Connecting Rod

Abstract

The connecting rod converts the reciprocating motion of piston into the rotating motion of the crankshaft. Generally, it can be seen in three parts, i.e., small end, shank and big end. The connecting rod motion is complex as the small end is reciprocating along cylinder axis and big end is rotating along with the crankpin. The Loads on a connecting rod are categorized as three types namely, Firing load, Inertia load and other loads. The analysis of loads on Connecting Rod by classical method must be carried out for sizing and shaping before going for detailed analysis using the finite element method for both static and dynamic loads. The examples for the classical method are available in the appendix. The analysis for the four load cases namely, Bolt Preload and Bearing and Bush Interference, Gas Pressure Loading, Inertia Loading and Combined Loading is presented. Enhancing the yield strength and fatigue strength is achieved by choice of Materials and heat treatment. Some practical aspects during design like Weight grouping of connecting rods, Push-out force test and Testing of the connecting rod are given. The fracture splitting method for connecting rods is becoming popular as an exercise in cost reduction. The manufacturing process of connecting rod is described in brief. At the end of the chapter various failure modes are described which are borne in mind while designing the connecting rod.

Appendices

Appendix 1

Distribution of connecting rod mass at small end and big end (refer Figs. 13.9 and 13.10)

Fig. 13.9

Location of C.G. for the connecting rod

Fig. 13.10

Distribution of reciprocating and rotating masses for con rod

Let

$$ \begin{aligned} L & = centre\,distance\,between\,Small\,end\,and\,big\,end \\ L_{1} & = distance\,of\,the\,Centre\,of\,gravity\,(CG)\,from\,the\,small\,end \\ L_{2} & = distance\,of\,CG\,from\,big\,end \\ L & = L_{1} + L_{2} \\ \end{aligned} $$

$$ \begin{aligned} W_{assly} & = weight\,of\,connecting\,rod\,assembly \\ W_{recip} & = reciprocating\,portion\,of\,the\,connecting\,rod\,assembly \\ W_{rot} & = rotating\,portion\,of\,the\,connecting\,rod\,assembly \\ W_{assly} & = W_{recip} + W_{rot} \\ \end{aligned} $$

Also, the moment of forces at CG is zero.

$$ W_{recip} \times L_{1} - W_{rot} \times L_{2} = 0 $$

Therefore,

$$ \begin{aligned} W_{rot} & = \left( {\frac{{L_{1} }}{{L_{2} }}} \right)W_{recip} \\ W_{assly} & = W_{recip} \left( {\frac{{L_{1} + L_{2} }}{{L_{2} }}} \right)\,{\text{Or}}\,W_{recip} = W_{assly} \left( {\frac{{L_{2} }}{{L_{1} + L_{2} }}} \right) \\ \end{aligned} $$

$$ \begin{aligned} L & = centre\,distance\,between\,the\,small\,end\,and\,the\,big\,end \\ R & = crank\,radius \\ \theta & = crank\,angle \\ \varphi & = obliquity\,angle \\ x & = displacement\,of\,piston\,assembly\,from\,the\,TDC\,at\,\theta \\ \end{aligned} $$

From the Fig. 13.11,

Fig. 13.11

Sketch of connecting rod at crank angle θ

$$ \begin{aligned} L + R & = x + L\cos \varphi + R\cos \theta \\ \therefore\,x & = (L + R) - (L\cos \varphi + R\cos \theta ) \\ \end{aligned} $$

But, $L\sin \varphi = R\sin \theta$

$$ \frac{x}{R} = \left( {1 + \frac{1}{\lambda }} \right) - \cos \theta - \frac{1}{\lambda }\sqrt {1 - \lambda^{2} sin^{2} \theta } $$

Where

$$ \lambda = \frac{R}{L} $$

Fourier development of the equation gives

$$ \frac{x}{R} = A_{0} + A_{1} \cos \theta - \frac{{A_{2} }}{4}\cos 2\theta + \frac{{A_{4} }}{16}\cos 4\theta - \frac{{A_{6} }}{36}\cos 6\theta + \cdots $$

Getting the values of A₀, A₁, A₂… in terms of λ and neglecting higher order terms of λ, λ being less than 1, we get,

i.e.

$$ \begin{aligned} \frac{x}{R} & = 1 + \frac{\lambda }{4} - \cos \theta - \frac{\lambda }{4}\cos 2\theta \\ velocity,v & = \dot{x} = \omega R\left( {\sin \theta + \frac{\lambda }{2}\cos 2\theta } \right) \\ acceleration,a & = \dot{v} = \ddot{x} = \frac{dv}{d\theta }\frac{d\theta }{dt} = \frac{dv}{d\theta }\omega \\ & = R\omega^{2} \left( {\cos \theta + \frac{\cos 2\theta }{n}} \right) \\ \end{aligned} $$

Where

$$ \begin{aligned} & n = \frac{1}{\lambda } \\ & \therefore Inertiaload = m_{recip} \times acceleration = m_{recip} R\omega^{2} \left( {\cos \theta + \frac{\cos 2\theta }{n}} \right) \\ \end{aligned} $$

Appendix 2

Inertia of I - section

The connecting rod small end and big end form a hinged joint in one plane where bend is tested. The other plane where twist is tested, the rod has more resistance to bending against the compressive load. In the latter case, the equivalent length is given below calculating the slenderness ratio.

$$ l = \frac{{l_{actual} }}{2} $$

Hence to have the connecting rod equally strong about the both the axes, the critical buckling load, F_i should be such that

$$ F = \frac{{f_{c} A}}{{1 + \alpha \left( {\frac{l}{{R_{xx} }}} \right)^{2} }} = \frac{{f_{c} A}}{{1 + \alpha \left( {\frac{l}{{2R_{yy} }}} \right)^{2} }} $$

i.e.

$$ \left( {\frac{1}{{R_{xx} }}} \right)^{2} = \left( {\frac{1}{{R_{yy} }}} \right)^{2} $$

or $I_{xx} = I_{yy}$ since $I = AR^{2}$ we get

$$ I_{xx} = 4I_{yy} $$

With the I-section defined by 4 t. 5 t. t as shown in the Fig. 13.12.

