A polystrene material has a resistivity to the impurity pi...

A polystrene material has a resistivity to the impurity (pi) of 2.5*10^-5 ohm m, to the plastic deformation (pd) of 3.5*10^-6 ohm m and a total (pt) of 4.0*10^-5 ohm m. Calculate its total resistivity (ohm mm) at 1173 \Kappa . 2.0*10^-5 1.2*10^-5 1.2*10^-2 3.5*10^-6

Transcript

00:01 So let's start with the problem over here.

00:02 We have an axle load of 22 kiloons and 45 mm wide and 15 m m m this is 45 mm wide and 15 m thick okay and this elongates 3m while the bar width okay this elongates that is this thing it elongates by 3mmm and this bar okay it's with the contracts okay contracts by 0 .25 okay and bar is 200 m.

00:40 Long okay the bar is 200 mm long at the 22 kilo newton load the stress is a proportional to is less than at 22 kilo newton load stress is less than proportional limit okay and we have to find the model of elasticity poisons ratio and change in bar thickness okay so first of all let's try to find the area of the bar okay so i'll write cross -sectional area of bar okay okay a is equal to 15 that is what well this cross -sectional area means see when you you will see the bar it will be like this i'll draw it better okay from one side if you will see this will be like this okay that's how the bar will be there in front of you if you see okay so this cross -sectional area is what we have to find or else i'll show you in a different way just one second okay so this is how the bar was okay so you can see uh the length is uh 45 okay and uh the width is, see, width is how much? 15, okay? and this total length was 200.

02:15 Okay, that's what was given.

02:16 Okay, so let's find out the area.

02:18 It is 15 m m into 45 m.

02:21 So this becomes 675 mm square.

02:25 Okay.

02:26 And now let us try to find the normal stress corresponding to corresponding 22 kilo newton axle load.

02:40 Okay you have to find the normal stress at this stress point okay so it will be 22 kilo newton okay uh it is to be multiplied with a thousand okay in order to convert it to newtons and just write the conversion okay and then we have to divide it using the area okay 675 m m square so this comes out to be 32 .59 mega pascal so, longitudinal strain, okay, in bar will be what? in just one second, in bar will be, okay, epsilon long, this will be delta by l, okay? this is a change, okay? how much change was there? it was extended by 3mm, that is it elongates by 3m, okay? so 3 we divided by the total length, okay? that is 200 okay so this becomes what this is 3 mm okay this is 200 mm so this becomes 0 .0150 m m.

03:56 Okay and this we will represent and modulus of elasticity therefore will be e is equal to sigma by epsilon long okay and this will be equal to 32...

Question

A polystrene material has a resistivity to the impurity (pi) of 2.510^-5 ohm m, to the plastic deformation (pd) of 3.510^-6 ohm m and a total (pt) of 4.010^-5 ohm m. Calculate its total resistivity (ohm mm) at 1173 \Kappa . 2.010^-5 1.210^-5 1.210^-2 3.5*10^-6

Question

A polystrene material has a resistivity to the impurity (pi) of 2.5*10^-5 ohm m, to the plastic deformation (pd) of 3.5*10^-6 ohm m and a total (pt) of 4.0*10^-5 ohm m. Calculate its total resistivity (ohm mm) at 1173 \Kappa . 2.0*10^-5 1.2*10^-5 1.2*10^-2 3.5*10^-6

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A polystrene material has a resistivity to the impurity (pi) of 2.510^-5 ohm m, to the plastic deformation (pd) of 3.510^-6 ohm m and a total (pt) of 4.010^-5 ohm m. Calculate its total resistivity (ohm mm) at 1173 \Kappa . 2.010^-5 1.210^-5 1.210^-2 3.5*10^-6