Question 6 of 6: Bolted Tension Joint — Bolt Size, Preload and Safety Factors
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 16-Mec-B1 Advanced Machine Design, December 2017. Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; only three of the four Part II problems (Problems 3–6) are required. All six problems are solved as a study resource. “State all assumptions clearly… assume any missing data and properly state it” is an explicit exam instruction.
Given. A single bolt clamps two members of total grip $l=2$ in (each member $l_1=l_2=1$ in), joint width $D=1$ in, carrying an external tensile load $P=2000$ lb (applied as $P/2$ to each side). Steel members, $E=30\times10^{6}$ psi.
Given data
Quantity
Value
Grip $l$
2 in
Joint width $D$
1 in
External load $P$
2000 lb
Member material
steel, $E=30\times10^6$ psi
Find. A suitable bolt size, the safety factors against yielding and joint separation, and the optimum preload (as a percentage of proof strength) that maximizes both.
Bolted tension joint: a preloaded bolt of stiffness $k_b$ in parallel with the clamped members of stiffness $k_m$; the external load $P$ splits by the joint constant $C$.
Approach. Choose a trial bolt, compute the bolt and member stiffnesses to get the joint stiffness constant $C=k_b/(k_b+k_m)$, then use the standard tension-joint load and separation factors. Setting the yielding and separation factors equal gives a closed-form optimum preload; at that preload both safety factors equal $F_p/P$.
Trial bolt. Select a $\tfrac12$–13 UNC SAE grade 5 bolt: tensile-stress area $A_t=0.1419\ \text{in}^2$, proof strength $S_p=85$ ksi, so proof load $F_p=S_pA_t=12.06\times10^{3}\ \text{lb}$.
Stiffnesses. Bolt: $k_b=\dfrac{A_dA_tE}{A_dl_t+A_tl_d}$ with the shank/thread split over the 2-in grip. Members (Shigley’s frustum fit for steel): $k_m=E\,d\,A\,e^{B\,d/l}$ with $A=0.78715$, $B=0.62873$. These give the joint constant
$$C=\frac{k_b}{k_b+k_m}=0.157.$$
Only about 16 % of the external load reaches the bolt; the rest unloads the clamped members.
Optimum preload (equal safety factors). The yielding load factor $n_p=\dfrac{S_pA_t-F_i}{CP}$ and the separation factor $n_0=\dfrac{F_i}{P(1-C)}$ are equal when
$$F_i=(1-C)\,S_pA_t=(1-C)F_p.$$
Hence the optimum preload as a percentage of proof strength is
$$\boxed{\ \frac{F_i}{F_p}=1-C=0.843\ \Rightarrow\ 84.3\%\ \text{of proof}\ }$$
i.e. $F_i=(0.843)(12\,060)=1.02\times10^{4}\ \text{lb}.$
Safety factors at the optimum. Substituting back, both factors collapse to the same value,
$$n_p=n_0=\frac{F_p}{P}=\frac{12\,060}{2000}=6.0.$$
The chosen $\tfrac12$-in grade-5 bolt is more than adequate; even a smaller bolt would suffice, but the $\tfrac12$-in size gives a robust, standard, easily-torqued joint.
Check: $D=1$ in is taken as the joint (member) width; it comfortably exceeds the standard washer-face frustum base for a $\tfrac12$-in bolt, so no frustum truncation is needed and Shigley’s exponential $k_m$ fit applies. Preload $F_i$ would be set in practice as a torque $T\approx0.2\,F_i\,d$.
