Question p.21.3: Calculate the shear flow distribution and the stringer and f...

The beam section at a distance of 1.5 m from the built-in end is shown in Fig. S.21.3.
The bending moment, M, at this section is given by
M = −40 × 1.5 = −60 kN m

Since the x axis is an axis of symmetry I_{xy} = 0; also M_y = 0. The direct stress distribution is then, from Eq. (16.18)

\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}{I_{x x}} y  (i)

in which I_{x x}=2 \times 1000 \times 112.5^2+4 \times 500 \times 112.5^2=50.63 \times 10^6 \mathrm{~mm}^4. Then, from Eq. (i), the direct stresses in the flanges and stringers are

Therefore

and
P_{z, 3}=P_{z, 5}=-P_{z, 4}=-P_{z, 6}=-133.3 \times 500=-66650 \mathrm{~N}
From Eq. (21.9)

P_{y, r}=P_{z, r} \frac{\delta y_r}{\delta z}  (21.9)

and

Thus the total vertical load in the flanges and stringers is
2 × 3332.5 + 4 × 1666.3 = 13 330.2 N
Hence the total shear force carried by the panels is
40 × 10³ − 13 330.2 = 26 669.8 N

The shear flow distribution is given by Eq. (20.11) which, since I_{x y}=0, S_x=0 and t_D = 0 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)

i.e.

or

q_s=-5.27 \times 10^{-4} \sum_{r=1}^n B_r y_r+q_{s, 0}  (ii)

From Eq. (ii)

The remaining distribution follows from symmetry. Now taking moments about the point 2 (see Eq. (17.17))

S_x \eta_0-S_y \xi_0=\oint p q_{\mathrm{b}} \mathrm{d} s+2 A q_{s, 0}  (17.17)

26 669.8 × 100 = 59.2 × 225 × 500 + 29.6 × 250 × 225 + 2 × 500 × 225_{q_{s,0}}

from which
q_{s,0} = −36.9 N/mm (i.e. clockwise)
Then

Finally