STRESS AND STRAIN. II 



very considerable lateral expanding force we know would be 

 necessary in order to prevent the bar from contracting later- 

 ally. This lateral compression is probably produced through 

 the medium of the lateral diagonal shears resulting from the 

 applied direct stress, these producing in turn the lateral strain 

 and its accompanying direct stress. 



If we consider a bar subjected to an applied longitudinal 

 stress and at the same time to various combinations of lateral 

 applied stresses, it is evident that each applied stress increases 

 or decreases the other applied stress at right angles to it. If 

 we assume proper values for Young's Modulus and Poisson's 

 Ratio it is not difficult to determine the actual stresses. As an 

 example, consider a steel bar 2" X 3" X 24" subjected to a 

 longitudinal pull of 100,000 lbs., and at the same time to a 

 total compressive force on the 3" face of 100,000 lbs., and a 

 total expansive force of 100,000 lbs. on the 2" face. Assume 

 e = coefficient of lateral strain = 1/3, and E = Young's 

 Modulus = 30,000,000. The apparent unit stresses will be : — 



100,000 



in the longitudinal direction Si = H ^ + 16,666 lbs. 



6 



per sq. in. ; in a direction perpendicular to the flat face, 



100,000 . , . ,. • 



Sf ^ ^= — 1389 lbs. ' sq. m., and m a direction 



72 



100,000 



perpendicular to the narrow face Sn ^ + = +2083 



48 



lbs. / sq. in. The corresponding lateral unit stresses would 



be equal to one-third of these values, and would be added or 



subtracted as circumstances required. The actual unit 



stresses in the three directions would be 



Sl = Si — 1/3 . Sn + 1/3 . Sf = 16,666 — 1/3 . 2083 + 



1/3 . 1389 = + 16,435 lbs. / sq. in. 

 Sf = — Sf — 1/3 . Sl — 1/3 . Sn = — 1389 — 1/3 • 16,666 — 



1/3 . 2083 = — 7638 lbs. / sq. in. 

 Sn == Sn — 1/3 . Sl -f 1/3 • Sf = 2083 — 1/3 . 16,666 + 

 1/3 . 1389 = — 3009 lbs. / sq. in. 



