Subsections

• Pressure
• Von Mises Stress
• Plastic Yield
• Principal Stresses
• Principal Strains
• Principal Shears
• Factor of Safety

Pre-defined Stress Calculations

  

For the general 3D case:

$$\mbox{Stress} = \left( \begin{array} {ccc} Sxx & Sxy & Sxz \\ Sxy & Syy & Syz \\ Sxz & Syz & Szz \\ \end{array} \right)$$

$$= - Pressure \left( \begin{array} {ccc} 1 & 0 & 0 \\ 0 & 1... ...{\prime} & Syz^{\prime} & Szz^{\prime} \\ \end{array} \right)$$

Where:

$$Sxx^{\prime} = (2*Sxx - Syy - Szz)/3.0$$

$$Syy^{\prime} = (2*Syy - Szz - Sxx)/3.0$$

$$Szz^{\prime} = (2*Szz - Sxx - Syy)/3.0$$

$$Sxy^{\prime} = Sxy$$

$$Sxz^{\prime} = Sxz$$

$$Syz^{\prime} = Syz$$

Pressure

 

$$\mbox{Pressure} = - (Sxx + Syy + Szz)/3.0$$

Von Mises Stress

 

Von Mises $$= \sqrt{1.5S_{ij^{\prime}}S_{ij^{\prime}}}$$

where:

$$S_{ij^{\prime}}S_{ij^{\prime}} = Sxx^{\prime^{2}} + Syy^{\pr... ... + 2.0(Sxy^{\prime^{2}} + Syz^{\prime^{2}} + Sxz^{\prime^{2}})$$

Plastic Yield

 

$$\mbox{Plastic Yield} = S_{ij^{\prime}}S_{ij^{\prime}} - 2Y^{2}/3.0$$

Where:

Y = Yield strength

Principal Stresses

 

Principal stresses P1, P2 and P3 are the ordered roots of the equation defined by

$$\left\vert \begin{array} {ccc} Sxx - \lambda & Sxy & Sxz \\... ... \\ Sxz & Syz & Szz - \lambda \\ \end{array} \right\vert = 0$$

Such that P1 $\gt$ P2 $\gt$ P3.

Principal Strains

 

For mathematical strains, the principal strains P1, P2 and P3 are the ordered roots of the equation defined by

$$\left\vert \begin{array} {ccc} Exx - \lambda & Exy & Exz \\... ... \\ Exz & Eyz & Ezz - \lambda \\ \end{array} \right\vert = 0$$

Such that P1 $\gt$ P2 $\gt$ P3.

Engineering strain is defined as $\gamma$ ij = 2*Eij.

So for engineering strains, the principal strains P1, P2 and P3 are the ordered roots of the equation defined by

$$\left\vert \begin{array} {ccc} Exx - \lambda & 0.5*Exy & 0.... ...*Exz & 0.5*Eyz & Ezz - \lambda \\ \end{array} \right\vert = 0$$

Such that P1 $\gt$ P2 $\gt$ P3.

It is this second equation that is used by RESULTS CALCULATE P-ESTRAIN

Principal Shears

 

Principal shears Q1, Q2, Q3 and QMAX

These are such that:

Q1 = 0.5 (P1 - P3)

Q2 = 0.5 (P1 - P2)

Q3 = 0.5 (P2 - P3)

QMAX = Q1

= Q3 ...(if absolute value of P1 = 0.0)

= Q2 ...(if absolute value of P3 = 0.0)

Factor of Safety

FS(t) = Fatigue Strength at temperature `t'

UTS(t) = Ultimate Tensile Strength at temperature `t'

P-Mean = Principal stress calculated from mean stress

$$\Vert P-Alt \Vert$$ = Absolute value of principal stress calculated from alternating stress

If $-UTS(t) < P-Mean < (-UTS(t) + FS(t))$ then: $FOS = \frac{UTS(t) + P-Mean}{\Vert P-Alt \Vert}$

If $(-UTS(t)+FS(t)) \leq P-Mean \leq 0$ then: $FOS = \frac{FS(t)}{\Vert P-Alt \Vert}$

If $0 < P-Mean < UTS(t)$ then: $FOS = \frac {FS(t) * (UTS(t) - P-Mean)} {(UTS(t) * \Vert P-Alt \Vert)}$

If $\Vert P-Mean \Vert \geq UTS(t)$ then: FOS = 0