LIVE Course for free

Rated by 1 million+ students
Get app now
JEE MAIN 2023
JEE MAIN 2023 TEST SERIES
NEET 2023 TEST SERIES
NEET 2023
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
45 views
in General by (54.0k points)
closed by

The power transmitted by a shaft 60 mm diameter at 180 RPM, if the permissible stress is 85 N/mm2


1. 68 kW
2. 650 kW
3. 1200 kW
4. 7 kW

1 Answer

0 votes
by (30.0k points)
selected by
 
Best answer
Correct Answer - Option 1 : 68 kW

Concept:

Power transmitted by shaft subjected to torque ‘T’ is given by,

\(P = \frac{{2\pi NT}}{{60}}\;Watt\)

Where, N = number of revolutions per minute, T = Applied torque N-m

Maximum shear stress induced in the shaft is given by,

\(\frac{{{τ _{max}}}}{R} = \frac{T}{{{I_P}}}\)

Where, R = Radius of shaft (m), IP = Polar moment of inertia (m4) and τmax = Maximum shear stress N/m2

Polar moment of inertia for solid circular shaft,

\({I_p} = \frac{\pi }{{32}} \times {D^4}\)

Calculation:

The diameter of the shaft, D = 60 mm

Number of revolutions per minute, N = 180 rpm

permissible shear stress = 85 N/mm2

The maximum torque that can be applied on the shaft ‘T’ is given by,

\(T = {τ _{max}} \times {Z_P}\)

\( = 85 \times \frac{\pi }{{16}} \times {D^3} = 85 \times \frac{\pi }{{16}} \times {60^3}\)

\( = 3604977.56\;Nmm \approx 3604.977\;Nm\)

Power Transmitted by shaft when subjected to torque ‘T’ is given by,

\(P = \frac{{2\pi NT}}{{60}}\)

\( = \frac{{2\pi \times 180 \times 3604.977}}{{60}}\)

\( = 67952.21\;W \approx 67.95\;kW\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...