STRESS CONCENTRATION FACTORS FOR ECCENTRIC SINGLE CIRCULAR HOLE IN FINITE-WIDTH PLATE


Theoretical stress concentration factors (Kt) of eccentric single circular hole in finite width plate can be calculated by this calculator for tension and bending loads. The maximum stress values at the edge of plate and hole are calculated if the loading parameters known. See footnotes of the "Results" table for the formulas for the stress calculations.

There exist some validity conditions for the equations which are used in the calculations. If input parameters don't satisfy validity conditions of equations, a warning message is given by the calculator.

Eccentric Single Circular Hole in Finite Width Plate :

Stress concentration factors for central single circular hole in finite-width plate
 INPUT PARAMETERS
Parameter Value
Plate width [D]
Hole diameter [d]
Plate thickness [t]
Edge distance [c]
Distributed tension force [P]
Bending moment [M]

Note: Use dot "." as decimal separator.

 


 RESULTS
LOADING TYPE - TENSION
Stress concentration factors for central single circular hole in finite-width plate under tension
Parameter Value
Stress concentration factor [Kt] * --- ---
Nominal tension stress [σnom] o ---
Maximum tension stress (at Point-B) [σmax] ---
LOADING TYPE - BENDING
Stress concentration factors for central single circular hole in finite-width plate under bending
Parameter Value
At Edge of Plate
Stress concentration factor at point - A  [KtA] * --- ---
Nominal tension stress [σnom ] + ---
Maximum tension stress (at Point-A) [σmax ] ---
At Edge of Hole
Stress concentration factor at point - B [KtB] * --- ---
Nominal tension stress [σnom] x ---
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)

Supplements:

Link Usage
Central Single Circular Hole in Finite Width Plate Stress concentration factors (Kt) of central single circular hole in finite width plate.

List of Equations:


Stress concentration factors for central single circular hole in finite-width plate
Tension
Stress concentration factors for central single circular hole in finite-width plate under 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
Stress concentration factors for central single circular hole in finite-width plate under 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
$${ C' }_{ 1 }=1.0286-0.1638(d/2c)+2.702{ \left( d/2c \right) }^{ 2 }$$
$${ C' }_{ 2 }=-0.05863-0.1335(d/2c)-1.8747{ \left( d/2c \right) }^{ 2 }$$
$${ C' }_{ 3 }=0.18883-0.89219(d/2c)+1.5189{ \left( d/2c \right) }^{ 2 }$$
$${ K }_{ tA }={ C' }_{ 1 }+{ C' }_{ 2 }\frac { c }{ e } +{ C' }_{ 3 }{ (\frac { c }{ e } ) }^{ 2 }$$
$${ \sigma }_{ nom }={ 6M }/{ t{ D }^{ 2 } }$$
$${ \sigma }_{ A }={ K }_{ tA }{ \sigma }_{ nom }$$

Reference: