# ECCENTRIC SINGLE CIRCULAR HOLE IN FINITE-WIDTH PLATE

Eccentric single circular hole in a plate. Stress concentration factors calculator (Kt) for tension and bending loads.

 INPUT PARAMETERS Parameter Value Plate width [D] mm cm m inch ft Hole diameter [d] Plate thickness [t] Edge distance [c] Distributed 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 [σnom] o --- MPa psi ksi Maximum tension stress (at Point-B) [σmax] --- LOADING TYPE - BENDING Parameter Value At Edge of Plate Stress concentration factor at point - A  [KtA] * --- --- Nominal tension stress [σnom ] + --- MPa psi ksi Maximum tension stress (at Point-A) [σmax ] --- At Edge of Hole Stress concentration factor at point - B [KtB] * --- --- Nominal tension stress [σnom] x --- MPa psi ksi Maximum tension stress (at Point-B) [σmax ] ---

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

Note 2: o For the formula, check List of Equation section.

Note 3: + σnom  = 6M/[tD2] (Nominal tension stress at the edge of plate due to bending)

Note 4: x σnom = 6M/[tD2] (Nominal tension stress at the edge of hole due to bending)

Note 5: KtA  = (σmaxnom) Theoretical stress concentration factor at point A in elastic range

Note 6: KtB  = (σmaxnom) Theoretical stress concentration factor at point B in elastic range

### 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 $${ K }_{ t }=3.000-3.140\frac { d }{ 2c } +3.667{ \left( \frac { d }{ 2c } \right) }^{ 2 }-1.527{ \left( \frac { d }{ 2c } \right) }^{ 3 }$$ $${ \sigma }_{ nom }=\frac { P\sqrt { 1-{ \left( d/2c \right) }^{ 2 } } }{ Dt(1-d/2c) } \frac { 1-c/D }{ 1-(c/D)\left[ 2-\sqrt { 1-{ (d/2c) }^{ 2 } } \right] }$$ $${ \sigma }_{ max }={ \sigma }_{ B }={ K }_{ t }{ \sigma }_{ nom }$$ Bending For Point B $$0\le d/2c\le 0.5,\quad 0\le c/e\le 1.0$$ $${ C }_{ 1 }=3.000-0.631(d/2c)+4.007{ \left( d/2c \right) }^{ 2 }$$ $${ C }_{ 2 }=-5.083+4.067(d/2c)-2.795{ \left( d/2c \right) }^{ 2 }$$ $${ C }_{ 3 }=2.114-1.682(d/2c)-0.273{ \left( d/2c \right) }^{ 2 }$$ $${ K }_{ tB }={ C }_{ 1 }+{ C }_{ 2 }\frac { c }{ e } +{ C }_{ 3 }{ (\frac { c }{ e } ) }^{ 2 }$$ $${ \sigma }_{ nom }={ 6M }/{ t{ D }^{ 2 } }$$ $${ \sigma }_{ B }={ K }_{ tB }{ \sigma }_{ nom }$$ For Point A $${ C' }_{ 1 }=1.0286-0.1638(d/2c)+2.702{ \left( d/2c \right) }^{ 2 }$$ $${ C' }_{ 2 }=-0.05863-0.1335(d/2c)-1.8747{ \left( d/2c \right) }^{ 2 }$$ $${ C' }_{ 3 }=0.18883-0.89219(d/2c)+1.5189{ \left( d/2c \right) }^{ 2 }$$ $${ K }_{ tA }={ C' }_{ 1 }+{ C' }_{ 2 }\frac { c }{ e } +{ C' }_{ 3 }{ (\frac { c }{ e } ) }^{ 2 }$$ $${ \sigma }_{ nom }={ 6M }/{ t{ D }^{ 2 } }$$ $${ \sigma }_{ A }={ K }_{ tA }{ \sigma }_{ nom }$$