Evidently each reaction equals one-half the sum of the load and weight of the beam, that is, (1,600+400) -=-1,000 pounds. To the left of a section 2 feet from the left end, the forces acting on the beam consist of the left reaction, the load on that part of the beam, and the weight of that part ; then since the load and weight of the beam per foot equal 200 pounds, V,-- 1,000-200 X 2 = 600 pounds.
To the left of a section four feet from the left end, the forces are the left reaction, the load on that part of the beam, and the weight ; hence V,= 1,000-200 X 4 =200 pounds.
Without further explanation, the student should see that V, =1,000-200X 6 =-200 pounds, V, = 1,000-200 X 8 = -600 pounds, = 1,000-200 X 10 = -1,000 pounds, V." 1,000-200 X 10+ 1,000 = 0.
3. Compute the values of the shear in example 1, taking into account the weight of the beam (400 pounds). (The right and left reactions are then 3,900 and 2,500 pounds respectively; see example 3, Art. 33.) We proceed just as in example 1, except that in each compu tation we include the weight of the beam to the left of the section (or to the right when computing from forces to the right). The weight of the beam being 40 pounds per foot, then (computing from the left) pounds, ,' 2,500-40= +2,460, ," V, = + 2,500-1,000-40 X 2 = +1,420, V, =--- 1- 2,500-1,000-40 X 3 = +1,380, V, =+2,500-1,000-40X4=+1,340, =+ 2,500-1,000 -40 x 5 = +1,300, Vg 2,500-1,000-40 X 6 = +1,260, V," =42,500-1,000-40 X 6-2,000 = -740, V, + 2,500-1,000-2,000-40 X 7 =-780, V,' =+ 2,500-1,000-2,000-40X 8 = -820, V," = +2,500-1,000-2,000-40 X 8-3,000 = -3,820, V, = + 2,500-1,000-2,000-3,000-40 X 9 = -3,860, = 2,50014,000-2,000-3,000-40 X 10 = -3,900, +2,500-1,000-2,000-3,000-40 X 10 + 3,900=0.
Computing from the right, we find, as before, that V, =- (3,900-3,000-40 X 3)=-780 pounds, V,' =-( 3,900-3,000-40 X 2)=-820, V," =-(3,900-40 X 2)=-3,820, etc., etc. 1. Compute the values of the shear -for sections of the beam represented ,in Fig. 10, neglecting the weight of the beam. (The right and left reactions are 3,300 and 4,000 pounds respectively; see example 2, Art. 33.) Ans. V, pounds, V, +1,000, / 2. Solve the preceding example, taking into account the weight of the beam, 42 pounds per foot. (The right and left reactions are 3,780 and 4,360 pounds respectively; see example 4, Art. 33.) r AT," = - 2,100 lbs. V, = +1,966 lbs. AT = -1,928 lbs.