46 Mechanics of Materials §3.3 S.F.)is then the reaction of 10 kN upwards,i.e.positive,and the bending moment reaction ×zero distance=zero. The following characteristics of the two diagrams are now evident and will be explained later in this chapter: (a)between B and C the S.F.is zero and the B.M.remains constant; (b)between A and B the S.F.is positive and the slope of the B.M.diagram is positive;vice versa between E and F; (c)the difference in B.M.between A and B =20kN m area of S.F.diagram between A and B. 3.3.S.F.and B.M.diagrams for uniformly distributed loads Consider now the simply supported beam shown in Fig.3.8 carrying a u.d.1.w=25 kN/m across the complete span. 25 kN/m A D E G eocoxcccoooxxcoooceeexxoeeeeccccco -12 m- 150 S.F.diagram (kN) 50 B.M.diagram (kN m) 450 400 400 250 250 Fig.3.8. Here again it is necessary to evaluate the reactions,but in this case the problem is simplified by the symmetry of the beam.Each reaction will therefore take half the applied load, i.e. RA=RB=- 2=150kN 25×12 The S.F.at A,using the usual sign convention,is therefore +150kN. Consider now the beam divided into six equal parts 2m long.The S.F.at any other point C is,therefore, 150-load downwards between A and C =150-(25×2)=+100kN The whole diagram may be constructed in this way,or much more quickly by noticing that the S.F.at A is +150kN and that between A and B the S.F.decreases uniformly,producing the required sloping straight line,shown in Fig.3.7.Alternatively,the S.F.at A is +150kN and between A and B this decreases gradually by the amount of the applied load (i.e.by 25×12=300kN)to-150 kN at B.46 Mechanics of Materials 53.3 S.F.) is then the reaction of 10 kN upwards, Le. positive, and the bending moment = reaction x zero distance = zero. The following characteristics of the two diagrams are now evident and will be explained later in this chapter: (a) between B and C the S.F. is zero and the B.M. remains constant; (b) between A and B the S.F. is positive and the slope of the B.M. diagram is positive; vice (c) the difference in B.M. between A and B = 20 kN m = area of S.F. diagram between A versa between E and F; and B. 3.3. S.F. and B.M. diagrams for uniformly distributed loads Consider now the simply supported beam shown in Fig. 3.8 carrying a u.d.1. w = 25 kN/m across the complete span. 25 kN/rn A C D E FG 0 RA R.2 I” 150 I50 0.M. dmgrorn (kN rn) 450 Fig. 3.8. Here again it is necessary to evaluate the reactions, but in this case the problem is simplified by the symmetry of the beam. Each reaction will therefore take half the applied load, i.e. 25 x 12 RA= Rs= ~ - 150 kN 2 The S.F. at A, using the usual sign convention, is therefore + 150kN. is, therefore, Consider now the beam divided into six equal parts 2 m long. The S.F. at any other point C 150 - load downwards between A and C = 150 - (25 x 2) = + 100 kN The whole diagram may be constructed in this way, or much more quickly by noticing that the S.F. at A is + 150 kN and that between A and B the S.F. decreases uniformly, producing the required sloping straight line, shown in Fig. 3.7. Alternatively, the S.F. at A is + 150 kN and between A and B this decreases gradually by the amount of the applied load (Le. by 25 x 12 = 300kN) to - 150kN at B