Question 16.4: A thin-walled cantilever with walls of constant thickness t ......
A thin-walled cantilever with walls of constant thickness t has the cross-section shown in Fig. P.16.4. It is loaded by a vertical force W at the tip and a horizontal force 2W at the mid-ection, both forces acting through the shear center. Determine and sketch the distribution of direct stress, according to the basic theory of bending, along the length of the beam for the points 1 and 2 of the cross-section. The wall thickness t can be taken as very small in comparison with d in calculating the sectional properties I_{x x},I_{x y}, and so forth.
Answer: \sigma_{z,1}\,(\mathrm{mid – point})=-0.05\;W l/t d^{2},\;\mathrm{\sigma}_{z,1}\,(\mathrm{built – in~end})=-1.85\;W l/t d^{2},
\sigma_{z,2}\:(\mathrm{mid – point})=-0.63\:W l/t d^{2},\;\mathrm{\sigma}_{z,2}\:(\mathrm{built – in~end})=0.1\,\,W l/t d^{2}
Referring to Fig. S.16.4.
In DB,
M_{x}=-W(l-z) (i)
M_{y}=0In BA,
M_{x}=-W(I-z) (ii)
M_{y}=-2W\left({\frac{l}{2}}-z\right) (iii)
Now referring to Fig. P.16.4, the centroid of area, C, of the beam cross-section is at the center of antisymmetry. Then
I_{x x}=2\left[t d\left({\frac{d}{2}}\right)^{2}+{\frac{t d^{3}}{12}}\right]={\frac{2t d^{3}}{3}}I_{y y}=2\left[td\left(\frac{d}{4}\right)^{2}+\frac{t d^{3}}{12}+t d\left(\frac{d}{4}\right)^{2}\right]=\frac{5t d^{3}}{12}
I_{x y}=t d\left({\frac{d}{4}}\right)\left({\frac{d}{2}}\right)+t d\left(-{\frac{d}{4}}\right)\left(-{\frac{d}{2}}\right)={\frac{t d^{3}}{4}}
Substituting for I_{x x},\,I_{y y},\,\mathrm{and}\ I_{x y} in Eq. (16.17) gives
\sigma_{z}=\left(\frac{M_{y}I_{x x}-M_{x}I_{x y}}{I_{x x}I_{y y}-I_{x y}^{2}}\right)x+\left(\frac{M_{x}I_{y y}-M_{y}I_{x y}}{I_{x x}I_{y y}-I_{x y}^{2}}\right)y (16.17)
\sigma_{z}=\frac{1}{t d^{3}}[(3.10M_{y}-1.16M_{x})x+\left(1.94M_{x}-1.16M_{y})y\right) (iv)
Along the edge l, x = 3d/4, y=d/2. Equation (iv) then becomes
\sigma_{z,1}={\frac{1}{t d^{2}}}{\big(}1.75M_{y}+0.1M_{x}{\big)} (v)
Along the edge 2, x = –d/4, y=d/2. Equation (iv) then becomes
\sigma_{z,2}={\frac{1}{t d^{2}}}{\left(-1.36M_{y}+1.26M_{x}\right)} (vi)
From Eqs (i)–(iii), (v), and (vi),
In DB,
In BA,
\sigma_{z,1}={\frac{W}{t d^{2}}}(3.6z-1.85l)\ \ \mathrm{whence}\,\sigma_{z,1}(\mathrm{A})=-{\frac{1.85Wl}{t d^{2}}}\sigma_{z,2}={\frac{W}{t d^{2}}}(-1.46z+0.1l)~{\mathrm{~whence}}\,\sigma_{z,2}({A})={\frac{0.1W l}{t d^{2}}}
