Impact and Shock Load Factor Calculator
Impact and Shock Load Factor Calculator
The impact factor from energy — exactly 2 for a suddenly applied load, 1 + √(1 + 2h/δ) for a dropped one — with the same member computed stiffer two ways, because one helps and the other makes it worse.
Impact and shock load factor
A 1 kN weight — about 100 kg — dropped 25 mm onto the end of a steel rod 200 mm² in section and one metre long
One line of energy balance, and everything on the page falls out of it
- the derivation
- the weight falls h, then a further δ as the member gives way, so it does work W(h + δ). All of it goes into the member’s strain energy, ½ k δ². Substitute k = W/δ_st, and you have a quadratic in δ whose positive root is δ = δ_st(1 + √(1 + 2h/δ_st)). The impact factor is that bracket. Nothing else is needed and nothing else is assumed
- the δ in W(h + δ)
- the term people drop, and dropping it is what makes the factor come out as √(2h/δ_st) instead of the correct form. Keeping it is what makes n = 2 rather than 0 when h = 0. That check — put the drop height to zero and see whether you get two — is the fastest way to tell whether an impact formula has been written down correctly
- n = 2
- a suddenly applied load doubles the stress. Exactly two, for every member, every material and every load, because δ_st cancels completely at h = 0. The reason is that the load is at full value from the first instant while the member’s resistance builds up from zero, so the load does twice the work that a gradually applied one would, and the member has to store it
- δ_st in the denominator
- the consequence almost nobody states. A stiffer member has a smaller δ_st, which makes 2h/δ_st larger and the factor larger. Stiffening a member to make it stronger can raise its peak stress. Whether it does depends on how you stiffened it, and this page computes both ways
- U = σ²V/2E
- the strain energy an axial member can hold at a given stress is proportional to its VOLUME. That is why a long bolt survives a shock that snaps a short one of the same area: the same energy spread over more length is less strain everywhere. It is also why waisted bolts exist — reducing the shank below the thread root makes the whole shank strain together instead of concentrating it in the threads
Worked example
A 1 kN weight — about 100 kg — dropped 25 mm onto the end of a steel rod 200 mm² in section and one metre long
THE MEMBER FIRST, and only one number about it is needed. k = AE/L = 200 × 206,000 / 1,000 = 41,200 N/mm, so the static deflection under 1 kN is δ_st = 0.02427 mm. Twenty-four microns. That tiny number is about to do all the work
THE FACTOR. n = 1 + √(1 + 2h/δ_st) = 1 + √(1 + 2 × 25 / 0.02427) = 1 + √2,061 = 46.40. Check the formula by putting the drop height to zero: √1 = 1 and n = 2, exactly, which is the suddenly applied case and is the test that the algebra is right
WHAT THAT MEANS IN FORCE. The rod sees 46.40 × 1,000 = 46,398 N at the peak — forty-six kilonewtons from a hundred-kilogram weight falling one inch. The stress goes from a comfortable 5.0 MPa to 232 MPa, and the rod stretches 1.126 mm instead of 0.02427
THE ENERGY, as a check. The weight fell 25 + 1.126 mm and did 26.13 J of work. The rod stored ½ k δ² = 26.13 J. Same number, as it has to be. And the same energy written as σ²V/2E with V = 200,000 mm³ gives 26.13 J again
NOW THE PART THAT SURPRISES PEOPLE. Make the rod twice as stiff by HALVING ITS LENGTH. δ_st halves, the factor rises to 65.2, the static stress is unchanged at 5.0 MPa, and the peak stress becomes 326 MPa — 41 per cent WORSE. Stiffening the member made it fail sooner
