Load Shear Moment Relations

38 Let us derive the basic relations between load, shear and moment. Considering an infinitesimal length dx of a beam subjected to a positive load5 w(x), Fig. 12.6. The infinitesimal section must also be in equilibrium.

39 There are no axial forces, thus we only have two equations of equilibrium to satisfy XFy = 0 and XMz = 0.

40 Since dx is infinitesimally small, the small variation in load along it can be neglected, therefore we assume w(x) to be constant along dx.

41 To denote that a small change in shear and moment occurs over the length dx of the element, we add the differential quantities dVx and dMx to Vx and Mx on the right face.

4 Note that this sign convention is the opposite of the one commonly used in Europe!

5 In this derivation, as in all other ones we should assume all quantities to be positive.

12.3 Sh

Figure 12.6: Free Body Diagram of an Infinitesimal Beam Segment

42 Next considering the first equation of equilibrium

The slope of the shear curve at any point along the axis of a member is given by the load curve at that point.

43 Similarly dx

(+J) £M0 = 0 ^M— + V— dx — w—dx- — (Mx + dMx) = 0

Neglecting the dx2 term, this simplifies to dM V ( ) — = V (x)

The slope of the moment curve at any point along the axis of a member is given by the shear at that point.

44 Alternative forms of the preceding equations can be obtained by integration

The change in shear between 1 and 2, AV21, is equal to the area under the load between x1 and x2.

The change in moment between 1 and 2, AM21, is equal to the area under the shear curve between x1 and x2.

45 Note that we still need to have V1 and M1 in order to obtain V2 and M2 respectively.

46 It can be shown that the equilibrium of forces and of moments equations are nothing else than dTij + = P dVi Ox, * P°i — p dt the three dimensional linear momentum dr + Pbi = Pdt and moment of momentum (rXt)dS +

i (rXpb)dV = df (rXpv)dV equations satisfied on the averse over the cross sector,

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