Question 20.7: A rectangular section thin-walled beam of length L and bread......
A rectangular section thin-walled beam of length L and breadth 3b, depth b, and wall thickness t is built-in at one end (Fig. P.20.7). The upper surface of the beam is subjected to a pressure which varies linearly across the breadth from a value p_{0} at edge AB to zero at edge CD. Note that the complete upper surface of the beam is subjected to this pressure. Find the vertical deflection of point A.
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
T_0=3 b^2 p_0(L-z) / 4 \quad T_1=3 b / 2Also, 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)
J=\frac{4 A^2}{\oint d s / t}=\frac{4\left(3b^2\right)^2}{8 b / t}=\frac{9 b^3 t}{2}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,
M_{x, 0}=3 b p_0(L-z)^2 / 4 \quad M_{x, 1}=1(L-z)Then
\int_L \frac{M_{x, 1} M_{x, 0}}{E I_{x x}} dz=\int_0^L \frac{3 b p_0}{4 E I_{x x}}(L-z)^3 d zin which
I_{x x}=2 \times 3 b t \times(b / 2)^2+2 t b^3 /12=5 b^3 t / 3Thus,
\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,
S_{y, 0}=-\frac{3 b p_0}{2}(L-z) \quad S_{y, 1}=-1Taking 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)
q_s=-\frac{S_y}{I_{x x}} \int_0^s t y d sThen
q_{12}=-\frac{3 S_y}{5 b^3 t} \int_0^{s_1}t\left(\frac{b}{2}\right) d s_1i.e.,
q_{12}=-\frac{3 S_y}{10 b^2} s_1from which
q_2=-\frac{9 S_y}{20 b}Also,
q_{23}=-\frac{S_y}{I_{x x}} \int_0^{s_2}t\left(\frac{b}{2}-s_2\right) d s_2-\frac{9S_y}{20 b}i.e.,
q_{23}=-\frac{3 S_y}{5 b^3}\left(\frac{b}{2} s_2-\frac{s_2^2}{2}\right)-\frac{9 S_y}{20 b}Then
\int_{\text {sect }} \frac{q_0 q_1}{G t} d s=4\int_0^{3 b / 2} \frac{3 b p_0(L-z)}{2 Gt}\left(\frac{3}{10 b^2}\right)^2 s_1^2 d s_1 +2\int_0^b \frac{3 b p_0(L-z)}{2 G t}\left(\frac{3}{20b}\right)^2\left(2 \frac{s_2}{b}-2 \frac{s_2^2}{b^2}+3\right)^2 d s_2which gives
\int_{\text {sect }} \frac{q_0 q_1}{G t} ds=\frac{1359 p_0}{1000 G t}(L-z)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
\Delta=\frac{p_0 L^2}{8 G t}+\frac{9 p_0 L^4}{80 E b^2 t}+\frac{1359 p_0 L^2}{2000 G t}Thus,
\Delta=\frac{p_0 L^2}{t}\left(\frac{9 L^2}{80 Eb^2}+\frac{1609}{2000 G}\right)