Question 2 of 8: Maximum bending and shear stresses in a cantilever channel

Question 2: Maximum bending and shear stresses in a cantilever 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. A cantilever channel loaded at the free end by two mutually perpendicular forces acting at the shear centre, so the section carries bending and bending-induced shear only — no torsion. The requested section lies 500 mm from the root of a 2500 mm beam, i.e. $x = 2000$ mm from the loaded end.

Find. The largest bending (direct) stress and the largest shear stress acting on the cross-section 500 mm from the built-in end, together with the points at which each occurs.

Approach. Reduce the walls to their median lines, compute the three section constants about the centroid, solve the unsymmetrical-bending pair for the stress field $\sigma = aZ + bY$, evaluate it at the four extreme fibres, then differentiate the same field along the span to obtain the shear flow and divide by each local thickness.

1. Median-line section constants. Treating each wall as a line of area gives $A = 300 + 120 + 120 = 540$ mm², and taking first moments about the web and the lower flange places the centroid at $\bar{Z}=20.0$ mm from the web and $\bar{Y}=75.0$ mm above the lower flange. About that centroid, $$I_{ZZ} = 5.041\times10^{5},\qquad I_{YY} = 1.9126\times10^{6},\qquad I_{YZ} = -2.700\times10^{5}\ \text{mm}^{4}$$The section is 3.8 times stiffer about the horizontal axis than about the vertical one, which is why the modest 500 N side load ends up dominating the answer.
2. Bending moments carried at the section. With the free-end loads $2000$ mm away, the internal moments about the centroidal axes are $M_Y = \int \sigma Z\,dA = 500(2000) = 1.00\times10^{6}$ and $M_Z = \int \sigma Y\,dA = -1000(2000) = -2.00\times10^{6}$ N·mm. Substituting into the unsymmetrical-bending pair $$a I_{ZZ} + b I_{YZ} = M_Y,\qquad a I_{YZ} + b I_{YY} = M_Z$$and solving the two-by-two system gives $$\boxed{a = 1.5401\ \text{MPa/mm},\qquad b = -0.8283\ \text{MPa/mm}}$$so the direct stress anywhere on the section is $\sigma = 1.5401\,Z - 0.8283\,Y$ with $Z$ and $Y$ measured from C.
3. Direct stress at the four extreme fibres. Evaluating the field corner by corner: the lower-flange tip ($Z=+100$, $Y=-75$) gives $154.0 + 62.1$; the lower-flange/web corner ($Z=-20$, $Y=-75$) gives $-30.8+62.1$; the upper-flange/web corner ($Z=-20$, $Y=+75$) gives $-30.8-62.1$; and the upper-flange tip ($Z=+40$, $Y=+75$) gives $61.6-62.1$. The extreme value is therefore $$\boxed{\sigma_{\max} = +216.1\ \text{MPa (tension) at the tip of the 120 mm lower flange}}$$with a maximum compressive value of $-92.9$ MPa at the upper-flange/web corner. Note that the 500 N horizontal load supplies 154.0 MPa of the 216.1 MPa peak — about 71 per cent — because the section is so much weaker about the vertical axis.
4. Neutral axis. Setting $\sigma = 0$ gives $Y = -(a/b)Z = 1.859\,Z$, a line through the centroid inclined $61.7^{\circ}$ above the $+Z$ direction. The extreme-fibre search above is really just a check of which corner lies furthest from this line, and the lower-flange tip wins comfortably.
5. Shear flow from the same stress field. For a cantilever with tip loads, $a$ and $b$ grow linearly with the distance from the free end, so $\partial\sigma/\partial x = a'Z + b'Y$ with $a' = a/x$ and $b' = b/x$. Longitudinal equilibrium of a wall element gives $$q(s) = -\int_0^{s}\frac{\partial\sigma}{\partial x}\,t\,ds$$walked from the open free edge. Starting at the lower-flange tip, where $q=0$, and integrating along the 1.0 mm flange, the flow builds to $q = 7.42$ N/mm at the web. Continuing up the 2.0 mm web it rises to a peak of $8.02$ N/mm about 38 mm above the lower flange, falls to $2.80$ N/mm at the upper-flange corner, and runs out to zero at the upper-flange tip — the closure that confirms the walk.
6. Statical check on the shear flow. Resolving the computed flow around the whole median line returns $(S_Z, S_Y) = (-500,\ +1000)$ N, exactly the applied tip loads. This check is worth the thirty seconds it takes: it catches a dropped thickness or a sign slip that no amount of formula-matching will reveal.
7. Maximum shear stress. Shear stress is $q/t$, so the peak flow and the peak stress need not coincide. Dividing wall by wall, the 1.0 mm lower flange reaches $7.42/1.0$ at its junction with the web, while the larger web flow of 8.02 N/mm is spread over 2.0 mm and gives only 4.01 MPa. Hence $$\boxed{\tau_{\max} = 7.42\ \text{MPa, in the 1.0 mm lower flange where it meets the web}}$$The bending stress is 29 times larger than the shear stress, which is the expected proportion for a slender beam and confirms that this section is bending-critical.