--- title: "Piston effort and crank-pin turning moment" subject: "Theory of Mechanism and Machine II" chapter: 1 key: "semester_7/theory_of_mechanism_and_machine_ii:engine-piston-effort-turning-moment" source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/engine-piston-effort-turning-moment site: BCT Engineering --- # Piston effort and crank-pin turning moment Theory of Mechanism and Machine II · Chapter 1: Engine Force Analysis - Asked in: year not given, 8 marks - Video: [Dynamic Force Analysis of Slider crank mechanism - Piston effort & Turning moment on crank shaft|DOM](https://www.youtube.com/watch?v=9Xz2Gw-a8Xs) — Mohith Dadu - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/engine-piston-effort-turning-moment - Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-1 ## Question The crank radius of a horizontal single-cylinder engine is 150 mm and the connecting rod is 600 mm long. The cylinder bore is 200 mm and the engine runs at 360 rpm. The mass of the reciprocating parts is 120 kg. When the crank is $40^\circ$ from the inner dead centre on the expansion (out) stroke, the gas pressure on the piston is 1.0 MPa (gauge, back pressure neglected). Neglecting friction and the weight of the parts, determine: (i) the inertia force, (ii) the net force on the piston (piston effort), (iii) the thrust in the connecting rod, (iv) the side thrust on the cylinder wall, and (v) the turning moment on the crankshaft. ## Short answer At 360 rpm, ω=37.70 rad/s, gas force 31416 N. Inertia force = 20708 N, net piston effort = 10708 N, connecting rod thrust = 10849 N, side thrust = 1743 N, turning moment = 1233 N·m. ## Answer **Data:** $r = 0.15$ m, $l = 0.6$ m, $n = l/r = 4$, $N = 360$ rpm, $m = 120$ kg, $\theta = 40^\circ$, $p = 1$ MPa, $d = 0.2$ m. #### Angular speed and gas force $$ \omega = \frac{2\pi N}{60} = \frac{2\pi \times 360}{60} = 37.70\ \text{rad/s} $$ $$ F_P = p\,\frac{\pi}{4}d^2 = 1\times10^{6}\times\frac{\pi}{4}(0.2)^2 = 31416\ \text{N} $$ #### (i) Inertia force $$ \begin{aligned} F_I &= m r\omega^2\left(\cos\theta + \frac{\cos 2\theta}{n}\right) \\ &= 120 \times 0.15 \times (37.70)^2 \left(\cos 40^\circ + \frac{\cos 80^\circ}{4}\right) \\ &= 25582.0 \times 0.8095 = 20708\ \text{N} \end{aligned} $$ #### (ii) Net force on piston During the out stroke the piston is accelerating, so the inertia force opposes the gas force: $$ F = F_P - F_I = 31416 - 20708 = 10708\ \text{N} $$ #### (iii) Thrust in the connecting rod $$ \sin\phi = \frac{\sin\theta}{n} = \frac{\sin 40^\circ}{4} = 0.1607 \Rightarrow \phi = 9.25^\circ $$ $$ F_Q = \frac{F}{\cos\phi} = \frac{10708}{0.9870} = 10849\ \text{N} $$ #### (iv) Side thrust on the cylinder wall $$ F_N = F\tan\phi = 10708 \times 0.1628 = 1743\ \text{N} $$ #### (v) Turning moment on the crankshaft $$ T = F_Q \cdot r\sin(\theta+\phi) = 10849 \times 0.15 \times \sin(49.25^\circ) = 1233\ \text{N·m} $$ **Answer:** Inertia force = 20708 N; piston effort = 10708 N; connecting-rod thrust = 10849 N; side thrust = 1743 N; turning moment = 1233 N·m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What expression gives the inertia force of the reciprocating parts at crank angle theta (with obliquity ratio n)? A. m r w^2 (cos theta + cos 2theta / n) B. m r w^2 (sin theta + sin 2theta / 2n) C. m r w^2 cos theta only D. m r w (cos theta + cos 2theta / n) Answer: A. F_I = m a with a = r w^2 (cos theta + cos 2theta / n). (Key point: Inertia force formula) 2. During the out (expansion) stroke when the piston is accelerating, how is the net force on the piston found? A. Gas force plus inertia force B. Gas force minus inertia force C. Inertia force minus gas force D. Gas force only; inertia is ignored Answer: B. The piston is accelerating, so the inertia force opposes the gas force: F = F_P - F_I = 31416 - 20708 = 10708 N. (Key point: Piston effort) 3. With piston effort F and rod angle phi, the rod thrust and the side thrust on the cylinder wall are: A. F cos(phi) and F sin(phi) B. F/cos(phi) and F tan(phi) C. F/sin(phi) and F cos(phi) D. F tan(phi) and F/cos(phi) Answer: B. Resolving: F_Q = F/cos(phi) along the rod, and F_N = F tan(phi) normal to the cylinder wall. (Key point: Connecting rod thrust and side thrust) 4. The turning moment on the crankshaft in this problem is obtained as: A. F_Q x r x sin(theta + phi) B. F x r x cos(theta) C. F_N x r x sin(theta) D. F_Q x l x sin(phi) Answer: A. The rod thrust F_Q acts at perpendicular distance r sin(theta+phi) from the crank centre, giving T = 10849 x 0.15 x sin(49.25 deg) = 1233 N.m. 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