22-Mec-A4 Design and Manufacture of Machine Elements · May 2015

Question 4 of 6: Safe Load on a Bolted Cantilever Bracket

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

Notes on this paper

Paper format. National Examinations, May 2015 — 07-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator permitted. Six questions in two parts: Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine-element design). The rubric asks for two questions from each part, and all questions carry equal value (25 % each). All six are solved here, because this set is a study resource rather than an examination script.

Reference texts.

S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology, 8th ed. — sheet-metal forming, deep drawing, abrasive machining.
M. P. Groover, Fundamentals of Modern Manufacturing, 7th ed. — stretch forming and the flow-curve treatment of forming loads.
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, 11th ed. — bolted and welded joints (Ch. 8, 9) and brakes (Ch. 16).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — net-section bending and bearing stress.

Units. The paper mixes systems deliberately: Q1, Q5 and Q6 are in US customary units (inch, pound, psi) and Q4 is metric. Each question is solved in the units in which it is set, as the exam intends.

Question 4: Safe Load on a Bolted Cantilever Bracket (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.

Quantity	Symbol	Value
Channel face (bolt row centred on it)	—	152 mm × 76 mm
Channel web thickness	$t_c$	6.4 mm
Channel yield strength (AISI 1006 HR)	$S_{y,c}$	170 MPa
Bar thickness × depth (from the figure)	$t_b \times h$	12 mm × 50 mm
Bar yield strength (AISI 1015 HR)	$S_{y,b}$	190 MPa
Bolts	—	three M12 × 1.75 ISO 5.8 shoulder bolts at 50 mm pitch
Bolt proof and yield strengths	$S_p,\ S_{y}$	380 MPa, 420 MPa
Load position	—	125 mm beyond the channel face
Design factor	$n_d$	2.8

Find. The largest force F that may be applied at the free end of the cantilever such that no failure mode — bolt shear, bearing on any member, or bending of the bar — reaches yield with a factor of 2.8 in hand.

Figure 3 — the joint as dimensioned on the exam drawing. The three shoulder bolts are centred on the 152 mm channel face, so the bolt-group centroid O lies 76 mm inside the face and the load F acts 76 + 125 = 201 mm from it.

Check: the moment arm. The 125 mm dimension on the exam drawing is struck from the channel face, not from the outer bolt: its left-hand extension line coincides with the edge of the channel outline, and scaling the drawing against the 50 mm bolt pitch puts that edge 76 mm (half of 152 mm) from the centre bolt. The moment arm about the bolt-group centroid is therefore $76 + 125 = 201$ mm. Had the dimension been taken to bolt B instead, the arm would be 175 mm, the critical bolt load would fall from $2.343F$ to $2.083F$, and the answer would rise from 1.99 kN to 2.24 kN — a 12 % difference, with the same failure mode governing.

Approach. Reduce the eccentric load to a force plus a couple at the bolt-group centroid, superpose the direct and moment-induced shears to find the worst-loaded bolt, then check that bolt load against every mode that can yield — shear of the bolt, bearing on the bolt, on the channel web and on the bar — and separately check bending of the bar at its net section, taking the smallest allowable load as the answer.

