# STRESS CONCENTRATION FACTORS FOR OPPOSITE SHOULDER FILLETS IN A STEPPED BAR

Opposite shoulder fillets in stepped flat bar stress concentration factors (Kt) can be calculated by this calculator for tension, bending and torsion loads. See footnotes of the "Results" table for the necessary equations for the stress calculations.

Calculated theoretical stress concentration factors can be used to predict maximum tension and shear stresses for the bar.

### Opposite Shoulder Fillets in Stepped Flat Bar:

 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)

### List of Equations:

 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 $$\frac { L }{ D } >-2.05(\frac { r }{ d } -0.025)+2$$ $${ \sigma }_{ nom }={ 6M }/{ t{ d }^{ 2 } }$$ $${ \sigma }_{ max }={ K }_{ t }{ \sigma }_{ nom }$$