Page 417 - Aircraft Stuctures for Engineering Student

398 Stress analysis of aircraft components

The actual shear flows are obtained by factoring the final shear flows by the ratio of
the applied torque to the torque corresponding to the final shear flows, Le.

Example 10.10
Solve Example 10.7 using the method of successive approximations.

From Example 10.7
61 = 1542 + 250 = 1792

611 = 250 + 725 + 233 + 725 = 1933

6111 = 736 + 233 + 736 + 368 = 2073
61,~ = 250

611,111 = 233

Hence, from Eqs (10.34)
250
CI,II = - 0.129
=
1933
250
CII!J = - 0.140
=
1792
233
C,,,J,, = - 0.1 12
=
2073
233
CII1,II = - 0.121
=
1933
Also
258000 = 287.9N/mm
‘I = 1792
355 Oo0 = 367.3 N/mm
‘I1 = 1933

The remainder of the solution is completed in Table 10.6. Note that the assumed
values of qI, qII and qIII are rounded off since the method is approximate.
The solution is virtually identical to that of Example 10.7. Note that the variation
in shear modulus is treated in the same way as that in Example 10.7, i.e. a reference
value is chosen and the actual thicknesses converted to modulus weighted thicknesses
t*; 6 is then Jds/t*.