The elasticity and resistance of the materials of engineering . body is subjected to compression, the edges of thecube become: <^(i â/j), a(\-\-rj^ and a(\ -\- rj^\ while thevolume of the parallelopiped takes the value: ^3(1 _ /^) (I + rj^f = a\i + ll2r, - i)] . . (2) As before, the higher powers of /^ are omitted. If the vol-ume of the cube is decreased, ^j must be found between oand 5^. Art. 4.âRelation between the Coefficients of Elasticity for Shearing andDirect Stress in a Homogeneous Body. A body is said to be homogeneous when its elasticity, of agiven kind, is the same in all directi
The elasticity and resistance of the materials of engineering . body is subjected to compression, the edges of thecube become: <^(i â/j), a(\-\-rj^ and a(\ -\- rj^\ while thevolume of the parallelopiped takes the value: ^3(1 _ /^) (I + rj^f = a\i + ll2r, - i)] . . (2) As before, the higher powers of /^ are omitted. If the vol-ume of the cube is decreased, ^j must be found between oand 5^. Art. 4.âRelation between the Coefficients of Elasticity for Shearing andDirect Stress in a Homogeneous Body. A body is said to be homogeneous when its elasticity, of agiven kind, is the same in all directions. Let Fig. I represent a body subjected to tension parallel toCD. That oblique section on which the shear has the greatesty^ E B intensity will make an angle of 45° with either of those faceswhose traces are CD or BD ; forif « is the angle which anyoblique section makes with BD,â q P the total tension on BD, andA the area of the latter surface,the total shear on any section whose area is A^ sec a, will beP sin a. Hence the intensity of shear is :. P sin a P -r, = -IT, stn a cos a. A sec a A (0 The second member of Eq. (i) evidently has its greatest Art. 4.] SHEARING AND DIRECT STRESS. 7 value for a = 45°. Hence, if the tensile intensity on BD is prepresented by -^ =/, the greatest intensity of shear will be : S = l (2) Then by Eq. (3) of Art. 2: ⢠^-A ⢠⢠⢠(3) In Fig. I EK and KG are perpendicular to each other, whilethey make angles of 45° with either AB or CD. After stress,the cube EKGH is distorted to the oblique parallelopipedEKGH. Consequently EKGH and EKGH correspond toABCD and AEFD, respectively, of Fig. i. Art. 2. The angu-lar difference EKG â EKG is then equal to cp; and EKE = GKG = ^. Also EKE = 45° _ ^. 2 -^2 Using, then, the notation of the preceding Arts., there willresult, nearly : ^«;. (45° - f) =-^^ = I -/(I + r); . (4) remembering that FKâ FK(\ + /) ; and that EF = FK{i - rl). From a trigonometrical formula, there is ob
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