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Problem 1 on Design of Shaft – Design of Machine

October 19, 2019

Hello Friends here in this video we will see a problem on the design of shaft and for that purpose here we have the question a shaft made up of mild steel is required to transmit 100 kilowatts at 300 rpm the supported length of the shaft is 3 meters it carries two pulleys each weighing 1500 Newton supported at a distance of one meter from the ends respectively assuming safe value of stress as 60 mega Pascal determine the diameter of the shaft now this is the question which we have on the design of shaft whatever is given here I will write that in the form of data first so here I am writing the data here it is given that a shaft is made up of mild steel which is the material of the shaft is required to transmit 100 kilowatts at 300 rpm so power is 100 kilowatts which we can say that it is also equal to 100 into 10 raised to 3 watts and speed is given here 300 rpm next it is given that the supported length of the shaft is 3 meters so here I can explain the supported length by an example that the shaft is placed along its length and then it is supported on both the sides like this and this supported length of the shaft is 3 meter given this I can say it is point a and point B next it carries two pulleys each weighing 1,500 Newton supported at a distance of one meter from the ends respectively so here it is given that this shaft it carries two pulleys and their weight is given that is weight of the pulley I will denote it by W and that is equal to 1500 Newton so it means this shaft is going to carry two pulleys and they are one meter distance from each end next assuming safe value of stress is 60 mega Pascal so this is the value of safe stress given safe stress is equal to 60 mega Pascal that is 60 Newton per mm square determine the diameter of the shaft so this is the question for us we have to find out how much would be the diameter of shaft so that I’ll denote it by capital D now as we have seen in this problem anywhere here it is not given that the shaft is hollow so I can say that the type of shaft here it is a solid shaft now you’re at both the ends the pulley are there they are weight is 1500 Newton on both the sides so now this is the question with us let us try to get the solution for this problem so in the solution part I will say that first of all I will explain this with a diagram more clear than this here I would be drawing a 3d diagram this shaft it carries two pulley and the weight of the pulley is same on both the sides their distances are also given in the problem so here I have drawn this diagram to explain it further now as I have C drawn in this diagram here this is the shaft which is supported at two ends means at here at this end it would be supported and at this end at the extreme ends this shaft it carries two pulleys as we can see here these are I can say that this is also a pulley and here also we have a pulley so what will happen in this case is that when the shaft is supported between a length of 3 meter and then the weight of the pulley would be acting downward so because of that the shaft is subjected to bending because of the weight of the pulley this shaft will Bend because weight will try to pull the shaft in the downward direction so there is bending of the shaft at the same time by the definition of shaft we know that it is a dating machine element so the shaft is constantly rotating and at the same time when it is rotating it is also subjected to bending because of the weight of the pulley on both the sides so I can see that here this shaft is subjected to both that is it is subjected to twisting torsion because of the rotation and it is subjected to bending that is because of the weight of the pulley so I can say that here the given shaft is subjected to both twisting and bending that is the given shaft essence I will read on since the given shaft is subjected to both twisting and bending so therefore here the shaft will be designed based on twisting moment and bending moment so while designing this I will consider both twisting and bending now since in this question they have given power they have given rpm so from that we can get the value of torque so I will say that since power transmitted by shaft is given by power as we know it is P is equal to 2pi and T upon 60 this is the formula of power next here I can see that therefore torque T will