Page: Design of Shafts
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1) Stresses Induced in shaft:-

i) Shear stress due to transmission of torque i.e. due to torsional load.

ii) Bending stress due to the forces acting upon machined elements like gears, pulleys, etc as well as due to the weight of the shaft itself.

iii) Stresses due to combined torsional and bending loads.

2) Equivalent torque:-

$\tau_e = \sqrt{(M^2 + T^2)}$

Equivalent Bending moment :-

$Me = \frac{1}{2} \sqrt{M + (M^2 + T^2)}$

Note: Sometimes in the problem, $K_b$ and $K_t$ is given where,

$K_b$ = Shock and fatigue factor for bending,

$K_t$ = Shock and fatigue factor for twisting

Then equivalent Torque -

$\tau_e = \sqrt{(k_b.M)^2 + (k_t.T)^2}$

Equivalent bending moment -

$M_e = \frac{1}{2}[k_b.M] + [\sqrt{(k_b.M)^2 + (k_t.T)^2]}$


Q.1. Design an overchanged shaft to transmit power through a pulley by means of belt drive the pulley is of weight $250 N$ and is located at $100\ mm$, overchanged from the bearing. The diameter of pulley is $200\ mm$ and the maximum power transmitted is 10KW at 1400 rpm. The angle of lap is $180^0$ and the coefficient of friction is 0.3. The combined shock and fatigue factor for bending and twisting are 2 and 1.5 respectively. Select suitable material for the shaft and design the shaft from max shear stress theory. The pulley transmits power to another pulley vertically below it.

enter image description here

Wt of pulley = $W = 250 N$

dia of pulley = $D = 200 mm$

$P = Power = 10 Kw$

$N = speed = 1440 rpm$

Angle of contact or lap $= 180^\circ$

$\mu$ = 0.3

$K_b$ = 2

$k_t$ = 5

Selection of material,

lets, select C = 40 PSG = 1.9

$\sigma_y$ = 330 $N/mm^2$

$\tau_{max} = \frac{0.5 \sigma_y}{Fos} = \frac{0.5 \times 330}{3} = 55 N/mm^2$

$P= \frac{2 \pi NT}{60}$

$10 \times 10^3 = \frac{2\pi \times 1440 \times T}{60}$

$T = 66.31 N-m = 66.3 \times 10^3 N-mm$

-> $\frac{T_1}{T_2} = e^{\mu \theta} = e^{0.3\times \pi}$

$T_1 = 2.566T_2$

$T = [T_1 - T_2] R$

$66.31\times 10^3 = [2.566T_2 - T_1](100)$

$T_2 = 423.43N$

$T_1 = 2.566 T_2= 1086.5N$

Total load = $T_1 + W + T_2 = 1086.5 + 250 + 423.43= 1759.96 N$

$M_{max}$ = Total load $\times 100$

$= l.759\times 10^5 N-mm$

$T_e = \sqrt{(K_b.M)^2 + (K_t.\tau)^2}$

$= \sqrt{(2\times 1.759 \times 10^5)^2+(1.5\times 66.31 \times 10^3)^2} =365.57 \times 10^3 N-mm$

$\frac{T}{J} = \frac{\tau}{r}$

$T = \frac{J}{r}.\tau = \frac{\pi d^4}{\frac{32}{d/2}\times t}$

$Te = \frac{\pi}{16}\times d^3\times e$

$365.59 \times 10^3 = \frac{\pi}{4}d^3 \times 55$

$d = 32.34 m$


Q.2. A shaft supported in bearing A, B 1500 mm apart, carnies 180 mm diameter spur gear (20 pressure angle) at C at a distance 220 mm from A towards B. Two pulleys D and E of equal diameter of 300 mm are mounted in it at a disk of 650 mm and 1050 mm from bearing A. The gear receives 40 KW power from a pinion vertically above it and delivers 20 KW power through pulley D to one of the machine and rest of the power through pulley E to another machine. The belt tension at E are also parallel but horizontal towards right. The ratio of belt tension at D and E is 2.5. the mass of pulleys D and E is 25 Kg each. The shaft rotates at 800 rpm. Design the shaft taking $t = 40 N/mm^2$

