Question 20.7: A rectangular section thin-walled beam of length L and bread......

The pressure loading is equivalent to a shear force/unit length of $3b p_{0}/2$ acting in the vertical plane of symmetry together with a torque $=3 b p_0(3 b / 2-b) / 2=3 b^2 p_0 / 4$ as shown in Fig. S.20.7. The deflection of the beam is then, from Eqs (20.14), (20.17), and (20.19),

$\Delta_T=\int_L \frac{T_0 T_1}{G J} d z$           (20.14)

$\Delta_M=\frac{1}{E} \int_L\left(\frac{M_{y, 1}M_{y, 0}}{I_{y y}}+\frac{M_{x, 1} M_{x, 0}}{I_{xx}}\right) d z$              (20.17)

$\Delta_S=\int_L\left(\int_{\text {sect }} \frac{q_0 q_1}{G t} d s\right) d z$              (20.19)

$\Delta=\int_L \frac{T_0 T_1}{G J} d z+\int_L\frac{M_{x, 1} M_{x, 0}}{E I_{x x}} dz+\int_L\left(\int_{\text {sect }} \frac{q_0 q_1}{G t} ds\right) d z$        (i)

Now

Also, from Eqs (3.12) and (18.4),

$T=G J{\frac{\mathrm{d}\theta}{\mathrm{d}z}}$          (3.12)

$\frac{ d \theta}{ d z}=\frac{T}{4 A^2} \oint \frac{ d s}{G t}$          (18.4)

Thus,

$\int_L \frac{T_0 T_1}{G J} d z=\int_0^L\frac{p_0}{4 G t}(L-z) d z=\frac{p_0 L^2}{8G t}$        (ii)

Also,

Then

in which

Thus,

$\int_L \frac{M_{x, 1} M_{x, 0}}{E I_{x x}} dz=\frac{9 p_0}{20 E b^2 t} \int_0^L(L-z)^3 d z=\frac{9p_0 L^4}{80 E b^2 t}$        (iii)

Further,

Taking the origin for s at 1 in the plane of symmetry where $q_{s,0}=0{\mathrm{~and~}}\mathrm{~since~}I_{x y}=0{\mathrm{~and~}}S_{x}=0,$ Eq. (17.15) simplifies to

$q_s=-\left(\frac{S_x I_{x x}-S_y I_{x y}}{I_{x x}I_{y y}-I_{x y}^2}\right) \int_0^s t x d s-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{xy}^2}\right) \int_0^s t y d s+q_{s, 0}$         (17.15)

Then

i.e.,

from which

Also,

i.e.,

Then

which gives

Hence

$\int_0^L\left(\int_{\text {sect }} \frac{q_0 q_1}{Gt} d s\right) d z=\frac{1359 p_0}{1000 G t}\int_0^L(L-z) d z=\frac{1359 p_0 L^2}{2000 G t}$        (iv)

Substituting in Eq. (i) from Eqs (ii)–(iv) gives

Thus,