Question 2 of 6: Bolted-Joint Design — Size, Preload and Safety Factors
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 22-Mec-B1 Advanced Machine Design — National Exams, May 2015. Open book, 3 hours, 100 marks. Part I (Problems 1–2) compulsory; answer any three of the four Part II problems (3–6). All six problems are solved here as a complete study resource.
Reference texts. R. G. Budynas & J. K. Nisbett, Shigley’s Mechanical Engineering Design (10th ed.) — deflection §4, fatigue §6, shafts §7, bolted joints §8, journal bearings §12, brakes & clutches §16; R. C. Juvinall & K. M. Marshek, Fundamentals of Machine Component Design; R. L. Norton, Machine Design: An Integrated Approach; R. C. Hibbeler, Mechanics of Materials.
Check (assumed data). The figure is dimensionless—it labels only the bolt diameter $d$, the member grip $l=l_1+l_2$ and the load $P$—so, as the exam permits (“assume any missing data and state it”), the design is carried out for a representative separating load $P=20\text{ kN}$ carried by a single bolt through two steel members of total grip $l=40\text{ mm}$ ($l_1=l_2=20\text{ mm}$), $E=207\text{ GPa}$. The method is general; only the numbers depend on these stated assumptions.
Given. A single bolt clamps two steel plates that an external load $P$ pulls apart; grip $l=40$ mm, members steel ($E=207$ GPa), assumed load $P=20$ kN.
Given / assumed data
Quantity
Symbol
Value
External separating load
$P$
20 kN
Grip (both members)
$l=l_1+l_2$
40 mm
Member / bolt modulus
$E$
207 GPa
Trial bolt
—
M12×55, Class 8.8
Find. A bolt size and preload $F_i$; the factors of safety against yielding and against joint separation; and the preload (as a percentage of proof strength) that maximizes them.
Bolted joint: one bolt of diameter $d$ through two steel members of combined grip $l=l_1+l_2=40$ mm; the external load $P$ (shown split as $P/2+P/2$) pulls the members apart.
Approach. Compute the bolt and member stiffnesses to get the joint stiffness constant $C$; the bolt then carries $F_b=F_i+CP$ while the members shed $(1-C)P$. Write the yielding and separation load factors as functions of preload, then equate them to find the preload that maximizes the smaller of the two.
Select a trial bolt and list its areas. Take an M12 coarse-thread bolt, Class 8.8 ($S_p=600$ MPa proof, $S_y=660$, $S_{ut}=830$ MPa). Its shank and tensile-stress areas are
$$A_d=\frac{\pi}{4}d^{2}=\frac{\pi}{4}(12)^2=113.1\text{ mm}^2,\qquad A_t=84.3\text{ mm}^2.$$
Bolt stiffness $k_b$. For an M12×55 bolt the threaded length is $L_T=2d+6=30$ mm, so the unthreaded shank is $l_d=55-30=25$ mm (all within the 40 mm grip) and the threaded portion in the grip is $l_t=40-25=15$ mm:
$$k_b=\frac{A_d A_t E}{A_d\,l_t+A_t\,l_d}=\frac{(113.1)(84.3)(207000)}{(113.1)(15)+(84.3)(25)}=5.19\times10^{5}\text{ N/mm}.$$
Member stiffness $k_m$. Using Shigley’s frustum fit for a two-plate steel joint (constants $A=0.78715$, $B=0.62873$):
$$k_m=E\,d\,A\,e^{B d/l}=(207000)(12)(0.78715)\,e^{0.62873(12/40)}=2.361\times10^{6}\text{ N/mm}.$$
Joint stiffness constant $C$. The fraction of external load felt by the bolt is
$$\boxed{C=\frac{k_b}{k_b+k_m}=\frac{5.19\times10^{5}}{5.19\times10^{5}+2.361\times10^{6}}=0.180.}$$
So the bolt sees $CP=0.180(20)=3.60$ kN of the external load and the clamped members lose $(1-C)P=16.4$ kN of clamping force.
Proof load and preload window. The proof load is $F_p=S_p A_t=(600)(84.3)=50.6$ kN. Good practice keeps the preload in $F_i=0.75$–$0.90\,F_p$; the exact optimum is found next.
Yielding and separation load factors. The bolt reaches proof when $F_i+n_y\,CP=F_p$, and the joint separates when $F_i=n_0(1-C)P$:
$$n_y=\frac{F_p-F_i}{CP},\qquad n_0=\frac{F_i}{(1-C)P}.$$
$n_y$ falls and $n_0$ rises as the preload increases, so the two curves cross at the preload that maximizes the smaller factor.
Optimum preload. Setting $n_y=n_0$ and solving,
$$\frac{F_p-F_i}{CP}=\frac{F_i}{(1-C)P}\;\Rightarrow\;F_i=(1-C)F_p.$$
$$\boxed{F_i=(1-C)F_p=(0.820)(50.6)=41.5\text{ kN}\;=\;82.0\%\ \text{of proof}.}$$
This sits neatly inside the recommended $0.75$–$0.90\,F_p$ band, so the M12 Class 8.8 bolt tightened to $\approx82\%$ of proof is the design.
Resulting safety factors. At the optimum both factors take the common value
$$\boxed{n_y=n_0=\frac{F_p}{P}=\frac{50.6}{20}=2.53.}$$
The joint is safe against both bolt yielding and separation with a comfortable factor of about $2.5$.
