1.1Convert Power and Speed to Torque
A shaft transmits (4\ \mathrm{kW}) at (1200\ \mathrm{rpm}). Find its torque.
Solution
Use
$$
T=\frac{9550P}{N}=\frac{9550(4)}{1200}=31.8\ \mathrm{N\,m}.
$$
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1.2Find a Gear Ratio
A 20-tooth pinion drives a 60-tooth gear. What are the speed ratio and output speed if the input is (1500\ \mathrm{rpm})?
Solution
$$
i=\frac{z_2}{z_1}=\frac{60}{20}=3,
\qquad N_2=\frac{1500}{3}=500\ \mathrm{rpm}.
$$
The output turns one-third as fast, with ideal torque multiplied by three.
1.3Compute Gear Tooth Force
A gear has (T=45\ \mathrm{N\,m}) and pitch diameter (d=90\ \mathrm{mm}). Find its tangential tooth force.
Solution
$$
F_t=\frac{2T}{d}=\frac{2(45)}{0.090}=1000\ \mathrm N.
$$
1.4Estimate Bearing Life
A ball bearing has (C=12\ \mathrm{kN}) and equivalent load (P=3\ \mathrm{kN}). Estimate (L_{10}) in million revolutions.
Solution
For a ball bearing, (p=3):
$$
L_{10}=\left(\frac{C}{P}\right)^3=\left(\frac{12}{3}\right)^3=64
$$
million revolutions.
1.5Combine Springs in Parallel
Two linear springs with stiffnesses (k_1=400\ \mathrm{N/m}) and (k_2=600\ \mathrm{N/m}) are installed in parallel. Find the equivalent stiffness and force at (x=0.05\ \mathrm m).
Solution
$$
k_{eq}=k_1+k_2=1000\ \mathrm{N/m},
\qquad F=k_{eq}x=1000(0.05)=50\ \mathrm N.
$$
1.6Separate Mean and Alternating Stress
A component experiences (sigma_{max}=120\ \mathrm{MPa}) and (sigma_{min}=20\ \mathrm{MPa}). Find (sigma_a) and (sigma_m).
Solution
$$
\sigma_a=\frac{120-20}{2}=50\ \mathrm{MPa},
\qquad
\sigma_m=\frac{120+20}{2}=70\ \mathrm{MPa}.
$$
1.7Count Planar Mechanism Mobility
A planar mechanism has (n=4) links and (j_1=4) one-degree-of-freedom joints, with no higher pairs. Find its mobility.
Solution
$$
M=3(n-1)-2j_1=3(4-1)-2(4)=1.
$$
It has one independent input, as expected for a conventional four-bar linkage.
1.8Compute Screw Lead Angle
A single-start screw has pitch (p=4\ \mathrm{mm}) and mean diameter (d_m=20\ \mathrm{mm}). Find its lead angle.
Solution
For a single-start screw, (l=p=4\ \mathrm{mm}). Thus
$$
\lambda=\tan^{-1}\left(\frac{4}{\pi(20)}\right)=3.64^\circ.
$$