Fig. 13.12

Typical dimensions of I section

This selection of cross section is taken as a starting guide line.

Appendix 3

Calculation of the connecting rod Big end (Bremi 1971)

Stresses at the big end is arrived at by the theory of curved beams. The big end is pulled by the inertia forces. Half connecting rod split by the plane passing through the centre line of the cylinder can be imagined as a hook. The shear and normal forces and the bending moment at the split. Since the split is in equilibrium the two halves of the connecting rod the respective forces on the faces are equal.

The reader is referred to the work done by P. Bremi (Bremi 1971). The formulae involved in brief are shown below. Refer Figs. 13.13, 13.14, 13.15 and 13.16 for simplified connecting rod model.

Fig. 13.13

Big end of connecting rod as a curved beam (Bremi 1971)

Fig. 13.14

Simplified rectangular cross section (Bremi 1971)

Fig. 13.15

Location of general cross section (Bremi 1971)

Fig. 13.16

Sketch showing various parameters of big end (Bremi 1971)

Bending stress,

$$ \sigma (z) = \frac{N}{F} + \frac{M}{rF} + \frac{M}{r\lambda F}\frac{z}{r + z} $$

(13.1)

where

$$ \lambda F = - \int_{{z_{1} }}^{{z_{2} }} {\frac{z}{r + z}b(z)} dz $$

(13.2)

$$ I(z)(r + z)b(z) = - \frac{\Delta \lambda F}{\lambda F}Q + \int_{{z_{1} }}^{z} {I\left( {z^{\prime } } \right)b(z)} dz^{\prime } $$

(13.3)

where

$$ \Delta \lambda F = - \int_{{z_{1} }}^{z} {\frac{{z^{\prime } }}{{r + z^{\prime } }}b\left( {z^{\prime } } \right)} dz^{\prime } $$

(13.4)

If

$$ \Delta F = \int_{{z_{1} }}^{z} {b\left( {z^{\prime } } \right)} dz^{\prime } $$

(13.5)

Then

$$ I(z) = \frac{Q}{b(z)(r + z)}\left[ {\frac{\Delta F}{F} - \frac{\Delta \lambda F}{\lambda F}} \right] $$

(13.6)

Deformation energy of the curved beam

$$ \overline{U} = \frac{{\sigma^{2} }}{2E} $$

(13.7)

$$ U = \int_{{\phi_{1} }}^{{\phi_{2} }} {\int_{{z_{1} }}^{{z_{2} }} {\frac{{\sigma^{2} }}{2E}} (r + z)b(z)} dzd\phi $$

(13.8)

Substituting from (13.1)

$$ U = \int_{{\phi_{1} }}^{{\phi_{2} }} {\frac{1}{2E}\left[ {\frac{{N_{r}^{2} }}{F} + 2\frac{NM}{F} + \frac{{M^{2} }}{rF}\left( {1 + \frac{1}{\lambda }} \right)} \right]} d\phi $$

(13.9)

$$ I(z) = \frac{Q}{{6h^{3} }}\left( {\frac{{h^{2} }}{4} - z^{2} } \right) $$

(13.10)

For comparison, a still simpler equation

$$ J^{{\prime }} (z) = \frac{Q}{bh} $$

(13.11)

$$ \overline{U}_{Q} = \frac{{J^{2} }}{2G}, $$

(13.12)

where G = shear modulus

$$ G = \frac{E}{2(1 + \gamma )}, $$

(13.13)

where γ = Poisson ratio

Deformation energy of a beam piece of length, l

$$ U_{Q} = \int_{0}^{l} {\int_{ - h/2}^{h/2} {\frac{{J^{2} }}{2G}b\,dz} } \,dx $$

(13.14)

Substituting (10) and (11) in the above equation,

$$ Deformation\,energy\,due\,to\,Q,U_{Q} = \int_{0}^{l} {\frac{3}{5}\frac{{Q^{2} }}{Gbh}} dx $$

(13.15)

Or

$$ U_{Q}^{{\prime }} = \int_{0}^{l} {\frac{1}{2}\frac{{Q^{2} }}{Gbh}} dx $$

(13.16)

When the beam is loaded with moment M,

$$ Deformation\,energy\,due\,to\,M,U_{M} = \int_{0}^{l} {\frac{{6M^{2} }}{{Ebh^{3} }}} dx $$

(13.17)

If the beam is loaded at the end with a force P,

$$ Q = P $$

(13.18)

and

$$ M = Px $$

(13.19)

$$ U = \frac{{2P^{2} l^{3} }}{{Ebh^{3} }}\left[ {1 + \frac{3}{5}(1 + \nu )\frac{{h^{2} }}{{l^{2} }}} \right] $$

(13.20)

Or

$$ U^{{\prime }} = 2\frac{{P^{2} l^{3} }}{{Ebh^{3} }}\left[ {1 + \frac{1}{2}(1 + \nu )\frac{{h^{2} }}{{l^{2} }}} \right] $$

(13.21)

In general case of a connecting rod,

If

$$ \nu = \frac{1}{3}\,and\,l = 3h $$

Then,

$$ U = \frac{{2P^{2} l^{3} }}{{Ebh^{3} }}[1 + 0.09] $$

(13.22)

And

$$ U^{{\prime }} = 2\frac{{P^{2} l^{3} }}{{Ebh^{3} }}[1 + 0.06] $$

(13.23)

The shear stress is not negligible since its contribution to the distortion energy is about 10% of that of the normal stress. Since Uis within 1% of U’ is only 1% Eq. 13.11 is used without loss of accuracy, instead of Eq. 13.10 while calculating the distortion energy.

