114 



PRINCIPAL ES IN liEAMS. 



Art. 76. 



rreme ilber and 100 Ibs. in shearing with the grain, the beam is 

 overstrained in shear. (See Cambria p. 361). The shear is the 

 same near the load as at the end and therefore the maximum re- 

 sultant unit stress may be greater than either of the above. How 

 to calculate this principal stress will now be shown. This is of 

 importance only in exceptional cases and for theoretical con- 

 siderations. 



It has been shown that on a vertical section of a beam 

 there are normal and tangential stresses, and on a horizontal sec- 

 tion there are tangential stresses. If equations are written giv- 

 ing the stresses on a plane making an angle < with the vertical 

 section, in terms of these stresses on vertical and horizontal 

 planes, the value of < which will make the normal stress on th<^ 

 inclined plane a maximum can be easily found. 



Fig. 86 shows the face of a triangular block, 

 taken indefinitely small, so that the unit stresses 

 on the three surfaces whose traces are AB, AC. 

 and BC may be taken constant. The plai 

 whose trace is AC is taken in a vertical section 

 on the tension side of a beam. The unit stresses 

 and not the total stresses are shown in the figure. Fig. 86. 

 It should be remembered that there is no variation in stress per- 

 pendicular to the plane of the paper, that there are no stresses 

 on the front and rear faces of the block, and that the unit shear- 

 ing stresses on AB and AC are equal (17). Fig. 86 is a repre- 

 sentation of the various stresses in a beam. 1 The problem is to 

 find that value of <f> which will make s t ' a maximum (Also s s '). 



Taking dA as the area of the plane whose trace is BC, the 

 equilibrium of the block requires the sum of the horizontal and 

 vertical components of the forces to be separately equal to zero. 

 The area of the plane AB = dA sin < and of AC, dA cos #. 



From the sum of the horizontal components 

 sfdA cos $ 8,'dA sin <f> sjlA sin $ s^dA cos tj>= 0. 

 Canceling the factor dA. this equation becomes, 



s! cos 4* / sin <j> s t sin <j> 8 cos <= 0. (a) 



In like manner from the vertical components, 



/ sin <H-*/ cos <f> s.eos<=0. (b) 



ir This is a special case of the general problem in which there is also 

 a normal stress on AB. 



