Error in finding shear force -shear strain curve

LAMMPS LAMMPS General Discussion
1

Respected Lammps users, I am using lammps version 2023 mpi on windows 10. I am facing problem of shear force doesnt graphically starts from zero rather than it starts increasing from 2.5 MPa. i am doing it with applying certain commands. I am solving the shear strength of molten and then solidified copper laminate on diamond surface. The way I apply the shear is by using velocity set but translate the whole copper block as whole and not by creating top and bottom velocity sets seperately as one of the example in lammps for shearing. Doing this way the chunk of copper is displaced as a bulk and sheared at copper-diamond interface where diamond block is stationery. Kindly assist me solving this problem along with how to calculate shear strain as I am calculating it using shear displacement/the interatomic gap between the copper and diamond which just 2 angstrom. Is this way right? In addition some lammps issues on the platform used xy tilt/normal length. What kind of xy tilt quantifies in terms of strain variables? I am using these commands in manuscript to find it.

compute p all reduce sum c_peratom[1] c_peratom[2] c_peratom[3]
#variable press atom -(c_stress[1]+c_stress[2]+c_stress[3])/(3.0*vol*${fraction})


variable p1 equal “-pxx/10000”
variable p2 equal “-pyy/10000”
variable p3 equal “-pzz/10000”
variable p4 equal “-pxy/10000”
variable p5 equal “-pyz/10000”
variable p6 equal “-pxz/10000”


variable vonM atom sqrt(((v_p1-v_p2)^2+(v_p2-v_p3)^2+(v_p3-v_p1)^2+6*(v_p4^2+v_p5^2+v_p6^2))/2)
variable p7 equal “v_vonM”
#variable von atom sqrt(((c_stress[1]-c_stress[2])^2+(c_stress[2]-c_stress[3])^2+(c_stress[3]-c_stress[1])^2+6*(c_stress[4]^2+c_stress[5]^2+c_stress[6]^2))/2)


#fix		3 mobile temp/rescale 10 300.0 300.0 10.0 1.0
#fix_modify	3 temp new2d

fix def_print all print 100 "${p1} ${p2} ${p3} ${p4} ${p5} ${p6}" file mg001.defo.txt screen no
fix statdump all print 1 "${p1} ${p2} ${p3} ${p4} ${p5} ${p6}" screen no file dump.stat

dump 1 all custom 5 dump.indent_loading.*.custom id type x y z  c_peratom[1] c_peratom[2] c_peratom[3]
dump 1 all custom 5 dump.indent_loading.*.custom id type x y z  v_vonM

dump shearstress all atom 100 m.facetshearstress id type  v_vonM

dump shear all atom 100 m.facetshearcontour

run		100000```

Thanks for cooperation
2

Impossible to answer with this limited information. But note that 2.5 MPa does not seem a high stress. Furthermore, when equilibrating with NPT and target pressure 0.0 also the normal stresses will not be zero in an individual timestep. Instead, the (time) average stress will be close to zero.

1 Like