be equal to P into 60 divided by 2 pi n power it is given as 100 into 10 raised to 3 watts into 60 divided by 2 pi into n is 300 that is the speed 300 rpm so from this I will get the value of torque and my answer is it is 3183 Newton meter now here we are getting torque in terms of Newton meter I’ll also write it in terms of Newton mm so it is 3 1 8 3 into 10 raised to 3 Newton millimeters after reading the torque now I will go on to find the bending moment because as I have told here that here we would be designing the shaft based on both twisting moment as well as bending moment so twisting moment we have got just now now I will calculate the bending moment and for that purpose I will say that since the shaft is supported at a and B so here at a we would be getting one reaction euler’s reaction at a you are right B also I would be getting one reaction or less are B now how to get this are NR BC it is very simple as you look at this beam it is exactly symmetrical that is the loading we can say the reactions the loads they are located at same distance from both the sides and even the load value are same so if the total load if I add this that is 1500 and 1500 that gives me 3000 Newton that 3000 Newton would be shared equally by both the supports so at each support I am getting the reaction as 1500 Newton so after getting our NRV now what i can do i can calculate moment either either at this that is I will say this is Point C or I can get the moment at this point which is point D so next I will say that therefore reactions at each supported end that would be our a is equal to RB and that is 1500 Newton I can write down since the beam is symmetric now after getting this I can say that therefore bending moment at Point C I can calculate either at Point C or at Point D both the values would be same so if I take the moment of this 1500 it would be 1500 into 1 because as we know moment is force in two perpendicular distance so it is 1500 into one so therefore the bending moment at Point C is equal to 1500 Newton into 1 meter so hence moment at C is 1500 Newton meter or I can write this as 1500 into 10 raised to 3 Newton millimeters now once I know the value of twisting moment and bending moment I can get the equivalent torque I can say that therefore equivalent twisting moment is given by the equivalent twisting moment is T equivalent that is equal to square root of T square plus M square that is it is the square root of torque and bending moment making square of that so finally I will say that therefore T equivalent will be equal to square root of the torque value which I am getting that is I have calculated it using the power and the torque was 3 1 8 3 Newton meter so it is 3 1 8 3 Plus bending moment is 1,500 so from this I will get equivalent torque in hence T equivalent that comes out to be three thousand five hundred and nineteen Newton meter or it is three 5 1 9 into 10 raised to 3 Newton millimeters so after getting equivalent torque now I can easily calculate the diameter of shaft I can say that therefore the equivalent wasting moment is also equal to PI by 16 D cube into tau here this formula I can say that it has come from by strength criteria for shaft design this formula has come from the strength criteria why because here we have stress so now I will go on putting the values T equivalent is 3 5 1 9 into 10 raised to 3 I will put the value in Newton mm so that I get the answer of diameter in terms of mm it is equal to PI by 16 D cube instead of tau I can write the value it is given in the problem as 60 mega Pascal so it is into 60 so finally D cube will be equal to 3 5 1 9 into 10 raise to 3 this 16 will go on to one side divided by PI into 60 and from this if I calculate I would be getting the cube of D and if I take the cube root of this answer therefore I will get my answer of capital D as it is sixty six point eight and since I am getting the answer is sixty eight point eight sixty six point eight I can make that as close to 70mm so this is my answer and the meaning of this answer is that here we should be using a shaft whose diameter should be equal to 70 mm for the given conditions it means whatever conditions they have given in the problem regarding the pulley and the kind of load the pulley has and then the shaft is subjected to the amount of power and speed they have given for that the diameter of shaft should be 70 mm so I hope this design of shaft is understood in this video

