Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — December 2014 — 07-Mec-B1, Advanced Machine Design. Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; answer any THREE of Part II (Problems 3–6). All six problems are solved here as a complete study resource.
Given. A confined-gasket bolted cylinder head; both members steel; grip length $l=D+E=45$ mm.
Given data
Sealing diameter
$D_g=150$ mm
Internal pressure
$p=6$ MPa
Number of bolts
$N=8$, M12 ($A_t=84.3\ \text{mm}^2$)
Grip
$l=D+E=20+25=45$ mm
Separation / yielding FoS
$n_0\ge1.5$, $\;n_p\ge2$
Steel modulus
$E=207$ GPa
Find. The required bolt preload $F_i$ and a suitable metric bolt grade.
Bolted cylinder head with a confined gasket seal (one representative M12 bolt shown; 8 equally spaced).
Approach. Find the pressure load carried per bolt, compute the joint stiffness constant $C$ from the bolt and member (frustum) stiffnesses, set the preload from the separation criterion, then pick the smallest bolt grade whose proof strength satisfies the yielding factor of safety.
External load per bolt. The gas pressure acting over the sealed area is shared by the 8 bolts: $$P=\frac{p\,(\pi/4)D_g^2}{N}=\frac{(6\times10^6)(\pi/4)(0.150)^2}{8}=\frac{106029}{8}=13.25\times10^{3}\ \text{N}.$$
Bolt stiffness. With grip $l=45$ mm and threaded length in grip $l_t=l-l_d$ (using $L_T=2d+6=30$ mm), the bolt stiffness is $$k_b=\frac{A_d A_t E}{A_d l_t+A_t l_d}\approx 2.9\times10^{8}\ \text{N/m}.$$
Member stiffness (frustum) and joint constant. Using Shigley’s exponential fit for steel, $k_m=E\,d\,(0.78715)\,e^{0.62873\,d/l}\approx1.6\times10^{9}$ N/m, so $$C=\frac{k_b}{k_b+k_m}=0.155.$$ The bolts carry only $C=15.5\%$ of any load change — a stiff-member, soft-bolt joint, as intended for a pressure seal.
Required preload from the separation criterion. Separation occurs when the member load reaches zero; the factor of safety against separation is $n_0=F_i/[P(1-C)]$. Setting $n_0=1.5$: $$\boxed{\;F_i=n_0\,P\,(1-C)=(1.5)(13254)(1-0.155)=1.68\times10^{4}\ \text{N}\;}$$ i.e. a preload of about 16.8 kN per bolt.
Grade selection from the yielding criterion. The bolt first reaches its proof strength at a load factor $n_p=(S_p A_t-F_i)/(C\,P)$. Requiring $n_p\ge2$ gives a minimum proof strength $$S_p\ge\frac{F_i+2CP}{A_t}=\frac{16800+2(0.155)(13254)}{84.3\times10^{-6}}=248\ \text{MPa}.$$ The lowest metric grade meeting this is 4.8 ($S_p=310$ MPa); grade 4.6 ($S_p=225$ MPa) fails. Selecting the readily-available grade 5.8 ($S_p=380$ MPa) gives ample margin: $$n_p=\frac{(380\times10^6)(84.3\times10^{-6})-16800}{(0.155)(13254)}=7.4\;(\ge2).$$ The preload also stays within the proof load, $F_i=16.8\ \text{kN}\le0.75\,S_p A_t=24.0\ \text{kN}$.
