582 Prof. E. H, Barton on the Lateral 



point beyond them, is determinate and equals their sum 

 although the actual value of M is unknown unless the 

 positions of all the forces are given also. 



It is obvious that this sum F of all the forces applied 

 between the free end at the origin and the point x, is also the 

 shearing force on the bar at x. 



Now suppose forces to be continuously applied between the 

 origin, and x and beyond x also. Then the broken straight- 

 line graph gives place to a continuous curve, and F becomes 

 the integral of the forces acting between the origin and x. 

 But we still have dM./dx=¥; and, since F is now itself a 

 continuous function of x, we may differentiate it and so obtain 



d 2 M/dx- = dF/dx = f) say (3) 



Further, this rate of increase of F (per unit length of x) 

 denoted by /, is obviously the value per unit length at x of 

 the impressed forces to whose sum that shearing force F is due. 



Hence, though the bending moment itself depends upon 

 the magnitudes and positions of forces acting elsewhere, and 

 the first derivative of the bending moment at any point 

 equals the sum of all the forces acting up to that point, the 

 second derivative of the bending moment at any point equals 

 the applied force per unit length acting there, and is inde- 

 pendent of all other considerations. 



This is the relation sought, for the equation of motion is a 

 relation between displacement and acceleration, and the latter 

 is proportional to the force on the element. Thus we need to 

 express the force in terms of the displacement, and by means 

 of the bending moment, this is now possible. 



Differential Equation of Motion. — We have hitherto dealt 

 with a bar at rest and bent by certain impressed forces and 

 the clamp or other arrangement at the opposite end. Hence 

 the element of unit length at x was in equilibrium owing to the 

 applied force / and the difference of the shearing forces at 

 its extremities, which must accordingly have had a re- 

 sultant — /. Now let all the impressed forces be suddenly 

 removed, then at x there must be an acceleration proportional 

 to the remaining force — /'. We may thus write 



copdy/dt=-f, . (4) 



where p is the density of the material. 

 Then equations (1), (3), and (4) give 



-f ■+«*&-«. ® 



where b=\/g/p is the speed of longitudinal waves in the bar. 



