22-Mec-A4 Design and Manufacture of Machine Elements · December 2017

Question 5 of 6: Boom and Tie-Rod — Tension Versus Compression

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 5: Boom and Tie-Rod — Tension Versus Compression (25 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.

Find. The factor of safety of the 12 mm tie-rod with the load acting downward, then with the load reversed to act upward, and the design conclusion those two numbers support.

Approach. Treat the boom as a rigid body pinned at the wall and take moments about that pin to get the tie-rod force, which is the same magnitude in both load cases. Then apply the appropriate failure criterion to each case: direct tensile yielding when the rod is in tension, and elastic column buckling when it is in compression, after checking the slenderness ratio to confirm which column formula applies.

1. Establish the geometry. The boom is pinned to the wall at the lower pin and runs straight to its tip, which the dimensions place 1.0 m out and 1.0 m up — a 45-degree member. The horizontal tie-rod is pinned to the wall 0.7 m above the boom's wall pin and meets the boom 0.7 m out, which is consistent, since a 45-degree line through the origin passes through the point (0.7, 0.7). The 6 kN load hangs from the boom tip on a vertical rod, so its line of action is 1.0 m from the wall pin.
2. Take moments on the boom about the wall pin. Only two forces have a moment about that pin: the applied load, at a horizontal arm of $a = 1.0\ \text{m}$, and the horizontal tie-rod force, at a vertical arm of $h = 0.7\ \text{m}$. $$\sum M_{O} = 0:\qquad P a = F_{t} h$$ $$F_{t} = \frac{P a}{h} = \frac{(6\ \text{kN})(1.0\ \text{m})}{0.7\ \text{m}}$$ $$\boxed{F_{t} = 8.571\ \text{kN}}$$ The sign works out so that the rod pulls the boom back toward the wall, that is the rod is in tension, which is what the arrangement is designed for. Note that the mechanical advantage is unfavourable: the tie-rod force exceeds the applied load by the factor $a/h = 1.43$, because the rod is attached well below the boom tip.
3. (a) Tensile stress and the safety factor against yielding. The rod is a simple two-force member in axial tension, so the stress is uniform over the full cross-section: $$A = \frac{\pi d^{2}}{4} = \frac{\pi (12)^{2}}{4} = 113.1\ \text{mm}^{2}$$ $$\sigma = \frac{F_{t}}{A} = \frac{8571\ \text{N}}{113.1\ \text{mm}^{2}} = 75.78\ \text{MPa}$$ Comparing that with the yield strength, $$n = \frac{S_{y}}{\sigma} = \frac{400}{75.78}$$ $$\boxed{n_{a} = 5.28}$$ A generous margin: the rod carries just under a fifth of the load that would yield it.
4. (b) Reverse the load and identify the new failure mode. Rotating the hanger through 180 degrees makes the 6 kN act upward. The moment equation is unchanged in form, so the magnitude of the tie-rod force is identical at 8.571 kN — but its sign reverses, and the rod is now in compression. A slender member in compression does not fail by yielding; it fails by buckling, so the tensile-yield calculation of part (a) is no longer the governing check and must be replaced.
5. Classify the column by slenderness ratio. For a solid circular section the radius of gyration is $k = \sqrt{I/A} = d/4 = 3.0\ \text{mm}$. The rod is pinned at both ends, so its effective length is its actual length of 700 mm and $$\frac{L}{k} = \frac{700}{3.0} = 233$$ The Euler and Johnson formulas meet at the tangency slenderness $$\left(\frac{L}{k}\right)_{1} = \sqrt{\frac{2\pi^{2}E}{S_{y}}} = \sqrt{\frac{2\pi^{2}(207\,000)}{400}} = 101$$ Since $233 \gg 101$, this is a long column and the Euler formula applies.
6. Apply the Euler buckling formula. With $I = \pi d^{4}/64 = 1018\ \text{mm}^{4}$ and pinned ends, $$P_{cr} = \frac{\pi^{2} E I}{L^{2}} = \frac{\pi^{2}(207\,000)(1018)}{(700)^{2}}$$ $$P_{cr} = 4244\ \text{N} = 4.244\ \text{kN}$$ The safety factor against buckling is therefore $$n = \frac{P_{cr}}{F_{t}} = \frac{4.244}{8.571}$$ $$\boxed{n_{b} = 0.50}$$ A factor of safety below unity means the member has already failed: the rod buckles under half the applied load, and the arrangement collapses. Note how far the buckling load is below the yield load — yielding would need $S_{y}A = 45.2\ \text{kN}$, more than ten times the buckling load, so material strength never enters the problem.
7. (c) Draw the design conclusion. The same rod, the same material, the same load magnitude and the same joints have gone from a comfortable safety factor of 5.28 to a failed member at 0.50 — a loss of a factor of 10.7 — purely because the direction of the load reversed. That asymmetry is the general rule, not an artefact of these numbers, and it has three parts.

   First, a tension member's capacity depends only on its cross-sectional area and its yield strength. Length, shape and end conditions are irrelevant; the stress is uniform, and every fibre works at full value.

   Second, a compression member's capacity depends on $EI/L_{e}^{2}$. It falls with the square of the length, is exquisitely sensitive to how the material is distributed about the axis rather than to how much of it there is, and depends on the end fixity, which is often uncertain in a real machine. Nothing about it is improved by choosing a stronger alloy, because $E$ is essentially the same for all steels.

   Third, buckling is unforgiving in a way yielding is not. Yielding is progressive, local and visible, and a slightly overloaded tension member simply stretches; buckling is a sudden instability that gives no warning and goes to complete collapse, and it is highly sensitive to initial crookedness, load eccentricity and imperfect pin alignment, so the real capacity is usually below the ideal Euler value.

   Conclusion: wherever the designer has a choice of load path, arrange the slender members of a machine to work in tension and let the short, stocky members take the compression. This is exactly why cranes, derricks, bicycle wheels, cable-stayed structures and this very boom arrangement are built as a short stubby strut plus a long slender tie. Where compression in a long member is unavoidable, do not try to fix it with a stronger material — increase $I$ for the same area by using a tube instead of a solid bar, shorten the effective length with intermediate bracing, or improve the end fixity.