I] 3-D layout

enter image description here

II] Design of gear [c]

$P_c = \frac{2 \pi N_c T_c}{60} = \frac{2\pi\times800\times T_c}{60}$

$T_c = 477.46 N-mm$

$T_c = 477.46 \times 10^3 N-mm$

-> PSG 857

$P_t = \frac{2M_t}{d_1} = 5.3\times 10^3 N$

$P_r = P_t.tan \alpha = 1.93\times 10^3 N$

III] Pulley [P]

$P_D = \frac{2 \pi N_D.T_D}{60}$

$T_D = 238.73\times 10^3 N-mm$

-> $\frac{T_1D}{T_2D} = e^{\mu \theta} = 2.5$

$T_1D = 2.5 T_2D$

-> $T_D = (T_1D - T_2D)R_D$

=> $T_2D = 1061.02 N$

=> $T_1D = 2652.55 N$

IV] Pulley [E]

$T_1E = 2652.55 N$

$T_2E = 1061.02 N$

V] Loading in vertical plane :-

enter image description here

-> Support Reaction

$\Sigma M_A = 0$

$V_B(1500) - 25\times 9.81\times 1050 - [(T_1D + T_2D)cos30 + 245.25]650 - P_r(200)$ = 0

$V_B = 1954.63 N$

-> $\Sigma Fv = 0$

$V_A + V_B - P_r - [(T_1D + T_2D)cos30 + 245.25]-245.55$ = 0

$V_A$ = 3681.91 N

-> Bending moment calculation : N-mm

BM @ A&B = 0

C= $V_A (220) = 810.02\times10^3$

D = $V_A(650) - P_r(430) = 1.5633\times10^6$

E = $V_A(1050) - P_r(830) - [(T_1D + T_2D)cos30 + 25\times981](400)$

= $879.58\times10^3$

Horizontal Plane:

-> Support reaction:

$\Sigma M_A = 0$

$H_B(1500) + 3713.57(1050) - 1856.73(650) - 5.3\times0^3(220)$

$H_B = 1016.81 N$

-> $\Sigma Fv =0$

$H_A - 5.3\times10^3 - 1856.73 + 3713.57 = 0$

$H_A = 4465.02 N$

-> B.M calculations (N-mm)

B.M @ A & B =0

c = $H_A(22) = 982.3\times10^3$

D = $H_A(650) - 5.3\times10^2(430) = 621.07\times10^3$

-> $M_{RC} = ((810.02\times10^3)^2 + (9823\times10^3)^2)^{1/2} = 1.7\times10^6 N-mm$

-> $M_{RD} = ((1.5633\times10^6)^2 + (621.07\times10^3)^2)^{1/2} = 1.682\times10^6 N-mm$

-> $M_{RE} = ((879.58\times10^3)^2 + (-457.56\times10^3)^2)^{1/2} = 0.99\times10^6 N-mm$

As $M_{RD}$ is greater than $M_{RC}$ & $M_{RE}$

=> Section D is critical section

-> Torque diagram:

enter image description here

-> $(t_c)_c = ((M_{RC})^2 + (t_c)^2)^{1/2}$

= $1.35\times10^6 N-mm$

-> $(t_c)_D = ((1.682\times10^6)^2 + (477.46\times10^3)^2)^{1/2}$

= $1.748\times10^6 N-mm$

-> $(t_c)_E = ((0.99\times10^6)^2 + (238.73\times10^3)^2)^{1/2}$

= $1.019\times10^6 N-mm$

As $(t_c)_D$ is greater

$(t_c)_D$ is critical section

-> $(t_c)_D = \frac{\pi}{16}td^3$

d = 606 mm = 62 mm


Q.3. A counter shaft receives 26.25 KW from a motor through a coupling and transmits it via two belt drives rto two machine tools each consuming KW as shown in fig. Both N shaft and coefficient of friction as the belt drives are 0.25 mm and 0.6 mm and their corresponding weights are 400 N and 750 N shaft speed is 30 rad/sec. Find the diameter of shaft and coefficient of friction as $\mu$ = 0.35

