54 STRENGTH OF MATERIALS 



Differentiating M with respect to x, 



dM 



= K l 



dx 



Therefore 



that is to say, the shear at any point of a beam is the first div 

 coefficient of the bending moment at that point. 



If the beam is uniformly loaded, as in (D) of the preceding 



nyj-y** 



article, Q = R^ - wx and M = R^x - -> from which equation (26) 



^ 



results as before. 



From equation (26) it follows that if the bending moment is ron- 

 stant the shear is zero; and conversely, if the shear is zero tin* U-nd- 



ing moment is constant. But - - = is the condition that ilu- 



dx 



bending moment shall be either a maximum or a minimum. Conse- 

 quently, at a point where the bending moment passes through a maxi- 

 mum or minimum value the shear is zero ; and conversely. 

 theorem is illustrated by the bending moment and shear diagran 



the preceding paragraph. 



54. Designing of beams In design inu brains tin* ] .ml, Inn is to 

 find the transverse dimensions of a beam <f nivt-n length ft! 

 material, so that it shall bear a uivcn load with safety. 



In order to solve this problem, the formula M = pS is \s rittM 



*-* 



P 



Then, from the given loading, the maximum value of M is determi 

 and by dividing the ultimate strength of the material l.y tin- j.i 

 factor of safety the safe unit stress p becomes known. Tin- <ju-. 

 of these two gives the section modulus of tin- ivjuiivd s<-.-ti..n. 



In the handbooks issued by the various structural imn and steel 

 companies, the section moduli of all the standard sections are tabu- 

 lated. If, then, the beam is to be of a standard shape, its si/ 

 found by simply looking in the tables for the value of S which corre- 



sponds most closely to the calculated value > the value chosen 



P 