For a bent beam,

$$ J(z) = \frac{Q}{F} $$

(13.24)

$$ UQ = \int_{{\phi_{1} }}^{{\phi_{2} }} {\frac{1}{2G}\frac{{Q^{2} }}{F}} rd\varphi $$

(13.25)

This energy can have the energy as Eq. 13.9 superimposed on it so that the following formula is valid for the deformation energy of the bent beam.

$$ U = \int_{{r_{1} }}^{{r_{2} }} {\frac{1}{2EF}\left[ {rN^{2} + 2NM + \frac{{1 + \frac{1}{\lambda }}}{r}m^{2} + r\frac{E}{G}Q^{2} } \right]} d\phi $$

(13.26)

Displacement due to N^l = n^l

Displacement due to Q^l = q^l

Displacement due to M^l = m^l

Ref Fig. 13.16

We define

$$ \alpha = \pi - \beta $$

(13.27)

And

$$ K_{\phi } = \left\{ {0\,for\,0 \le \phi \le \frac{\pi }{2},K\,for\,\frac{\pi }{2} \le \phi \le \alpha } \right\} $$

(13.28)

for the split beam shown in figure.

In the split beam, there will be no load N, φ and C, moment created in the section of the connecting rod. The following relationships are obtained.

$$ N_{\phi } = N\cos \phi + Q\sin \phi - K_{\phi } \cos \phi $$

(13.29)

$$ Q_{\phi } = - N\sin \phi + Q\cos \phi + K_{\phi } \sin \phi $$

(13.30)

Moment of inertia of all forces at the centre of connecting rod,

$$ M_{\phi } = N\left( {r_{0} - r_{\phi } \cos \phi } \right) - Qr_{\phi } \sin \phi + K_{\phi } \sin \phi $$

(13.31)

Substituting in Eq. 13.26, the equations for deformation energy of half big end are as follows:

$$ \begin{aligned} U & = \frac{1}{2}\left[ {a_{NN} N^{2} + a_{NQ} NQ + a_{NM} NM + a_{NK} NK + a_{QN} QN + a_{QQ} Q^{2} } \right. \\ & \quad + a_{QM} QM + a_{QK} QK + a_{MN} MN + a_{MQ} MQ + a_{MQ} MQ + a_{MM} M^{2} + a_{MK} MK \\ & \quad \left. { + a_{KN} KN + a_{KQ} KQ + a_{kM} KM + a_{KK} K^{2} } \right] \\ \end{aligned} $$

(13.32)

$$ \begin{aligned} a_{NN} & = \int_{0}^{\alpha } {\frac{{r_{\phi } }}{EF}} \left[ {\cos^{2} \phi + 2\frac{{r_{\phi } }}{r}\left( {\frac{{r_{0} }}{{r_{\phi } }} - \cos \phi } \right) + \left( {1 + \frac{1}{\lambda }} \right)\frac{{r_{\phi }^{2} }}{{r^{2} }}\left( {\frac{r}{{r_{\phi } }} - \cos \phi } \right)^{2} } \right. \\ & \quad \left. { + \frac{E}{G}\sin^{2} \phi } \right]d\phi \\ \end{aligned} $$

(13.33)

$$ a_{QQ} = \mathop \smallint \limits_{0}^{\alpha } \frac{{r_{\phi } }}{EF}\left[ {\sin^{2} \phi - 2\frac{{r_{\phi } }}{r}\sin^{2} \phi + \left( {1 + \frac{1}{\lambda }} \right)\frac{{r_{\phi }^{2} }}{{r^{2} }}\sin^{2} \phi + \frac{E}{G}\cos^{2} \phi } \right]d\phi $$

(13.34)

$$ a_{MM} = \mathop \smallint \limits_{0}^{\alpha } \frac{{r_{\phi } }}{EF}\left[ {\left( {1 + \frac{1}{\lambda }} \right)\frac{{r_{\phi }^{2} }}{{r^{2} }} - \frac{1}{{r_{\phi }^{2} }}} \right]d\phi $$

(13.35)

$$ a_{QM} = a_{MQ} = \mathop \smallint \limits_{0}^{\alpha } \frac{{r_{\phi } }}{EF}\left[ {\frac{{r_{\phi } }}{r}\sin \phi \frac{1}{{r_{\phi } }} - \left( {1 + \frac{1}{\lambda }} \right)\frac{{r_{\phi }^{2} }}{{r^{2} }}\sin \phi \frac{1}{{r_{\phi } }}} \right]d\phi $$

(13.36)

$$ a_{MN} = a_{NM} = \mathop \smallint \limits_{0}^{\alpha } \frac{{r_{\phi } }}{EF}\left[ {\frac{{r_{\phi } }}{r}\cos \phi \frac{1}{{r_{\phi } }} + \left( {1 + \frac{1}{\lambda }} \right)\frac{{r_{\phi }^{2} }}{{r^{2} }}\left( {\frac{r}{{r_{\phi } }} - cos\phi } \right)\frac{1}{{r_{\phi } }}} \right]d\phi $$

(13.37)

$$ \begin{aligned} a_{NQ} = a_{QN} & = \int_{0}^{\alpha } {\frac{{r_{\phi } }}{EF}\left[ {\cos \phi \sin \phi - \frac{{r_{\phi } }}{r}\cos \phi \sin \phi + \frac{{r_{\phi } }}{r}\sin \phi \left( {\frac{r}{{r_{\phi } }} - \cos \phi } \right)} \right.} \\ & \quad - \left. {\left( {1 + \frac{1}{\lambda }} \right)\frac{{r_{\phi }^{2} }}{{r^{2} }}\left( {\frac{r}{{r_{\phi } }} - cos\phi } \right)\sin \phi - \frac{E}{G}\sin \phi \cos \phi } \right]d\phi \\ \end{aligned} $$

