Specific compression energy

Demolded Proctor body after the compressive strength test (material: drainable, mineral, trass-modified bedding mortar ) .

Demolded Proctor body after the compressive strength test (material: drainable, mineral bedding mortar).

The specific compression energy W (or volume-related compression work) is a common specification in the manufacture of Proctor bodies (see Proctor density ). It is given in MNm / m³.

Explanations

The production of a Proctor body takes place under standardized (DIN 18127 [2.93]) conditions, so the method of operation is specified and must be strictly adhered to. In the following, the specific compression energy in the manufacture of a Proctor body is derived and calculated as an example. With a simple Proctor density ρ _{Pr, there} are 25 beats per layer filled, with 3 layers that is 75 beats. A test cylinder with a diameter d = 100 mm and a height h ₁  = 120 mm is assumed . The mass of the falling weight of the compactor is G = 2.5 kg . The path (free fall) of the weight is h ₂  = 30 cm .

calculation

${\displaystyle W_{\rho _{\Pr }}\approx (Schlaege)\cdot \left({\frac {G(kg)\cdot h_{2}(m)\cdot 9,81(m/s^{2})}{V_{\dim }(m^{3})}}\right)\cdot 10^{-6}\;[MNm/m^{3}]}$

The following specific energy results when manufacturing a Proctor body:

${\displaystyle W_{\rho _{\Pr }}\approx 75\cdot \left({\frac {2,5\cdot 0,3\cdot 9,81}{\pi \cdot 0,05^{2}\cdot 0,12}}\right)\cdot 10^{-6}=0,585\approx {\underline {\underline {0,6MNm/m^{3}}}}}$

This can be converted at will. For example, if you want to calculate the volume-related compaction work during the incorporation of a paving stone into the bedding layer (see paving ), you can proceed as follows:

Assumptions made

• Bedding height is 5 cm (when compacted)
• The area of ​​the paving stone is 225 cm² (a / b = 15/15)
• 3 blows from a height of 22 cm with a 2.0 kg hammer (your own strength - when striking - is neglected here, as if the hammer were falling!).

${\displaystyle W\approx 3\cdot \left({\frac {2,0\cdot 0,22\cdot 9,81}{0,05\cdot 0,15^{2}}}\right)\cdot 10^{-6}\approx {\underline {\underline {0,004MNm/m^{3}}}}}$

conversion

As an example, the conversion from MNm / m³ to ft-lbf / ft³ should be carried out:

${\displaystyle {\begin{array}{l}0,6MNm/m^{3}=600.000Nm/m^{3}\\1,356Nm\cong 1,0\;ft-lbf\\\end{array}}}$
such as
${\displaystyle 1,0ft^{3}\cong 0,02832m^{3}}$

calculation

${\displaystyle W\approx {\frac {600.000\cdot 0,02832}{1,356}}\approx {\underline {\underline {12500ft-lbf/ft^{3}}}}}$

0.6 MNm / m³ therefore corresponds to approximately 12500 ft-lbf / ft ³ (ft-lbf = " foot-pound- force")

Web links

• ASTM International

Individual evidence

1. ↑ DIN EN 13286-2: 2013-02 Appendix A // Unbound and hydraulically bound mixtures - Part 2: Laboratory test method for determining the reference dry density and the water content - Proctor test