written 5.7 years ago by | modified 4.6 years ago by |

The twist should not be more than 1^0 in a shaft of length 2m and The maximum shear stress should not exceed 30 N/mm^2 Take modulus of rigidity =1x10^5 N/mm^2

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Determine the diameter of a solid shaft, which will transmit 300 kW at 250 rpm and the working conditions to be satisfied are:

written 5.7 years ago by | modified 4.6 years ago by |

The twist should not be more than 1^0 in a shaft of length 2m and The maximum shear stress should not exceed 30 N/mm^2 Take modulus of rigidity =1x10^5 N/mm^2

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written 4.7 years ago by | • modified 4.6 years ago |

**Data:** (P) Power = 300 KW = 300 \times 10^3 W, (d) Diameter = ?

N = 250 rpm.

(L) Length = $2 \ m = 2 \ \times 10^3 \ mm$

$(\theta)$ Angle of twist = 1° = $ 1 \times \frac{\pi}{180} \ (radians)$

$(\zeta)$ maximum shear stress = $30 \ N/mm^2$

(G) = Modulus of RIGIDITY = $1 \times 10^5 \ N/mm^2$

$\rightarrow$ **1] Power Developed by a shaft.**

$= \frac{2 \ \pi \ N \ T}{60}$

$300 \times 10^3 = \frac{2 \pi \times 250 \ T}{60}$

$\therefore$ $T = 1145.15 \ Nm = 11.459 \times 10^6 \ Nmm.$

**2] For strength:**

$\frac{T}{J_p} = \frac{\zeta}{r}$

$\frac{11.459 \times 10^6}{\frac{\pi}{32} \times d^4} = \frac{30}{(\frac{d}{2})}$

$\therefore$ d = 124.83 mm.

**3] For stiffness:**

$\frac{T}{J_p} = \frac{G \theta}{l}$

$\frac{11.459 \times 10^6}{\frac{\pi}{32} \times d^4} = \frac{1 \times 10^5 \times}{2 \times 10^3} (1 \times \frac{\pi}{180})$

$\therefore$ d = 107.54 mm.

Now, value of diameter of shaft such that it must satisfies condition for strength and stiffness for that we are taking greater values of two **i.e. d = 124.83 mm.**

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