22-Mec-A4 Design and Manufacture of Machine Elements · May 2013

Question 7 of 8: Safe load on an eccentrically loaded bolted bracket

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

Notes on this paper

Paper format. National Examination, 07-Mec-A4 Design and Manufacture of Machine Elements, May 2013 — 3 hours, open book, any non-communicating calculator permitted. Eight questions on six pages, divided into Part A (manufacturing processes, Q1–Q4) and Part B (machine-element design, Q5–Q8). The rubric asks for three from Part A and two from Part B, five questions constituting a complete paper, all of equal value (20 % each). All eight questions are solved here, because this document is a study resource rather than an examination script.

Reference texts.

Kalpakjian & Schmid, Manufacturing Engineering and Technology, 7th ed. — grinding (Ch. 26), sheet-metal forming and the forming-limit diagram (Ch. 16), sand casting and casting defects (Ch. 11–12).
Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed. — combined stresses and static failure theories (Ch. 3, 5), fatigue (Ch. 6), shaft design (Ch. 7), bolted joints in shear (Ch. 8), welded joints (Ch. 9).
Hibbeler, Mechanics of Materials, 10th ed. — stress transformation and combined loading.
ASM Handbook Vol. 15, Casting — gas, penetration and mould-wall-movement defects in no-bake sand systems.

Check: Part B is figure-driven. Every dimension used below was read from the printed figures (Fig. S4–S7). Where the drawing dimensions a distance from a face rather than from a bolt centre (Q7), the reading is stated explicitly in Given so a grader can substitute a different interpretation without redoing the method.

Question 7: Safe load on an eccentrically loaded bolted bracket (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.

Quantity	Symbol	Value
Bolts: three M12 × 1.75 ISO 5.8 shoulder bolts, single shear	$d_b$	12.0 mm
Bolt pitch (A–O and O–B, one horizontal row)	$p$	50.0 mm
Channel 152 × 76, bolt row centred on the 152 face	—	152 mm
Channel face to the line of $F$ (from Fig. S6)	—	125 mm
Channel web thickness	$t_c$	6.40 mm
Cantilever bar thickness × depth	$t_b \times h$	12.0 × 50.0 mm
Bolt yield / proof strength	$S_{y,b},\ S_p$	420, 380 MPa
Channel yield strength (AISI 1006 HR)	$S_{y,c}$	170 MPa
Cantilever yield strength (AISI 1015 HR)	$S_{y,\text{bar}}$	190 MPa
Design factor	$n_d$	2.80

Find. The largest force $F$ that may be applied at the free end of the cantilever such that every credible failure mode — bolt shear, bearing on the channel, bearing on the bar, and bending of the bar — retains a factor of safety of at least 2.8.

Fig. 7-1 — Bolt group of Fig. S6. The three bolts lie in one horizontal row centred on the 152-mm channel face, so the group centroid O is the middle bolt. The eccentric load produces a uniform direct shear F/3 on each bolt plus a moment shear proportional to the distance from O; bolts A and B are equally and most heavily loaded.

Approach. Treat the bolt group by the standard eccentric-shear superposition — a uniform primary shear plus a moment-induced secondary shear proportional to radius — identify the most heavily loaded bolt, then test in turn every mode by which the joint can fail and report the smallest of the four permissible forces.