(13.38)

$$ \begin{aligned} a_{NK} = a_{KN} & = \int_{0}^{\alpha } {\frac{{r_{\phi } }}{EF}\frac{{K_{\phi } }}{K}\left[ { - \cos^{2} \phi + \frac{{r_{\phi } }}{r}\cos^{2} \phi - \frac{{r_{\phi } }}{r}\cos \phi \left( {\frac{r}{{r_{\phi } }} - \cos \phi } \right)} \right.} \\ & \quad \left. { - \left( {1 + \frac{1}{\lambda }} \right)\frac{{r_{\phi }^{2} }}{{r^{2} }}\left( {\frac{r}{{r_{\phi } }} - cos\phi } \right)\cos \phi - \frac{E}{G}\sin^{2} \phi } \right]d\phi \\ \end{aligned} $$

(13.39)

$$ \begin{aligned} a_{QK} = a_{KQ} & = \int_{0}^{\alpha } {\frac{{r_{\phi } }}{EF}\frac{{K_{\phi } }}{K}\left[ { - \sin \phi \cos \phi + \frac{{r_{\phi } }}{r}\sin\upphi\cos \phi } \right.} \\ & \quad \left. { - \left( {1 + \frac{1}{\lambda }} \right)\frac{{r_{\phi }^{2} }}{{r^{2} }}\sin\upphi\cos \phi - \frac{E}{G}\sin\upphi\cos \phi } \right]d\phi \\ \end{aligned} $$

(13.40)

$$ a_{MK} = a_{KM} = \mathop \smallint \limits_{0}^{\alpha } \frac{{r_{\phi } }}{EF}\frac{{K_{\phi } }}{K}\left[ { - \frac{{r_{\phi } }}{r}\cos \phi \frac{1}{{r_{\phi } }} + \left( {1 + \frac{1}{\lambda }} \right)\frac{{r_{\phi }^{2} }}{{r^{2} }}\cos \phi \frac{1}{{r_{\phi } }}} \right]d\phi $$

(13.41)

$$ a_{KK} = \mathop \smallint \limits_{0}^{\alpha } \frac{{r_{\phi } }}{EF}\frac{{K_{\phi }^{2} }}{{K^{2} }}\left[ {\cos^{2} \phi - 2\frac{{r_{\phi } }}{r}\cos^{2} \phi + \left( {1 + \frac{1}{\lambda }} \right)\frac{{r_{\phi }^{2} }}{{r^{2} }}\cos^{2} \phi + \frac{E}{G}\sin^{2}\upphi} \right]d\phi $$

(13.42)

Stresses on the big end

$$ n = \frac{\partial U}{\partial N} = a_{NN} N + a_{NQ} Q + a_{NM} M + a_{NK} K $$

(13.43)

$$ q = \frac{\partial U}{\partial Q} = a_{QN} N + a_{QQ} Q + a_{QM} M + a_{QK} K $$

(13.44)

$$ m = \frac{\partial U}{\partial M} = a_{MN} N + a_{MQ} Q + a_{MM} M + a_{MK} K $$

(13.45)

To these we add the following quantity

$$ k = \frac{\partial U}{\partial K} = a_{KN} N + a_{KQ} Q + a_{KM} M + a_{KK} K $$

(13.46)

For the above calculations, the connecting rod is assumed to be split. In actual case, it is not. Now,

$$ N^{l} = N^{r} $$

(13.47)

$$ Q^{l} + Q^{r} = P $$

(13.48)

$$ M^{l} = M^{r} $$

(13.49)

$$ n^{l} = - n^{r} $$

(13.50)

$$ q^{l} = q^{r} $$

(13.51)

$$ m^{l} = - m^{r} $$

(13.52)

A set of linear equations for the six unknowns N^l, Q^l, m^l, N^r, Q^r and M^r is obtained using Eqs. 43–45 for each half of the big end. This system can be easily converted back to the system

$$ \left( {a_{NN}^{l} + a_{NN}^{r} } \right)N^{l} + \left( {a_{NQ}^{l} - a_{NQ}^{r} } \right)Q^{l} + \left( {a_{NM}^{l} + a_{NM}^{r} } \right)M^{l} = - a_{NQ}^{r} P - a_{NK}^{l} K^{l} - a_{NK}^{r} K^{r} $$

(13.53)

$$ \left( {a_{QN}^{l} - a_{QN}^{r} } \right)N^{l} + \left( {a_{QQ}^{l} + a_{NQ}^{r} } \right)Q^{l} + \left( {a_{QM}^{l} - a_{NM}^{r} } \right)M^{l} = a_{QQ}^{r} P - a_{QK}^{l} K^{l} + a_{QK}^{r} K^{r} $$

(13.54)

$$ \left( {a_{MN}^{l} + a_{MN}^{r} } \right)N^{l} + \left( {a_{MQ}^{l} - a_{MQ}^{r} } \right)Q^{l} + \left( {a_{MM}^{l} + a_{MM}^{r} } \right)M^{l} = - a_{MQ}^{r} P - a_{MK}^{l} K^{l} - a_{MK}^{r} K^{r} $$

(13.55)

This selection of cross section is taken as a starting guide line. The calculations are iterated for the modified configuration, depending on the magnitude of the stresses. A typical stress distribution is shown in Fig. 13.17.

Fig. 13.17

Stresses in big end of the connecting rod (Bremi 1971)

Appendix 4

Connecting rod Big end bolt sample calculation

The calculations are performed for a connecting rod having two bolts. The side force Q is not considered as the rod assumed with horizontal split.

