Question

Computer Problem # 3 The common configuration of an internal combustion engine is represented by a slider crank mechanism as shown in Figure. The connecting rod AB weighs 1.21b with mass center of gravity at G. The piston and piston pin at A weigh 1.81b. The connecting rod has radius of gyration about G of 1.12 in. The engine runs at a speed range of 2000 - 3000 rpm. a) b) Plot the velocity and acceleration of the center of gravity G of the connecting rod versus time for two revolutions of the crank (at three different engine speeds). For the same engine speeds in (a), plot the velocity and acceleration of pin A versus time for two revolutions of the crank. Plot the magnitude of the force on the piston pin A as function of crank angle 0. c) Hint: Neglect the force exerted by the gas in the cylinder compared with the dynamic forces generated. Also, to simplify the problem you might neglect the weights of other components. 89

Answer 1

.. Piston Motion are simple Harmonic Motion DAS K=1.7") 0,= 90 = 0 = 0 = 0, 4 then repeat: o 90 180 270 360 450 540 636 726 1068tings linis Piston Velocity, o= r. Ao sing V max = r. To If N=2000rpm o &o=90 - if w= 2A4, = LAN - 25x2000 60 60 w=209.44 rad/s. velocity diagram taigi - 2 Rev. of Crank 890 - = Umax Red Black Blue 1.7xł x 209.44 Colour = N=2000 rpm Colour =N=2500 rpm colour N = 3000 rpm 671940 = 7/2.1 inls. 466070 in/52 in ing L-298280 if 0 = 0 & 180 va o diagram of o=270 u=-712.1 in/s. a o=360 Acceleration .-298280 in/82 _466070 & then again same as start oo to 360 reached 720- in Two revolution of -671140 in/82 crank.

af N = 2500 rpm Acceleration, a = te. Two cos(AO) = rtw sinko ih N = 2000 rpm v= 0 w2261-8 rad's O=0, xa=298280 iner 0 = 90 0=90, aa=0 v= - 17x AX2R x 2500 . a = - 29820 in/12 0=180 1-7X Сох, 10=270, a= 0 & = 890. 1 inls. o=360 , a=29820. 14182 if 0 = 180 as same repeated. U=0 If N=2500 rpm Ho = 270 0=0, a =466070 [n/82 v=-890.1 0=90, a = o wn yo=360 0 = 780 a = -466070 ingh as same repeated. = 890 lins 9f N=3000 soon as we same as repeat q N=3000 spm wote wo 25N RX 3000 well 0=90, a=0 over 60 60 10=180 a=-671140 inch w = 314.16 radis B=60 o zo 0 -90, = 1068.1 iuls 0 = 1.80 v= o 1 0=270 v= - 10 68. I indd o=360, u=0 same as repeated. I 70=450 20=0, a=671140 in 152

Odry w Weigth of Riston A = 1.2 lb force, fam. a for 40400 SO N=2000 rpm Fmax = mx amx fax = 1.2 x 298280 for = 357940 lbf. Sokmin = 1.2xt-a) min=-357940 for = mx (90) co N= 2500 force diagram f=0 Fmx=559280 lbf. As same for other speed but or values chang (hP) as thy). &min=-559280 lbf. I N=3000 rpm KX720