56 Mechanics of Materials §3.11 finite areas and the S.F.must change gradually from one value to another across these areas. The vertical line portions of the S.F.diagrams are thus highly idealised versions of what actually occurs in practice and should be replaced more accurately by lines slightly inclined to the vertical.All sharp corners of the diagrams should also be rounded.Despite these minor inaccuracies,B.M.and S.F.diagrams remain a highly convenient,powerful and useful representation of beam loading conditions for design purposes. Examples Example 3.1 Draw the S.F.and B.M.diagrams for the beam loaded as shown in Fig.3.17,and determine (a)the position and magnitude of the maximum B.M.,and(b)the position of any point of contraflexure. 5KN 2kN 4kN/m A 3m Im- -Im 15 S.F.Diagram B.M.Diagram 17.5 13 3 Fig.3.17. Solution Taking the moments about A, 5RB=(5×1)+(7×4)+(2×6)+(4×5)×2.5 R,=5+28+12+50 5 19kN56 Mechanics of Materials 43.1 1 finite areas and the S.F. must change gradually from one value to another across these areas. The vertical line portions of the S.F. diagrams are thus highly idealised versions of what actually occurs in practice and should be replaced more accurately by lines slightly inclined to the vertical. All sharp corners of the diagrams should also be rounded. Despite these minor inaccuracies, B.M. and S.F. diagrams remain a highly convenient, powerful and useful representation of beam loading conditions for design purposes. Examples Example 3.1 Draw the S.F. and B.M. diagrams for the beam loaded as shown in Fig. 3.17, and determine (a) the position and magnitude of the maximum B.M., and (b) the position of any point of contraflexure. L I S.F. Diagram / + \ Fig. 3.17. Solution Taking the moments about A, 5RB= (5 x 1)+(7 x 4)+(2 x 6)+(4 x 5) x 2.5 .. 5+28+12+50 = 19kN 5 R, =