Elasticity/Stress example 4

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Example 4

Given:

The octahedral plane is the plane that is equally inclined to the directions of the three principal stresses. For any given stress of state there are eight such planes.

Show:

  1. The normal traction on an octahedral plane is given by
𝐭noct=13(σ1+σ2+σ3)=13Iσ.
  1. The projected shear traction on an octahedral plane is given by
𝐭soct=13(σ1σ2)2+(σ2σ3)2+(σ3σ2)2=132Iσ26IIσ.

Here (σ1,σ2,σ3) are the principal stresses and (Iσ,IIσ) are the first two invariants of the stress tensor (σ).

Solution

Let us take the basis as the directions of the principal stresses 𝐧^1, 𝐧^2, 𝐧^3. Then the stress tensor is given by

[σ]=[σ1000σ2000σ3]

If 𝐧^o is the direction of the normal to an octahedral plane, then the components of this normal with respect to the principal basis are no1, no2, and no3. The normal is oriented in such a manner that it makes equal angles with the principal directions. Therefore, no1=no2=no3=no. Since no12+no22+no32=1, we have no=1/(3).

The traction vector on an octahedral plane is given by

𝐭o=𝐧^o[σ]=noσ1,noσ2,noσ3

The normal traction is,

N=𝐭o𝐧^o=no2σ1+no2σ2+no2σ3

Now, Iσ=(σ1+σ2+σ3). Therefore,

N=(1/3)(σ1+σ2+σ3)=(1/3)Iσ

The projected shear traction is given by

S=𝐭o𝐭oN2

Therefore,

S=no2σ12+no2σ22+no2σ32(1/9)(σ1+σ2+σ3)2

Also,

IIσ=σ1σ2+σ2σ3+σ3σ1

If you do the algebra for S, you will get the required relations.

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