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5.2 Loads between nodal points Generally, structures must resist loads applied between joints or natural nodal points of the system. One way to treat intermediate concentrated loads is to insert artificial nodes, such as p and q in Fig. 5.4(a). Additional nodes may also be used when the load is distributed as in Figs. 5.4(b) and 5.4(c). The distributed loads can be “lumped” as concentrated loads at suitably selected arbitrary nodes, and the degrees of freedom at these and the actual joints are treated as the unknowns of the problem. The lumped loads must be statically equivalent to the distributed loads they replace. It is desirable to have recourse to a rigorous method that does not require “artificial” joints. The approach that is of most general use with the displacement method is one that employs the related concepts of fictitious joint restraint, fixed-end forces, and equivalent nodal loads. This approach will be explained by considering an example. But first, the approach already used is reviewed by illustrating its application to the continuous beam of Fig. 5.5(a). The solution of this problem of a system loaded only at a natural node would start with eqn (3.6), P K . (3.6)

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The support and the remaining degrees of freedom are grouped as in eqn (3.7),

ff fsf f

sf sss s

K KPK KP

, (3.7)

where all quantities related to the supports are assigned subscript s and those relating to the remaining degrees of freedom have the subscript f . For this particular loading and support conditions,

000

0000

a

a

b

ff fs c

sf ssyb

yc

yd

md

P v

RRRR

θθθ

−

=

K KK K

. (5.19)

Solution for the unknown displacements, reactions, and internal forces proceeds in the usual way. The moment diagram and the deflected structure are shown in Fig. 5.5(c) and 5.5(d). Now consider the same structure but with a uniformly distributed load of intensity q in the center span (see Fig. 5.6). This loading, which is between nodal points, will be treated in two stages and the results summed (see Fig. 5.6(a)-(c)). First, as in Fig. 5.6(a), assume the existence of fictitious external constraints capable of reducing the nodal degrees of freedom to zero (clamping the joints). The constraining forces, which in this

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case consist of two direct forces and two moments, are shown in their positive sense under the sign convention used. It should be clear that F

mcP must actually be negative in order to constrain the rotation at point c . It should also be accepted that these forces are completely independent of the system – they are not being supplied by the real beam or its real supports. The solution for the fixed-end forces are assumed to be known (see Table 5.1). Knowing the fixed-end forces, the internal forces and deformations corresponding to the assumed constraints can be calculated, as shown in the bending moment diagram and deflected structure of Fig. 5.6(a). Now it is necessary to remove the fictitious constraints by applying to the nodes loads that are equal and opposite to the fixed-end forces and permitting the system to deform under the action of these likewise fictitious loads. The reversed fixed-end forces are called equivalent nodal loads. They are indicated in Fig. 5.6(b) by symbol EP and appropriate subscripts. Solving in the usual way for the displacements, reactions, and the internal forces caused by these loads results in the bending moment diagram and deflected structure of Fig. 5.6(b). Last, the total solution is obtained by summing the two parts. In arriving at the free body diagram of Fig. 5.6(c), the two sets of fictitious forces have canceled each other and what is left is the real system, in which all the requirements of equilibrium and compatibility are satisfied. Represent the above physical process algebraically. The internal forces and displacements of the fixed-end problem (Fig. 5.6(a)), must be obtained by some means not detailed here, and reserved for addition to the solution of the nodal displacement problem (Fig. 5.6(b)). One way to formulate the solution to the displacement problem is after the fashion of eqn (3.6), but with the addition of the fixed-end forces to the right-hand side of the equation; thus, FP K P . (3.6a) Physically, this states that, in the absence of any nodal displacement, i.e., 0 , { }P would be equal to the vector of fixed-end forces { }FP . Conceptually, this formulation is useful because it helps keep clear the distinction between any real nodal loads and reaction components of { }P and the fictitious components that comprise { }FP .

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The support and the remaining degrees of freedom of eqn (3.6a) may be grouped as before,

F

ff fsf f fF

sf sss s s

K KP PK KP P

. (3.7a)

For the particular loading in Fig. 5.6, the above may be written as,

2

2

0 00 00 /120 /12

0 / 20 / 20 00 0

a

a

b

ff fs c

sf ssyb

yc

yd

md

v

qLqL

R qLR qLRR

θθθ

− = +

K KK K

. (5.20)

Transference of the { }FP vector in any of the above formulations (eqns 3.6a, 3.7a, or 5.20) to the left hand side of the equation is the algebraic equivalent of applying the reversed fixed-end forces as nodal loads in Fig. 5.6(b). Hence, F EP P P P K , (5.21) where { }EP is the vector of equivalent nodal loads defined in the discussion of Fig. 5.6(b). For the illustrative example of Fig. 5.6, the statement of the displacement problem is therefore,

2

2

00

/12/12

/ 2 0/ 2 0

00

a

a

b

ff fs c

sf ssyb

yc

yd

md

v

qLqL

R qLR qL

RR

θθθ

−

= − −

K KK K

. (5.22)

Solution for the unknown nodal displacements, the real reactions ( , , ,yb yc yd mdR R R R ), and the

internal forces proceeds in the usual way, but with appropriate accounting for the loads between nodes. Thus from eqn (3.7a) for s 0 ,

1 F

f ff f fK P P , (3.7b)

Fs sf f sP K P . (3.7c)

In determining forces and displacements within the loaded members, add the results of eqn (5.22) to the solution of the fixed-end problem. Formally, this part of the problem can be symbolized by augmenting the element eqns (3.11) in the same way that eqns (3.6) and (3.7) were modified to obtain the above global equations. Thus, FF k F . (3.11a)