Page 483 - Aircraft Stuctures for Engineering Student

464 Structural constraint

Also substituting for PSI and PB2 from Eqs (1 1.40), we obtain

PGt

or
a2PA ~2p - PGt (11.44)
--
a9 A--- dEB
where X2 = Gt(2B + A)/dEAB. The solution of Eq. (11.44) is of standard form
and is

PA
PA = CcoshAz+DsinhXz+- (1 1.45)
2B+A
.The boundary conditions are PA = 0 when z = 0 and q1 = q2 = 0 = dPA/az at the
built-in end (no shear loads are applied). Hence

PA =- (1 - cosh XZ + tanh XL sinh Xz)
2B+A
or, rearranging

P -- [l- cash X(L - Z) (1 1.46)
A-2B+A cosh XL
Hence
cash X(L - Z)
CA = - (1 1.47)
[l-
2B+A cosh XL
Substituting for PA in the second of Eqs (1 1.40), we have

4B + A cash X(L - Z) 1 (1 1.48)
P ” - 2(2B + A) [-+ A coshXL
-
whence
4B + A cash X(L - Z) 1
nB1 = 2B(t+ A) [ A+ (1 1.49)
coshXL
Also from Eqs (1 1.40)

so that
1
P- cash X(L - Z) (1 1.50)
cosh AL
and
-PA [ - cash X(L - z)]
nB2 = (1 1.51)
2B(2B + A) cosh XL