2 find the diameter of a shaft to transmit twisting moments...

2 Find the diameter of a shaft to transmit twisting moments varying from 800 N-m to 1600 Nm. The ultimate tensile strength for the material is 600 MPa and yield stress is 450 MPa. Assume the stress concentration factor = 1.2, surface finish factor = 0.8 and size factor = 0.85

Transcript

00:01 Hello guys i am here to show this problem so find the diameter of the sub to transmit twisting movement varying from 800 nanometer to 600 nanometer the ultimate tensile strength is for the material is 600 and the yield is still is 540 450 megapaths so first of all you write the given data so the maximum tensile strength so it is a twisting movement so the maximum is t maximum is equal to 600 and 1 ,600 nanometer and the minimum is 6800 it is the twisting movement and the ultimate tensile strength of the material means sigma ultimate is equal to 600 megapascal and the yield strength sigma y is equal to 450 megapascal and the value of stress concentration factor means k is equals to 1 .2 and the finish factor k they it is the finish factor which is 0 .8 and the size factor which is denoted by kb is equal to 0 .8 5 so these data which is given in the now we start the solst pollution so first of all we know that t m what is the t m t m is the mean torque how to calculate the mean tortum plus t minimum divided by two so it is sixteen hundred plus eight hundred divided by two and after solving this it is twelve hundred you know so the value of t m it is the mean torque which is equal to twelve hundred now we sold the t a so what is t a t a is the variable torque is calculated by t max minus t minimum divided by so it is 600 1600 sorry minus 800 divided by 2 and it is equal to 400 nano so t a equals to 400 now the means here stress so what is the means here stress mean serious what is the formula of the means here stress means here stress means tau m is equal to 16 tm divided by pi dq and the tm the value of tm is 1200 so it is tau m means here stress is equal to 16 multiplied by the t m which is 16 1200 divided by so tau m is equal to 6 1 1 .55 divided by t 2 and so it is not nanometer it is newton meter and so it is not nanometer it is newton meter and the variable here stress tau so what is the problem of the tau it is 16 t a divided by 5 dq so now put the values here so it is 2 037 .18 divided by t newton per meter square so it's in meter square meter so we earlier calculated first thing is main shear stress and variable shear stress and we know that the endurance limit of this year so what is the endurance limit endurance limit of shear so it is denoted by tau e which is 0 .55 is equal to 0 .5 multiplied sigma ultimate strength so it is the cr and the sigma is equals to sigma e is the endurance limit of the shear which is so sigma the value of sigma e means endurance stress which is denoted by 0 .5 ultimate stress so ultimate stress is 0 .5 multiplied and 600 it is equal to 300 megapascal so now the tau is equal to 0 .55 multiplied now we put the value 300 in so the value of tau is equal to 165 megapascal so it is the endurance limit of the now the equivalent shear stress so what is the equivalent shear stress so to calculate the equivalent series stress first of all equivalent shear stress so tau yes is equal to tau m plus tau a multiplied by tau by divide by tau e plus sorry multiplied by k so now put the values here so tau equivalent is equal to the value of tm which is 6111 1 .55 divided by dq plus 2037.

04:51 1 8 divided by tq multiplied by the value of tau wise is 2 55 multiplied by 10 x the power minus 6 divided by the value of tau e which is 165 multiplied 10 at the power minus 6 multiplied by kb which is 0 .5 by 0 .8 multiplied 0 .85 so now solely the value of tau equivalent is equaled is 6 1 1 .5 divided by tq plus t 85 .5...

Question

2 Find the diameter of a shaft to transmit twisting moments varying from 800 N-m to 1600 Nm. The ultimate tensile strength for the material is 600 MPa and yield stress is 450 MPa. Assume the stress concentration factor = 1.2, surface finish factor = 0.8 and size factor = 0.85

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