Locate the bolt-group centroid and the moment arm. The three bolts A, O and B are equally spaced at 50 mm on a row centred on the 152 mm channel face, so the centroid is at the middle bolt O. The load acts $$r_F = \frac{152}{2} + 125 = 76 + 125 = 201\ \text{mm}$$ from O, and the joint carries a direct shear $V = F$ together with a couple $M = 201F$ N·mm.
Direct (primary) shear. Shared equally among the three bolts, $$F' = \frac{V}{N} = \frac{F}{3} = 0.3333F$$
Moment (secondary) shear. With $r_A = r_B = 50$ mm and $r_O = 0$, $$\textstyle\sum r_i^2 = 2(50)^2 = 5000\ \text{mm}^2$$ $$F'' = \frac{M\,r}{\sum r_i^2} = \frac{201F(50)}{5000} = 2.010F$$ The centre bolt sits at the centroid and carries no moment shear at all.
Identify the critical bolt. The bolt row is horizontal and the load is vertical, so both the primary and the secondary shears are vertical. At bolt B, on the loaded side, they act in the same sense and add; at bolt A they oppose: $$F_B = 0.3333F + 2.010F = 2.343F \qquad F_A = |2.010F - 0.3333F| = 1.677F$$ $$\boxed{F_B = 2.343F \ \text{governs}}$$
Bolt shear. These are shoulder bolts, so the unthreaded shank of nominal diameter 12 mm lies in the shear plane: $$A_s = \frac{\pi(12)^2}{4} = 113.1\ \text{mm}^2, \qquad S_{sy} = 0.577S_y = 0.577(420) = 242.3\ \text{MPa}$$ Setting $2.343F/A_s = S_{sy}/n_d$, $$F = \frac{242.3}{2.8}\cdot\frac{113.1}{2.343} = 4177\ \text{N}$$
Bearing on the channel web. The projected bearing area is the thinner member's thickness times the bolt diameter, $6.4(12) = 76.8$ mm², and the channel is the weakest material at 170 MPa: $$F = \frac{170}{2.8}\cdot\frac{76.8}{2.343} = 1990\ \text{N}$$
Bearing on the bar and on the bolt. The bar is thicker (12 mm) and stronger (190 MPa), and the bolt material is stronger still (420 MPa against the 6.4 mm channel): $$F_{bar} = \frac{190}{2.8}\cdot\frac{12(12)}{2.343} = 4170\ \text{N}, \qquad F_{bolt} = \frac{420}{2.8}\cdot\frac{76.8}{2.343} = 4916\ \text{N}$$
Bending of the bar at its net section. The bar is a cantilever in its own right, and the critical section is the one through the outer bolt B, where the bending moment is still large and the section is weakened by the hole: $$M_B = (201 - 50)F = 151F\ \text{N}\!\cdot\!\text{mm}$$ $$I_{net} = \frac{t_b\,(h^3 - d^3)}{12} = \frac{12\,(50^3 - 12^3)}{12} = 123{,}272\ \text{mm}^4, \qquad c = 25\ \text{mm}$$ $$F = \frac{190}{2.8}\cdot\frac{123{,}272}{151(25)} = 2216\ \text{N}$$ Direct shear of the bar over the same net section allows 17.9 kN and is nowhere near critical.
Select the safe load. Collecting the five checks, the smallest governs: $$\boxed{F_{safe} = 1.99\ \text{kN}\ \text{(bearing on the channel web)}}$$

The result is worth a comment. The bolts themselves are the strongest element in the joint by a factor of more than two; what limits this bracket is the soft, 6.4 mm channel web crushing against the shank of the outer bolt. Doubling the bolt size would buy almost nothing, whereas a filler plate or a washer plate behind the channel web, or spreading the three bolts to a wider pitch to cut the secondary shear, would each raise the capacity substantially.

Failure mode	Allowable F (N)
Bearing on the channel web (170 MPa, 6.4 mm)	1990 — governs
Bending of the bar at the net section through B	2216
Bearing on the bar (190 MPa, 12 mm)	4170
Shear of the M12 shoulder bolt	4177
Bearing on the bolt (420 MPa)	4916
Shear of the bar at the net section	17,853
Safe force F	1.99 kN

Topic. Shear joints loaded eccentrically in the plane of the fastener pattern.

Key relations. Primary shear $F' = V/N$; secondary shear $F'' = Mr_i/\sum r_i^2$ acting perpendicular to $r_i$; vector sum on each fastener; bearing stress $\sigma = F_{bolt}/(t\,d)$ on the projected area; shear yield $S_{sy} = 0.577S_y$; net-section bending $\sigma = Mc/I_{net}$ with $I_{net} = t(h^3 - d^3)/12$.

Why this works. Moving the load to the centroid of the fastener pattern and adding the couple that this move omits is just statics, but it separates the problem into two effects that can be superposed on each fastener. The secondary shear grows linearly with distance from the centroid, so the outermost fastener on the side where the two contributions align is always the critical one — here that is bolt B, carrying 2.34 times the whole applied load. The second half of the method matters just as much: a bolted joint has several independent ways to yield, and the design factor must be satisfied by all of them, so the honest answer is the minimum over a list of checks rather than the result of a single formula.

Common pitfall. Measuring the moment arm from the outer bolt or from the channel face instead of from the bolt-group centroid, which is the single most common error on this problem. A close second is checking only the bolts. Here the bolts are more than twice as strong as the joint, and stopping after the shear check would overstate the capacity by 110 %. Finally, remember to use the gross depth minus the hole when computing $I$ for the bar — the critical section passes straight through a bolt hole.

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

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