Question 27.4: A uniform, four-boom beam, built in at one end, has the rect......
A uniform, four-boom beam, built in at one end, has the rectangular cross-section shown in Fig. P.27.4. The walls are assumed to be effective only in shear, the thickness and shear modulus being the same for all walls, while the booms, which are of equal area, carry only direct stresses. Assuming that the cross-section remains undistorted by the loading, calculate the twist at the free end due to a uniformly distributed torque loading T = 20 Nm/mm along its entire length. Take G = 20,000 N/mm² and G/E = 0.36.
From Fig. P.27.4, the torque at any section of the beam is given by
T=20\times10^{3}\left(2500-z\right)\mathrm{Nm}\mathrm{m} (i)
Eq. (27.16) for the warping distribution along boom 4 then becomes
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+\frac{20\times10^{3}(2500-z)}{8a b G}{\left(\frac{b}{t_{b}}-\frac{a}{t_{a}}\right)} (ii)
where
\mu^2=\frac{8 G t_b t_a}{B E\left(b t_a+a t_b\right)}Comparing Figs 27.6 and P.27.4, t_{b}=t_{a}=1.0\;\mathrm{mm},a=500\;\mathrm{mm},b=200\;\mathrm{mm},\,\mathrm{and}\,B=800\;\mathrm{mm}^{2}. Then
\mu^{2}=\frac{8\times0.36\times1.0\times1.0}{800(200\times1.0+500\times1.0)}from which
\mu=2.27\times10^{-3}Eq. (ii) then becomes
w=C\cosh2.27\times10^{-3}z+D\sinh2.27\times10^{-3}z-3.75\times10^{-4}(2500-z) (iii)
When z = 0, w = 0, hence, from Eq. (iii), C = 0.9375. At the free end the direct stress in boom 4 is zero so that the direct strain ∂w/∂z = 0 at the free end. Hence, from Eq. (iii), D = –0.9386 and the warping distribution along boom 4 is given by
w=0.9375\cosh2.27\times10^{-3}z-0.9386\sinh2.27\times10^{-3}z-3.75\times10^{-3}(2500-z) (iv)
Substituting for w from Eq. (iv) and T from Eq. (i) in Eq. (27.11),
\frac{ d \theta}{ d z}=\frac{4 w\left(b t_a-at_b\right)}{a b\left(b t_a+a t_b\right)}+\frac{2 T}{a bG\left(b t_a+a t_b\right)} (27.11)
{\frac{\mathrm{d}\theta}{\mathrm{d}z}}=-10^{-5}[1.6069\cosh2.27\times10^{-3}z-1.6088\sinh2.27\times10^{-3}z -3.4998\times10^{-3}(2500-z)] (v)
Then
When z = 0, θ = 0 so that, from Eq. (vi),
F=-10^{-5}\times{\frac{1.6088}{2.27\times10^{-3}}}and
At the free end where z = 2500 mm, Eq. (vii) gives
\theta=0.1036\,\mathrm{rad}=5.9^{\circ}