Stress Drop in Earthquake Source Calculator
Estimate the static stress drop of an earthquake from its seismic moment and source radius, or from its corner frequency.
🌍 What is Stress Drop in an Earthquake Source?
Stress drop is the average amount by which the shear stress on a fault falls when an earthquake happens. Before the quake, tectonic loading builds up stress on the fault. During the rupture, the two sides of the fault slip past each other and part of that stored stress is released. The difference between the stress before and after is the static stress drop, and for most earthquakes it is only a few megapascals, roughly 30 times atmospheric pressure.
The simplest model of an earthquake source is a circular crack that slips by an average amount over a flat disc of radius r. For this model Eshelby showed in 1957 that the stress drop is 7/16 times the seismic moment divided by the cube of the radius. The seismic moment M0 is the rigidity times the fault area times the average slip, and it can be found from the moment magnitude Mw. This calculator accepts either one.
Seismologists use stress drop to compare earthquakes of very different sizes, to judge whether a fault is strong or weak, to calibrate ground-motion models, and to study induced seismicity from fluid injection. Stress drop controls how much high-frequency shaking an earthquake radiates, which matters for engineering design and hazard maps.
The radius of a small earthquake cannot be measured directly. It is usually inferred from the corner frequency of the S-wave spectrum, the frequency above which the spectral amplitude falls off. Brune (1970) related the two with r = 2.34 β / (2π fc). The second tab of this calculator applies that relation, then computes the stress drop from the result.
A common misconception is that large earthquakes release more stress. In fact, stress drop is almost independent of magnitude: a Mw 3 and a Mw 8 earthquake typically have a similar stress drop, and the larger one simply ruptures a far bigger area. Because radius is cubed, the result is also very sensitive to the radius estimate, so published stress drops carry large uncertainties of a factor of two or more.
📐 Formula
📖 How to Use This Calculator
Steps
💡 Example Calculations
Example 1 — Moderate Crustal Earthquake
Mw 6.0 with a 5 km source radius
Example 2 — Moment Given Directly in N·m
M0 = 1×1019 N·m with a 12 km source radius
Example 3 — Microearthquake from Corner Frequency
Mw 3.0, S-wave corner frequency 4 Hz, shear-wave speed 3.5 km/s
Example 4 — Moderate Event from Spectral Analysis
Mw 5.0, S-wave corner frequency 0.6 Hz, shear-wave speed 3.5 km/s
❓ Frequently Asked Questions
🔗 Related Calculators
What is stress drop in an earthquake?
Stress drop is the average difference between the shear stress on a fault before and after an earthquake. For a circular crack it equals 7/16 times the seismic moment divided by the cube of the source radius, and it is usually reported in megapascals (MPa) or bar.
How do you calculate earthquake stress drop from seismic moment and radius?
Use the Eshelby circular crack formula, stress drop = (7/16) * M0 / r^3, with M0 in newton metres and r in metres. A Mw 6.0 earthquake (M0 = 1.12e18 N m) with a 5 km radius gives about 3.93 MPa.
How do you convert moment magnitude to seismic moment?
Use M0 = 10^(1.5 * Mw + 9.05) in newton metres, the Hanks and Kanamori definition. Mw 6.0 corresponds to 1.12e18 N m, and every increase of one magnitude unit multiplies the moment by about 31.6.
How is source radius found from corner frequency?
Brune's relation gives r = 2.34 * beta / (2 * pi * fc), where beta is the shear-wave speed and fc is the S-wave corner frequency of the source spectrum. With beta = 3.5 km/s and fc = 0.6 Hz the radius is about 2.17 km.
What is a typical stress drop for an earthquake?
Most earthquakes have stress drops between about 0.1 and 10 MPa (1 to 100 bar), with a median near 3 MPa. Some events, especially intraplate and deep earthquakes, reach tens of MPa.
What is the difference between stress drop and apparent stress?
Stress drop is the change in static stress on the fault. Apparent stress is the radiated seismic energy divided by the seismic moment times the rigidity, and it measures how much energy was radiated per unit slip. They are related but not equal.
What is the difference between MPa and bar for stress drop?
One megapascal equals 10 bar, so a 3.93 MPa stress drop is 39.3 bar. Seismologists use both units, with MPa now more common in recent papers.
Why is the stress drop of small and large earthquakes so similar?
Seismic moment grows as radius cubed for earthquakes of similar stress drop. Observations show that moment and rupture size scale together over many orders of magnitude, so stress drop stays near 0.1 to 10 MPa while magnitude changes enormously.
What assumptions does the circular crack model make?
It assumes a flat circular rupture in an infinite uniform medium, with uniform stress drop across the crack and slip that is greatest at the centre. Real faults are rectangular, rough and heterogeneous, so the result is an average estimate.
Why does stress drop depend so strongly on source radius?
Radius enters the formula cubed. If a source radius estimate is 30 percent too small, the computed stress drop is about 2.9 times too large. This is the main source of scatter in published stress drop catalogs.
Does a high stress drop mean a more dangerous earthquake?
High stress drop earthquakes radiate more high-frequency energy for their size, which tends to increase shaking at short periods near the source. It does not by itself make an earthquake larger, since magnitude depends on moment.