Formulas FormulasThe infinite beam extends infinitely along the x axis (positive and negative values). The procedure to compute the deflection produced by a load column (distributed load) on the beam is as follows:
1. Use the flexural rigidity (D) to express the "strength" of the elastic lithosphere (i.e. elastic beam):
D = E*T3/(12*(1-ν2)) (1)
where E is Young´s modulus, ν is Poisson ratio, and T is the thickness of the elastic lithosphere.
2. Invent a term D(α,x):
D(α,x) = exp(-x/α)*cos(x/α) (2)
where x is the horizontal coordinate and α is a length parameter given by:
α = ((4*D)/(ρf*g))^(1/4) (3)
where ρf is the density of the foundation. This is the difference between the density of the mantle and the density of the material filling the basin. g is gravity.
3. The deflection (u) at any point (c) along the elastic lithosphere (i.e. elastic beam) is equal to:
a. If point (c) is under the load column:
u = (q/(2*k))*(2-D(α,a)-D(α,b)) (4)
b. If point (c) is to the left of the load column:
u = (q/(2*k))*(D(α,a)-D(α,b)) (5)
c. If point (c) is to the right of the load column:
u = (-q/(2*k))*(D(α,a)-D(α,b)) (6)
where (q/k) = h*(ρ/ρf), h is the load height, and ρ is the load density. a and b are measured as absolute values of distances from point (c) to the left and right borders of the load column, respectively.
The semi-infinite beam has a free end at x = 0 and extends infinitely in the positive x direction.The deflection of this beam is obtained by first considering the forces that are generated during bending of an infinite beam, and then "nulling" the bending moment and shear force at the free-end (x = 0). The procedure is described in detail by Hetenyi (1946). Here I will enumerate the equations to compute the deflection of a semi-infinite beam due to a load colum (distributed load).
1. Introduce new terms A(α,x), B(α,x), C(α,x):
A(α,x) = exp(-x/α)*(cos(x/α) + sin(x/α)) (7)
B(α,x) = exp(-x/α)*sin(x/α) (8)
C(α,x) = exp(-x/α)*(cos(x/α) - sin(x/α)) (9)
2. Compute the moment (MO) and shear force (QO) at required free-end (x = 0):
MO = ((q*α2)/4)*(B(α,a') - B(α,b')) (10)
QO = ((q*α)/4)*(C(α,a') - C(α,b')) (11)
Where a' and b' are the distances from the required free end (x = 0) to the left and right side of the load column, respectively.
3. Compute the moment (MC) and shear force (QC) required to cancel the moment and shear force at x = 0:
MC = -((q*α2)/2)*[2*(B(α,b') - B(α,a')) + (C(α,a') - C(α,b'))] (12)
QC = (q*α)*[(B(α,b') - B(α,a')) + (C(α,a') - C(α,b'))] (13)
To simplify the equations let's call the second terms of equation 12 and 13 aa and bb:
aa = 2*(B(α,b') - B(α,a')) + (C(α,a') - C(α,b')) (14)
bb = (B(α,b') - B(α,a')) + (C(α,a') - C(α,b')) (15)
4. The deflection at any point (with coordinates x) along the semi-infinite beam is equal to:
a. If point is under the load column:
u = (q/(2*k))*[(2-D(α,a)-D(α,b)) + bb*A(α,x) - aa*B(α,x)] (16)
b. If point is to the left of the load column:
u = (q/(2*k))*[(D(α,a)-D(α,b)) + bb*A(α,x) - aa*B(α,x)] (17)
c. If point is to the right of the load column:
u = (q/(2*k))*[(D(α,b)-D(α,a)) + bb*A(α,x) - aa*B(α,x)] (18)
For both, infinite and semi-infinite beams, the displacement profile is computed for each load column. The total displacement profile is computed by adding the displacement profiles of the load columns.
For elastic beams of variable thickness, the solutions for infinite and semi-infinite beams are given by Bodine (1981). These solutions are numerical and are based on finite differences. The solution for an infinite (unbroken) beam is given by Bodine in his program FMDEPYE, and for a semi-infinite (broken) beam in his program FMDEPYEC. The reader is referred to Bodine (1981) for further details about his programs.
For the same constant elastic thickness and same loads distribution, the Hetenyi and Bodine solutions don't give exactly the same results. This is because Hetenyi's solution is analytical while Bodine's solution is numerical, and also because in Hetenyi's solution loads are entered as columns, while in Bodine's solution loads change continuously along the profile. This difference will be less marked as the width of the columns is reduced.