Question 20.5: A uniform beam with the cross-section shown in Fig. P.20.5(a......
A uniform beam with the cross-section shown in Fig. P.20.5(a) is supported and loaded as shown in Fig. P.20.5(b). If the direct and shear stresses are given by the basic theory of bending, the direct stresses being carried by the booms and the shear stresses by the walls, calculate the vertical deflection at the ends of the beam when the loads act through the shear centers of the end cross-sections, allowing for the effect of shear strains. Take E = 69,000 N/mm² and G = 26,700 N/mm². Boom areas:
1, 3, 4, 6 = 650 mm²; 2, 5 = 1,300 mm².
Referring to Fig. S.20.5(a), the x axis of the beam cross-section is an axis of symmetry so that I_{x y}=0. Further, S_{{y}} at the end A is equal to –4450 N and S_{x}=0. The total deflection, Δ, at one end of the beam is then, from Eqs (20.17) and (20.19),
\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{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)
in which q_{0}, from Eqs (20.20) and (20.11), is given by
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)\left(\int_0^s t_{ D } x ds+\sum\limits_{r=1}^n B_r x_r\right)-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right)\left(\int_0^s t_{ D } y d s+\sum\limits_{r=1}^n B_r y_r\right)+q_{s, 0} (20.11)
q_0=-\left(\frac{S_{x, 0} I_{x x}-S_{y, 0} I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right)\left(\int_0^s t_{ D } x ds+\sum\limits_{r=1}^n B_r x_r\right) -\left(\frac{S_{y, 0} I_{yy}-S_{x, 0} I_{x y}}{I_{x x} I_{y y}-I_{xy}^2}\right)\left(\int_0^s t_{ D } y d s+\sum\limits_{r=1}^n B_r y_r\right) (20.20)
q_{0}=-\frac{S_{y,0}}{I_{x x}}\sum\limits_{r=1}^{n}B_{r}y_{r}+q_{s,0} (ii)
and
q_{1}={\frac{q_{0}}{4450}}Since the booms carry all the direct stresses, I_{x x} in Eq. (i) is, from Fig. S.20.5(a),
I_{x x}=2\times650\times100^{2}+2\times650\times75^{2}+2\times1300\times100^{2}=46.3\times10^{6}~\mathrm{mm}^{4}Also, from Fig. S.20.5(b) and taking moments about C,
R_{\mathrm{B}}\times500-4450\times1750-4450\times1250=0from which
R_{\mathrm{B}}=26\,700\,\mathrm{N}Therefore in AB,
M_{x,0}=4450z\;\;M_{x,1}=zand in BC,
M_{x,0}=33.4\times10^{6}-22250z\;\;M_{x,1}=7500-5zThus the deflection, \Delta_{M}, due to bending at the end A of the beam, is, from the first term on the righthand side of Eq. (i),
\Delta_M=\frac{1}{E I_{x x}}\left\{\int_0^{1250}4450 z^2 d z+\int_{1250}^{1500}4450(7500-5 z)^2 d z\right\}i.e.,
\Delta_M=\frac{4450}{69000 \times 46.3 \times 10^6}\left\{\left[\frac{z^3}{3}\right]_0^{1250}-\frac{1}{15}\left[(7500-5 z)^3\right]_{1250}^{1500}\right\}from which
\Delta_{M}=1.09\,\mathrm{mm}Now ‘cut’ the beam section in the wall 12. From Eq. (20.11), i.e.,
q_{s}=-{\frac{S_{y}}{I_{x x}}}\sum\limits_{r=1}^{n}B_{r}y_{r}+q_{s,0} (iii)
The remaining distribution follows from symmetry. The shear load is applied through the shear center of the cross-section so that dθ/dz = 0 and q_{s,0} is given by Eq. (17.28), i.e.,
q_{s, 0}=-\frac{\oint q_{ b } d s}{\oint d s} (17.28)
q_{s, 0}=-\frac{\oint q_{ b } d s}{\oint d s}\quad(t=\text { constant })in which
\oint d s=2 \times 300+2 \times 250+2 \times 100+2\times 75=1450 mmi.e.,
q_{s,0}=-\frac{2S_{y}}{1450I_{x x}}(-130000\times250-195000\times100+48750\times75)from which
q_{s,0}=66\,681S_{y}/I_{x x}Then
\begin{aligned}& q_{12}=66681 S_y / I_{x x} \\& q_{23}=-63319 S_y / I_{x x} \\& q_{34}=-128319 S_y /I_{x x} \\& q_{16}=-115431 S_y / I_{x x}\end{aligned}Therefore the deflection, \Delta_{S}, due to shear is, from the second term in Eq. (i),
\Delta_S=\int_L\left(\int_{\text {sect }} \frac{q_0 q_1}{G t} d s\right) d zi.e.,
Thus,
\Delta_S=\int_L 2 \frac{S_{y, 0} S_{y, 1} \times 4.98\times 10^{12}}{26700 \times 2.5 \times\left(46.3\times 10^6\right)^2} d z=\int_L 6.96\times 10^{-8} S_{y, 0} S_{y, 1} d zThen
\Delta_S=\int_0^{1250} 6.96 \times 10^{-8} \times 4450 \times 1 d z+\int_{1250}^{1500} 6.96 \times 10^{-8}\times 22250 \times 5 d zfrom which
\Delta_{S}=2.32\,{\mathrm{mm}}The total deflection, Δ, is then
\Delta=\Delta_{M}+\Delta_{S}=1.09+2.32=3.41\,{\mathrm{mm}}