"Example Pr Example Problemoblem 3 3.Draw SFD and BMD for the single side overhanging beamsubjected to loading as shown below. Determine theabsolute maximum bending moment and shear forcesand mark them on SFD and BMD. Also locate points ofcontra flexure if any. 5kN 2 kN 10kN/m A D C B 4m 1m 2m5kN 2 kN 10kN/m A B R R 4m A 1m B 2m Solution : Calculation of Reactions: SM = 0 A - R × 5 + 10 × 4 × 2 + 2 × 4 + 5 × 7 = 0? R = 24.6 kN B B SF = 0 y R + 24.6– 10 x 4 – 2 + 5 = 0? R = 22.4 kN A A5kN 2 kN 10kN/m 5 2 4 3 7 0 1 6 5 4 0 2 3 7 6 1 R =22.4kN A 4m 1m 2m R =24.6kN B Shear Force Calculations: V =0;V = 22.4 kNV = - 19.6 + 24.6 = 5 kN 0-0 1-1 5-5 V = 22.4– 10 × 4 = -17.6kN V = 5 kN 2-2 6-6 V = - 17.6 – 2 = - 19.6 kN V = 5 – 5 = 0 (Check) 3-3 7-7 V = - 19.6 kN 4-45kN 2 kN 10kN/m A C B D R =22.4kN A 4m 1m 2m R =24.6kN B 22.4kN 5 kN 5 kN x = 2.24m 17.6kN 19.6kN 19.6kN SFD5kN 2 kN 10kN/m X A D x C B X R =22.4kN A 4m 1m 2m R =24.6kN B Max. bending moment will occur at the section where the shear force is zero. The SFD shows that the section having zero shear force isavailablein the portion AC. Let that section be X-X, considered at a distance xfrom support A as shown above.The shear force at that section can be calculated as Vx-x = 22.4- 10. x = 0 ? x = 2.24 m5kN 2 kN 10kN/m A D C B R =22.4kN A 4m 1m 2m R =24.6kN B Calculations of Bending Moments:M = M = 0 A D M = 22.4 × 4 – 10 × 4 × 2 = 9.6 kNm CM = 22.4 × 5– 10 × 4 × 3– 2 × 1 = - 10kNm (Considering Left portionB of the section) Alternatively M = -5 × 2 = -10 kNm (Considering Right portion of the section) BAbsolute Maximum Bending Moment is atX- X ,Mmax = 22.4 × 2.24 – 10 × (2.24)2 / 2 = 25.1 kNm5kN 2 kN 10kN/m X A D C B x = 2.24m X R =22.4kN A 4m 1m 2m R =24.6kN B Mmax = 25.1 kNm 9.6kNm Point ofcontra flexure 10kNm BMD5kN 2 kN 10kN/m X A D x = 2.24m C B X R =22.4kN A 4m 1m 2m R =24.6kN B 22.4kN 5 kN 5 kN x = 2.24m 17.6kN 19.6kN 19.6kN SFD Point ofcontra flexure 9.6kNm 10kNm BMD"