Sandstone porosity, cementation and strength

A pile of sand has no strength, and can be deformed with the brush of a hand. Sandstones have a great variety of strength. Certain horizons in the local Cretaceous Dakota sandstones, can be easily broken and crumbled by hand. Other horizons require a hammer and a good strong blow. A coloration difference in this case provides insight as to the cause of the difference - the amount of cementation. The more iron oxide cement the darker and the stronger the sandstone. From sandstone to sandstone the type of cementation will also be a significant factor. So it is useful to understand that more than one factor is at play. As cementation increases the porosity will decrease. A sand may have a porosity from 35-40% depending on a variety of factors, such as how the sand grains are arranged, the range of sizes, and the shapes of the grains. The source below provided some data to test the relationship between porosity and sandstone strength.

Original data taken from: Hale, P. A. & Shakoor, A., 2003, A laboratory investigation of the Effects of Cyclic Heating and Cooling, Wetting and Drying, and Freezing and Thawing on the Compressive Strength of Selected Sandstones: Environmental and Engineering geoscience, vol IX, p. 117-130. Table 3: "Mean values of engineering properties measured... "

x=porosity | y=unconfined strength | z scores for x | z score for y |

12.32 | 2636 | 1.0387907185144 | -1.04472102928152 |

13.94 | 3162 | 1.34550535350711 | -0.981827324710856 |

6.94 | 7580 | 0.020195202304047 | -0.453568034229502 |

4 | 16899 | -0.53643506120124 | 0.66070275074011 |

2.94 | 23739 | -0.737124884097704 | 1.47856004971984 |

0.86 | 14224 | -1.13093132902661 | 0.340853587761924 |

The above plot shows the linear regression done in Excel. The r-squared value indicates that porosity can predict 72% of the variance in strength. However, the statistical significance of this is still questionable, since there is an n of only 6. With a lower n your chances of getting a relationship with a high r-squared value by accident are increased.

The above plot shows the z values of porosity and strength plotted against each other. The slope of the line is -.849. This is a visual measure of the goodness of fit. Note that if you square it you have the same value as the R-squared computed by Excel and shown in the graph above. However, the regression line above suggests that when porosity is equal to about 15% the strength should be zero. Does this make sense. At what value of porosity might you expect the strength to approach 0? One simple conclusion is that the relationship is not a linear one. We can try a transformation and see if that helps.

x=porosity | y=unconfined strength | log strength |

12.32 | 2636 | 3.42094540592197 |

13.94 | 3162 | 3.49996186559619 |

6.94 | 7580 | 3.87966920563205 |

4 | 16899 | 4.22786100595501 |

2.94 | 23739 | 4.37546242044717 |

37 | 1 | 0 |

Did the transformation help? Note the much higher r-squared value. Remember that a standard deviation and mean work well to capture a normal distribution. It may be that by taking the log we normalized the population of y values. Note that letting y = to zero results in an x intercept of 54% porosity. This is not a reasonable initial porosity value for most sandstones.

In the above graph we have included a zero value of strength for a sand of typical porosity. the relationship improves further, but how does this influence the statistical interpretations you can make? Excel does give you the option of forcing the line to go through the origin.