# LARGE CIRCUMFERENTIAL GROOVE ON A CIRCULAR SHAFT

Large circumferential groove on a circular shaft. Stress concentration factors (Kt) calculator for tension, bending and torsion loads.

 INPUT PARAMETERS Parameter Value Diameter of larger shaft section [D] mm cm m inch ft Diameter of smaller shaft section [d] Radius [r] Tension force [P] N kN lbf Bending moment [M] N*m lbf*in lbf*ft Torque [T]

Note:Use dot "." as decimal separator.

 RESULTS LOADING TYPE - TENSION Parameter Value Stress concentration factor[Kt] * --- --- Nominal tension stress at shaft [σnom ] o --- MPa psi ksi Maximum tension stress due to tension load (at Point-A) [σmax ] --- LOADING TYPE - BENDING Parameter Value Stress concentration factor [Kt] * --- --- Nominal tension stress at shaft [σnom ] + --- MPa psi ksi Maximum tension stress due to bending (at Point-A) [σmax ] --- LOADING TYPE - TORSION Parameter Value Stress concentration factor [Kt] ** --- --- Nominal shear stress at shaft [τnom ] x --- MPa psi ksi Maximum shear stress due to torsion (at Point-A) [τmax ] ---

Note 1: Maximum stress is occured at point A

Note 2: * Geometry rises σnom by a factor of Kt.  ( Kt = σmaxnom)

Note 3: ** Geometry rises τnom by a factor of Kt.  ( Kt = τmaxnom)

Note 4: o σnom= 4P/(πd2) (Nominal tension stress occurred due to tension load)

Note 5: + σnom = 32M/(πd3) (Nominal tension stress occured due to bending)

Note 6: x τnom = 16T/(πd3) (Nominal shear stress occureed due to  torsion)

### Definitions:

Stress Concentration Factor: Dimensional changes and discontinuities of a member in a loaded structure causes variations of stress and high stresses concentrate near these dimensional changes. This situation of high stresses near dimensional changes and discontinuities of a member (holes, sharp corners, cracks etc.) is called stress concentration. The ratio of peak stress near stress riser to average stress over the member is called stress concentration factor.

Kt: Theoretical stress concentration factor in elastic range = (σmaxnom)

### Formulas:

 Tension $$0.3\quad \le \quad \frac { r }{ d } \le \quad 1.0$$ , $$1.005\quad \le \quad \frac { D }{ d } \le \quad 1.10$$ $${ C }_{ 1 }=-81.39+153.10(D/d)-70.49{ (D/d) }^{ 2 }$$ $${ C }_{ 2 }=119.64-221.81(D/d)+101.93{ (D/d) }^{ 2 }$$ $${ C }_{ 3 }=-57.88+107.33(D/d)-49.34{ (D/d) }^{ 2 }$$ $${ K }_{ t }={ C }_{ 1 }+{ C }_{ 2 }\frac { r }{ d } +{ C }_{ 3 }{ (\frac { r }{ d } ) }^{ 2 }$$ $${ \sigma }_{ nom }=4P/\pi { d }^{ 2 }$$ $${ \sigma }_{ max }={ \sigma }_{ A }={ K }_{ t }{ \sigma }_{ nom }$$ Bending $$0.3\quad \le \quad \frac { r }{ d } \le \quad 1.0$$ , $$1.005\quad \le \quad \frac { D }{ d } \le \quad 1.10$$ $${ C }_{ 1 }=-39.58+73.22(D/d)-32.46{ (D/d) }^{ 2 }$$ $${ C }_{ 2 }=-9.477+29.41(D/d)-20.13{ (D/d) }^{ 2 }$$ $${ C }_{ 3 }=82.46-166.96(D/d)+84.58{ (D/d) }^{ 2 }$$ $${ K }_{ t }={ C }_{ 1 }+{ C }_{ 2 }\frac { r }{ d } +{ C }_{ 3 }{ (\frac { r }{ d } ) }^{ 2 }$$ $${ \sigma }_{ nom }=32M/\pi { d }^{ 3 }$$ $${ \sigma }_{ max }={ \sigma }_{ A }={ K }_{ t }{ \sigma }_{ nom }$$ Torsion $$0.3\quad \le \quad \frac { r }{ d } \le \quad 1.0$$ , $$1.005\quad \le \quad \frac { D }{ d } \le \quad 1.10$$ $${ C }_{ 1 }=-35.16+67.57(D/d)-31.28{ (D/d) }^{ 2 }$$ $${ C }_{ 2 }=79.13-148.37(D/d)+69.09{ (D/d) }^{ 2 }$$ $${ C }_{ 3 }=-50.34+94.67(D/d)-44.26{ (D/d) }^{ 2 }$$ $${ K }_{ t }={ C }_{ 1 }+{ C }_{ 2 }\frac { r }{ d } +{ C }_{ 3 }{ (\frac { r }{ d } ) }^{ 2 }$$ $${ \tau }_{ nom }=16T/\pi { d }^{ 3 }$$ $${ \tau }_{ max }=\tau _{ A }={ K }_{ t }{ \tau }_{ nom }$$