46 RESISTANCE OF MATERIALS 



4. Take a section between the origin and the first concentration, let 

 x denote the distance of this section from the origin, and find the gen- 

 eral expressions for the moment and shear at this section in terms of x. 



5. Proceed in the same way for a section between each pair of 

 consecutive concentrations. 



6. Plot these equations, checking the work by means of the 

 general relations stated in the preceding article. 



7. Plot the shear diagram first. In plotting this diagram it is 

 convenient to follow the direction in which the forces act. Thus, in 

 Fig. 45 the shear at the left end is equal to the reaction and may 

 be laid off in the same direction, that is, upwards. Proceeding to 

 the right, drop the shear diagram by an amount equal to each load 

 as it is met, until the reaction at the right end is reached, which will 

 bring the shear diagram back to the base line. By following this 

 method the shear diagram will always begin and end on the base 

 line, which serves as a check on the work. 



8. Note that as long as the shear diagram lies above the base line the 

 shear is positive and therefore A M is also positive ; that is to say, the 

 moment is increasing. Where the shear diagram crosses the axis, 

 the moment diagram must attain its highest or lowest point. When 

 the shear diagram lies below the base line, the moment is decreasing. 



9. Compute numerical values of the moment and shear at the 

 critical points of the diagrams, and indicate these numerical values 

 on the diagrams. 



A sample set of diagrams as they should be drawn by the student 

 is shown in Fig. 45. 



APPLICATIONS 



76. A beam 16 ft. long is supported at the left end and at a point 4 ft. from the 

 right end, and carries a uniform load of 200 Ib./ft. over its entire length and a 

 concentrated load of 1 ton at a point 4 ft. from the left end. Sketch the shear 

 and moment diagrams and note the maximum shear and maximum moment. 



Solution. On cross-section paper indicate the loading as shown in Fig. 45. 



To find either reaction, take moments about the other point of support. Thus, 

 for the left reaction E l we have 



R l . 12 - 3200 4 + 2000 -8 = 0, whence E l = 2400 Ib. 

 Similarly, for R 2 , R 2 . 12 - 3200 . 8 - 2000 -4 = 0, whence E 2 = 2800 Ib. 



As a check on the correctness of these results, sum of loads is 3200 + 2000 = 5200, 

 and sum of reactions is 2400 + 2800 = 5200. 



