Question 27.5: Figure P.27.5 shows the doubly symmetrical idealized cross-s......
Figure P.27.5 shows the doubly symmetrical idealized cross-section of a uniform box beam of length l. Each of the four corner booms has area B and Young’s modulus E, and they constitute the entire direct stress carrying area. All the thin walls have the same shear modulus G. The beam transmits a torque T from one end to the other, and at each end, warping is completely suppressed. Between the ends, the shape of the cross-section is maintained without further restriction of warping. Obtain an expression for the distribution of the end load along the length of one of the corner booms. Assuming b t_{1}\gt a t_{2}, indicate graphically the relation between torque direction and tension and compression in the boom end loads.
The warping distribution along the top right-hand corner boom is given by Eq. (27.16), i.e.,
w=C \cosh \mu z+D \sinh \mu z+\frac{T}{8 a bG}\left(\frac{b}{t_b}-\frac{a}{t_a}\right) (27.16)
w=C\cosh\mu z+D\sinh\mu z+w_{0} (i)
where
\mu^{2}=\frac{8G t_{2}t_{1}}{B E(b t_{1}+a t_{2})}\;\;\mathrm{and}\;\;w_{0}=\frac{T}{8a b G}\left(\frac{b}{t_{2}}-\frac{a}{t_{1}}\right)At each end of the beam the warping is completely suppressed, i.e., w = 0 at z = 0 and z=l. Thus, from
Eq. (i),
i.e.,
\mathrm{C}=-w_{0}and
0=C\cosh\mu l+D\sinh\mu l+w_{0}which gives
D={\frac{w_{0}}{\sinh\mu l}}(\cosh\mu l-1)Hence, Eq. (i) becomes
w=w_{0}\left[1-\cosh\mu z+{\frac{(\cosh\mu l-1)}{\sinh\mu l}}\sinh\mu z\right] (ii)
The direct load, P, in the boom is then given by
P=\sigma_{z}B=B E{\frac{\partial w}{\partial z}}Thus, from Eq. (ii),
P=\mu B E w_{0}\left[-\sinh\mu z+{\frac{(\cosh\mu l-1)}{\sinh\mu l}}\cosh\mu z\right] (iii)
or, substituting for w_{0} from above,
P={\frac{\mu B E T}{8a b G t_{1}t_{2}}}(b t_{1}-a t_{2})\left[-\sinh\mu z+{\frac{(\cosh\mu l-1)}{\sinh\mu l}}\cosh\mu z\right] (iv)
For a positive torque – i.e., T is counterclockwise when viewed along the z axis to the origin of z – the term in square brackets in Eq. (iv) becomes, when z = 0,
\frac{\cosh\mu l-1}{\sinh\mu l}which is positive. Thus at z = 0, the load in the boom is tensile. At z = l, the term in square brackets in Eq. (iv) becomes
\frac{1-\cosh\mu l}{\sinh\mu l}which is negative. Thus at z=l, the load in the boom is compressive. Also, from Eq. (iv), ∂P/∂z = 0 at z=l/2, and the distribution of boom load is that shown in Fig. S.27.5. The reverse situation occurs for a negative, i.e., a clockwise, torque.