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  • Reply Manasa Gopu April 29, 2017 at 3:52 pm

    thank u so much sir

  • Reply ॐVeerbhadra Gamer April 30, 2017 at 5:22 pm

    provide design procedure of shafts

  • Reply Panchal Anand May 4, 2017 at 9:20 am

    nice teaching sir

  • Reply Panchal Anand May 4, 2017 at 9:22 am

    cotter joint and knuckle joint example ke video Upload kro na sir plz

  • Reply Rehan Khan May 9, 2017 at 6:34 pm

    Hi sir
    your performance is very well.
    But last topic sound recording is not well.

  • Reply Mohamad Imam Wicaksono May 11, 2017 at 5:23 am

    Thank you so much sir, really helpful! i wanna ask, if we have a shaft and pulley exactly the same with your question but its vertically positioned. how do we solve the problem?

  • Reply Nicole Tara May 13, 2017 at 12:39 am

    Awesome!! Thank you, sir

  • Reply Explain To All May 19, 2017 at 8:08 pm

    ty sir and try to cover problem which difficult rather this types of simple ones….

  • Reply lavesh kumar May 21, 2017 at 1:00 pm

    sr I have some problem in shaft in combined loading

  • Reply shaharin babu May 25, 2017 at 7:09 am

    such an excellent class sir

  • Reply malli smarttrak May 29, 2017 at 1:04 pm

    sir easy to understand that sir

  • Reply malli smarttrak May 29, 2017 at 1:04 pm

    thanks sir

  • Reply SkyNight June 9, 2017 at 4:49 pm

    Thank you very much. This is very helpful

  • Reply amit kumar June 19, 2017 at 1:47 am

    It is so nice

  • Reply amit kumar June 19, 2017 at 1:48 am

    Sir please upload material science video.

  • Reply praveen kumar July 14, 2017 at 6:08 am

    good teaching

  • Reply Chaitu Harsha July 20, 2017 at 8:11 am

    helpful….thanks for sharing

  • Reply Prateek A. July 26, 2017 at 4:51 pm

    Thank you sir for this beautifully explained problem.
    Only one query though.
    Why did you select 70mm as the approximate answer and not 68, as it is the next even whole number?

  • Reply Prasad Sardeshmukh August 7, 2017 at 2:50 pm

    Ekeeda is doing a great job. Keep up the good work 🙂

  • Reply Brian Ahirekar August 13, 2017 at 7:39 pm

    U. Explained the problem with such great precession Sir.. Thanks.

  • Reply C G GURUMANJUNATH September 5, 2017 at 8:00 am

    P = (2*pi*N*T) / (60 *1000) kW , N – rpm, T – Nm . And. T=9.55*10^6 * P/ n Nmm, P – kW , n – rpm Can anyone prove that above 2 equations are equal with proper steps and proof?

  • Reply Rahul Mangate September 10, 2017 at 5:24 pm


  • Reply SANDEEP kumar September 15, 2017 at 3:45 pm

    Sir problem combined pully and gear ki

  • Reply AJEY SINGH September 29, 2017 at 12:14 pm

    Why no consideration of Critical frequency ? Did I missed that part ???

  • Reply Vineeth Bhogireddy October 9, 2017 at 4:22 pm

    Can u add problems……. on desgin on cotter joint, knuckle joint.

  • Reply abhiyender sharma October 22, 2017 at 2:19 am

    Sir,You r another best teacher for me, Thank you

  • Reply Farooq Khan October 26, 2017 at 6:55 am


  • Reply Countdown FUN October 29, 2017 at 8:37 am

    At 9:15 who else got deaf on right ear (headphone users)

  • Reply Manveer Jaan November 16, 2017 at 2:12 am


  • Reply Moe Mesh November 24, 2017 at 1:32 pm

    His reaction at 10:44 = 1500 N

  • Reply Mandy singh November 26, 2017 at 2:51 pm

    Your T value is 3183 but this is not a correct value. Correct value is 3.141

  • Reply universal education December 8, 2017 at 1:28 pm

    Hello friends now you can learn Engineering subjects in simple way watch this video and go to the channel for more videos.

  • Reply universal education December 8, 2017 at 2:38 pm

  • Reply Akash shet December 8, 2017 at 3:49 pm

    Excellent superb teaching in easy way

  • Reply Mustaq Honyal December 16, 2017 at 10:25 am

    Sri plz upload spur gear and helical gear
    And bevel gear

  • Reply Rahul Naik January 21, 2018 at 2:50 pm

    Please sir make in hindi

  • Reply Technical courses January 27, 2018 at 6:42 pm

    Sir aap ek WhatsApp group bnaayei or sab students Ko aad kr diji ye jis sa jyada se jyada log aap k video ka advantage le sake

  • Reply rajesh kannan March 11, 2018 at 1:20 pm

    Sir why we take bending moment from c (or)d

  • Reply vivekananda reddy March 12, 2018 at 12:18 pm

    instead of pulleys if two bearings are connected, will then the bending moment acts on the system. NO, then what will be the equation since bending moment not actcing

  • Reply Akshay Patil. March 24, 2018 at 4:11 pm

    Nice sir..thank you so much

  • Reply sachin ramesh April 3, 2018 at 9:12 am

    too slow

  • Reply Ayush Chauhan April 7, 2018 at 12:44 pm

    Why bending moment at point C was only taken and not from point D , while calculating total bending moment on the shaft ????

