1 12 MECHANICS 



point a directly under the first load. The moment here, 

 on the left-hand part of the beam, is ^X3 = 5,600X3 

 = 16,800 ft.-lb. 



Then find the moment on the left-hand portion of the beam 

 around point b. Here the positive moment is R\ X 10 2, or 

 5,600X10^ = 58,800 ft.-lb. There are, however, two neg- 

 ative moments acting, and these must be subtracted from 

 the positive moment in order to get the resultant moment. 

 One of these moments is that due to the concentrated load 

 of 1,200 lb., and the other is that due to the part of the dis- 

 tributed load to the left of the point b. The negative moment 

 of the concentrated load is therefore 1,200X7^ = 9,000 ft.-lb. 

 The portion of the distributed load considered is 800 X4 

 =3,600 lb. Its moment arm may be considered to extend 

 from the point b to the center of the portion under consid- 

 eration, or 4-*-2=21 ft. Its moment is therefore 3,600X2} 

 = -8,100 ft.-lb. The resultant moment of the left-hand 

 section about b is therefore 58,800-9,000-8,100= +41,700 

 ft.-lb. 



The bending moment about c is 5,600X18-1,200X15 

 - (800 X 9) X 7^ = 100,800 -18,000 -54,000 = +28,800 ft.-lb. 



The bending moment of the forces acting on a beam, 

 tending to break it at any section, may be calculated from 

 either end of the beam. It is customary to call the bending 

 moment positive if it tends to turn the left-hand part in the 

 direction of the hands of a clock, and negative if it tends to 

 turn it in the opposite direction. However, the moment 

 may be calculated from either end, remembering that if it 

 is calculated from the right-hand end the bending moment 

 acts in an opposite direction and will therefore receive a sign 

 opposite to that given the left-hand end of the beam. 



As has been stated, the reason a beam does not break is 

 because its strength at any transverse section is sufficient to 

 resist the moment of the forces about that section. It has 

 also been shown that the bending moment is different at 

 different points along a beam. It is therefore important to 

 find out around what point the forces acting on a beam 

 exert their maximum moment, and then, by the method 

 already given, to find this maximum moment. 



