Consider the Euler-Bernoulli equations in the following form:

E I ( x ) d 2 w ( x ) d x 2 = − M ( x ) (1)

and

d 2 M ( x ) d x 2 = − q ( x ) (2)

where w denotes the transverse deflection and M(x) represents the bending moment at point x. The coefficient EI(x), called the flexural rigidity, is the product of the modulus of elasticity E and the moment of inertia I of the beam’s cross-section. The function q(x) represents the transversely distributed load (see Figure 1). We will demonstrate the discretization process based on the mixed form of the beam bending equations, Equations (1) and (2). Here, we begin by assuming that w and M are approximated in the usual manner by appropriate shape functions and unknown parameters:

w = N w e ,   M = N M e (3)

where N represents element shape functions and w_e and M_e are the nodal parameters to be determined. Higher order functions can be used for shape functions, but the same linear shape functions were used for w_e and M_e to simplify the calculation. For the Galerkin solution, the original shape functions are simply used as weighting functions. If we consider the Galerkin form for an element of length l , we obtain:

− 1 E I ∫ 0 l N T N d x M e + ∫ 0 l d N T d x d N d x d x w e = [ N T θ ] 0 l (4)

and

∫ 0 l d N T d x d N d x d x M e = ∫ 0 l N T q d x + [ N T V ] 0 l (5)

where θ and V denote the slope and shear force, respectively. Here N is the linear shape function, which is:

N 1 = 1 − ξ ,     N 2 = ξ ,     ξ = x / l (6)

With the above approximations, Equations (4) and (5) yields the following

[ k 11 k 12 k 21 0 ] { M e w e } = { θ F } (7)

where:

k 11 = − 1 E I ∫ 0 l N T N d x ,   k 12 = k 21 T = ∫ 0 l d N T d x d N d x d x (8)

{ M e w e } = { M 1 M 2 w 1 w 2 } ,     { θ F } = { θ 1 θ 2 V 1 + Q 1 V 2 + Q 2 }

We consider the inverse analysis of the flexural rigidity of a beam as applied to a finite-element solution of Equation (7). Suppose that the force acting on the beam domain, the boundary conditions, and the deflection at each point on the beam are given. Under these conditions, the bending moment at each point can be easily obtained from the second equation of Equation (7). The first equation of Equation (7) is written as follows:

− l 6 E I [ 2 1 1 2 ] { M 1 M 2 } + 1 l [ 1 − 1 − 1 1 ] { w 1 w 2 } = { θ 1 θ 2 } (9)

By transforming Equation (9) into an equation for unknown quantity 1/EI_e of flexural rigidity, the following equation is obtained.

[ A ] e { X } e = { B } e (10)

with:

[ A ] e = [ l 3 M 1 + l 6 M 2 l 6 M 1 + l 3 M 2 ] ,   { X } e = { 1 E I e } ,   { B } e = − { θ 1 θ 2 } + [ 1 l − 1 l − 1 l 1 l ] { w 1 w 2 } (11)

When a beam is divided into m elements, the number of flexural rigidity values to be considered as unknown quantities will become m, i.e., the same as the number of elements. The next step is the assembly of the final equations of the type given by Equation (12). This is accomplished, according to the rule of Equation (10), by simple addition of all the numbers in the appropriate space of the global matrix:

[ A ] { X } = { B } (12)

in which [A] is an n × m matrix, for which m is the total number of unknowns and n is the number of equations. For this problem, singular value decomposition is applied to [A], which yields:

[ A ] = [ U ] [ Λ ] [ D ] T (13)

where [ U ] , [ Λ ] and [ D ] each denote the presence of n × n , n × m and m × m matrices, respectively. Matrix [ Λ ] is given as follows, while the matrix rank [ A ] is expressed by p:

[ Λ ] = [ λ 1 λ 2 0 ⋱ λ p 0 0 ⋱ ] ,     λ 1 ≥ λ 2 ≥ ⋯ ≥ λ p > 0 , p ≤ min ( m , n ) (14)

where λ i is the singular value of [ Λ ] . This immediately yields:

{ X } = [ D ] [ Λ ] − 1 [ U ] T { B } (15)

If we consider the system of equations in Equation (7) to be a mixed system, in which M and w, the solution can proceed by eliminating M from the first equation and substituting it into the second to obtain:

{ F } = − [ K 21 ] [ K 11 ] − 1 [ K 12 ] { w } + [ K 21 ] [ K 11 ] − 1 { θ } (16)

As the initial slope θ → 0 , the equation above changes to:

{ F } = − [ K 21 ] [ K 11 ] − 1 [ K 12 ] { w } (17)

The vector {F} is termed the equivalent nodal force. The distributed forces q using interpolations of the form:

q = N q e (18)

where q e represents the distributed force parameters to be determined. For linearly distributed forces q over the element, the force vector by

l 6 [ 2 1 1 2 ] { q 1 q 2 } = { Q 1 Q 2 }     or     [ H ] e { q } e = { Q } e (19)

The next step is the assembly of the final equation of the type given by:

[ H ] { q } = { F } (20)

Finally, the unknown distributed forces vector q can be obtained by solving the resulting equation system.

We consider a simply supported beam subject to a concentrated load p applied at the center of the beam. Since any unit system may be used, provided it is unified throughout, dimensionless quantities are used here. Thus the length of the beam is l = 10.

E I ( x ) = { 10 ,       ( 0 ≤ x ≤ 3 , 7 ≤ x ≤ 10 ) 20 ,                   ( 3 ≤ x ≤ 7 )

In Figure 2, the solid line shows the distribution of flexural rigidity assumed in solving the problem, and the circles show the results of identification obtained using the proposed method. For the monitoring point, at which deflection is determined, the beam is divided into ten equal parts in Figure 2(a) and forty equal parts in Figure 2(b). In Figure 2, this method reproduces the behavior of a stepped beam almost exactly.

The next example is a beam with small defects in close proximity, as shown in Figure 3, where the flexural rigidity is expressed as:

E I ( x ) = { 10 ,     ( 0 ≤ x ≤ 3 ,   3.2 ≤ x ≤ 4.2 ,   4.8 ≤ x ≤ 10 ) 9 ,                                     ( 3 ≤ x ≤ 3.2 ) 8 ,                                   ( 4.2 ≤ x ≤ 4.8 )

In this example, it is assumed that the reduction in flexural rigidity might be dependent on the level of damage. The beam is divided into fifty elements for identification. The results obtained are shown by plotting of the circles. It can be observed in Figure 3 that reductions in rigidity are identified with high accuracy. Figure 3 indicates that the flexural rigidity was evaluated precisely, except around the simple support, where the beam deflection was very small.

We show the numerical results for the first example the simply supported beam subject to two concentrated loads, p = 1. In this example, we assume that the deflection is measured at 21 points. The circles at x = 0 and x = 10 in Figure 4 denote the vertical reactions at the supporting point. In Figure 4, the method reproduces the loads acting on a beam almost exactly.

In the next example, we show the numerical results for identification of a simply supported beam subjected to a uniform load acting over a portion of the beam, as indicated in Figure 5. Although a significant overshoot occurred around x = 3.5 and 8.5, on the whole, good identification results were obtained. In this figure, circles at x = 0 and x = 10 denote the vertical reactions at the supporting point. Although there was a part disturbed by overshoot, its amount was small and good one was obtained as an estimated value.