Stress Concentration Factor Calculator

What a stress concentration factor is

A stress concentration factor, written Kt, is the ratio between the highest stress at a change in shape and the stress from an average load on the same section. In short, Kt = sigma max / sigma nom, and it depends on geometry alone. Double the hole and Kt changes; change the alloy and Kt does not.

The classical case was solved in 1898 by Ernst Kirsch, who worked out the elastic stresses around a round hole in an infinitely large plate. His answer is a peak of three times the far field stress at the hole edge, so Kt = 3, which is the number quoted for a small hole in a wide plate. The University of Washington stress concentrations notes walk through the Kirsch solution, its elastic assumptions and the handbooks that tabulate the finite width corrections.

Three words in the definition carry weight. The factor is theoretical, so it assumes elastic material with no local yielding. The stress it multiplies is nominal, the average on a reference section you choose, and different choices give different Kt values for the same part. The peak it predicts is local, which is why a part can survive a Kt of 3 without failing.

Handbook values come from Peterson's Stress Concentration Factors and from Roark's Formulas for Stress and Strain, built on decades of photoelastic and numerical work.

A central hole in a plate of finite width

Real plates have edges, and the edges change the stress field, so the factor shifts as the hole takes up more of the width. The common fit for a central hole in a plate in tension uses the ratio x = d / W, with d the hole diameter and W the plate width:

Kt = 3 - 3.14 x + 3.667 x squared - 1.527 x cubed

The fit returns 3.00 as x goes to zero, so it agrees with the Kirsch result for a small hole, and it falls as the hole grows because the net section carries the same load over less material. It is used for x up to 0.5, where the plate has lost half its width to the hole. The formula and its use appear in the University of Washington finite element course material, which pairs it with the peak stress relation sigma x,max = Kt p w / (w - d), and in the fracture mechanics reference that derives the net section nominal stress from the same geometry.

At d/W = 0.1 the net factor is 2.721, at 0.3 it is 2.347, at 0.5 it is 2.156. The drop is slow because the edges only matter once the hole is a fair share of the width.

Net section or gross section: pick one and stay with it

The same hole can be quoted two ways, and mixing them up is the most common error in this family of calculations. The net section nominal stress divides the load by the material left at the hole, so sigma nom = F / ((W - d) t). The gross section nominal stress divides by the full width, so sigma nom = F / (W t).

The net section stress is always the larger of the two, so the net section factor is always the smaller one. They are linked by a single line: Kt gross = Kt net / (1 - d/W). At d/W = 0.3 the net factor of 2.347 becomes 3.353 gross, and at 0.5 it becomes 4.312. The peak stress is the same either way, which is the check to apply to any number in this family.

Design codes quote hole factors on the net section, since that is the stress that has to stay below yield, while older tabulations often use the gross section. Note which one a number refers to before comparing it.

Worked example: a 30 mm hole in a 100 mm plate

Take a 100 mm wide plate, 5 mm thick, pulled by 12,000 N, with a 30 mm hole in the middle. The ratio is x = 30 / 100 = 0.3, so Kt = 2.347. The remaining material is 70 mm across, the net area is 350 square mm, and the nominal stress is 12,000 / 350 = 34.29 MPa. The peak at the hole edge is 2.347 x 34.29 = 80.46 MPa.

On the gross section the same load gives 12,000 / (100 x 5) = 24.00 MPa, and the factor that pairs with it is 3.353, which lands on the same 80.46 MPa. That is the check that the two conventions agree.

The imperial version is worth doing once by hand: ratio 0.3, net area 2.8 x 0.2 = 0.56 square inches, nominal stress 2,700 / 0.56 = 4,821 psi, peak 2.347 x 4,821 = 11,315 psi. In metric that peak is 78.01 MPa, and the gap against 80.46 MPa is rounding in the inch figures.

Opposite semicircular edge notches

The second geometry comes from Table 6-1 of Roark's Formulas for Stress and Strain, the case for opposite single U-shaped notches in a finite width plate in tension. The notch is semicircular when the depth h equals the root radius r, and the handbook gives a closed form with the ratio x = 2h / D, where D is the plate width and 2h is the depth removed from both edges:

Kt = 3.065 - 3.472 x + 1.009 x squared + 0.405 x cubed

The reference stress is the net section, so sigma nom = F / ((D - 2h) t). As x goes to zero the factor returns 3.065, the value for a shallow semicircular notch, and it falls as the notches grow because the ligament narrows.

Worked through: an 80 mm wide plate, 6 mm thick, pulled by 20,000 N, with 8 mm deep semicircular notches on both edges. The ratio is x = 16 / 80 = 0.2, so Kt = 2.414. The ligament is 80 - 16 = 64 mm, the net area is 64 x 6 = 384 square mm, the nominal stress is 20,000 / 384 = 52.08 MPa, and the peak at the notch root is 2.414 x 52.08 = 125.74 MPa. On the gross section the same part reads 3.018 and 41.67 MPa.

Two limits matter. The closed form assumes semicircular notches, so h = r exactly; a sharper notch needs the general coefficient set in the same table. And the calculator holds 2h/D at or below 0.5, which keeps the ligament at least as wide as the notch depth.

What Kt does not tell you

The biggest gap is material behavior. Kt assumes elastic material, so a ductile steel that yields locally at the hole sheds the peak into the surrounding material. The peak is a starting point, not a failure load, and for static loads on ductile metals the notch sensitivity q is often taken as zero.

Fatigue is different. The effective factor is Kf = 1 + q(Kt - 1), with q running from 0 for a material that ignores notches to 1 for full sensitivity. With the worked example factor of 2.347 and q = 0.8, Kf = 1 + 0.8 x 1.347 = 2.077. Sharper notches lower q, because a small radius produces a steep gradient and the material has less room to redistribute.

Thickness adds a third layer. The formulas here are two dimensional, which means a thin plate in plane stress. NASA Technical Paper 3192, published in 1992, ran three dimensional finite element cases for straight-shank and countersunk holes over hole radius to thickness ratios from 0.1 to 2.5, and found that the peak under remote tension can sit about 37 percent above the classical two dimensional value, with Kt = 3 recovered for much thicker or thinner plates. A thick block with a drilled hole can be worse than the flat plate formula suggests.

Surface finish, residual stress, weld toes and corrosion pits all act like a hole and none appear in these equations.

Where the calculator stops

Two geometries are implemented, the two the sources give as closed forms with a stated range: a central circular hole in a plate of finite width, and opposite semicircular edge notches in tension. Anything else has to come from a handbook chart or finite element analysis, and the calculator will not guess.

Not implemented here: shoulder fillets, stepped shafts, keyways, elliptical holes, a single edge notch, bent or torsion loaded parts, rows of holes, bearing stresses under pins and any thick section effect. Most have charts in Peterson's or in Roark's Table 6-1, and checking one against this page catches a misread axis.

Input ranges keep the numbers inside the published fits. The hole ratio d/W must be 0.5 or less, the notch ratio 2h/D must be 0.5 or less, and width, thickness and load must be positive. When a limit is reached the calculator says so instead of printing a precise looking number.

What the calculator returns

The default case is the worked example: a 100 mm wide, 5 mm thick plate with a 30 mm hole, pulled by 12,000 N, with the central hole geometry selected.

Switching to the notched plate gives the second example: an 80 mm wide, 6 mm thick plate with 8 mm deep notches and a 20,000 N load, for Kt = 2.414, 52.08 MPa nominal and 125.74 MPa peak.

Every value is computed in your browser and nothing is sent anywhere. The page is a study aid for two published formulas, not a replacement for the handbook charts.