AND THE OTHER WAY. Make it twice as stiff by DOUBLING ITS AREA instead. The factor rises to exactly the same 65.2, because the stiffness change is the same — but the static stress halves, and the peak stress becomes 163 MPa, 30 per cent BETTER. Identical stiffness, opposite outcome
THE RULE THAT RECONCILES THEM. Once the drop is much larger than the static deflection, the peak stress is √(2hWE/AL), and A and L appear only as their PRODUCT — the volume of metal that is straining. Here that limit gives 227 MPa against the exact 232. Shortening the rod removed volume; fattening it added volume. Stiffness was never the thing that mattered. If you want a member to survive impact, give it more material to spread the energy through, or give it more length — and if you can, put something soft in the load path, because raising δ_st is the only lever that acts on the factor itself
The impact factor against drop height, in units of the static deflection
| h / δ_st | n = 1 + √(1 + 2h/δ_st) | δ_max on this page’s default member (mm) | The √(2h/δ) approximation | How far that approximation is out (%) |
|---|---|---|---|---|
| 0.0 | 2.000 | 0.0485 | 0.000 | — |
| 0.5 | 2.414 | 0.0586 | 1.000 | -58.6 |
| 1.0 | 2.732 | 0.0663 | 1.414 | -48.2 |
| 2.0 | 3.236 | 0.0785 | 2.000 | -38.2 |
| 5.0 | 4.317 | 0.1048 | 3.162 | -26.7 |
| 10.0 | 5.583 | 0.1355 | 4.472 | -19.9 |
| 50.0 | 11.050 | 0.2682 | 10.000 | -9.5 |
| 100.0 | 15.177 | 0.3684 | 14.142 | -6.8 |
| 500.0 | 32.639 | 0.7922 | 31.623 | -3.1 |
| 1,000.0 | 45.733 | 1.1100 | 44.721 | -2.2 |
| 5,000.0 | 101.005 | 2.4516 | 100.000 | -1.0 |
The same member made stiffer, two different ways, on the page’s own default
| Stiffness × | δ_st (mm) | Impact factor n | Peak stress if stiffened by SHORTENING (MPa) | … relative | Peak stress if stiffened by ADDING AREA (MPa) | … relative |
|---|---|---|---|---|---|---|
| 0.25 | 0.09709 | 23.72 | 118.6 | 0.511 | 474.3 | 2.045 |
| 0.50 | 0.04854 | 33.11 | 165.5 | 0.714 | 331.1 | 1.427 |
| 1.00 | 0.02427 | 46.40 | 232.0 | 1.000 | 232.0 | 1.000 |
| 2.00 | 0.01214 | 65.20 | 326.0 | 1.405 | 163.0 | 0.703 |
| 4.00 | 0.00607 | 91.78 | 458.9 | 1.978 | 114.7 | 0.495 |
| 8.00 | 0.00303 | 129.38 | 646.9 | 2.788 | 80.9 | 0.349 |
Shock and fatigue factors for rotating shafts — a different tool for a different job
| Nature of the loading | K_m, bending | K_s, torsion |
|---|---|---|
| Gradually applied or steady | 1.5 | 1.0 |
| Suddenly applied, minor shock | 1.5 to 2.0 | 1.0 to 1.5 |
| Suddenly applied, heavy shock | 2.0 to 3.0 | 1.5 to 3.0 |
Twice, for nothing — and a stiffer member is worse, not better
A suddenly applied load produces exactly twice the static stress. Not approximately, not for steel, not under some conditions: exactly twice, for any member, any material and any load. It falls out of one line. The load is at full value from the instant it is released, while the member’s resistance builds up from zero as it deflects, so the load does twice the work a gradually applied one would. The member has to store all of it, and storing twice the energy in a linear spring means deflecting by twice as much and carrying twice the force. Put a drop height of zero into the formula on this page and you get 2, which is the check that the algebra is right — and it is a check that catches the commonest way of writing the formula down wrong, which is to forget that the weight keeps falling while the member gives way.
