Shear flow and bending stresses in a closed trapezoidal beam: 22-Mec-B9 Advanced Engineering Structures, May 2018, Q6

Question 6: Shear flow and bending stresses in a closed trapezoidal beam (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A closed trapezoidal thin-wall beam of uniform 2 mm wall thickness. Taking the bottom-left corner as the origin with z to the right and y upward, the four corners of the median line are A(0, 0), B(0, 400), C(600, 150) and D(600, 0), all in mm. A vertical force of 15,000 N acts downward in the plane of the right-hand (150 mm) web.

Wall	From → to	Length (mm)
Bottom	A → D	600
Right web	D → C	150
Sloping top	C → B	650
Left web	B → A	400

Find. (a) the shear flow all the way round the section, and (b) the bending stress at each of the four corners of a section 500 mm behind the loaded one.

Question 6 — the closed trapezoidal section, its centroid and the neutral axis of bending. Because the section has no axis of symmetry the neutral axis is not horizontal, even though the load is purely vertical.

Approach. The walls are effective in bending as well as shear, so the section properties come from the walls themselves and the product of inertia does not vanish. Compute the centroid and the three second moments, walk the open-section shear flow round a cut, close it by taking moments about a convenient corner, and finally apply the unsymmetrical-bending stress formula at the four corners.

Section area and centroid. Each wall is a thin strip of area t L at its own mid-point: $$A = 2(600+150+650+400) = 3600\ \text{mm}^2, \qquad \bar z = 258.3\ \text{mm}, \qquad \bar y = 150.0\ \text{mm}$$
Second moments about the centroid. For a straight thin wall running from $(z_1,y_1)$ to $(z_2,y_2)$ in centroidal coordinates, $$I_{xx} = \sum \frac{tL}{3}\left(y_1^2+y_1y_2+y_2^2\right), \qquad I_{yy} = \sum \frac{tL}{3}\left(z_1^2+z_1z_2+z_2^2\right)$$ $$I_{xy} = \sum \frac{tL}{6}\left(2z_1y_1+z_1y_2+z_2y_1+2z_2y_2\right)$$ giving $I_{xx} = 6.90\times10^{7}$, $I_{yy} = 1.678\times10^{8}$ and $I_{xy} = -3.50\times10^{7}$ mm⁴. The product term is negative because the deep end of the section sits at low z.
Open-section shear flow. Cut the box at corner A and integrate anticlockwise, using the general form that admits $I_{xy} \neq 0$ with $S_z = 0$: $$q_b(s) = -\frac{S_yI_{yy}}{I_{xx}I_{yy}-I_{xy}^2}\int_0^s ty\,\mathrm{d}s + \frac{S_yI_{xy}}{I_{xx}I_{yy}-I_{xy}^2}\int_0^s tz\,\mathrm{d}s$$ The walk gives $q_b = -41.23$ N/mm at D, $-41.50$ at C, $+0.76$ at B and closes at zero back at A.
Close the cell by taking moments. The unknown constant $q_{s,0}$ is fixed by requiring the shear flows to have the same moment about A as the applied load, which acts 600 mm to the right of A: $$\sum q_b(2A_{\text{swept}}) + q_{s,0}\,(2A_{\text{cell}}) = S_y \times 600, \qquad 2A_{\text{cell}} = 330\,000\ \text{mm}^2$$ $$q_{s,0} = +1.46\ \text{N/mm}$$ The enclosed area, 165,000 mm², is the trapezoid’s own area $600(400+150)/2$ — a free check on the sweep.
Shear flow round the section (part a). Adding the constant to the open flows: $$q_A = 1.46,\quad q_D = -39.77,\quad q_C = -40.04,\quad q_B = 2.22\ \text{N/mm}$$ The flow varies parabolically along each wall, and its largest value is $$\boxed{|q|_{\max} = 41.3\ \text{N/mm in the right-hand web, giving } \tau_{\max} = 20.6\ \text{MPa}}$$ The peak sits 78.7 mm above the bottom of that web — precisely where the wall crosses the neutral axis, which is where the first moment stops accumulating. The flows resolve to 15,000 N vertically and zero horizontally, as they must.
Bending moment at the section of interest (part b). Treating the beam as a cantilever with the load at the free end, the section 500 mm inboard carries $$M_x = 15\,000 \times 500 = 7.5\times10^6\ \text{N}\cdot\text{mm}$$ hogging, so fibres above the centroid go into tension.
Unsymmetrical bending stress. With $M_y = 0$ but $I_{xy} \neq 0$, $$\sigma = \frac{M_x\left(I_{yy}\,y - I_{xy}\,z\right)}{I_{xx}I_{yy}-I_{xy}^{2}}$$ in which y and z are measured from the centroid. Evaluating at the four corners: $$\sigma_A = -24.8,\quad \sigma_B = +23.8,\quad \sigma_C = +8.7,\quad \sigma_D = -9.6\ \text{MPa}$$ $$\boxed{\sigma_{\max} = 24.8\ \text{MPa compression at corner A}}$$
Where the neutral axis lies. Setting $\sigma = 0$ gives $y = (I_{xy}/I_{yy})z = -0.209z$, a line through the centroid inclined 11.8° below the horizontal. Integrating the stress field over the walls returns zero axial force and exactly $7.5\times10^6$ N·mm about the horizontal axis, which confirms the whole calculation.