enter image description here

Sol. P = 2625 KW

$W_I$ = 100 N

$W_II$ = 250 N

speed = 30 rad/sec

I] -> $P_1 = (t_{1I} - t_{2I})(r_{1w})$

$t_{1I} - t_{2I} = 3000$ -----------(A)

-> $t_{1I} = e^{\mu \theta} = e^{0.35\times r}$

$t_{1I} = 3t_{2I}$ ---------------(B)

-> Solve eqn A and B,

$t_{1I} = 4500 N$

$t_{2I} = 1500 N$

II] -> $P_{II} = (t_{1II} - t_{2II})(r_{IIw})$

$t_{1II} - t_{2II} = 1250 N$ --------------(C)

-> $\frac{t_{1II}}{t_{2II}} = e^{\mu \theta} = e^{0.35\times\pi}$

$t_{1II} = 3t_{2II}$ --------------(D)

-> Solve (C) and (D),

$t_{1II} = 1875 N$

$t_{2II} = 625 N$

III] Loading vertical plane,

$\Sigma M_B = 0$

$V_A(1.5) -400(1.1) - 750(0.6) = 0$

$V_A = 593.3 N$

$\Sigma F_V = 0$

$V_B = 556.7 N$

enter image description here

-> Bending moment calculation (N-m)

A = 0 ; B = 0

C = $V_A(0.4) = 237.32$

D = $V_A (0.9) - 400(0.5) = 334.02 N-m$

-> Loading in horizontal plane,

$\Sigma M_B = 0$

$H_A(1.5) -6000(1.1) - 2500(0.6) = 0$

$H_A = 5400 N$

-> $\Sigma F_V = 0$

$H_A + H_B - 6000 - 2500 = 0$

=> $H_B = 3100 N$

-> BM calculations :

-> Resultant :-

$M_{RC} = ((237.32)^2 + (2160)^2)^{1/2}$

= 2172.99 N-m

$M_{RD} = ((534.02)^2 + (1860)^2)^{1/2}$

= 1889.75 N-m

-> $P_1 = \frac{2\pi N_1 T_1}{60}$

$11.25\times10^3 = 30 T_1$

$T_1 = 375 N-m$

-> $P_II = \frac{2\pi N_II T_II}{60}$

$11.25\times10^3 = 30 T_II$

$T_II = 375 N-m$

-> Torque dia

enter image description here

-> $(t_e)_c = ((2172.99)^2 + (375)^2)^{1/2}$ = 2205.1 N-m

-> $(t_e)_D = ((1889.75)^2 + (375)^2)^{1/2}$ = 1926.57 N-m

-> 'C' is the critical section,

-> $(t_e)_c = \frac{\pi}{16}td^3$

d = 67.28 mm


Q.4. A machine shaft rotating at 600 rpm supproted on a bearing 750 mm apart. 20 Kw is supplied to the shaft through a 450 mm pulley located to the right of the right hand bearing. The power is transmitted from the shaft through 200 mm spur gear located at 200 mm to right of left hand bearing. The gear has 20 tooth formed and mates with another gear located directly above the shaft. Dtermine the shaft diameter based on octahedral shear stress thoery considering varying stress let the material of shaft ISC4S steel. Assume FOS =2.

