SF – BM Relationship F Load M Rl B E C F G SFD A D M BMD M = reaction moment - Σ load moments = Area (ABCD) – Area (CGFE) = Area under the SFD (to one side of the position of M) Thus:- Between any 2 points on the beam span, The increase in BM value = area under the SFD (integration principle) ∴ the BM value reduces with –ve Shear Force values ∴ the position of maximum BM is at the position of zero Shear Force. 2 Standard maximum BM values. W Central point load:- Mmax = WL/4 L UDL:- Mmax = wL^2 w kN/m 8

E.G.8 (Canti) 20 20 10 10 25 60 45 SFD (kN) 80 95 0 17.5 70 BMD (kN m) 157.5 Note: The curved form of the BMD corresponding to the sloping lines of the SF diagram. E.G.9 60 kN Rl Rr 2m 6m 1m 115.9 79.9 Part SFD (NTS) 19.9 D Distance D = 19.9/18 = 1.11 m ∴ Maximum BM occurs at 3.11 m (2 + 1.11) from the LHS. Max BM.value = (115.9*3.11) –(60*1.11) –(18*3.11)3.11/2 = 207 kNm But; SAG HOG BM at RHS = -18*1*1/2 = -9 kNm ∴ A Point of Contraflexure occurs.

15 kN/m

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18 kN/m

Bending Moment – Shear Force – Loading E.G. 60 kN 40 kN 4m 1m 60 SFD (kN) D D = 60/25 = 2.4 m 40 72 max

40 BMD (kNm) Max BM value = 60*2.4 – (2.4)^2*25 = 72 kNm 2 P x At random position P, BM(P) = 60x – 25*x^2 (for x ≤ 4m) 2 d BM(P) = 60 – 25x d x = SF(P) d^2 BM(P) = -25 d x^2 = Loading intensity Note:- Maximum moment value occurs at position of zero shear force. i.e d M = 0, so, 60 – 25x = 0 ⇒ x = 2.4 m d x & Max BM = 60x – 25x^2/2 at the position of x = 2.4 m

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Parabola Straight line

25 kN/m

25 kN/m