written 7.2 years ago by | modified 2.2 years ago by |
Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II
Marks: 10 Marks
Year: Dec 2015
written 7.2 years ago by | modified 2.2 years ago by |
Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II
Marks: 10 Marks
Year: Dec 2015
written 7.2 years ago by |
Given data:-
$Z_A=52 \\ Z_B=56 \\ Z_E=Z_F=36$
To find:-
Step 1:- Calculating the no. of teeth in C & D
$r_d= D_E+ r_b \\ D_d=2D_E+D_b$
Similarly,
$r_c=D_F +r_a \\ D_c=2D_F+D_a$
But the no. of teeth are proportional to the P.C.D, for same module
$\therefore Z_D=Z_b+2*Z_{DE} \\ =56+2*36 \\ =56+72 \\ Z_D=128$ $\hspace{3cm}$ $Z_c=2_{ZF}+Z_A \\ =72+52 \\ Z_c=124$
Calculate the speed of ‘c’ if arm is rotated by 200 rpm and D is fixed.
It is given that $N_g=200 rpm$ and $N_D$ is fixed
$\therefore N_D=x+ y=0 \\ \therefore N_G=y=200 rpm \\ \therefore x=-200 rpm$
Speed of gear $C= x. Z_D/Z_B*Z_A/Z_C +y$
=-200128/15652/124+200
=8.29 rpm (clockwise)
Step III- Calculate the speed of ‘c’ at $N_g=200 rpm$ and $N_D= -20 rpm$
$N_g=200 rpm \\ N_D=-20 \\ N_A= x + y=-20 \\ N_G=y=200 \\ X=-220 rpm$
$\therefore$ Speed of gear $‘c’= x. Z_D/Z_B.Z_A/Z_C +y$
=-220128/5652/124+200
=-210+200
=-10.87 in opposite direction of arm.