--- title: "Inertia force and turning moment in an engine" subject: "Theory of Mechanism and Machine I" chapter: 8 key: "semester_6/theory_of_mechanism_and_machine_i:reciprocating-engine-inertia-force-torque" source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/reciprocating-engine-inertia-force-torque site: BCT Engineering --- # Inertia force and turning moment in an engine Theory of Mechanism and Machine I · Chapter 8: Force Analysis of Mechanisms - Asked in: year not given, 8 marks - Video: [Numerical on Inertia Force Calculation 1](https://www.youtube.com/watch?v=YtsSmPxCaS0) — Neeraj Gautam - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/reciprocating-engine-inertia-force-torque - Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-8 ## Question Derive an expression for the inertia force of the reciprocating parts of a horizontal engine. The following data relate to a horizontal single-cylinder engine: cylinder bore 100 mm, crank radius 100 mm, connecting rod length 400 mm, speed 600 rpm, mass of reciprocating parts 1.5 kg. When the crank is 30 degrees from the inner dead centre on the working stroke, the gauge pressure on the piston is 0.8 MPa. Neglecting the mass of the connecting rod and friction, find the net force on the piston, the thrust on the cylinder walls, the force in the connecting rod and the turning moment on the crankshaft. ## Short answer Inertia force F_I = m ω² r (cosθ + cos2θ/n). For the given data, net piston force is 5696.3 N, cylinder wall thrust 717.7 N, connecting rod force 5741.4 N, turning moment 347.0 N·m. ## Answer #### Inertia force of reciprocating parts Let $r$ = crank radius, $l$ = connecting rod length, $n = l/r$, $\theta$ = crank angle from the inner dead centre (IDC), $\omega$ = crank speed (constant), $m$ = reciprocating mass. Displacement of the piston from IDC: $$ x = r(1 - \cos\theta) + l(1 - \cos\phi), \qquad \sin\phi = \frac{\sin\theta}{n} $$ Using $\cos\phi = \sqrt{1 - \sin^2\theta/n^2} \approx 1 - \dfrac{\sin^2\theta}{2n^2}$ and $\sin^2\theta = \dfrac{1 - \cos2\theta}{2}$, the displacement from the crank centre is $$ x_c \approx r\cos\theta + \frac{r}{4n}\cos2\theta + \text{constant} $$ Differentiating twice with $\theta = \omega t$, acceleration of the piston: $$ a = -\omega^2 r\left(\cos\theta + \frac{\cos2\theta}{n}\right) $$ The inertia force is equal and opposite to $ma$ (D'Alembert's principle). Taking the working-stroke direction (towards the crank) as positive, its magnitude is $$ F_I = m\omega^2 r\left(\cos\theta + \frac{\cos2\theta}{n}\right) $$ A positive value means the inertia force opposes the gas force (as near IDC); it is negative (aids motion) in the later part of the stroke. The net force on the piston is $F = F_P - F_I$. #### Numerical Data: $D = 100$ mm, $r = 0.1$ m, $l = 0.4$ m so $n = 4$, $N = 600$ rpm, $m = 1.5$ kg, $\theta = 30^\circ$, $p = 0.8$ MPa. $$ \omega = \frac{2\pi \times 600}{60} = 62.832\ \text{rad/s}, \quad \omega^2 = 3947.8 $$ Gas force on piston: $$ F_P = p\times\frac{\pi}{4}D^2 = 0.8\times10^6 \times 0.00785 = 6283.2\ \text{N} $$ Inertia force: $$ F_I = 1.5 \times 3947.8 \times 0.1\,(\cos 30^\circ + \tfrac{\cos 60^\circ}{4}) = 1.5 \times 3947.8 \times 0.1\,(0.8660 + 0.1250) = 586.9\ \text{N} $$ #### (a) Net force on piston (along the line of stroke) $$ F = F_P - F_I = 6283.2 - 586.9 = 5696.3\ \text{N} $$ #### (b) Thrust on cylinder walls $$ \sin\phi = \frac{\sin30^\circ}{4} = 0.125 \Rightarrow \phi = 7.18^\circ $$ $$ F_N = F\tan\phi = 5696.3 \times \tan7.18^\circ = 717.7\ \text{N} $$ #### (c) Force in connecting rod $$ F_Q = \frac{F}{\cos\phi} = \frac{5696.3}{\cos 7.18^\circ} = 5741.4\ \text{N} $$ #### (d) Turning moment Tangential component of the rod force on the crank pin: $F_T = F_Q\sin(\theta + \phi)$, with $\theta + \phi = 37.18^\circ$. $$ F_T = 5741.4 \times \sin 37.18^\circ = 3469.7\ \text{N} $$ $$ T = F_T\times r = 3469.7 \times 0.1 = 347.0\ \text{N m} $$ The radial component on the crank pin is $F_R = F_Q\cos(\theta + \phi) = 4574.3$ N, which loads the shaft bearings. **Answer: $F = 5696.3$ N, $F_N = 717.7$ N, $F_Q = 5741.4$ N, $T = 347.0$ N m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The inertia force of the reciprocating parts is: A. m w^2 r (cos(theta) + cos(2 theta)/n) B. m w r (sin(theta) + sin(2 theta)/n) C. m w^2 r sin(theta) D. m w^2 l cos(theta) Answer: A. Differentiating the piston displacement twice gives a = -w^2 r (cos(theta) + cos(2theta)/n); the inertia force has magnitude m w^2 r (cos(theta) + cos(2theta)/n). (Key point: Inertia force expression) 2. How is the net force on the piston found? A. F = F_P + F_I B. F = F_P x F_I C. F = F_I only D. F = F_P - F_I Answer: D. The inertia force opposes the gas force near IDC, so F = F_P - F_I = 6283.2 - 586.9 = 5696.3 N. (Key point: Net force on piston) 3. With sin(phi) = 0.125 and net force F, the side thrust and rod force are: A. F_N = F cos(phi), F_Q = F tan(phi) B. F_N = F tan(phi), F_Q = F/cos(phi) C. F_N = F sin(phi), F_Q = F cos(phi) D. F_N = F/tan(phi), F_Q = F sin(phi) Answer: B. Resolving at the piston: F_Q cos(phi) = F, and F_N = F tan(phi). (Key point: Thrust and rod force) 4. How is the turning moment on the crankshaft obtained? A. T = F_Q cos(theta + phi) x r B. T = F x l C. T = F_Q sin(theta + phi) x r D. T = F_N x r Answer: C. The tangential component on the crank pin is F_Q sin(theta + phi); multiplying by r gives T = 347 N m. 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