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M'

= 0

(zero shear force).

We can now write the homogeneous part of the equations of equilibrium as e" - —w'" + ^-¡9cos2(A) - sin(A) cos(A) = 0,

GJ GJ GJ

K — /// qcci , qcci— -, w""--6 + — u/sin(A)cos(A) - — <9cos2(A) = 0.

EI EI EI

Differentiating the first equation with respect to y and transforming the set of equations so that the highest derivative terms, 6 and w"", have coefficients of unity, we obtain

Multiplying the first equation by cos( A ) and the second equation by sin( A ) and subtracting the second equation from the first, we obtain a single equation in terms of 6—6 cos( A ) — w' sin(A) as

EI GJ qeccil2 ,

EIGJ-K2 G J

0 0

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