[30 pts] In the gear system shown in the picture the motor applies a torque of 4

Question

[30 pts] In the gear system shown in the picture the motor applies a torque of 400 Nm to the gear at A. A torque of Te 700 Nm is removed from the shaft at gear C, and the remaining torque is removed at gear D. Shafts (1) and (2) are made of solid steel [G-80 GPa]. Determine the torque in shaft (1) and shaft (2). diameters for shafts (1) and (2). diameter that can be used for the shaft if the maximum shear stress must not exceed 40 MPa b) If the maximum shear stress must not exceed 40 MPa, determine the minimum permissible c) If the same diameter is to be used for shafts (1) and (2), determine the minimum permissible and the rotation angle of gear D relative to gear B must not exceed 3 degrees. 00 m diameier INCLUDE units when appropriate: (1D 300-mm diameter 15 m c) dmin =

Explanation / Answer

from the given data in the question

applied torque, T = 400 Nm

Tc = 700 Nm

Ta = 400 Nm

G = 80 GPa

a. let torque in shaft 1 be T1

then

from force balance at the junction of gears B and A

T1/300 mm = T/100 mm

hence

T1 = 3T = 1200 Nm

now let torque in shaft 2 be T2

then again from torque balance

T1 - Tc = T2

T2 = 1200 - 700 = 500 Nm

b. maximum shear stress, Sm = 40*10^6 Pa

maximum possible diameter of shaft 1 = d1

for shaft 2 = d2

then

for shaft maximum shearing stress is 16T/pi*d^3

hence

40*10^6 = 16*T1/pi*d1^3

d1 = 0.05346 m

and similiarly

40*10^6 = 16*T2/pi*d2^3

d2 = 0.039929 m

c. for same diameter on both shafts , dmin

then

for maximum relative twist of 3 deg = 3*pi/180 rad

Smax = 40 Mpa

3*pi/180 = L1*Smax/G*dmin + L2*Smax/G*dmin

3*pi/180 = Smax(L1 + l2)/G*dmin

dmin = 0.02387 m

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