Question 20.6: A cantilever, length L, has a hollow cross-section in the fo......
A cantilever, length L, has a hollow cross-section in the form of a doubly symmetric wedge, as shown in Fig. P.20.6. The chord line is of length c, wedge thickness is t, the length of a sloping side is a/2, and the wall thickness is constant and equal to t_{0}. Uniform pressure distributions of the magnitudes shown act on the faces of the wedge. Find the vertical deflection of point A due to this given loading. If G = 0.4E, t/c = 0.05, and L = 2c, show that this deflection is approximately 5.600p_{0}c^{2}/E t_{0}.
At any section of the beam the applied loading is equivalent to bending moments in vertical and horizontal planes, to vertical and horizontal shear forces through the shear center (the center of symmetry C) plus a torque. However, only the vertical deflection of A is required so that the bending moments and shear forces in the horizontal plane do not contribute directly to this deflection. The total deflection is, 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)
Referring to Fig. S.20.6, the vertical force/unit length on the beam is
1.2 p_0 \frac{c}{2}+p_0 \frac{c}{2}+0.8 p_0\frac{c}{2}-p_0 \frac{c}{2}=p_0 c \quad \text {(upward) }acting to the right and at a distance 0.2t above the horizontal axis of symmetry. Thus, the torque/unit length on the beam is
p_{c}c\times0.2c-p_{0}t\times0.2t=0.2p_{0}\left(c^{2}-t^{2}\right)acting in a counterclockwise sense. Then, at any section, a distance z from the built-in end of the beam,
T_{0}=0.2p_{0}(c^{2}-t^{2})(L-z)~~~~T_{1}=-1\frac{c}{2} (unit load acting upward at A)
Comparing 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 \frac{ d s}{t}}i.e.,
J=4\left(2 \frac{t}{2} \frac{c}{2}\right)^2 / \frac{2a}{t_0}=\frac{t^2 c^2 t_0}{2 a}Then
\int_0^L \frac{T_0 T_1}{G J} d z=-\int_0^L\frac{0.1 p_0\left(c^2-t^2\right) c}{G t^2 c^2t_0 / 2 a}(L-c) d z=\frac{0.1 p_0 a L^2\left(t^2-c^2\right)}{G t^2 t_0 c} (ii)
The bending moment due to the applied loading at any section a distance z from the built-in end is given by
M_{x, 0}=-\frac{p_0 C}{2}(L-z)^2 \text { also }M_{x, 1}=-1(L-z)Thus,
\int_0^L \frac{M_{x, 1} M_{x, 0}}{E I_{x x}} dz=\frac{p_0 c}{2 E I_{x x}} \int_0^L(L-z)^3 d zin which
I_{x x}=2 \frac{(a)^3 t_0 \sin ^2 \alpha}{12}=\frac{a^3 t_0}{6}\left(\frac{t / 2}{a /2}\right)^2=\frac{a t^2 t_0}{6}Then
\int_0^L \frac{M_{x, 1} M_{x, 0}}{E I_{x x}} dz=\frac{3 p_0 c}{E a t^2 t_0}\left[-\frac{1}{4}(L-z)^4\right]_0^L=\frac{3 p_0 c L^4}{4 E a t^2 t_0} (iii)
The shear load at any section a distance z from the built-in end produced by the actual loading system is given by
S_{y, 0}=p_0 c(L-z) \text { also } S_{y, 1}=1From Eq. (17.15), in which I_{x y}=0{\mathrm{~and~}}S_{x}{=}0,
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 s+q_{s, 0} (iv)
If the origin of s is taken at the point l, q_{s,0}=0, since the shear load is applied on the vertical axis of symmetry, Eq. (iv) then becomes
q_s=-\frac{S_y}{I_{x x}} \int_0^s t y d sand
q_{12}=-\frac{6 S_y}{a t^2 t_0} \int_0^st_0\left(-\frac{t}{2}+s \sin \alpha\right) d si.e.,
q_{12}=\frac{6 S_y}{a t^2}\left(\frac{t}{2} s-\frac{t}{a} \frac{s^2}{2}\right)Thus,
q_{12}=\frac{3 S_y}{a t}\left(s-\frac{s^2}{a}\right)The remaining distribution follows from symmetry. Then
\int_{\text {sect }} \frac{q_0 q_1}{G t} d s=4\times \frac{9 p_0 c(L-z)}{G a^2 t^2 t_0} \int_0^{a /2}\left(s-\frac{s^2}{a}\right)^2 d si.e.,
\int_{\text {sect }} \frac{q_0 q_1}{G t} d s=\frac{3p_0 c a(L-z)}{5 G t^2 t_0}Then
\int_0^L\left(\int_{\text {sect }} \frac{q_0 q_1}{Gt} d s\right) d z=\frac{3 p_0 c a}{5 G t^2t_0}\int_0^L(L-z) d z=\frac{3 p_0 c a L^2}{10 Gt^2t_0} (v)
Now substituting in Eq. (i) from Eqs (ii), (iii), and (v),
\Delta=\frac{0.1 p_0 a L^2\left(t^2-c^2\right)}{Gt^2 t_0 c}+\frac{3 p_0 c L^4}{4 E a t^2 t_0}+\frac{3p_0 c a L^2}{10 G t^2 t_0}i.e.,
\Delta=\frac{p_0 L^2}{t^2t_0}\left[\frac{a\left(t^2-c^2\right)}{10 Gc}+\frac{3 c L^2}{4 E a}+\frac{3 c a}{10 G}\right]Substituting the given values and taking a \bumpeq c
\Delta=\frac{p_0(2 c)^2}{(0.05 c)^2t_0}\left[\frac{c\left[(0.05 c)^2-c^2\right]}{4E}+\frac{3 c(2 c)^2}{4 E c}+\frac{3 c^2}{4 E}\right]Neglecting the term (0.05 c)^2 \text { in}\left[(0.05 c)^2-c^2\right] gives
\Delta=\frac{5600 p_0 c^2}{E t_0}