Problem on Calculation of Diameter of Solid Shaft – Torsion – Strength of Materials


Let us take the second question a solid shaft in a rolling mill transmits 20 kilowatts at 2 Hertz Determine the diameter of the shaft if shear stress is not to exceed 40 mega Pascal and angle of twist is limited to 6 degree in a length of 3 meter full stop take capital G is equal to 83 GPA that is modulus of rigidity now the question which we have in front of us we will write the data for this first here it is given that a solid shaft in a rolling mill transmits 20 kilowatts so power is given it is in kilowatts we are converting it into watts by multiplying it by thousand and two Hertz this 2 Hertz is nothing but the speed here it is 2 Hertz means it is in 2 revolutions per second so this we can convert it into revolutions per minute by multiplying it with 60 so therefore I have rpm as 120 next determine the diameter of the shaft this is the question here we have to calculate the diameter if shear stress is not to exceed 40 mega Pascal FS is 40 mega Pascal that is 40 Newton per mm square and angle of twist is limited to 6 degree theta 6 degree so I will convert it into Radian by multiplying it by PI and dividing it by 180 in a length of 3 meter the length of the shaft is 3 meter so it is 3000 mm take capital G that is modulus of rigidity as 83 GPA means 83 into 10 raised to 3 Newton per millimeter square so here I have to calculate the diameter and this is the question which we have as we can see your FS is also there and modulus of rigidity so we would be using strength criteria as well as the rigidity criteria but before that since power and speed are known let us try to get torque first so we will start with the solution since power transmitted by shaft is given by P is equal to two pi n T by 6 T therefore T is equal to P into 60 divided by 2 pi n power is given in the question it is 20 into 10 raised to 3 watts divided by 2 pi into n n is 120 rpm therefore I will be getting the answer of torque as 1.5 9 into 10 raised to 3 Newton meter now I have to convert it into Newton mm so multiplying it by thousand so this becomes 1 point 5 9 into 10 raise to 6 Newton mm this is the value of torque which we are getting now let us write first the strength criteria based on strength criteria the formula is T by J is equal to FS by R therefore T it is 1.5 9 into 10 raise to 6 J is PI by 32 D raise to 4 for solid shaft FS it is given for T radius is diameter by 2 so from left hand side and right hand side D gets cancelled so here I have D cube this becomes 16 so therefore D cube will be equal to 1.5 9 into 10 raise to 6 divided by PI by 16 into 40 so here if I calculate numerator upon denominator and then take the cube root I will be getting the answer of diameter as 58.7 4mm this becomes equation number 1 so based on strength criteria my diameter is 58.7 4mm for solid shaft now based on rigidity criteria the formula is T by J is equal to G theta by L therefore T it is 1.5 9 into 10 raise to 6 J is PI by 32 D raise to 4 modulus of rigidity is given in the question it is 83 into 10 raised to 3 theta is 6 PI by 180 divided by the length of the shaft is 3000 mm so from this if I calculate if I shift D raise to 4 onto right hand side and this term in the denominator here I would be getting first d raise to 4 as 5 point 6 0 into 10 raise to 6 now if I want to calculate the diameter of shaft I will have to take the square root twice so if I take square root twice of this number I would be getting my diameter of solid shaft as 48 point 6 4 mm this becomes my second equation now if I compare from equation number 1 and equation number 2 in equation 1 the diameter is fifty eight point seven four in equation 2 it is 48 point six four so I can say that we will be selecting the greater diameter hence 58.7 for mm will be the answer for us that is we are having a solid shaft and its diameter will be equal to 58 point 7 4 mm and with this we complete the question

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