KKLATloXS BETWEEN STKESS AND DEFORMATION 23 



"Eliminating q f , 



p' = p x sin 2 + p v cos 2 a + 2 q sin a cos a. 

 From trigonometry, 



1 cos 2 a 



2 1 -f cos 2 a . 

 > cos 3 a = 2 sin a cosa = sui 2 a. 



2 2 



Therefore, by substituting these values, 



(2) p' = $* PV + Py P* cos 2 a + # ski 2 a. 



Similarly, by eliminating j/ from equations (1), 



(3) j' = ^-* sin 2 a + # cos 2 a. 



Problem 23. At a certain point in a vertical cross section of a beam the unit 

 normal stress is 300 lb./in. 2 , and the unit shear is 100 lb./in. 2 . Find the normal 

 stress and the shear at this point in a 

 plant- inclined at 30 to the horizontal. 



Solution. Suppose a small cube cut 



i he beam at the point J\ 

 10). Then, by the theorem in Article 23, 

 there will also be a unit shear of inten- 

 sity q on the top and bottom faces of 

 ibe. In the present case, there- 

 fore, p x = 300 lb./in. a . p, = 0, and 

 q = 100 lb./in. 2 . Substitutin. 



values in equations (2) ami (3). and 

 tg a = 80, the unit normal stress 



I i.;. 10 



and m on a plane through N inclined at 80 to the horizontal are 



p'= 1IH.5 lb./in.-, 7'= 179.8 lb./in. 2 . 



26. Maximum normal stress. The condition that p' shall be a 



: mi in or a minimum is that -j- 

 to equation 1 1 



0. Applying this condition 



whence 



(5) 

 and consequently 



tan2a = 



2? 



- 

 P V -P X 



ATT 



