Page 302 - Physical Principles of Sedimentary Basin Analysis

284 Flexure of the lithosphere

The length of the fiber is l = (R − z)φ when φ is the angle. The strain in the fiber is the
elongation of the fiber divided by its original length,
dl z
xx =− = (9.1)
l R
where a minus sign is added to ensure that compression gives positive stress. If we for a
moment assume that we are bending a beam instead of a plate then the fiber stress σ xx
follows directly from Hooke’s law,
z
σ xx = E xx = E (9.2)
R
where E is Young’s modulus. We will now compute the bending moment along a vertical
cross-section at an arbitrary lateral position in a beam. The contribution to the bending
moment (or torque) from a fiber of thickness dz and a distance z from the neutral surface
is then
2
dM = zdF = z σ xx dz = (E/R)z dz (9.3)
where the force in the 2D fiber is dF = σ xx dz. The total bending moment in the beam is
found by integration over the entire thickness of the plate, and we get
E h/2 2 Eh 3
M = z dz = . (9.4)
R −h/2 12 R
The deflection is in the general case represented by a function of distance, w(x), which is
not exactly a circular arc (see Figure 9.2b). Nevertheless, the local curvature of a function
w(x) at a point x is found from the circle that fits the curve locally. The radius of this circle
is approximated by
2
1 d w
≈ (9.5)
R dx 2
as shown in Note 9.1, and the bending moment for the beam is
2
3
Eh d w
M = . (9.6)
12 dx 2
We have so far assumed that bending a 2D plate is similar to bending a beam. The stress
state of a slightly bent beam can be represented by just one non-zero stress component,
σ xx , and we can apply Hooke’s law (9.2). The stress in a bent plate is a little different. It
is first assumed that there is no stress in the plate in the y-direction, σ yy = 0. Instead of
assuming that σ zz is zero too, which is the assumption for a beam, it is instead assumed that
the strain in the vertical direction ( zz ) is zero. Then it follows from equation (3.90) that
1
zz = σ zz − νσ xx − νσ yy = 0. (9.7)
E
Using that σ yy = 0 we get that σ zz = νσ xx , and when σ yy = 0 and σ zz = νσ xx are
inserted into equation (3.88) for lateral strain we get
1
1
2
xx = σ xx − νσ yy − νσ zz = 1 − ν σ xx . (9.8)
E E