CENTRAL SINGLE CIRCULAR HOLE IN FINITE-WIDTH PLATE


Central single circular hole in finite width plate. Stress concentration factors (Kt) for tension, in-plane and simple transverse bending loads.

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]
Tension force [P]
In-plane bending moment [M]
Transverse bending moment [M1]

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 the edge of hole [σmax ] ---
LOADING TYPE - IN-PLANE BENDING
Stress concentration factors for central single circular hole in finite-width plate under bending
Parameter Value
At Edge of Hole
Stress concentration factor at point - A [KtA] * --- ---
Nominal tension stress [σnom ] + ---
Maximum tension stress (at Point-A) [σmax ] ---
At Edge of Plate
Stress concentration factor at point - B [KtB] * --- ---
Nominal tension stress [σnom] x ---
Maximum tension stress (at Point-B) [σmax ] ---
LOADING TYPE - SIMPLE TRANSVERSE BENDING
Stress concentration factors for central single circular hole in finite-width plate under simple transverse bending Stress concentration factors for central single circular hole in finite-width plate under simple transverse bending
Stress concentration factor at point A[KtA] * --- ---
Nominal tension stress [σnom ] # ---
Maximum tension stress (at Point-A) [σmax ] ---


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

Note 2: o σnom= P/[t(D-d)] (Nominal tension stress at the plate cross section due to tension load)

Note 3: + σnom  = 6Md/[t(D3-d3)] (Nominal tension stress at the edge of hole due to bending)

Note 4: x σnom = 6MD/[t(D3-d3)] (Nominal tension stress at the edge of plate due to bending)

Note 5: # σnom = 6M1/[t2(D-d)] (Nominal tension stress at the edge of plate due to bending)

Note 6: α=30°

Note 7: KtA  = (σmaxnom) Theoretical stress concentration factor at point A in elastic range

Note 8: 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:

Formulas:

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
For $$0\le \frac { d }{ D } \le 1$$
$$3.000-3.140\frac { d }{ D } +3.667{ \left( \frac { d }{ D } \right) }^{ 2 }-1.527{ \left( \frac { d }{ D } \right) }^{ 3 }$$
$${ \sigma }_{ nom }=P/[(D-d)t]$$
$${ \sigma }_{ max }={ K }_{ t }{ \sigma }_{ nom }$$
In-Plane Bending
Stress concentration factors for central single circular hole in finite-width plate under bending
At the edge of hole
Kta = 2 (independent of d/D)
$${ \sigma }_{ nom }=6Md/[({ D }^{ 3 }-{ d }^{ 3 })t]$$
σmax@A = Ktσnom
At the edge of plate
$${ K }_{ tb }=\frac { 2d }{ D } (\alpha ={ 30 }^{ \circ })$$
$${ \sigma }_{ nom }=6MD/[({ D }^{ 3 }-{ d }^{ 3 })t]$$
σmax@B = Ktσnom
Simple Transverse  Bending
Stress concentration factors for central single circular hole in finite-width plate under simple transverse bending
For $$0\le \cfrac { d }{ D } \le 0.3$$ and $$1\le d/t\le 7$$
$${ K }_{ t }=\left[ 1.793+\frac { 0.131 }{ d/t } +\frac { 2.052 }{ { \left( d/t \right) }^{ 2 } } -\frac { 1.019 }{ { \left( d/t \right) }^{ 3 } } \right] \times \left[ 1-1.04(d/D)+1.22{ (d/D) }^{ 2 } \right]$$
$${ \sigma }_{ nom }=6{ M }_{ 1 }/[(D-d){ t }^{ 2 }]$$
σmax@A = Ktσnom

Reference: