Question 3 of 8: Unsymmetrical bending of a thin-walled channel

Question 3: Unsymmetrical bending of a thin-walled channel (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.

Set up axes in the plane of the section with $z$ horizontal, positive towards the flange tips, and $y$ vertical, positive upward, with the origin at the foot of the web. In that frame the wall mid-lines are: the web from $(0,0)$ to $(0,120)$; the upper flange from $(0,120)$ to $(50,120)$; the lower flange from $(0,0)$ to $(100,0)$. Point A is the lower-flange tip at $(100,0)$.

Find. The direct bending stress at point A on the section 1500 mm from the loaded end.

Approach. Because the section has no axis of symmetry, $I_{yz} \neq 0$ and the stress must come from the general unsymmetrical bending formula; the loads act at the shear centre, so there is no torsion to superpose.

1. Locate the centroid. Treating each wall as a line of area $t\,L$ concentrated on its mid-line, the three wall areas are $3.0\times120 = 360$, $3.0\times50 = 150$ and $1.5\times100 = 150$ mm$^2$, totalling $A = 660$ mm$^2$. Then$$\bar z = \frac{\sum A_i z_i}{A} = \frac{0+150(25)+150(50)}{660} = 17.045\ \text{mm},\qquad\bar y = \frac{360(60)+150(120)+0}{660} = 60.0\ \text{mm}$$The two flanges have equal areas — $50\times3.0 = 100\times1.5$ — which is why the centroid falls at exactly mid-web depth. That is worth noticing, because it removes a whole class of arithmetic slips from the rest of the calculation.
2. Compute the second moments about the centroidal axes. For a straight thin wall running between centroid-relative end points $(a_1,b_1)$ and $(a_2,b_2)$ the exact three-point formulae are$$I = \frac{tL}{3}\big(b_1^2+b_1b_2+b_2^2\big),\qquad I_{yz} = \frac{tL}{6}\big(2a_1b_1+a_1b_2+a_2b_1+2a_2b_2\big)$$Summing over the three walls gives$$I_{zz} = 1.5120\times10^{6}\ \text{mm}^4,\qquad I_{yy} = 4.3324\times10^{5}\ \text{mm}^4,\qquad I_{yz} = -2.2500\times10^{5}\ \text{mm}^4$$The product of inertia comes almost entirely from the lower flange, which sits far from both centroidal axes; the web contributes nothing because it straddles the horizontal axis symmetrically.
3. Find the two bending moments at the section. Take the free-end portion of length $a = 1500$ mm as a free body. With $F_y = +900$ N and $F_z = -500$ N applied at its tip, moment equilibrium about the centroid of the cut gives$$M_z \equiv \int \sigma y\,\mathrm{d}A = -a F_y = -1.350\times10^{6}\ \text{N}\cdot\text{mm},\qquad M_y \equiv \int \sigma z\,\mathrm{d}A = -a F_z = +7.50\times10^{5}\ \text{N}\cdot\text{mm}$$The negative $M_z$ is the familiar cantilever result — an upward tip load puts the top fibres into compression — while the offset of the shear centre from the centroid contributes only torque, not bending moment, and so does not enter here.
4. Write the stress as a plane and invert. With no axial force the stress must be linear in position, $\sigma = \mathcal{A}y + \mathcal{B}z$. Substituting that into the definitions of $M_z$ and $M_y$ gives two simultaneous equations whose solution is$$\mathcal{A} = \frac{M_zI_{yy}-M_yI_{yz}}{I_{zz}I_{yy}-I_{yz}^2},\qquad\mathcal{B} = \frac{M_yI_{zz}-M_zI_{yz}}{I_{zz}I_{yy}-I_{yz}^2}$$With $I_{zz}I_{yy}-I_{yz}^2 = 6.0426\times10^{11}$ mm$^8$ this evaluates to $\mathcal{A} = -0.68845$ MPa/mm and $\mathcal{B} = +1.37360$ MPa/mm.
5. Evaluate the stress at point A. Point A lies at $(z,y) = (100,0)$, so relative to the centroid $z_A = 82.955$ mm and $y_A = -60$ mm:$$\sigma_A = \mathcal{A}y_A + \mathcal{B}z_A = (-0.68845)(-60) + (1.37360)(82.955)$$The two terms are $+41.31$ MPa and $+113.95$ MPa. These are the $\mathcal{A}y$ and $\mathcal{B}z$ terms, not the separate load contributions: because $I_{yz}\neq0$, each coefficient mixes both moments. Their sum gives$$\boxed{\sigma_A = +155.3\ \text{MPa (tensile)}}$$
6. Audit the result. Two checks cost nothing. First, integrating the assumed linear field over the three walls must return zero net axial force and reproduce both applied moments — it does, to four significant figures. Second, the neutral axis is the line $\mathcal{A}y+\mathcal{B}z = 0$ through the centroid, of slope $\mathrm{d}y/\mathrm{d}z = -\mathcal{B}/\mathcal{A} = 1.995$, i.e. inclined at $63.4^{\circ}$ to the horizontal. Point A is the section point furthest from that line on the tension side, which is precisely why the examiner asks for it.

The split between the two load contributions is the engineering lesson here. The 500 N horizontal load is only 56 per cent of the 900 N vertical one, yet it supplies 89 per cent of the stress at A. Evaluated one load at a time, the 900 N load alone gives $+16.37$ MPa at A and the 500 N load alone gives $+138.88$ MPa. This is because it bends the channel about its weak axis — $I_{yy}$ is only 29 per cent of $I_{zz}$. Open thin-walled sections are extremely sensitive to loads applied out of the plane of the web, and a designer who sizes such a member for the vertical load alone will underestimate the peak stress by a factor of more than nine (155.3 against 16.4 MPa). The steep neutral axis is the geometric statement of the same fact: the section bends about an axis nowhere near the one the loads suggest.