Bolt sizeM12 × 1.5

Bolt Quality10.9

m_recip. 1.720 kg.

m_rod without cap1.305 kg.

$$ \begin{aligned} r & = crank\,radius = 0.054\,{\text{m}} \\ & \varpi = \frac{2\pi n}{60}where\,n = 3150\,rpm\,330\,{\text{rad/sec}}. \\ L & = con.\,rod\,centre\,distance\,0.187\,{\text{m}} \\ Fmax & = (m\,recip + m\,rod)r\varpi 2\left( {1 + \frac{r}{L}} \right)22925\,{\text{N}} \\ \end{aligned} $$

Bolt torque10 kgm with oil (80% yield)

Preload as per VDI 2230 (VDI 2230) is estimated as 54004 N per bolt

Total preload108008 N

The force F_interference due to bearing overstand-

Assuming 25 kg/mm² stress on bearing,

$$ \begin{array}{*{20}l} {{\text{The}}\,{\text{dimensions}}\,{\text{of}}\,{\text{bearings}}{:}\,\quad \quad 2\,{\text{mm}}\,{\text{thick}} \times 23\,{\text{mm}}\,{\text{length}}} \hfill \\ \begin{aligned} Area\,on\,one\,side & = 46\,{\text{mm}}^{ 2} \\ Force\,on\,each\,side\,25 \times 46 & = 1150\,{\text{kg}} \\ Force\,on\,both\,sides & = 2300\,{\text{kg}} \\ Force\,on\,each\,bolt & = 1150\,{\text{kg}} \\ \therefore Finterference & = 11280\,{\text{N}} \\ \end{aligned} \hfill \\ {\quad \quad \therefore Total\,load\,on\,bolt} \hfill \\ \begin{aligned} & = Fmax + Finterference \\ & = 22925 + 11280\,{\text{N}} \\ & = 34205\,{\text{N}} \\ \quad \quad \therefore Cover\,factor & = \frac{108008}{34205} \\ & = 3.16 > 2 \\ \end{aligned} \hfill \\ {\quad \quad Hence\,safe} \hfill \\ \end{array} $$

Connecting rod Big end bolt detailed calculation as per VDI 2230

The connecting rod big end joint as shown in Fig. 13.18 is example of eccentrically clamped eccentrically loaded joint. The analysis is to be done for 12.9 grade M9 × 1 bolt and grade-12 nut.

Fig. 13.18

Forces acting on the interface of a connecting rod bearing cap bolted joint

The bolts are tightened with a high precision tightening spindle. C45 was selected as the material for the clamped parts. For the rated engine speed (n = 4000 rpm), we have the following initial parameters.

Axial force at the interface

$$ Fa = 3.7 \times 10^{3} \,{\text{N}} $$

Bending moment at the interface

$$ M_{bA} = 30\,{\text{Nm}} $$

Transverse force at the interface

$$ F_{Q} = 420\,{\text{N}} $$

From the bending moment M_bA and the axial load F_A, the lever arm of the eccentric load application is determined as

$$ a = \frac{{M_{bA} }}{{F_{A} }} = 8.1\,{\text{mm}} $$

Assuming for simplicity that the bending moment is constant over the clamping length ${\text{l}}_{\text{K}}$

Calculation procedure

The calculations are made following the calculations steps R1 to R10 given in Sect. 4.1 VDI 2230.

R1Rough determination of the bolt diameter d and the clamping length ratio l_K/d.

The bolt diameter is given as 9 mm from the design shown in Fig. 13.19. The clamping length ration is

Fig. 13.19

Dimensions(mm) of the clamping and clamped parts of the connecting rod bearing cap bolted joint and the nut and head bearing surfaces

$$ \frac{{l_{K} }}{d} = \frac{41.5}{9} = 4.6 $$

Rough determination of the surface pressure under the bolt head:

$$ p = \frac{{\frac{{F_{lA} }}{0.9}}}{{A_{P} }} \le P_{G} $$

Measurement of screw in mm

Place	l1	d1
	2.5	8.7
	3.0	9.2
	6.5	9.0
	10.0	9.2
	15.8	8.35
	3.7	7.68

Other Measurements

Thread		Clamped parts		Nut bearing		Head bearing
d2	8.27	DA	15.5	Dw, min	13.8	dhw, min	14
d3	7.66	db	9.25	Db, max	9.9	dho, min	9.65
d	9	lg	41.5	D1	7.9	dh	9.25
P	7	g	0.3	dh	9.43	df	8.7
Ad3	46 mm2	a	8.1	Dh0, max	10	db, max	9
		u	5.3	Dh0, min	14	dw, min	13.2
		v	6.8			dp	13.4
		w	1.4
		n	17.6

$F_{M} = 42.6 \times 10^{3} \,{\text{N}}\,for\,\mu_{G} = 0.12$ from Table 3 of VDI 2230.

With

$$ \begin{aligned} A_{P\,head,min} & = \frac{\pi }{4}\left( {d_{W,min}^{2} - d_{h\,a,max}^{2} } \right) \\ & = 64\,{\text{mm}}^{ 2} \\ \end{aligned} $$

and

$$ \begin{aligned} A_{Pnut,min} & = \frac{\pi }{4}\left( {D_{w,min}^{2} - D_{ha,max}^{2} } \right) \\ & = 71\,{\text{mm}}^{ 2} \\ \end{aligned} $$

follows

$$ P_{max} = \frac{{F_{M} }}{{0.9\,A_{Phead,min} }} = 740\frac{\text{N}}{{{\text{mm}}^{ 2} }} > P_{G} $$

where P_G = 700 N/mm² from Table 39 of VDI2230. Further examination in R10.

R2 Determination of the tightening factor α_A

The bolt is tightened using a high precision tightening spindle, which has been adjusted by measuring the elongation of the bolt (after pre-calibrating the bolt as the force measuring element).

