Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

21.3 Torsion 583

Fig.21.4

Alternative solution of Example 21.2.

Substitutingforq 32 ,q 43 ,andq 54 fromthepreceding,weobtain

4 ( 381 q 21 +18740.5)= 100 × 103

fromwhich

q 21 =16.4N/mm

and

q 32 =46.7N/mm, q 43 =69.9N/mm, q 54 =83.4N/mm,andsoon

Theshearflowdistributionduetotheappliedtorqueis,fromEq.(21.2)

q=

15 × 106

2 ×4.56× 105

=16.4N/mm

acting in an counterclockwise sense completely around the section. This value of shear flow is now
superimposedontheshearflowsproducedbytheshearload;thisgivesthesolutionshowninFig.21.3;
thatis,

q 21 =16.4+16.4=32.8N/mm

q 161 =16.4−16.4=0,andsoon