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 t0. Uniform pressure distributions

Question

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 [latex]t_0[/latex]. Uniform pressure distributions of 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 [latex]5600 p_0 c^2 / E t_0[/latex].

Accepted Answer

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 centre (the centre 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)

[latex]\Delta_T=\int_L \frac{T_0 T_1}{G J} \mathrm{~d} z[/latex] (20.14)

[latex]\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_{x x}} ight) \mathrm{d} z[/latex] (20.17)

[latex]\Delta_S=\int_L\left(\int_{ ext {sect }} \frac{q_0 q_1}{G t} \mathrm{~d} s ight) \mathrm{d} z[/latex] (20.19)

[latex]\Delta=\int_L \frac{T_0 T_1}{G J} \mathrm{~d} z+\int_L \frac{M_{x, 1} M_{x, 0}}{E I_{x x}} \mathrm{~d} z+\int_L\left(\int_{ ext {sect }} \frac{q_0 q_1}{G t} \mathrm{~d} s ight) \mathrm{d} z[/latex] (i)

Referring to Fig. S.20.6 the vertical force/unit length on the beam is

[latex]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[/latex] (upwards)

acting at a distance of 0.2c to the right of the vertical axis of symmetry. Also the horizontal force/unit length on the beam is

[latex]1.2 p_0 \frac{t}{2}+p_0 \frac{t}{2}+0.8 p_0 \frac{t}{2}-p_0 \frac{t}{2}=p_0 t[/latex]

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
[latex]p_0 c imes 0.2 c-p_0 t imes 0.2 t=0.2 p_0\left(c^2-t^2 ight)[/latex]
acting in an anticlockwise sense. Then, at any section, a distance z from the built-in end of the beam

[latex]T_0=0.2 p_0\left(c^2-t^2 ight)(L-z) \quad T_1=-1 \frac{c}{2}[/latex] (unit load acting upwards at A)

Comparing Eqs (3.12) and (18.4)

[latex]T=G J \frac{\mathrm{d} heta}{\mathrm{d} z}[/latex] (3.12)

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{T}{4 A^2} \oint \frac{\mathrm{d} s}{G t}[/latex] (18.4)

[latex]J=\frac{4 A^2}{\oint \frac{\mathrm{d} s}{t}}[/latex]

i.e.

[latex]J=4\left(2 \frac{t}{2} \frac{c}{2} ight)^2 / \frac{2 a}{t_0}=\frac{t^2 c^2 t_0}{2 a}[/latex]

Then

[latex]\int_0^L \frac{T_0 T_1}{G J} \mathrm{~d} z=-\int_0^L \frac{0.1 p_0\left(c^2-t^2 ight) c}{G t^2 c^2 t_0 / 2 a}(L-z) \mathrm{d} z=\frac{0.1 p_0 a L^2\left(t^2-c^2 ight)}{G t^2 t_0 c}[/latex] (ii)

The bending moment due to the applied loading at any section a distance z from the built-in end is given by

[latex]M_{x, 0}=-\frac{p_0 c}{2}(L-z)^2 \quad ext { also } M_{x, 1}=-1(L-z)[/latex]

Thus

[latex]\int_0^L \frac{M_{x, 1} M_{x, 0}}{E I_{x x}} \mathrm{~d} z=\frac{p_0 c}{2 E I_{x x}} \int_0^L(L-z)^3 \mathrm{~d} z[/latex]

in which

[latex]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} ight)^2=\frac{a t^2 t_0}{6}[/latex]

Then

[latex]\int_0^L \frac{M_{x, 1} M_{x, 0}}{E I_{x x}} \mathrm{~d} z=\frac{3 p_0 c}{E a t^2 t_0}\left[-\frac{1}{4}(L-z)^4 ight]_0^L=\frac{3 p_0 c L^4}{4 E a t^2 t_0}[/latex] (iii)

The shear load at any section a distance z from the built-in end produced by the actual loading system is given by

[latex]S_{y, 0}=p_0 c(L-z) \quad ext { also } S_{y, 1}=1[/latex]

From Eq. (17.15), in which [latex]I_{x y}=0 ext { and } S_x[/latex] = 0

[latex]q_s=-\left(\frac{S_x I_{x x}-S_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight) \int_0^s t x \mathrm{~d} s-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight) \int_0^s t y \mathrm{~d} s+q_{s, 0}[/latex] (17.15)

[latex]q_s=-\frac{S_y}{I_{x x}} \int_0^s t y \mathrm{~d} s+q_{s, 0}[/latex] (iv)

If the origin of s is taken at the point l, [latex]q_{s, 0}=0[/latex] since the shear load is applied on the vertical axis of symmetry, Eq. (iv) then becomes

[latex]q_s=-\frac{S_y}{I_{x x}} \int_0^s t y \mathrm{~d} s[/latex]

and

[latex]q_{12}=-\frac{6 S_y}{a t^2 t_0} \int_0^s t_0\left(-\frac{t}{2}+s \sin \alpha ight) \mathrm{d} s[/latex]

i.e.

[latex]q_{12}=\frac{6 S_y}{a t^2}\left(\frac{t}{2} s-\frac{t}{a} \frac{s^2}{2} ight)[/latex]

Thus

[latex]q_{12}=\frac{3 S_y}{a t}\left(s-\frac{s^2}{a} ight)[/latex]

The remaining distribution follows from symmetry. Then

[latex]\int_{ ext {sect }} \frac{q_0 q_1}{G t} \mathrm{~d} s=4 imes \frac{9 p_0 c(L-z)}{G a^2 t^2 t_0} \int_0^{a / 2}\left(s-\frac{s^2}{a} ight)^2 \mathrm{~d} s[/latex]

i.e.

[latex]\int_{ ext {sect }} \frac{q_0 q_1}{G t} \mathrm{~d} s=\frac{3 p_0 c a(L-z)}{5 G t^2 t_0}[/latex]

Then

[latex]\int_0^L\left(\int_{ ext {sect }} \frac{q_0 q_1}{G t} \mathrm{~d} s ight) \mathrm{d} z=\frac{3 p_0 c a}{5 G t^2 t_0} \int_0^L(L-z) \mathrm{d} z=\frac{3 p_0 c a L^2}{10 G t^2 t_0}[/latex] (v)

Now substituting in Eq. (i) from Eqs (ii), (iii) and (v)

[latex]\Delta=\frac{0.1 p_0 a L^2\left(t^2-c^2 ight)}{G t^2 t_0 c}+\frac{3 p_0 c L^4}{4 E a t^2 t_0}+\frac{3 p_0 c a L^2}{10 G t^2 t_0}[/latex]

i.e.

[latex]\Delta=\frac{p_0 L^2}{t^2 t_0}\left[\frac{a\left(t^2-c^2 ight)}{10 G c}+\frac{3 c L^2}{4 E a}+\frac{3 c a}{10 G} ight][/latex]

Substituting the given values and taking [latex]a \simeq c[/latex]

[latex]\Delta=\frac{p_0(2 c)^2}{(0.05 c)^2 t_0}\left[\frac{c\left[(0.05 c)^2-c^2 ight]}{4 E}+\frac{3 c(2 c)^2}{4 E c}+\frac{3 c^2}{4 E} ight][/latex]

Neglecting the term (0.05c)² in [(0.05c)² − c²] gives

[latex]\Delta=\frac{5600 p_0 c^2}{E t_0}[/latex]

Question p.20.6: A cantilever, length L, has a hollow cross-section in the fo...

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