Question 4 of 6: Mounting reactions on a four-wheel-drive transmission

Question 4: Mounting reactions on a four-wheel-drive transmission

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The vertical force carried by each of the four mounts A, B, C and D that holds the transmission and front-differential assembly in equilibrium.

Approach. Work out every shaft torque from speed ratios and power conservation, then sum the torques the rotating drivetrain applies to the housing about the vehicle’s longitudinal (X) and transverse (Y) axes, and let the two mount pairs react those two couples.

1. Fix the drive-shaft torque from the transmission ratio. With no losses, power is conserved through the gearbox, so torque rises in the same proportion as speed falls: $$N_{\text{drive}}=\frac{N_{\text{eng}}}{i}=\frac{2400}{2.0}=1200\ \text{rpm},\qquad T_{\text{drive,total}}=i\,T_{\text{eng}}=2.0\times100=200\ \text{lb}\cdot\text{ft}$$ That 200 lb·ft leaves the gearbox on two shafts turning at the same 1200 rpm. With all four wheels having full traction and an even centre split, each drive shaft carries $$T_{\text{front,ds}}=T_{\text{rear,ds}}=\tfrac{1}{2}(200)=100\ \text{lb}\cdot\text{ft}$$
2. Carry the front branch through the differential. The front drive shaft runs at 1200 rpm and the wheel axle shafts at 400 rpm, so the front differential ratio is $$i_{\text{diff}}=\frac{1200}{400}=3.0 \qquad\Rightarrow\qquad T_{\text{axle,total}}=3.0\times100=300\ \text{lb}\cdot\text{ft}$$ An open differential divides that equally, so each front wheel receives $$T_{\text{wheel}}=\tfrac{1}{2}(300)=\boxed{150\ \text{lb}\cdot\text{ft}}$$
3. Check the power balance before going further. Every torque above must be consistent with one input: $$T_{\text{eng}}N_{\text{eng}}=100(2400)=240{,}000$$ $$T_{\text{rear}}N_{\text{drive}}+2\,T_{\text{wheel}}N_{\text{axle}}=100(1200)+2(150)(400)=120{,}000+120{,}000=240{,}000\ \checkmark$$ in lb·ft·rpm, equivalent to P = 2πTN/33,000 = 45.7 hp.
4. Sum the torques the drivetrain applies to the housing about the longitudinal axis X. Take X pointing rearward along the engine and drive-shaft centreline and Y pointing toward the right-front wheel. The engine drives the input shaft in the direction of its own rotation, so it feeds +100 lb·ft about X into the box; the rear drive shaft, which turns in the same sense at half the speed, carries 100 lb·ft out about X and therefore reacts −100 lb·ft on the box: $$\sum M_X = T_{\text{eng}}-T_{\text{rear,ds}} = 100-100 = \boxed{0}$$ This cancellation is the crux of the problem. The 2.0 ratio doubles the output torque, but half of it is diverted to the front axle and leaves the free body about a different axis, so the longitudinal torque entering and leaving the housing are exactly equal.
5. Sum the torques about the transverse axis Y. The only shafts crossing the boundary along Y are the two front wheel axle shafts, and both turn in the same sense, so their reactions add: $$\sum M_Y = 2\,T_{\text{wheel}} = 2(150)=\boxed{300\ \text{lb}\cdot\text{ft}}$$ The housing must therefore be held against a pure couple of 300 lb·ft about the transverse axis, with no net force and no couple about the longitudinal axis.
6. React each couple with the mount pair that has an arm for it. A vertical force produces a moment about Y in proportion to its longitudinal offset, and a moment about X in proportion to its transverse offset. Mounts A and B are separated by 24 in along X, so they are the pair that can resist the Y-couple; C and D are separated by 12 in along Y and resist the X-couple. Converting the couples to pound-inches, $$F_A=F_B=\frac{\sum M_Y}{d_{AB}}=\frac{300\times12}{24}=\boxed{150\ \text{lb}}$$ $$F_C=F_D=\frac{\sum M_X}{d_{CD}}=\frac{0\times12}{12}=\boxed{0}$$ Because the mounts carry no net force, A and B form an equal and opposite pair: one pushes up 150 lb, the other pulls down 150 lb. With the engine rotation sense shown in the figure, the drivetrain reaction lifts the front of the housing, so the front mount A is loaded 150 lb downward and the rear mount B 150 lb upward. Reversing the assumed engine rotation reverses both signs but not the magnitudes.
7. Close the equilibrium check. Vertical force: −150 + 150 + 0 + 0 = 0 ✓. Moment about Y: 150 lb over a 24 in arm gives 3600 lb·in = 300 lb·ft, matching Step 5 ✓. Moment about X: zero from A and B (both on the centreline) plus zero from C and D ✓.