Page 503 - Aircraft Stuctures for Engineering Student

484 Structural constraint

column the solution of Eq. (iii) in Example 11.2 is
d8
- = Ccoshpz + D sinhp (i)
dz
At the base of the column warping of the cross-section is suppressed so that, from
Eq. (9.65), dO/dz = 0 when z = 0. Substituting in Eq. (i) gives C = 0. The moment
couple at the top of the column is obtained from Eq. (1 1.77) and is
Mr = P~AR = -100 x 2.5 x IO3 = -25 x lo5 kNmm2
Therefore, from Eq. (11.74) and noting that Jc uz2ARtds = Mr, we have

-_ d28 - 2'5 lo5 lo3 - 0.06 10-6/~~2
dz2 200000 x 2.08 x 1O'O -
at z = 3000mm. Substitution in the differential of Eq. (i) gives D = 0.04 x so
that Eq. (i) becomes

= 0.04 x sinh 0.54 x 10-32 (ii)
dz
Integration of Eq. (ii) gives
8 = 0.08 COS^ 0.54 x 10P3z + F

At the built-in end (z = 0) 8 = 0 so that F = -0.08. Hence
8 = 0.08(~0sh0.54 x 10-3~ - 1) (iii)

At the top of the column (z = 3000mm) the angle of twist is then
8(top) = 0.08cosh0.54 x x 3000 = 0.21 rad(l2.01")
The axial load is applied through the centroid of the cross-section so that no bending
occurs and Eq. (1 1.76) reduces to

At the base of the column

(see Eq. (1 1.74))

Therefore, from Eq. (ii)

(Mr)==o = -200000 x 2.08 x 10" x 0.02 x loP6 = -83.2 x 106Nmm'
The direct stress distribution at the base of the column is then, from Eq. (iv)
100 x io3 83.2 x io6
uz = - -
400 x 5 2.08 x
or
uz = -so - 4.0 x 10-~2~~