P df li di an axial load is placed on a rod of diameter di...

Name: p df li di an axial load is placed on a rod of diameter di and length of l after the load is placed the diameter of the rod is measured to be df find the magnitude of the load p given d 11137
Uploaded: 2022-02-10T08:33:06-08:00
Duration: 5 min 7 s
Channel: Ina R
Description: p df li di an axial load is placed on a rod of diameter di and length of l after the load is placed the diameter of the rod is measured to be df find the magnitude of the load p given d 11137

P D$_f$ L$_i$ D$_i$ An axial load is placed on a rod of diameter, D$_i$, and length of L. After the load is placed, the diameter of the rod is measured to be D$_f$. Find the magnitude of the load P given: D$_i$ = 10 mm D$_f$ = 9.997 mm L$_i$ = 100 mm $\nu$ = 0.3 E = 200 GPa Magnitude of load P = Number kN (input 4 significant figures)

Transcript

00:01 Hello students in this question we are given with values like length of the rod which is 100 mm which is equals to 0 .1 meters we are also given with the diameter of rod which is 15 millimeters which is equal to 0 .015 meters we are also given with some more data like p which is equal to 10 kiloons which can also be written as 10 raise to power 4 neurons we are also given with mu which is 0 .35 and we are also given with e which is 70 which is 70 g pascals which can also be written as 70 multiplied by 10x per 9 pascals and we have to find the change in diameter delta d on application of the force so let's calculate as we know that the longitudinal strain the longitudinal strain is given by the formula longitude strain is given by formula sigma upon e and we also know that the we also know that the normal stress is also given by another formula which is sigma is equals to p by a where p is the force on with p is force on the surface area or we can say not normal force on the surface area and a is the surface area okay so let's calculate the surface area first which is which can be calculated by pi by 4 multiplied by d square so let's put the value of d here so it will be pi by 4 multiplied by 0 .015 square so on solving this we get 1 .76 multiplied by 10 r to power minus 4 meter squares okay so so from here we can calculate the longitudinal stress, we can calculate the normal stress which is sigma, which will be p by a.

02:07 So the values of p are given in the equation which is 10 kiloons and so it can be written as per 4 neutrons and the value of a as we have calculated is 1 .76 multiplied by 10 s to par minus 4.

02:22 So on solving this we got 5 .56 multiplied by 56 .5 .5 .5.

02:28 Multiplied by 10x to par 6 pascal which can be written as 56 .6 megapascals.

02:34 Okay, so we will solve this question further...

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