5 octahedral shear stress theory

Octahedral Shear Stress theory

1. OCTAHEDRAL SHEAR STRESS THEORY Presented By- Laxman G Tanpure (152040014) Rohan Pendke (152040015) Manoj D Mahajan (152040016)

2. Octahedral Plane The octahedral plane is the plane which is equally inclined to all the three principal axes of reference Since Hydrostatic stress alone does not cause yeilding , we can find a material plane , called as octahydral plane, where the stress state can be decoupled into dilation strain energy and distortion strain energy

3. Octahedral Plane For octahedral planes which are equally inclined to the reference principal axes, they have the direction cosine of nx , ny and nz all equal. So nx = ny = nz =±(1/ squre root of 3)

4. T he stresses which act, the normal stress, the shearing stresses which act on these octahedral planes, we call those stresses as octahedral stresses. A t any plane which is having outward normal and the resulting stress in the x-direction, y-direction and in the z direction , can be represented in terms of the normal rectangular stress components : σ x , σ y ,σ z and the corresponding shear stresses: τxy , τyx , and τzx along with the direction cosines nx , ny , and nz .

5. Tn -resultant of vector σ n and vector τ n Tn = R x⒤+ R y ⒥+ R z ⒦ Here Rx= σ xnx + τ xyny + τ xznz Ry = τ yxnx + σ yny + τ yznz Rz = τ zxnx + τ zyny + σ znz

6. If we consider the plane perpendicular to x direction then, σ x= σ 1 and τ xy = τ xz =0 Similaraly σ y = σ 2 and τ yx = τ yz =0 for plane perpendicular to y direction Again σ z = σ 3 and τ z x = τ zy =0 for plane perpendicular to z direction As the planes are perpendicular x,y,z direction respectively so only principle stress acts.

7. As σ x= σ 1, σ y= σ 2 and σ z= σ 3 and τ xy = τ xz = τ yz =0 Therefore we get as Rx= σ 1nx R y = σ 2ny Rz = σ 3nz Now magnitude of Tn =square root of(( σ 12+ σ 2 2+ σ 3 2 )/3).

8. Vector Tn = σ 1nxi+ σ 2nyj+ σ 3nzk Which is resultant vector of σ n and τ n. Again σ n= Tn.n n is normal vector in the direction of σ n. Therefore σ n= σ 1*nx^2+ σ 2*ny^2= σ 3*nz^2 => σ oct =( σ 1+ σ 2+ σ 3)/ 3 As nx = ny = nz = =±(1/ squre root of 3)

9. From diagram, Tn2= σ n2+ τ n2 τ n2= Tn2- σ n2 And we get shear stress on octahedral plane as τ n= (1/3)*Square root of(( σ 1- σ 2 )2+( σ 1- σ 3 )2+( σ 2 - σ 3 )2)

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