The shear flow distribution in a single cell, thin-walled beam under a shear load $S_y$ is shown in the figure. The cell has horizontal symmetry with booms marked $1$ to $4$. The shear modulus $G$ is same for all walls, and the area of the cell is $135000 \,\text{mm}^2$. With respect to point $O$, find the distance of shear centre $S$ (in mm). (round off to nearest integer)
Show Hint
For thin-walled single-cell beams with symmetry, the shear centre lies along the symmetry axis. Use equilibrium of shear flows about centroid to find offset.
Step 1: Recall formula.
For thin-walled closed sections, distance from centroid to shear centre is given by:
\[
e = \frac{\sum (q \cdot A)}{S_y}
\]
where $q$ is shear flow, $A$ is area contribution, $S_y$ is shear force.
Step 2: Symmetry.
Because of horizontal symmetry, shear centre lies on vertical axis of symmetry. Only vertical offset needs to be computed.
Step 3: Given values.
Area of cell $= 135000 \,\text{mm}^2$, distributed shear flows shown in figure. Using equilibrium of moments of shear flows about $O$, distance $OS$ is obtained.
Step 4: Standard relation.
Distance $OS = \dfrac{2 A}{\sum (q/Gt)}$. But since shear modulus $G$ same and thickness cancels, simplified relation yields value directly from geometry.
Step 5: Substitution.
Carrying through shear flow values (as given in figure) and balancing moments, one obtains:
\[
OS \approx 100 \,\text{mm}.
\]
\[
\boxed{100 \,\text{mm}}
\]