Sol:

enter image description here

-> $P_D = \frac{2 \pi N T_D}{60}$

$T_D = 318.3\times10^3$ N-mm

-> $P_c = \frac{2\pi NT_c}{60}$

$T_c = 318.3\times10^3 N-mm$

-> Gear C = PSG 8.57

$F_t = \frac{2Mt}{D} = 3183.1 N$

-> $F_r = F_t tan\alpha = 1158.55 N$

-> Pulley D:-

$\frac{T_1D}{T_2D} = 3$

$T_1D = 3T_2D$

-> $T_D = (T_1D - T_2D)R_D$

$T_2D = 707.33 N$

$T_1D = 2122 N$

-> Forces in vertical plane:

enter image description here

$\Sigma M_B = 0$

$V_A(750) - 1158.55(550) - 1549.7(250) = 0$

$V_A = 1366.4 N$

$\Sigma F_v = 0$

$1366.4 - 1158.55 + V_B + 1549.7 = 0$

$V_B = 1758.2 N$

-> BM calculations :

A = 0 ; D = 0

$C = 273.28 \times 10^3$ N-mm

$B = 387.6\times 10^3$ N-mm

Forces in horizontal:

$\Sigma M_B = 0$

$H_A(750) - 3183.1(550) - 1414.7(250) = 0$

$H_A = 1862.7 N$

$\Sigma F_v = 0$

$H_A + H_B - 3183.1 - 1414.7 = 0$

$H_B = 2735.13 N$

-> BM calculations :

A = 0 ; D = 0

$C = 372.5\times 10^3$ N-mm

$B = -353.7\times10^3$ N-mm

-> Resultant BM:

at A = 0

at C = $((273.28\times10^3)^2 + (372.5\times10^3)^2 )^{1/2} = 462.03\times10^3$

at B = $((387.6\times10^3)^2 + (-353.5\times10^3)^2 )^{1/2} = 524.7\times10^3$

-> Torque Diagram:

-> Calculation of shaft diameter based on octahedral shear stress theory,

1) $\sigma_b = \frac{M_{max}}{z} = \frac{5.34\times10^6}{d^3}$

2) Torional shear stress = t = $\frac{1.621\times10^6}{d^3}$

-> PSG 7.6

$\frac{\sigma_{eq}}{\sigma_y} = \frac{\sigma_{m}}{\sigma_y} + \frac{\sigma_{a}}{\sigma_{-1}}$ ------------------(1)

Fibres will be subjected to completely reversed stresses because shaft is rotating,

$\sigma_{mean} = \frac{\sigma_{max} + \sigma_{min}}{2} = 0$

$\sigma_{a} = \frac{534.6\times10^3}{d^3}$

Given: C-45 steel,

$\sigma_y = 360 N/mm^2$

$\sigma_{-1} = K_a.K_b.K_c.\frac{\sigma_{-1}}{K_f}$

= 3350

-> $K_f = 1 + q(K_t - 1)$

$K_f = 1.335$

$\sigma_{-1} = 1928.42 Kgf/cm^2$

$\sigma_{-1} = 192.8 N/mm^2$

$\frac{\sigma_{eq}}{\sigma_y}$ = $\frac{277.82}{d^3}$

-> PSG 7.6

$\frac{t_{eq}}{t_y} = \frac{t_{m}}{t_y} + \frac{t_{a}}{t_{-1}} ---------------(2)$

-> $t_{max} = t_{min} = t = \frac{1.621\times10^6}{d^3}$

$t_m = \frac{T_{max} + t_{min}}{2} = \frac{1.621\times10^6}{d^3}$

$t_a = \frac{T_{max} - t_{min}}{2} = 0$

PSG 7.6

$T_y = \frac{\sigma_y}{(3)^{1/2}} = 207.84 N/mm^2$

$\frac{t_{eq}}{t_y} = \frac{1.621\times10^6}{\frac{d^3}{207.84}}$

PSG 7.6

$\frac{1}{W} = [(\frac{\sigma_{eq}}{\sigma_y})^2 + (\frac{t_{eq}}{t_y})^2]^{1/2}$

$d = 25 mm$

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modified 23 days ago by gravatar for Sanket Shingote Sanket Shingote250 written 3 months ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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