Question 21.2: The cantilever beam shown in Fig. 21.6 is uniformly tapered ...
The cantilever beam shown in Fig. 21.6 is uniformly tapered along its length in both x and y directions and carries a load of 100 kN at its free end. Calculate the forces in the booms and the shear flow distribution in the walls at a section 2 m from the built-in end if the booms resist all the direct stresses while the walls are effective only in shear. Each corner boom has a cross-sectional area of 900 mm² while both central booms have cross-sectional areas of 1200 mm².
the internal force system at a section 2 m from the built-in end of the beam is
S_y=100 \mathrm{kN} \quad S_x=0 \quad M_x=-100 \times 2=-200 \mathrm{kN} \mathrm{m} \quad M_y=0
The beam has a doubly symmetrical cross-section so that I_{xy} = 0 and Eq. (16.18) reduces to
\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 (16.18)
\sigma_z=\frac{M_x y}{I_{x x}} (i)
in which, for the beam section shown in Fig. 21.6(b)
Then
\sigma_{z, r}=\frac{-200 \times 10^6}{5.4 \times 10^8} y_r
or
\sigma_{z, r}=-0.37 y_r (ii)
Hence
P_{z, r}=-0.37 y_r B_r (iii)
The value of P_{z, r} is calculated from Eq. (iii) in column ② in Table 21.1; P_{x, r} and P_{y, r} follow from Eqs (21.10) and (21.9), respectively in columns ⑤ and ⑥. The axial load P_r, column ⑦, is given by [②² + ⑤² + ⑥²]^{ 1/2} and has the same sign as P_{z, r} (see Eq. (21.12)). The moments of P_{x, r} and P_{y, r} are calculated for a moment centre at the centre of symmetry with anticlockwise moments taken as positive. Note that in Table 21.1, P_{x, r} and P_{y, r} are positive when they act in the positive directions of the section x and y axes, respectively; the distances η_r and ξ_r of the lines of action of P_{x, r} and
Table 21.1
| ① | ② | ③ | ④ | ⑤ | ⑥ | ⑦ | ⑧ | ⑨ | ⑩ | ⑪ |
| P_{z,r} | δx_r/δz | δy_r/δz | P_{x,r} | P_{y,r} | P_r | ξ_r | η_r | P_{x,r}η_r | P_{y,r}ξ_r | |
| Boom | (KN) | (KN) | (KN) | (KN) | (m) | (m) | (KN m) | (KN m) | ||
| 1 | -100 | 0.1 | -0.05 | -10 | 5 | -101.3 | 0.6 | 0.3 | 3 | -3 |
| 2 | -133 | 0 | -0.05 | 0 | 6.7 | -177.3 | 0 | 0.3 | 0 | 0 |
| 3 | -100 | -0.1 | -0.05 | 10 | 5 | -101.3 | 0.6 | 0.3 | -3 | 3 |
| 4 | 100 | -0.1 | 0.05 | -10 | 5 | 101.3 | 0.6 | 0.3 | -3 | 3 |
| 5 | 133 | 0 | 0.05 | 0 | 6.7 | 177.3 | 0 | 0.3 | 0 | 0 |
| 6 | 100 | 0.1 | 0.05 | 10 | 5 | 101.3 | 0.6 | 0.3 | 3 | -3 |
P_{y, r} from the moment centre are not given signs since it is simpler to determine the sign of each moment, P_{x, r}η_r and P_{y, r} ξ_r, by referring to the directions of P_{x, r} and P_{y, r} individually.
P_{y, r}=P_{z, r} \frac{\delta y_r}{\delta z} (21.9)
P_{x, r}=P_{z, r} \frac{\delta x_r}{\delta z} (21.10)
P_r=P_{z, r} \frac{\left(\delta x_r^2+\delta y_r^2+\delta z^2\right)^{1 / 2}}{\delta z} (21.12)
From column ⑥
\sum_{r=1}^6 P_{y, r}=33.4 \mathrm{kN}From column ⑩
\sum_{r=1}^6 P_{x, r} \eta_r=0From column ⑪
\sum_{r=1}^6 P_{y, r} \xi_r=0From Eq. (21.15)
S_{x, w}=S_x-\sum_{r=1}^m P_{z, r} \frac{\delta x_r}{\delta z} \quad S_{y, w}=S_y-\sum_{r=1}^m P_{z, r} \frac{\delta y_r}{\delta z} (21.15)
S_{x, w}=0 \quad S_{y, w}=100-33.4=66.6 \mathrm{kN}
The shear flow distribution in the walls of the beam is now found using the method described in Section 20.3. Since, for this beam, I_{x y}=0 \text { and } S_x=S_{x, w}=0 \text {, }, Eq. (20.11) reduces to
\begin{aligned}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_{\mathrm{D}} x \mathrm{~d} s+\sum_{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_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r\right)+q_{s, 0}\end{aligned} (20.11)
q_s=\frac{-S_{y, w}}{I_{x x}} \sum_{r=1}^n B_r y_r+q_{s, 0} (iv)
We now ‘cut’ one of the walls, say 16. The resulting ‘open section’ shear flow is given by
q_{\mathrm{b}}=-\frac{66.6 \times 10^3}{5.4 \times 10^8} \sum_{r=1}^n B_r y_ror
q_{\mathrm{b}}=-1.23 \times 10^{-4} \sum_{r=1}^n B_r y_r (v)
Thus
\begin{aligned}&q_{\mathrm{b}, 16}=0 \\&q_{\mathrm{b}, 12}=0-1.23 \times 10^{-4} \times 900 \times 300=-33.2 \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 23}=-33.2-1.23 \times 10^{-4} \times 1200 \times 300=-77.5 \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 34}=-77.5-1.23 \times 10^{-4} \times 900 \times 300=-110.7 \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 45}=-77.5 \mathrm{~N} / \mathrm{mm}(\text { from symmetry) } \\&q_{\mathrm{b}, 56}=-33.2 \mathrm{~N} / \mathrm{mm}(\text { from symmetry) }\end{aligned}giving the distribution shown in Fig. 21.7. Taking moments about the centre of symmetry we have, from Eq. (21.16)
S_x \eta_0-S_y \xi_0=\oint q_{\mathrm{b}} p \mathrm{~d} s+2 A q_{s, 0}-\sum_{r=1}^m P_{x, r} \eta_r+\sum_{r=1}^m P_{y, r} \xi_r (21.16)
\begin{aligned}-100 \times 10^3 \times 600=& 2 \times 33.2 \times 600 \times 300+2 \times 77.5 \times 600 \times 300 \\&+110.7 \times 600 \times 600+2 \times 1200 \times 600 q_{s, 0}\end{aligned}from which q_{s, 0}=-97.0 N/mm (i.e. clockwise). The complete shear flow distribution is found by adding the value of q_{s,0} to the q_b shear flow distribution of Fig. 21.7 and is shown in Fig. 21.8.
