Discussion
Applying \( 90 \) kN m to that box, the rate of twist is nearest to
Answer: D
The rate is \( T/GJ = 90\,000/3.04 \times 10^6 = 0.0296 \) radians per metre, which is \( 0.0296 \times 57.3 = 1.70 \) degrees per metre. Over a ten-metre semi-span that would accumulate to seventeen degrees, which shows why wing torsional rigidity is watched so closely.
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