# OPPOSITE SHOULDER FILLETS IN A STEPPED BAR

Opposite shoulder fillets in stepped flat bar. Stress concentration factors (Kt) calculator for tension, bending and torsion loads.

 INPUT PARAMETERS Parameter Value Thickness of stepped section  [D] mm cm m inch ft Thickness of flat section [d] Radius [r] Length of stepped section [L] Width of bar[t] Tension force [P] N kN lbf Bending moment [M] N*m lbf*in lbf*ft

Note: Use dot "." as decimal separator.

 RESULTS LOADING TYPE - TENSION Parameter Value Stress concentration factor [Kt] * --- --- Nominal tension stress at flat bar [σnom ] o --- MPa psi ksi Maximum tension stress due to tension load [σmax ] --- LOADING TYPE - BENDING Parameter Value Stress concentration factor [Kt] * --- --- Nominal tension stress at flat bar [σnom ] + --- MPa psi ksi Maximum tension stress due to bending [σmax ] ---

Note 1: * Geometry rises σnom by a factor of Kt. (Kt = σmaxnom)

Note 2: o σnom= P/(td) (Nominal tension stress occurred due to tension load)

Note 3: + σnom = 6M/(td2) (Nominal tension stress occured due to bending)

### 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 $$0.1\le \frac { h }{ r } \le 2.0$$ $$2\le \frac { h }{ r } \le 20$$ $${ C }_{ 1 }=1.006+1.008\sqrt { h/r } -0.044h/r$$ $${ C }_{ 1 }=1.020+1.009\sqrt { h/r } -0.048h/r$$ $${ C }_{ 2 }=-0.115-0.584\sqrt { h/r } +0.315h/r$$ $${ C }_{ 2 }=-0.065-0.165\sqrt { h/r } -0.007h/r$$ $${ C }_{ 3 }=0.245-1.006\sqrt { h/r } -0.257h/r$$ $${ C }_{ 3 }=-3.459+1.266\sqrt { h/r } -0.016h/r$$ $${ C }_{ 4 }=-0.135+0.582\sqrt { h/r } -0.017h/r$$ $${ C }_{ 4 }=3.505-2.109\sqrt { h/r } +0.069h/r$$ $${ K }_{ t }={ C }_{ 1 }+{ C }_{ 2 }\frac { 2h }{ D } +{ C }_{ 3 }{ (\frac { 2h }{ D } ) }^{ 2 }+{ C }_{ 4 }{ (\frac { 2h }{ D } ) }^{ 3 }$$ where $$\frac { L }{ D } >-1.89(\frac { r }{ d } -0.15)+5.5$$ $${ \sigma }_{ nom }=P/td$$ $${ \sigma }_{ max }={ K }_{ t }{ \sigma }_{ nom }$$ Bending $$0.1\le \frac { h }{ r } \le 2.0$$ $$2\le \frac { h }{ r } \le 20$$ $${ C }_{ 1 }=1.006+0.967\sqrt { h/r } +0.013h/r$$ $${ C }_{ 1 }=1.058+1.002\sqrt { h/r } -0.038h/r$$ $${ C }_{ 2 }=-0.270-2.372\sqrt { h/r } +0.708h/r$$ $${ C }_{ 2 }=-3.652+1.639\sqrt { h/r } -0.436h/r$$ $${ C }_{ 3 }=0.662+1.157\sqrt { h/r } -0.908h/r$$ $${ C }_{ 3 }=6.170-5.687\sqrt { h/r } +1.175h/r$$ $${ C }_{ 4 }=-0.405+0.249\sqrt { h/r } -0.200h/r$$ $${ C }_{ 4 }=-2.558+3.046\sqrt { h/r } -0.701h/r$$ $${ K }_{ t }={ C }_{ 1 }+{ C }_{ 2 }\frac { 2h }{ D } +{ C }_{ 3 }{ (\frac { 2h }{ D } ) }^{ 2 }+{ C }_{ 4 }{ (\frac { 2h }{ D } ) }^{ 3 }$$ where $$\frac { L }{ D } >-2.05(\frac { r }{ d } -0.025)+2$$ $${ \sigma }_{ nom }={ 6M }/{ t{ d }^{ 2 } }$$ $${ \sigma }_{ max }={ K }_{ t }{ \sigma }_{ nom }$$