)ρ is the strain energy density (u/V) multiplied by the density (ρ), which is equal to the specific strain energy (u/mass):

se = (u/V)ρ

Multiplying both sides of the equation by the volume V, we get:

uρ = (1/2)E(L^(-2))(δ²)ρV

Dividing both sides by the mass (m = ρV), we get:

(u/m) = (1/2)E(L^(-2))(δ²)

Substituting the expression for work done (W = Fδ) into this equation, we get:

(u/m) = (1/2)(F/δ)(δ/L)^2

Simplifying, we get:

(u/m) = (1/2)(F/L)δ

Since F/L is the strain energy per unit volume (w), we can write:

(u/m) = (1/2)wδ

Multiplying both sides by the mass (m = ρV), we get:

u = (1/2)wmδ

Therefore, we can write the relationship between strain energy (u) and work done (W) as:

u = (1/2)wmδ = (1/2)Wδ = (1/2)Fd

where F is the applied force and d is the displacement.

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