Question

What is the relation between young's modulus bulk modulus and modulus of rigidity

Answer 1

I’ve altered the question slightly by changing “poison’s modulus” to “Poisson’s ratio,” since that’s what I think the questioner is referring to.

This could be answered by simply providing the equations relating these elastic constants. But that’s not as interesting as demonstrating how these equations arise. Thus I’ll do that starting with the equation tying together bulk modulus K, Poisson’s ratio ν, and elastic modulus E.

The generalized Hooke’s law for the x-, y- and z-directions is

ϵx=1E[σx−ν(σy+σz)]ϵx=1E[σx−ν(σy+σz)]

ϵy=1E[σy−ν(σz+σx)]ϵy=1E[σy−ν(σz+σx)]

and

ϵz=1E[σz−ν(σx+σy)]ϵz=1E[σz−ν(σx+σy)]

where ε is strain and σ is stress.

Adding up these three equations gives

(1) ϵx+ϵy+ϵz=1−2νE(σx+σy+σz)ϵx+ϵy+ϵz=1−2νE(σx+σy+σz)

Now note in the above equation that ϵx+ϵy+ϵzϵx+ϵy+ϵz is volume strain Δ and that σx+σy+σzσx+σy+σz is just 3σm3σm where σmσm is mean stress. Accordingly, Eq. (1) becomes

(2) Δ=1−2νE3σmΔ=1−2νE3σm

But since bulk modulus is defined as σmΔσmΔ, we have

(3) K=σmΔ=E3(1−2ν)K=σmΔ=E3(1−2ν)

which is the desired relation between K, E, and ν.

To develop the relation between G and E, imagine the two square elements in Fig. 1a below subjected to an axial load P in the x-direction. The square element with sides of unit length and oriented parallel to the x and y axes undergoes normal strain as depicted in Fig. 1b. As expected, the sides parallel to the x-direction elongate to length 1 + ϵxϵx whereas the sides parallel to the y-axis Poisson contract to length 1 – νϵxνϵx

The other square element, delineated by the dotted line, is oriented 45 degrees to the first square element. As such, it undergoes shear strain γ such that its right angles become 90° – γ and 90° + γ as shown in Fig. 1b. Therefore, axial loading creates not only normal strains but also shearing strains, which suggests a connection between shear modulus and elastic modulus.

To get that, consider the red-shaded element of Fig. 1b and reproduced in Fig. 1c. Comparing Figs. 1b and 1c reveals θ to be half of 90° – γ. Therefore, invoking the formula for the tangent of the difference of two angles and also that the tangent of a small angle is roughly equal to that angle (expressed in radians),

(4) tanθ=1−γ21+γ2tan⁡θ=1−γ21+γ2

But in Fig. 1c it’s also evident that

(5) tanθ=1−νϵx1+ϵxtan⁡θ=1−νϵx1+ϵx

Setting Eq. (4) equal to Eq. (5) and exploiting that ϵxϵx is much less than 1 then gives

(6) γ=(1+ν)ϵxγ=(1+ν)ϵx

Substituting into Eq. (6) γ=τGγ=τG, ϵ=σEϵ=σE, and σ=PAσ=PA (A being cross sectional area), plus also incorporating that a force balance gives τ=P2Aτ=P2A, yields the desired equation:

(7) G=E2(1+ν)G=E2(1+ν)

Equations (3) and (7) embody the interrelationship among G, E, K and ν. These equations contain all you need to solve for an elastic constant knowing the others. For example, say you were given G and K and wanted to know ν. You’d just solve Eqs. (3) and (7) for E and set them equal to each other, which then allows obtaining ν.