A stiffer structure is worse under impact, and that is the most useful thing on this page. The impact factor is 1 + √(1 + 2h/δ_st), and the static deflection is in the denominator. Halve the static deflection and the factor goes up. Stiffening a member to “make it stronger” therefore raises the factor it sees, every time. Whether the peak STRESS goes up or down depends entirely on how the stiffness was bought. Shorten the member and the static stress is unchanged while the factor rises, so the peak stress rises as roughly the square root of the stiffness. Fatten it and the static stress falls in proportion while the factor rises only as the square root, so the peak stress falls. Identical stiffness increase, opposite outcome, and the page computes both.
What reconciles them is volume. Once the drop is large compared with the static deflection, the peak stress in an axial member is √(2hWE/AL) — and the area and the length appear only as their product. The thing that absorbs energy is the volume of metal that is straining, and the stiffness is beside the point. That is also the cleanest statement of why a long bolt survives a shock that snaps a short one of the same area, why waisted bolts exist, and why a wire rope sling is kinder than a chain of the same rating. Strain energy per unit volume at a given stress is σ²/2E, so more volume at the same stress is more energy absorbed — and a lower modulus helps too, which is why aluminium and titanium absorb more energy per unit volume than steel at the same stress.
Service factors are a different tool and they are not interchangeable with this. The shock and fatigue factors published for rotating shafts run from 1.5 for a steady load to 3 for a heavy shock. Notice that a STEADY load already gets 1.5, which no energy calculation would ever produce: the factor is carrying a fatigue allowance for rotating bending as well as a shock allowance, and the bands are wide and overlapping because they encode judgement rather than mechanics. They are the right tool for “this drive is rough, how much margin should the shaft carry”. The energy method is the right tool for “this mass falls this far onto this member”. Do not multiply them together. The drive service factor page carries the drive families in full.
What the model assumes, and what it leaves out. It assumes the striking body is rigid and stays in contact; that no energy is lost in the blow, to sound, heat, local crushing or friction; that the member deflects in its static shape, so its own mass and inertia contribute nothing; that the material stays linear elastic; and that the whole member participates. Every one of those is wrong to some degree, and they mostly err on the safe side — energy that goes into noise, local denting or plastic deformation is energy the member does not have to store elastically. The one that does not err safely is the last. A real impact is a stress WAVE: for the first moments only the material near the struck end knows anything has happened, and the local stress there is far higher than the uniform value this page computes. So the closed form is a reasonable upper bound on the overall deflection and force, and it is NOT a prediction of local damage at the point of the blow.
Frequently asked questions
Why does a suddenly applied load double the stress?
Because the load is at full value from the first instant while the member’s resistance builds from zero as it deflects. The load does work Wδ; the member stores ½kδ². Setting them equal gives δ = 2W/k = 2δ_st, so both the deflection and the force are doubled. It is exactly two for every member and every material, because the stiffness cancels. The member then oscillates about the static position and settles there once damping has taken the energy out — but on its first swing it goes to twice the static deflection, and that is when it breaks if it is going to.
What is the impact factor for a dropped load?
n = 1 + √(1 + 2h/δ_st), with h the drop height and δ_st the deflection the member would have under the same load applied slowly. Multiply the static stress and the static force by it. Check the formula by putting h = 0: it gives exactly 2, the suddenly applied case, which is the fastest way to tell whether you have written it down correctly. The common simplification √(2h/δ_st) is within ten per cent once the drop is more than about fifty times the static deflection, and badly wrong below that.
Does making a part stiffer help under impact?
It raises the impact factor, always, because the factor depends on how far the member deflects and a stiff member deflects less. Whether the peak stress goes up or down depends on how you made it stiffer. Shortening it at the same section raises the peak stress by roughly the square root of the stiffness increase — halving the length costs 41 per cent more stress. Adding area at the same length lowers it by roughly the same factor. What actually matters is the volume of material that is straining: in the large-drop limit the peak stress is √(2hWE/AL) and A and L appear only as a product.
Why does a long bolt survive shock better than a short one?
