Art. 76. 



PRINCIPAL STRESSES IN BEAMS. 



115 



Multiplying the first equation by cosj> and the second by 

 sin <f>, and adding, s s ' is eliminated. 



/ sin $ cos , sin <f> cos <fr s, cos 1 < = 0. 



*! sin* ^+*/ sin <f> cos <f>s sin <fcos<j> = 



/ = s t cos* <+2 * * in <f>cos<j> which reduces to 



9^_L^i n 9^ (25) 



2 * (26) 



Ivjiiation (26) is gotten by multiplying (a) by sin<f>, (b) by 

 cos <j>, subtracting, and introducing the angle 2 . 

 :n (25), for a maximum or a minimum s t ' 



tan 2 <f> = 2 



(27) 



Tan 2 < has the same value for two angles differing from 

 each other by 180 ; therefore <#> has two values differing by 90, 

 one of which locates the plane upon which s t ' is a maximum, and 

 the other the plane upon which s t ' is a minimum. By substituting 

 the value of 2 <t> from (27) in (25) these values are determined. 

 From (c) or (27), 



2st *t 



These values in equation (25) give 



The larger value is the maximum principal stress and the 

 smaller value the minimum principal stress (12). These will be 

 of opposite sign, because the negative term is greater than the 

 positive term; this means that one is tension arid the other 

 compression. On the compression side of a beam s t is replaced 

 by s c and s/ by s/. 



If equation (28) gives the values of the principal stresses, 

 the value of ^ from equation (27) should make s s ' = in equa- 

 tion (26) ; that this is true is apparent by comparing equations 

 (c) and (26). 



When .?, is zero as it is in the extreme upper and lower fibres, 

 equation (28) becomes s t ' = s t or 0; the resultant stress is equal 

 to the bending stress tension in this case and the compressive 

 stress is zero. Now the principal stresses lie in the horizontal 

 and vertical planes, as is also shown by equation (27). 



When St = Q, as it does at the neutral axis, equation (28) 





