284 REPORTS ON THE STATE OP SCIENCE. — 1916. 



Guest's law does not apply to elastic limits as at present defined, but only 

 to the drop stresses. This is perhaps natural, for the drop or minimum 

 stress after the general breakdown is probably the natural resistance 

 of the material, whereas the elastic limits may have been affected by 

 preliminary strainings and by ageing effects. It should also be men- 

 tioned that the changes of curvature of the elastic linos are very much more 

 marked in the tension and compression cases T^ and C„ than in the shearing 

 (torsion) cases, for in these latter it is only the outer fibres of the samples 

 which are affected. 



Both in the tension and the compression experiments two strain 

 indicators were used and corrections were made in the final results for 

 eccentricity of pull. These corrections were less than 1 ton for the 

 tension tests. Duplicate tests on the same material demonstrated that 

 these corrections are necessary and that the methods adopted are fairly 

 correct. 



Note on Shearing Stress Strain Diagrams. 



The problem is to determine the shearing stress strain curve from the 

 torsion moment or stress strain curve. 



Assume that two similar cylindrical shells of the respective semi- 

 diameters X| and X and the thicknesses dx and dx are subjected to equal 

 circumferential shearing stresses S, then the respective torsion moments 

 rfM, and dM. stand in the relation 



dM,/dM. = x^'-'/x\ 



This relation holds good for a number of concentric cylindrical sheila 

 constituting a solid bar, provided of course that the stress distribution 

 is similar in each bar of the respective radii r^ and r. 



M,/M = riVr^ 

 Let r,=:r + ^r, then Ui=M{r + drjyr^=llfl+^'[''\. 



Assume that the smaller of the two rods of the radius ;• receives the 

 addition of a thin cylindrical shell of the thickness dr, then its diameter 

 will be the same as the other bar, and if this added cylinder be stressed 

 circumferentially with the stress S, which exists in its original outer 

 fibre, then the torsion moment M2 for this compounded bar is : — 



M2=M + 2n . Sr'' .dr. 



These two torsion moments M, and M, Avould be obtained with one and 

 the same bar if in the first one, the shear strain angle at the surface, 

 were a,, and if in the second it were : — 



dr\ ,, ,j dr 



a-2=ai{r + dr)/r=aJl + ^^j, then as Ja=a2 — a,=a' 



r 



we may in the above equations replace dr/r by da/a, and combine them 

 as follows : — 



M2-M, = dM = 2.nSr'--3M^, 



o a 



and we have : — 



