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In Shigley(5th Edition), in Chapter 14. Dynamics of Reciprocating Engine ,Section 14.7 Inertia Forces, the inertia torque exerted by the engine on the crankshaft is given as enter image description here $$ \mathbf{T}_{21}^{\prime \prime}=-\left(-m_{B} A_{B} \tan \phi\right) x \hat{\mathbf{k}} $$

Background : In this section they have assumed equivalent masses $m_A$ and $m_B$ for the crank and connecting rod (link 3) at the crank Pin A and wrist pin B respectively. (with $m_A$ rotating mass at A ($m_{A}=m_{2} \frac{r_{G}}{r}$ + $m_{3} \frac{l_{B}}{l}$) where $r_G$ is the CG of Crank from $O_2$ and $l_B$ is the CG of connecting rod from B and $m_B$ reciprocating mass at B, ($m_{B}=m_{3} \frac{l_{A}}{l}$ + $m_{4}$) , $l_A$ is the CG of connecting rod from A) . From the formulas for inertia force and torques or equivalently Frame Force and Frame Moment(Sec 12.7 Shaking Forces and Moments) as : $$ \begin{array}{l} \mathbf{F}_{s}=\sum\left(-m_{j} \mathbf{A}_{G_{j}}\right) \\ \mathbf{M}_{s}=\sum\left[\mathbf{R}_{G_{j}} \times\left(-m_{j} \mathbf{A}_{G_{j}}\right)\right]+\sum\left(-I_{G_{j}} \alpha_{j}\right) \end{array} $$ we can calculate the inertia forces : \begin{array}{l} -m_{A} \mathbf{A}_{A}=m_{A} r \omega^{2}(\cos \omega \hat{t} \mathbf{i}+\sin \omega t \hat{\mathbf{j}}) \\ -m_{B} \mathbf{A}_{B}=m_{B} r \omega^{2}\left(\cos \omega t+\frac{r}{l} \cos 2 \omega t\right) \hat{\mathbf{i}} \end{array} As per my understanding we incorporate Inertia Forces in dynamic systems to reduce the system to static system using d'Alembert Principle.

My doubt is although we have reduced the system to its equivalent mass system and the respective inertia forces for both of them pass through $O_2$

  1. How is the inertia force generating a torque about $O_2$ ? (like as mentioned in the text, it says $-m_A\mathbf{A}_A$ can be neglected because it does not have any moment arm about $O_2$ , so why isn't $-m_B\mathbf{\ddot{X}}$ also ignored along the same line of reasoning?
  2. The inertia force due to $m_B$ is resolved into components and only one of the component ($-m_B\mathbf{\ddot{X}}\:tan \phi$ ) is used to calculate the torque. Won't the net torque if we use both the component be zero, as the force $-m_B\mathbf{\ddot{X}}$ has zero moment arm in the first place and hence the net torque of its components about $O_2$ will also be zero.
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  • $\begingroup$ What is $A_B$ where is $T_{21}''$ applied? $\endgroup$ – John Alexiou Aug 23 at 0:35
  • $\begingroup$ $\textbf{A}_B$ is the acceleration of Body B i.e the piston. $T^{\prime\prime}_{21}$ is the torque applied to the crankshaft at $O_{2}$ . Upon reading further I came to the conclusion. The inertial force acting $-m_B\mathbf{\ddot{X}}$ on B can 'transmit' the force(acting along the connecting rod $F_{43}$ ) through the connecting rod, and the rod being mass-less will transfer the forces.We have $F_{34}=\frac{-m_B\mathbf{\ddot{X}}}{cos\phi}$ using force balance at point $B$ . $\endgroup$ – vanguard478 Aug 23 at 14:50
  • $\begingroup$ Torque at $O_2$ will be $r_\perp*F_{43}$ and $r_\perp=x*sin\phi$. Using these two equations we get $\mathbf{T}_{21}^{\prime \prime}=-\left(-m_{B} A_{B} \tan \phi\right) x \hat{\mathbf{k}}$ $\endgroup$ – vanguard478 Aug 23 at 14:50
  • $\begingroup$ Please let me know if my line of reasoning is wrong. $\endgroup$ – vanguard478 Aug 23 at 14:50
  • $\begingroup$ So $A_B$ is the scalar acceleration vector of point [4]? And $\boldsymbol{A}_B = A_B \boldsymbol{\hat{x}}$. I must admit the nomenclature is confusing me a lot, so I am having difficulty following the logic. $\endgroup$ – John Alexiou Aug 23 at 16:33
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Upon reading further I came to the conclusion. The inertial force acting $-m_B\mathbf{\ddot{X}}$ on B can 'transmit' the force(acting along the connecting rod $F_{43}$ ) through the connecting rod, and the rod being mass-less will transfer the forces.We have $F_{34}=\frac{-m_B\mathbf{\ddot{X}}}{cos\phi}$ using force balance at point $B$ . Torque at $O_2$ will be $r_\perp*F_{43}$ and $r_\perp=x*sin\phi$. Using these two equations we get $\mathbf{T}_{21}^{\prime \prime}=-\left(-m_{B} A_{B} \tan \phi\right) x \hat{\mathbf{k}}$

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