Page 306 - Physical Principles of Sedimentary Basin Analysis

288 Flexure of the lithosphere

x = l
^ x = 0 ^
^ x
q 0 q 0
q
0
q 0
q 0

F 0

Figure 9.5. The deflection of a loaded horizontal beam clamped at one end.

Exercise 9.3 Show that the neutral surface passes through the center of mass of the plate.

Exercise 9.4 Find the deflection of a horizontal beam of length l that is clamped at one
end and loaded at the free end with a force F 0 (see Figure 9.5). The beam is also loaded
uniformly along its length by a load q 0 , which could be the weight of the beam.
Solution: The total torque at the position x is
l 1

τ(x) = F 0 (l − x) + q 0 (x − x)dx = F 0 (l − x) + q 0 (l − x) 2 (9.24)
x 2
where the first term is the torque from the force F 0 , and the second term is the torque from
the uniform load q 0 . The internal bending moment of the beam must balance the external
moment τ(x), and we have the equation
2
d w 1 2
D = F 0 (l − x) + q 0 (l − x) (9.25)
dx 2 2
3
where D = Eh /12 is the flexural rigidity of the beam. The equation for the deflection can
be rewritten as
2
d w F 0 l 3 q 0 l 4 2
x
x
= (1 −ˆ) + (1 −ˆ) (9.26)
d ˆ x 2 D 2D
x
x
using the dimensionless coordinate ˆ = x/l. The deflection w(ˆ) is obtained after
integrating twice
4
3
x
w(ˆ) = w 1 (1 −ˆx) + w 2 (1 −ˆx) + c 1 ˆx + c 2 (9.27)
4
3
where w 1 = q 0 l /(24D) and w 2 = F 0 l /(6D), and where c 1 and c 2 are integration con-
x
x
stants. The two boundary conditions w = 0 and dw/d ˆ = 0 at the clamped end (ˆ = 0)
x
give the two integration constants. The first boundary condition w = 0at ˆ = 0 gives that
c 2 =−w 1 −w 2 and the second boundary condition dw/d ˆx = 0 gives that c 1 = 4w 1 +3w 2 .
The wanted deflection of the beam is then

4
3
w(ˆ) = w 1 (1 −ˆx) + 4ˆx − 1 + w 2 (1 −ˆx) + 3ˆx − 1 (9.28)
x