The tightening factor α_A = 1.6 according to Table 8of VDI 2230 (for large angles of rotation, fine threads and resilient joints).

R3 Determination of the required minimum clamp load, F_K erf

• 1. The requirement for friction grip at the interface (μ_Tr = 0.12) is: $F_{K\,erf1} = \frac{{F_{0} }}{{\mu_{Tr} }} = 3.5 \times 10^{3} \,{\text{N}}$

• 2. To avoid one-sided opening at the rated speed of the engine, F_K erf2 is calculated (from Eq. 3.52 of VDI 2230) using the dimensions given in Fig. 13.19.

  $$ F_{{K\,erf_{2} }} = \frac{{\left( {a - s} \right)u}}{{\frac{{I_{BT} }}{{A_{D} }} + su}}F_{A} = 8.1 \times 10^{3} \,{\text{N}} $$

  Where

  $$ \begin{aligned} & A_{D} = 121\,{\text{mm}}^{ 2} \\ & I_{BT} = 2474\,{\text{mm}}^{ 4} \\ \end{aligned} $$

R4 Determination of the load factor, Φ_en

The transmission of bending moments and normal forces at the interface leads to a bolted joint with eccentric load application. Moreover, the load is not introduced under the bolt head and the nut but inside the clamped parts. Since the greater part of the connecting rod joint surrounding the bolt can be considered to be a clamping sleeve, n = 1/3 is estimated in this case.

With

$$ A_{ers} = 122\,{\text{mm}}^{ 2} $$

Using Eq. (3.34 of VDI 2230)

$$ \Phi _{en} = n\frac{{\delta_{P} \left( {1 + \frac{{a\,s\,A_{ers} }}{{I_{B\,ers} }}} \right)}}{{\delta_{S} + \delta_{P} \left( {1 + \frac{{s^{2} A_{ers} }}{{I_{B\,ers} }}} \right)}} $$

The resilience of the bolt is calculated using Eq. 3.8 of VDI 2230

$$ \delta_{S} = \delta_{K} + \delta_{1} + \delta_{2} + \cdots + \delta_{GM} $$

Its components are found as follows

• 1. Thread elasticity in the nut:

  $$ \delta_{G} = \frac{0.5d}{{E_{S} A_{3} }} = 0.477 \times 10^{ - 6} \,{\text{mm/N}} $$

  Where

  $$ E_{S} = 205 \times 10^{3} \,{\text{N/mm}}^{ 2} $$

• 2. Elasticity due to the displacement of the nut:

  $$ \delta_{M} = \frac{{l_{M} }}{{E_{S} A_{N} }} = 0.276 \times 10^{ - 6} \,{\text{mm/N}} $$

  Where

  $$ l_{M} = 0.4\,d $$

• 3. Elasticity of the free and loaded part of the thread:

  $$ \delta_{6} = \frac{{l_{6} }}{{E_{S} A_{3} }} = 0.392 \times 10^{ - 6} \,{\text{mm/N}} $$

• 4. Elasticity of the shank:

  $$ \begin{aligned} \delta_{1 \cdots 5} & = \sum\limits_{i = 1 \cdots 5} {\frac{{l_{i} }}{{E_{S} A_{N} }}} \\ & = 3.064 \times 10^{ - 6} \,{\text{mm/N}} \\ \end{aligned} $$

• 5. Elasticity of the head:

  $$ \delta_{K} = \frac{0.4\,d}{{E_{S} A_{N} }} = 0.276 \times 10^{ - 6} \,{\text{mm/N}} $$

  We thus have

  $$ \begin{aligned} \delta_{S} & = \delta_{G} + \delta_{M} + \delta_{6} + \delta_{1 \cdots 5} + \delta_{K} \\ & = 4.49 \times 10^{ - 6} \,{\text{mm/N}} \\ \end{aligned} $$

Resilience of the clamped parts δ_P:

For the determination of the resilience of the clamped parts, the small eccentricity of the bolt (s = 0.3) is not allowed for. Thus, δ_P will be determined instead of δ ^* _P . The assembly preload which causes embedding acts concentrically. For a cross-section study of the clamped parts of a substitution body Fig. 13.5 of VDI 2230 should be referred. Since the joint has a clearance hole, with dimensions D_A and d_h, and because

$$ d_{w} \le D_{A} \le d_{w} + l_{K} $$

We find with Eq. (3.17 of VDI 2230)

$$ A_{ers} = 122\,{\text{mm}}^{ 2} $$

Where

$$ D_{A} = 15.5\,{\text{mm}};d_{w} = 13.2\,{\text{mm}};l_{K} = 41.5\,{\text{mm}};d_{h} = 9.25\,{\text{mm}} $$

With

$$ E_{P} = 205 \times 10^{3} \,{\text{N/mm}}^{ 2} $$

We find

$$ \delta_{P} = \frac{{l_{K} }}{{E_{P} A_{ers} }} = 1.66 \times 10^{ - 6} \,{\text{mm/N}} $$

Thus, with I_B ers = 2833 mm⁴ because D_ers = 15.5 mm determined from A_ers and d_h = 9.25 mm we obtain from Eq. (3.43) of VDI 2230

$$ \varPhi_{en} = 0.118 $$

R5Determination of the loss of preload due to F_Z embedding

From Fig. 50 of VDI 2230 we find the amount of embedding f_Z = 6.1 × 10⁻³ mm (rounded), for l_K/d = 4.6. Thus, the loss of preload due to embedding,

$$ F_{Z} = f_{Z} \frac{1}{{\delta_{P} + \delta_{S} }} = 0.976 \times 10^{3} \,{\text{N}} $$

R6 Determination of the required bolt size

The maximum assembly preload is calculated from:

$$ F_{M\,\hbox{max} } = \alpha_{A} \left[ {F_{K\,erf} + \left( {1 -\Phi _{en} } \right)F_{A} + F_{Z} } \right] $$

However, for the case of a bearing cap bolted joint, F_M min must be further increased by the amount F_L required for the elastic and plastic deformation which occurs in the shell bearing because of over sizing.