The inclined neutral axis is the feature to take away from this question. A purely vertical load on an unsymmetrical section produces bending about an axis that is not horizontal, so the extreme fibre is not simply the highest or lowest point: here corner A, at the bottom of the deep end, is more highly stressed than corner D at the bottom of the shallow end even though both lie on the same bottom wall. Ignoring $I_{xy}$ and using $M y / I_{xx}$ would give +27.2 MPa at B and −16.3 MPa at A — overstating the tension corner by 14% and understating the critical compression corner by 34%, and putting the extreme fibre at the wrong corner.

Quantity	Value
Centroid (from corner A)	z̄ = 258.3 mm, ȳ = 150.0 mm
I_xx / I_yy / I_xy	6.90 × 10⁷ / 1.678 × 10⁸ / −3.50 × 10⁷ mm⁴
Closing shear flow q_s,0	1.46 N/mm
(a) Shear flow at A / D / C / B	1.46 / −39.77 / −40.04 / 2.22 N/mm
(a) Peak shear flow and stress	41.3 N/mm, τ = 20.6 MPa, in the right-hand web
(b) σ_A (bottom left)	−24.8 MPa
(b) σ_B (top left)	+23.8 MPa
(b) σ_C (top right)	+8.7 MPa
(b) σ_D (bottom right)	−9.6 MPa
Neutral axis	through the centroid at 11.8° below the horizontal

Topic. Shear flow and unsymmetrical bending of a closed section whose walls carry direct stress.

Key relations. Thin-wall second moments $I_{xx}, I_{yy}, I_{xy}$ by the three-point formulae; open flow $q_b$ from the general shear equation with $I_{xy}\neq 0$; closure by moment equivalence about any point; $\sigma = M_x(I_{yy}y - I_{xy}z)/(I_{xx}I_{yy}-I_{xy}^2)$; neutral axis $y = (I_{xy}/I_{yy})z$.

Why this works. Once the walls are declared effective in bending, the section behaves like an ordinary beam cross-section, and the only complication is that neither centroidal axis is principal. The general bending formula handles that by coupling the two moments through $I_{xy}$; it reduces to the familiar $My/I$ the moment the section gains an axis of symmetry. The shear problem is the same one-redundancy argument as for any closed cell, except that the position of the load now matters, because it is the moment equation that fixes the closing flow.

Common pitfall. Assuming the maximum bending stress must occur at the deepest point, and dropping $I_{xy}$ — both errors are invisible in a symmetric section and both are large here. In the shear calculation, forgetting that the applied load has a line of action (600 mm from A) turns the moment equation into a different problem entirely.

Source. Megson, Aircraft Structures for Engineering Students, 6th ed., §16.2 (unsymmetrical bending) and §17.3 (shear of closed sections); Boresi & Schmidt, 6th ed., ch. 7 (shear centre and shear flow).

Question 6 of 8: Shear flow and bending stresses in a closed trapezoidal beam

Question 6: Shear flow and bending stresses in a closed trapezoidal beam (20 marks)