Question 27.3: A singly symmetrical, thin-walled, closed section beam is bu......

Referring to Fig. S.27.3, the shear flows in the walls 12 and 23 are constant, since the walls are straight (see Eq. (27.1)). Choosing O as the origin of axes, from Eq. (27.1),

q=G t\left(p \frac{ d \theta}{ d z}+\frac{ d u}{ d z}\cos \psi+\frac{ d υ}{ d z} \sin \psi\right)         (27.1)

i.e.,

q_{12}=G t(0.8666R\theta^{\prime}+0.866u^{\prime}+0.5υ^{\prime})          (i)

q_{23}=G t(0.866R\theta^{\prime}-0.866u^{\prime}+0.5υ^{\prime})          (ii)

q_{31}=G t(R\theta^{\prime}-u^{\prime}\cos\phi-υ^{\prime}\sin\phi)        (iii)

Resolving forces vertically,

i.e.,

q_{12}+q_{23}-\int_0^\pi q_{31} \sin \phi d\phi=\frac{5000}{R}          (iv)

Substituting in Eq. (iv) for q_{12},\,q_{23},\,\mathrm{and}\,\,q_{31} from Eqs (i)–(iii), respectively, gives

R \theta^{\prime}-9.59 υ^{\prime}=-\frac{18656.7}{G t R}        (v)

Resolving forces horizontally,

i.e.,

1.732 q_{12}-1.732 q_{23}-\int_0^\pi q_{31} \cos \phi d \phi=\frac{8660.3}{R}        (vi)

Substituting in Eq. (vi) for q_{12},\,q_{23},\,\mathrm{and}\,q_{31} from Eqs (i)–(iii), respectively, gives

u^{\prime}=\frac{1894.7}{G t R}        (vii)

Taking moments about O,

i.e.,

1.732 q_{12}+1.732 q_{23}+\int_0^\pi q_{31} d \phi=\frac{8660.3}{R}          (viii)

Substituting in Eq. (viii) for q_{12},\,q_{23},\mathrm{~and~}q_{31} from Eqs (i)–(iii), respectively, gives

R\theta^{\prime}-0.0444υ^{\prime}={\frac{1410.0}{G t R}}        (ix)

Now, subtracting Eq. (ix) from (v),

whence,

υ^{\prime}={\frac{2102.1}{G t R}}        (x)

Then, from Eq. (v),

R\theta^{\prime}={\frac{1502.4}{G t R}}          (xi)

Substituting for u′, υ′, and θ′ from Eqs (vii), (x), and (xi), respectively, in Eqs (i)–(iii) gives