    Or what happened is u calculated bending moment of point D about point C ? , I m confused

  • Reply Ravi Metra April 15, 2018 at 5:39 pm

    Very good sir …nice

  • Reply kaan d April 29, 2018 at 3:09 am

    give this man a cookie
    some advanced examples would be better tho

  • Reply babu kc April 30, 2018 at 7:23 am

    Sound is getting low in the vedio

  • Reply Gurvinder Singh May 1, 2018 at 10:00 am

    when i calculate torque formula i found wrong answer .how i calculate

  • Reply Mahi bai May 7, 2018 at 12:02 pm

    Thanks you sir

  • Reply Rahul Roy June 1, 2018 at 7:32 pm

    Is shaft taken as massless??

  • Reply Mohamad Hamad June 3, 2018 at 12:39 pm

    You have the worst hand in the history of hands. Nice video tho

  • Reply sujoy mondal June 9, 2018 at 3:36 pm


  • Reply BABU seamerzz August 5, 2018 at 11:52 am

    Nice sir plz upload with different load…

  • Reply Azam Jan August 20, 2018 at 6:07 pm

    Amazing. i never had this much good understanding of such problems.

  • Reply Shubham Verma September 3, 2018 at 4:07 pm

    Nice sir

  • Reply SEM MES September 13, 2018 at 3:28 am

    do you add a safety factor to your 70mm

  • Reply Sumeet Singh September 18, 2018 at 11:40 pm

    Sir , if shaft in step then daimeter is?

  • Reply N vishwakarma September 25, 2018 at 2:33 pm

    aap screen pr likhane k liye kis instrument ka use karte hai?

  • Reply HIMANSHU NAHARIYA September 26, 2018 at 5:58 pm

    Iske ans me 70 mm kha se aaya

  • Reply smoked buns October 6, 2018 at 8:29 am

    Good Sir, may you please help me with my homework?
    the problem goes like this.
    The following data apply to a motor and gear drive for a centrifugal pump. The shaft is supported by
    bearing A and B. A flexible coupling C connects the pump to the shaft.
    Pump output, 500 gpm
    Total head, 60 ft
    Pump speed, 500 rpm
    Efficiency of pump, 64 percent
    Motor speed, 1,800 rpm
    Efficiency of gears, 97 percent
    Power cost, 3 cents per kwhr
    Determine the following: (a) maximum horsepower required of motor; (b) horsepower of motor
    to be recommended, assuming a duty cycle of full load for 15 min and rest 15 min; (c) diametral
    pitch, pitch diameters, and face width of gears, assuming 14.5o
    teeth, 20 teeth for pinion, SAE 1045
    for pinion and SAE 1030 for gear; (d) loads on bearing A and B; (e) size of shaft to be recommended,
    assuming allowable stress in shear, ss
    = 5,500 psi; (f) power cost of operating pump, dollars per day.
    The electrical efficiency of the motor may be assumed as 87 percent.

  • Reply HARRY POTTER October 6, 2018 at 8:36 pm

    I say thank u sir by subscribe ur channel😋

  • Reply Ramu Maddi October 7, 2018 at 3:57 pm

    nice video sir

  • Reply Dilip Kumar October 9, 2018 at 4:23 pm

    Thank you sir

  • Reply Pulkit Prabhat October 11, 2018 at 4:57 pm

    make a video on shaft subjected to two bending at 90°.. vertically and horizontally….

  • Reply tim paulpartha October 22, 2018 at 4:43 pm

    sir which book should i use for mechanics of materials

  • Reply H r October 25, 2018 at 12:32 pm

    It's very useful to me….i learn Clearly for ur videos …Tq sir…

  • Reply IRFAN PATHAN FAN CLUB October 28, 2018 at 2:08 am

    it is vtu based right

  • Reply ABDUL FAHEEM October 30, 2018 at 9:57 am

    1 MPa = 10^6 N/(m^2)

  • Reply MUKESH KUMAR November 11, 2018 at 12:17 pm

    Sir bahut achha padate h

  • Reply Lovely Murugesh November 16, 2018 at 2:03 am

    Amazing sir wt.a teaching

  • Reply AKASH KUMAR November 20, 2018 at 3:25 pm

    sir aap all-rounder hai Kya
    aap sabhi subject padhate h

  • Reply Akhil A Oachira November 20, 2018 at 5:25 pm

    Nyz lecture 😊

  • Reply Ahmad Sandhu November 23, 2018 at 3:36 pm

    how you take moment at C by neglecting force at D and reaction of B ?????

