Question 21.2: The cantilever beam shown in Fig. 21.6 is uniformly tapered ...

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

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

From column ⑩

From column ⑪

From 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

or

q_{\mathrm{b}}=-1.23 \times 10^{-4} \sum_{r=1}^n B_r y_r  (v)

Thus

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)

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.