To compensate for over-sizing of the shell bearings, an axial load F_L of 6.2 × 10³ N per bolt is required.

Thus, in this case we have

$$ F_{M\,\hbox{max} } = \alpha_{A} \left[ {F_{K\,erf} + \left( {1 -\Phi _{en} } \right)F_{A} + F_{Z} + F_{L} } \right] $$

With

$$ \begin{aligned} & F_{K\,erf} = F_{{K\,{\text{erf}}\,2}} ,since\,F_{{K\,{\text{erf}}\,2}} > F_{{K\,{\text{erf}}\,1}} \\ & F_{M\,max} = 29.7 \times 10^{3} \,{\text{N}} \\ \end{aligned} $$

Assuming a coefficient of friction μ_G = 0.12 (Table 5 of VDI 2230), we obtain from Eq. (3.26 of VDI 2230) an assembly preload of

$$ F_{M} = \sigma_{M} A_{0} = 41.6 \times 10^{3} {\text{N}} $$

for the bolt M9 × 1 of strength grade 12.9. Therefore, A₀= A_s= 49.8 mm² and

$$ \begin{aligned} \sigma_{M} & = \frac{{\nu R_{p\,0.2\,min} }}{{\sqrt {1 + 3\left[ {\frac{{2d_{2} }}{{d_{0} }}\left( {\frac{P}{{\pi d_{2} }} + 1.155\mu_{G} } \right)} \right]^{2} } }} \\ & = 835\frac{\text{N}}{{{\text{mm}}^{ 2} }} \\ d_{0} & = d_{s} = 7.963\,{\text{mm}} \\ R_{p\,0.2} & = 1100\frac{\text{N}}{{{\text{mm}}^{ 2} }} \\ \end{aligned} $$

And with $\nu = 0.9$, follows

$$ F_{M} > F_{M\,max} $$

The dimensions of the bolt are correct.

With $\mu_{K} = 0.12$, we have a tightening torque of

$$ \begin{aligned} M_{A} & = F_{M} \left( {0.16P + \mu_{G} \times 0.58d_{2} + \frac{{D_{k\,m} }}{2}\mu_{K} } \right) \\ & = 60.3\,{\text{Nm}} \\ \end{aligned} $$

where

$$ \begin{aligned} \frac{{D_{k\,m} }}{2} & = \frac{{D_{w\,min} + D_{ha\,max} }}{4} = 5.95 \\ \mu_{G} & = 0.12 \\ \end{aligned} $$

R7 The check for l_K/d and Φ can be omitted since these values are already determined exactly.

R8 The check that the maximum permissible bolt force is not exceeded

$$ \begin{aligned} &\Phi _{en} F_{A} \le 0.1\,R_{p0.2\,\hbox{min} } A_{0} \\ & 440\,N < 5480\,N \\ \end{aligned} $$

R9 Determination of the alternating stress endurance of the bolt.

Because of the eccentric clamping and loading, the bolt is subject to tensile stresses and to bending stresses.

From Eq. (3.74 b of VDI 2230),

$$ \sigma_{SA\,b} = \left[ {1 + \left( {\frac{1}{{\Phi _{en} }} - \frac{s}{a}} \right)\frac{{l_{K} }}{{l_{ers} }}\frac{{E_{s} }}{{E_{P} }}\frac{{a\,\pi d_{3}^{3} }}{{8\overline{I}_{B\,ers} }}} \right]\frac{{\Phi _{en} F_{A} }}{{A_{d3} }} $$

It follows from

$$ \begin{aligned}\Phi _{en} & = 0.118,\frac{{l_{K} }}{{l_{ers} }} = 1,F_{A} = 3.7 \times 10^{3} N,\frac{{E_{s} }}{{E_{p} }} = 1,\overline{I}_{B\,ers} = I_{B\,ers} - \frac{{\pi d_{h}^{4} }}{64} \\ & = 2474\,{\text{mm}}^{ 4} \\ \end{aligned} $$

That on the side in tension

$$ \begin{aligned} \sigma_{SA\,b} & = (1 + 4.88) \times 9.49\frac{\text{N}}{{{\text{mm}}^{ 2} }} \\ & = 9.5\frac{\text{N}}{{{\text{mm}}^{ 2} }} + 46.5\frac{\text{N}}{{{\text{mm}}^{ 2} }} \\ & = 56\frac{\text{N}}{{{\text{mm}}^{ 2} }} \\ \end{aligned} $$

With strain gauges attached to the bolt shank in the vicinity of the interface, a tensile stress variation of 52 N/mm² and a compressive stress variation of –32 N/mm² relative to the pre-load were measured, after applying an axial force of F_A to a joint with minimum bolt pre-load $F_{V\,erf} = F_{{K\,{\text{erf}}\,2}} + F_{L}$. From this we obtain a tensile stress component of 10 N/mm² (9.5 N/mm² was calculated) and a bending stress component of 42 N/mm² (46.5 N/mm² was calculated). Figure 13.20 gives a comparison between the calculated and measured stress distribution in the interface plane of the bolt.