  • Reply Er. Anil Kumar November 24, 2018 at 12:42 pm

    Aap figure achha banate h

  • Reply Er. Anil Kumar November 24, 2018 at 12:46 pm

    Good teaching sir

  • Reply AMAN yadav December 9, 2018 at 4:58 am

    Thanku you sir

  • Reply LiamPhillips101 December 16, 2018 at 2:21 am

    "By strength criteria for shaft design" is the 'Maximum Torsional
    Resisting Moment' for a solid cylinder shaft
    – Tmax –

    (Nm, in lb)

  • Reply The Baraik December 16, 2018 at 3:26 am

    Why did you put 70mm instead of 66.8mm.. I didn't get it. Someone reply me. I'm confused about it!

  • Reply swami soundu December 17, 2018 at 4:43 am

    Sir can you please explain problem which containing both pulley and gears and inclined angle.

  • Reply Novendra Anon December 17, 2018 at 9:59 am

    why pick bending moment at c , how to pick the maximum bending moment ?

  • Reply amanyisye mwalulesa December 17, 2018 at 6:20 pm

    Sir which things should I consider during the design of screw shaft for squeezing tomato..?

  • Reply RAHUL SHINDE December 22, 2018 at 1:17 pm

    Why the moment is taken about point c

  • Reply RAHUL SHINDE December 22, 2018 at 1:17 pm

    Why only the moment of only one force is used not other forces

  • Reply Maruti Dodmani January 9, 2019 at 3:01 pm

    If pulley takes at 1mtr from left bearing toward left side ,is it right?

  • Reply abd ullah January 28, 2019 at 9:17 pm

    Please I need to shum solutions in design

  • Reply Krishna Raj February 16, 2019 at 1:54 pm

    Unit of speed is not N, it's a m/s right

  • Reply Ankit Pal March 2, 2019 at 2:29 am

    Nice explanation…thank u so much…

  • Reply Noor Khan March 6, 2019 at 12:55 pm

    sorry sir but your lecture will help more if you upload them in hindi or urdu rather than your current language

  • Reply Charles Karuru March 11, 2019 at 6:28 pm

    In this problem why did the lectirer not consider the effect of tension of the belt drives…if the shaft is rotating ot means it is driven by the belt drives therefore the effect of tension in tightside and slack side must be considered

  • Reply pratik shetty April 13, 2019 at 7:51 pm

    Just perfect ❤

  • Reply Bharath Kumar April 19, 2019 at 7:17 am

    Superb sir

  • Reply om patil May 1, 2019 at 12:16 pm

    Watch at 1.5x speed

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  • Reply mia khalifa May 22, 2019 at 10:33 am

    Ha bc 3-3 4-4 add dikha do bich me padne mat diyo

  • Reply Hasith Pereira June 1, 2019 at 8:09 am

    very useful thanks a lot ,upload more ,cheers

  • Reply Amit Kumar June 13, 2019 at 1:16 pm

    plz correcr the BM AT C by considering the forces at c and d….

  • Reply U WUT M8? June 17, 2019 at 9:28 pm

    I just had an exam on this 80% similar

  • Reply Chaithanya colan July 1, 2019 at 7:36 pm

    Thankyou sir and pls ceep up the good work for students like me

  • Reply ananth .... July 31, 2019 at 2:40 am

    Lecture was good but I stead of writing everything in the video which co dunes lot of time please keep everything ready and just explain things

  • Reply prakash gade August 1, 2019 at 6:27 pm

    Sir do your dme 1 lectures are based on pune University syllabus ?
    Do you make lectures of subjects heat transfer, turbo machines, and tom 2 ?
    Please reply

  • Reply Izzat Danial September 11, 2019 at 5:54 pm

    So many fucking ads man..?! Nice video tho

  • Reply UNIQUE Coaching Classes September 21, 2019 at 1:32 pm


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