A transmission shaft supporting a spur gears B and the pulley D is shown in Fig. The shaft is mounted on two bearings A and C. The diameter of the pulley and the pitch circle diameter of the gear are 450 mm and 300 mm respectively. The pulley transmits 20 kW power at 500 rpm to the gear. $P_1$ and $P_1$ are belt tensions in the tight and loose sides, while $P_t$ and $P_r$ are tangential and radial components of gear tooth force. Assume, $P_1 = 3P_2$ and $Pr = P_t tan (20°).$ The gear and pulley are keyed to the shaft. The material of the shaft is steel 50C4 (Sut = 700 and Syt = 460 $N/mm^2$). The factors kb and kt of the ASME code are 1.5 each. Determine the shaft diameter using the ASME code.

Subject: Machine Design -I

Topic: Shaft, Keys and Coupling

Difficulty: High

Step 1:Permissible shear stress:

$0.30S_yt=0.30(460)=138 N/mm^2$

$0.18S_yt=0.18(700)=126 N/mm^2$

The lower of the two values is 126 N/mm^2 and there are keyways on the shaft.

\tau_max=0.75(126)=94.5 N/mm^2

Step 2: Torsional moment The torque transmitted by the shaft is given by,

$M_T=\frac{60*10^6}{2 \pi n}=\frac{60*10^6(20)}{2 \pi 500}=381971.86 N-mm$

Step 3: Bending moment

$(P_1-P_2)*225=381971.86$

$(P_1-P_2)=1697.65 N$ (a)

$P_1=3P_2$ (b)

From (a) and (b),

$P_1=2546.48N and P_2=848.83N$

$(P_1+P_2)=3395.31N $

$P_T*150=381971.86$

$P_T-2546.48N$

$P_r=P_ttan(20 ^\circ)=(2546.48)tan(20 ^\circ)=926.84N $

At C,

$M_b=1358.124 N-mm$ $M_t=381971.86 N-mm$

Step 4: Shaft diameter:

$d^3=\frac{16}{\pi \tau_max}\sqrt{(k_bM_b)^2+(k_TM_T)^2} $

$=\frac{16}{\pi 94.5}\sqrt{(1.5*1358124)^2+(1.5*381971.86)^2}$

$d=48.5mm$