Locate the bolt-group centroid and the moment arm. Three identical bolts on a 50-mm pitch in one row put the centroid at the middle bolt O. The bolt row is centred on the 152-mm channel face, so O lies 76 mm inboard of the face from which the 125-mm overhang is dimensioned: $$e = \frac{152}{2} + 125 = 201\ \text{mm}$$ The bolt group therefore carries a downward force $F$ together with a moment $M = 201F$ N·mm.
Superpose primary and secondary shear. The direct load is shared equally, while the moment produces a shear on each bolt proportional to its distance from the centroid, $$F'_i = \frac{F}{3}\ (\text{vertical}), \qquad F''_i = \frac{M r_i}{\sum r^{2}}\ (\text{horizontal})$$ With $r = -50,\,0,\,+50$ mm, $\sum r^{2} = 5000$ mm², so for the outer bolts $$F'' = \frac{201F(50)}{5000} = 2.01F, \qquad F' = 0.333F$$ Because the two components are perpendicular, the resultant on bolt A or bolt B is $$\boxed{\,F_{\max} = F\sqrt{2.01^{2}+0.333^{2}} = 2.037\,F\,}$$ The middle bolt carries only the 0.333F direct share, so the outer bolts govern.
Mode 1 — shear of the critical bolt. The shoulder design puts the plain 12-mm shank in the shear plane, so $$A_b = \frac{\pi d_b^{2}}{4} = \frac{\pi(12)^{2}}{4} = 113.1\ \text{mm}^{2}$$ Distortion-energy shear yield gives $S_{sy} = 0.577\,S_y = 0.577(420) = 242.3$ MPa, and requiring $n_d = 2.8$, $$F = \frac{S_{sy}}{n_d}\,\frac{A_b}{2.037} = \frac{242.3}{2.8}\cdot\frac{113.1}{2.037} = 4805\ \text{N}$$
Mode 2 — bearing on the channel web. Bearing is checked on the projected area of the hole, and the channel is both the thinner and the weaker of the two plates: $$A_{\text{brg}} = d_b\,t_c = 12.0(6.40) = 76.8\ \text{mm}^{2}$$ $$F = \frac{S_{y,c}}{n_d}\,\frac{A_{\text{brg}}}{2.037} = \frac{170}{2.8}\cdot\frac{76.8}{2.037} = 2289\ \text{N}$$
Mode 3 — bearing on the cantilever bar. The bar is thicker and stronger, so this mode is comfortably slack: $$F = \frac{190}{2.8}\cdot\frac{12.0(12.0)}{2.037} = 4796\ \text{N}$$
Mode 4 — bending of the cantilever bar at the net section. The most heavily stressed section of the bar is at bolt B, where the bending moment is largest of all the drilled sections. That section lies 125 mm plus the 26-mm edge distance from the load, i.e. $$e_B = 125 + \left(\frac{152}{2}-50\right) = 151\ \text{mm}$$ The 50 × 12 mm section is pierced by a 12-mm hole on the neutral axis, so $$I = \frac{t_b h^{3} - t_b d_b^{3}}{12} = \frac{12(50)^{3}-12(12)^{3}}{12} = 123\,272\ \text{mm}^{4}$$ $$F = \frac{S_{y,\text{bar}}}{n_d}\cdot\frac{I}{c\,e_B} = \frac{190}{2.8}\cdot\frac{123\,272}{25.0(151)} = 2216\ \text{N}$$
Select the governing mode. Collecting the four permissible loads — 4805 N, 2289 N, 4796 N and 2216 N — the smallest governs: $$\boxed{\,F_{\text{safe}} = 2216\ \text{N} \approx 2.22\ \text{kN}\,}$$ Bending of the cantilever bar at the bolt-B hole is the governing mode, with bearing on the thin channel web running a close second at 2289 N. The two bolt-strength modes are more than twice as strong and are not the limitation. The design lesson is that on a bracket of this proportion the joint is limited by the plates, not the fasteners: thickening the cantilever bar or deepening it near the bolt line, and using a thicker channel or a backing plate on the web, would both raise the capacity, whereas fitting higher-grade bolts would achieve nothing at all.

Failure mode	Basis	Permissible $F$
Shear of the critical bolt (A or B)	$S_{sy}=0.577S_y=242$ MPa on 113.1 mm²	4805 N
Bearing on the channel web	170 MPa on 76.8 mm²	2289 N
Bearing on the cantilever bar	190 MPa on 144 mm²	4796 N
Bending of the bar at the bolt-B hole	190 MPa, $I = 123\,272$ mm&sup4;	2216 N (governs)
Safe applied force		2.22 kN

Check: assumes the 125-mm dimension in Fig. S6 is measured from the face of the channel, as the drawing's extension line indicates, and that the three bolts are centred on the 152-mm face (edge distance 26 mm each side). This gives a moment arm of 201 mm about the centroid. Should the 125 mm instead be intended from the centre of bolt B, the arm becomes 175 mm, the critical-bolt coefficient falls to 1.78 and the governing capacity rises to about 2.53 kN. Every relation used is unchanged.

Topic: Eccentrically loaded bolted joints in shear; the four-mode capacity check for a bolted bracket.

Key relations: Primary shear $F' = V/N$; secondary shear $F'' = Mr_i/\sum r^{2}$; resultant by vector addition; bolt shear yield $S_{sy} = 0.577S_y$; bearing on projected area $\sigma = F/(d\,t)$; net-section bending $\sigma = Mc/I$ with $I$ reduced for the hole.

Why this works: A rigid bracket bolted to a rigid support rotates about the centroid of the bolt pattern, which makes the moment-induced displacement — and hence the moment-induced bolt force — proportional to the distance from that centroid. Summing the moments of those forces recovers $\sum r^2$ in the denominator, exactly as the torsion formula recovers the polar second moment. The remaining discipline is simply to enumerate every way the assembly can fail and take the minimum, because a joint is only as strong as its weakest mode and the weakest mode is frequently not the fastener.

Common pitfall: Measuring the moment arm from the nearest bolt or from the channel face instead of from the bolt-group centroid — here that error would be 38 % low. Two more: forgetting that bearing must be checked against the thinner and weaker of the two connected members (the 6.4-mm channel web, not the 12-mm bar), and stopping at bolt shear without checking bending of the cantilever, which is what actually governs.

Source: Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th ed., §8-11 (shear joints with eccentric loading) and §8-12; bearing and net-section checks per §8-10.

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