Fig. 13.20

Stress distribution in the interface plane of the bolt

On the tensile side, the bolt alternating stress is

$$ \sigma_{B} = \frac{{\sigma_{SA\,b} }}{2} = \pm 28\,N/{\text{mm}}^{ 2} $$

Figure 48 of VDI 2230 gives

$$ \sigma_{A} = \pm 54\,N/{\text{mm}}^{ 2} $$

Thus, we have $\sigma_{B} < \sigma_{A}$

R10 Checking calculation of the surface pressure under the head bearing

The nut bearing is not examined.

$$ A_{P\,nut\,min} > A_{P\,head\,min,} $$

See R1

$$ \begin{aligned} & p = \frac{{F_{M} +\Phi _{en} F_{A} }}{{A_{P\,head\,\hbox{min} } }} = 660\frac{\text{N}}{{{\text{mm}}^{ 2} }} \\ & p < p_{G } = 700\,{\text{N/mm}}^{ 2} \\ \end{aligned} $$

$p_{G} \,value$ is taken from Table 9 of VDI 2230.

The checking calculation shows that the function of the bolted joint, under strength stipulations is fulfilled.

Symbols and notations (VDI 2230)

A	Cross sectional area
A D	Interface area minus the area of the hole for the bolt
A ers	Equivalent cross section area of a hollow cylinder with the same elastic resilience of the clamped parts
A P	bearing area of the bolt head or nut
D A	Outside diameter of the clamped sleeve- for the interface surfaces of a varying circular form (dA = the diameter of the inner circle)
D a	Inner diameter of bearing surface of the nut
D ha	Inner diameter of the bearing area of the clamped parts under the nut (at the start of the fillet of the clamped parts)
D hW	Outside diameter of the bearing area of the clamped parts under the nut (at the thread or at the start of the fillet of the outside diameter)
D W	Outside diameter of bearing area of the nut (at the start of the fillet of the outside of the nut)
D f	minor diameter of the thread in the nut
E P	Young’s modulus of the parts clamped
E s	Young’s modulus of the bolt
F A	Axial force calculated along the bolt axis or the components for a given working load, FB
F B	Working load in a joint in any direction
F K erf	required clamping load for sealing functions, friction grip and prevention of one-sided opening at the interface
F M	Initial clamping load (assembly preload); The values in tables of VDI 2230 are calculated with a 90% of the elastic limit using $\sigma_{red}$
F M max	Initial clamping load for which the bolt is designed considering lack of precision in the tightening technique the expected bedding in during operation, the minimum required clamping load
F M min	Minimum initial clamping load established at FM max because of lack of precision in the tightening technique
F Q	Transverse force normal to bolt axis
F V	Preload
F V erf	Minimum preload required to ensure sealing functions, friction grip and one sided opening at the interface by loss of the force at the interface
F Z	Loss of preload due to bedding in during operation
I B ers	Equivalent area moment of inertia
$\overline{I}_{B\,ers}$	$I_{Bers}$ minus area moment of inertia for the bolt hole
I BT	Area moment of inertia for the interface
M b	Bending moment at the bolting point due to the axial loads, FA and FS applied eccentrically
P	thread pitch
R p 0.2	0.2% proof stress as per DIN ISO 898 Part 1
a	Distance at which the load acts from the axis of the surface AB
d	Bolt diameter = outside diameter of the thread (nominal diameter)
d a	Inner diameter of the face of the bolt head (at the inlet of the transition radius of the shank)
d c	Core diameter of the face of the bolt head
d h	Bore diameter of the clamped parts; inner diameter of the equivalent cylinder
d ha	Inner diameter of the bearing surface of the bolt head (from the start of the fillet the bore)
d hW	Outer diameter of the bearing surface of the clamped parts with the bolt head (from the start of the fillet of the outer diameter of the clamped parts)
d W	Outside diameter of the plane head bearing surface (at the inlet of the transition radius of the head)
d 0	Diameter at the smallest cross section of the bolt shank
d 2	Pitch diameter of the bolt thread
d 3	Minor diameter of bolt thread
F 2	Plastic deformation by bedding in
l	Length, in general
li	Length of the bolt
l t	Clamping length
n, $\bar{n}$	A factor given by the ratio of the thickness of sections of clamped parts unloaded by the axial load, FA to the clamping length lK
p	Surface pressure
p G	maximum permissible pressure under a bolt head
s	Distance between the bolt and the axis of the surface AB
u	Distance between the edge of clamped part (prism) from the axis of the surface AB (in the direction opposite to A-A)
v	Distance between the edge in clamped part (prism) from the axis of the surface As (in the direction opposite to A-A)
$\alpha_{A}$	Tightening factor, $F_{Mmax} /F_{Mmin}$
$\delta$	Elastic resilience
$\delta_{G}$	Resilience of engaged thread
$\delta_{i}$	Resilience of any part, i
$\delta_{K}$	Resilience of bolt head
$\delta_{p}$	Resilience of the clamped parts for the concentric bolting and concentric loading
$\delta_{p*}$	Resilience of the clamped parts for eccentric loading
$\delta_{p**}$	Resilience of the clamped parts for eccentric bolting and eccentric loading
$\delta_{s}$	Resilience of the bolt
$\mu_{G}$	Coefficient of friction in the thread
$\mu_{K}$	Coefficient of friction for bolt head
$\gamma$	A fraction of the yield load to which the bolt is tightened
$\sigma_{A}$	Stress amplitude at the endurance limit
$\sigma_{a}$	Alternating stress acting on the bolt
$\sigma_{M}$	Tensile stress due to FM
$\sigma_{red}$	Equivalent stress, (relative stress)
$\sigma_{SA\,b}$	tensile stress (due to bending) in bolt thread caused by the axial load FSA and the bending moment Mb due to eccentric application of load
$\sigma_{SA\,d}$	Same as $\sigma_{SAb}$, but compression stress due to bending
$\phi$	Load factor, FSA/FA
$\phi_{e}$	Load factor for eccentric application of axial load FA
$\phi_{en}$	Load factor, $\phi_{e}$ for introduction of load through the clamped parts
$\phi_{n}$	Load factor for introduction of axial load FA concentrically at a distance = n lK