Because strain energy at a given stress is proportional to volume: U = σ²V/2E. A longer bolt of the same area has more volume, so it can hold the same energy at a lower stress. Concretely, doubling a bolt’s grip length halves its stiffness, doubles its static deflection, and drops the peak stress by about thirty per cent for the same blow. It is also why waisted bolts exist: reducing the shank diameter below the thread root makes the whole shank strain evenly instead of concentrating the strain in the first few threads, which both adds effective volume and moves the peak away from the stress raiser.
Is a service factor the same as an impact factor?
No, and combining them will give you a nonsense. A service factor is a duty allowance: the ASME-derived table for rotating shafts gives 1.5 for a steady load, 1.5 to 2.0 for a minor shock and 2.0 to 3.0 for a heavy one, and the fact that a STEADY load already gets 1.5 tells you the number is carrying a fatigue allowance as well. An impact factor is a mechanics result for a specific mass falling a specific distance, and it routinely exceeds 20. Use the table when the duty is described in words; use the energy method when the drop is described in millimetres.
How accurate is the energy method for impact?
It is a reasonable upper bound on the overall deflection and a poor predictor of local damage. It assumes no energy is lost in the blow, which is conservative, and it assumes the member deflects in its static shape with no inertia of its own, which is conservative when the striking mass is much heavier than the member and optimistic when it is not. What it cannot see at all is the stress wave: a real blow loads the material near the point of impact long before the far end knows anything has happened, and the local stress there is far above the uniform value. It also assumes elasticity, and a real ductile member will yield and absorb energy plastically, which usually saves it.
How do I reduce an impact load?
Raise the static deflection, which is the only quantity the factor depends on, and add volume, which is what reduces the stress at a given factor. In practice: make the load path longer, put something soft in it, use a spring or a rubber element, use a lower-modulus material, use a longer or waisted bolt, or lower the load instead of dropping it. Ten times the static deflection cuts the factor by about three. Note what does NOT help: a stronger material of the same size, which has the same modulus, the same deflection, the same factor and the same peak stress. Strength lets a part survive an impact; it does not reduce one.
Related calculators
References
- S. Timoshenko, Strength of Materials, Part II: Advanced Theory and Problems, the chapter on impact and stress waves. Cited for the energy method used here and for its limits, which Timoshenko states plainly: the method assumes the striking body is rigid, that no energy is lost at the blow, and that the struck member deflects in its static shape — so it ignores the inertia of the member itself and it cannot see a stress wave at all. It is an upper bound on deflection and a lower bound on local damage.
- Richard Budynas and Keith Nisbett, Shigley’s Mechanical Engineering Design, the impact and energy methods sections. Cited for the standard form of the impact factor used here and for the resilience relation U = σ²V/2E, which is the reason a long bolt survives a shock that breaks a short one of the same area. Neither is reproduced: both are derived on the page from the energy balance, and the closed form is checked against a Runge-Kutta integration of the equation of motion.
- Stress Analysis Manual of the U.S. Air Force Flight Dynamics Laboratory, section 10, Table 10-4, “Values of Shock and Fatigue Factors for Rotating Shafts (from ASME Code)”, as published by Engineering Library (engineeringlibrary.org, read 29 September 2026). The three-row table reproduced on this page: Km 1.5 and Ks 1.0 for a gradually applied or steady load, Km 1.5 to 2.0 and Ks 1.0 to 1.5 for a suddenly applied minor shock, Km 2.0 to 3.0 and Ks 1.5 to 3.0 for a heavy one. The underlying ASME code for the design of transmission shafting has been withdrawn; the table survives it, and it is quoted here for what it is — a duty allowance, not an impact factor.
- Hertz’s own worked problems and every reference above use consistent SI: newtons, millimetres and megapascals, with 1 MPa = 1 N/mm². That identity is used throughout these four pages and is worth stating once, because the single commonest arithmetic error in contact and impact work is a factor of a thousand between N/m² and N/mm².