Flexure Elements
4.3 FLEXURE ELEMENT
Similarly, the internal bending moment at a cross section must be equivalent to the resultant moment of the normal stress distribution, so
M(x) = −
A
yxdA= Ed2v dx2
A
y2dA (4.9)
The integral term in Equation 4.9 represents the moment of inertia of the cross- sectional area about the zaxis, so the bending moment expression becomes
M(x) =EIz
d2v
dx2 (4.10)
Combining Equations 4.6 and 4.10, we obtain the normal stress equation for beam bending:
x= −M(x)y
Iz = −yEd2v
dx2 (4.11)
Note that the negative sign in Equation 4.11 ensures that, when the beam is sub- jected to positive bending moment per the convention depicted in Figure 4.1c, compressive (negative) and tensile (positive) stress values are obtained correctly depending on the sign of the ylocation value.
v1 v2 (a)
Figure 4.3
(a) and (b) Beam elements with identical end deflections but quite different deflection characteristics. (c) Physically unacceptable discontinuity at the connecting node.
v1 v2
v 0
(b) (c)
elements, thus precluding the physically unacceptable discontinuity depicted in Figure 4.3c.
In light of these observations regarding rotations, the nodal variables to be associated with a flexure element are as depicted in Figure 4.4. Element nodes 1 and 2 are located at the ends of the element, and the nodal variables are the trans- verse displacements v1 andv2at the nodes and the slopes (rotations) 1and2. The nodal variables as shown are in the positive direction, and it is to be noted that the slopes are to be specified in radians. For convenience, the superscript (e) indicating element properties is not used at this point, as it is understood in con- text that the current discussion applies to a single element. When multiple ele- ments are involved in examples to follow, the superscript notation is restored.
The displacement function v(x)is to be discretized such that
v(x)= f(v1,v2,1,2,x) (4.12) subject to the boundary conditions
v(x=x1)=v1 (4.13)
v(x=x2)=v2 (4.14)
dv dx
x=x1
=1 (4.15)
dv dx
x=x2
=2 (4.16)
1
2
y
x
v1 v2
L
1 2
Figure 4.4 Beam element nodal displacements shown in a positive sense.
M1
M1
Mz F1L M2 M1
F1L M1 M2
x
F1 F2
1 2
Figure 4.5 Bending moment diagram for a flexure element. Sign convention per the strength of materials theory.
Before proceeding, we assume that the element coordinate system is chosen such thatx1=0andx2= L to simplify the presentation algebraically. (This is not at all restrictive, since L =x2−x1in any case.)
Considering the four boundary conditions and the one-dimensional nature of the problem in terms of the independent variable, we assume the displacement function in the form
v(x)=a0+a1x+a2x2+a3x3 (4.17) The choice of a cubic function to describe the displacement is not arbitrary.
While the general requirements of interpolation functions is discussed in Chapter 6, we make a few pertinent observations here. Clearly, with the specifi- cation of four boundary conditions, we can determine no more than four con- stants in the assumed displacement function. Second, in view of Equations 4.10 and 4.17, the second derivative of the assumed displacement function v(x) is linear; hence, the bending moment varies linearly, at most, along the length of the element. This is in accord with the assumption that loads are applied only at the element nodes, as indicated by the bending moment diagram of a loaded beam element shown in Figure 4.5. If a distributed load were applied to the ele- ment across its length, the bending moment would vary at least quadratically.
Application of the boundary conditions 4.13–4.16 in succession yields
v(x =0)=v1=a0 (4.18)
v(x =L)=v2=a0+a1L +a2L2+a3L3 (4.19) dv
dx
x=0
=1 =a1 (4.20)
dv dx
x=L
=2 =a1+2a2L+3a3L2 (4.21)
Equations 4.18–4.21 are solved simultaneously to obtain the coefficients in terms of the nodal variables as
a0=v1 (4.22)
a1=1 (4.23)
a2= 3
L2(v2−v1)− 1
L(21+2) (4.24)
a3= 2
L3(v1−v2)+ 1
L2(1+2) (4.25)
Substituting Equations 4.22–4.25 into Equation 4.17 and collecting the coeffi- cients of the nodal variables results in the expression
v(x)=
1− 3x2 L2 + 2x3
L3
v1+
x−2x2 L + x3
L2
1
+ 3x2
L2 −2x3 L3
v2+
x3 L2 − x2
L
2 (4.26)
which is of the form
v(x)=N1(x)v1+N2(x)1+N3(x)v2+N4(x)2 (4.27a) or, in matrix notation,
v(x)=[N1 N2 N3 N4]
v1
1
v2
2
=[N]{} (4.27b) where N1,N2,N3, and N4 are the interpolation functions that describe the dis- tribution of displacement in terms of nodal values in the nodal displacement vector{}.
For the flexure element, it is convenient to introduce the dimensionless length coordinate
= x
L (4.28)
so that Equation 4.26 becomes
v(x)=(1−32+23)v1+L(−22+3)1+(32−23)v2
+L2(−1)2 (4.29)
where 0≤ ≤1. This form proves more amenable to the integrations required to complete development of the element equations in the next section.
As discussed in Chapter 3, displacements are important, but the engineer is most often interested in examining the stresses associated with given loading conditions. Using Equation 4.11 in conjunction with Equation 4.27b, the normal
stress distribution on a cross section located at axial position xis given by x(x,y) = −y Ed2[N]
dx2 {} (4.30)
Since the normal stress varies linearly on a cross section, the maximum and min- imum values on any cross section occur at the outer surfaces of the element, where distance yfrom the neutral surface is largest. As is customary, we take the maximum stress to be the largest tensile (positive) value and the minimum to be the largest compressive (negative) value. Hence, we rewrite Equation 4.30 as
x(x) =ymaxEd2[N]
dx2 {} (4.31)
and it is to be understood that Equation 4.31 represents the maximum and mini- mum normal stress values at any cross section defined by axial coordinate x. Also ymaxrepresents the largest distances (one positive, one negative) from the neutral surface to the outside surfaces of the element. Substituting for the interpolation functions and carrying out the differentiations indicated, we obtain
x(x)= ymaxE 12x
L3 − 6 L2
v1+
6x L2 − 4
L
1+ 6
L2 −12x L3
v2
+ 6x
L2 − 2 L
2
(4.32) Observing that Equation 4.32 indicates a linear variation of normal stress along the length of the element and since, once the displacement solution is obtained, the nodal values are known constants, we need calculate only the stress values at the cross sections corresponding to the nodes; that is, at x =0andx= L. The stress values at the nodal sections are given by
x(x =0)=ymaxE 6
L2(v2−v1)− 2
L(21+2)
(4.33) x(x= L)= ymaxE
6
L2(v1−v2)+ 2
L(22+1)
(4.34) The stress computations are illustrated in following examples.