Hello friends here in this video we will see a problem on calculation of stress in each material here we have a question a copper wire 20 mm square in cross-section and steel wire 30 mm square in cross-section both 1 meter long are rigidly connected to plates on either side so I will write them in the data here two different materials are given copper wire so the first material is copper next and steel wire copper wire is 20 mm square in cross-section it means the area of copper wire is given a suffix C I’ll denote it by a suffix e next steel wire is 30 mm square in cross-section a suffix s next both 1 meter long so the length is 1 meter of both copper as well as steel so this is thousand mm are rigidly connected two plates on either side for that I will draw the diagram in the solution part they jointly share a load of 8 kilo Newton so capital P is 8 kilo Newton that is 8000 Newton then Young’s modulus for both steel and copper I have been given for steel it is 20 into 10 raise to 5 mega Pascal that is directly I can write it in terms of Newton per mm square and stress in copper also a Young’s modulus is 1 into 10 raise to 5 Newton per mm square now the question is to find the stresses produced in each material that is here we have to calculate how much is the stress in copper and how much is the stress in steel so this much is the question which we have now let us try to get the solution to this problem in the solution part I will explain it with a diagram first as it has been mentioned there are two plates now to these two plates it is given that copper wire and steel wire are rigidly connected to plates so here is the copper wire and then we have a steel wire so steel wire and copper wire respectively they have their individual areas as we have written in the data but both are having the same length that is 1 meter so the length is same that is 1 meter and they jointly share a load of 8 kilo Newton so over this assembly of copper and steel wire a joint load is acting the direction is not given suppose I am assuming that it is a tensile kind of a load which is applied and this has a value of 8 kilo Newton now the question is to find the stresses produced in each material that is stress in copper and stress in steel and since the area are given we’ll start the problem in this way that since cross section area of copper and steel are given therefore the cross-section area of steel is 30 mm square and that of copper is 20 mm squared it is mentioned next as we can see from the diagram the length of both of them are same and when they are rigidly connected with the help of plates and if we are applying a load the chances are that the strain in copper will be same as that of the strain in steel because they are rigidly connected the length is also same so they act as a single unit so here we can say that since strain in steel is equal to strain in copper because they are rigidly connected in the length is same and subjected to same kind of loading so therefore strain in steel is e suffix s strain in copper is e suffix E from Hookes law we have a relation that Young’s modulus is stress upon strain so therefore strain is equal to stress upon Young’s modulus so I will use this relation there for stress upon young’s modulus for steel is equal to stress upon young’s modulus for copper so therefore we have stress in steel is equal to Young’s modulus of Steel upon young’s modulus of copper into stress and copper so I will put the values the Youngs modulus for copper for steel first of all it is given as 20 into 10 raised to 5 and for copper it is 1 into 10 raised to 5 so from this and it is into Sigma C so stress in steel is equal to 20 into Sigma C so this will be the equation number one that is stress in steel is 20 times more the stress in copper next year as the load is applied to both copper and steel with the help of the plates so the load P is equally shared by copper and steel so I can say that since the copper and steel wire are rigidly connected so therefore the total load is shared equally by steel and copper so therefore total load is equal to load shared by steel plus load shared by copper so therefore the total load is capital P shall load shared by steel is piece of excess load shared by copper is P suffix C so therefore here I will put the values that is total load is given as 8 kilo Newton so it is 8 into 10 raised to 3 Newton load shared by steel load is equal to stress into area so stress in steel into area of steel plus stress in copper into area of copper I will go on putting the values stress in steel is 20 into Sigma C which is given in equation 1 next area of steel it is given in the problem that it is 30 plus stress in copper into area of copper is 20 so here I have 8 into 10 raise to 3 if I calculate this term that becomes 6 20 into Sigma C so Sigma C is 8 into 10 raised to 3 divided by 6 2005 a Sigma C value it comes out to be 12 point 9 Newton per mm square so this will be the first answer now after getting the first answer we can put this value of Sigma C in equation number 1 where we have Sigma s is equal to 20 into Sigma C so therefore put Sigma C is equal to 12 point 9 Newton per mm square in equation number 1 so hence stress in steel is 20 into 12 point 9 and this comes out to be 258 point zero 6 Newton per mm square this will be the second answer so if we look into this question they were telling us to calculate two things that to find the stresses produced in each material and there were two material copper and steel so we have to find the individual stresses and that we have calculated and with this we complete the question

i think the load should not be shared equally by both wires…is it?

Thank you Sir

Please sir upload mohr's circle graphical problem

Sir please correct it …you can't rigid the both wire as both the wires has different strain value.

Thanks sir

sir thank you very much for all your effort to reach students through wonderfull explantion in all subjects. dear sir its my sincere request to get accessed for private videos. kindly help

Sir pls can u solve more problems in this topic

Thanks sir

superb sir…

Thank you sir… Nice explanation…. Please give some more examples

A steel tube 2.4 cm external diameter and 1.8 cm internal diameter encloses a copper rod 1.5

cm diameter to which it is rigidly joined at each end. If at a temperature of 100C there is no

longitudinal stresses calculate the stresses in the rod and the tube when the temperature is

raised to 2000C. Es=210 kN/mm2

and Ec=1000 kN/mm2

. Coefficient of linear expansion for

steel is 11(10^-6)/°C and for copper 18(10^-6)/°C

Please explain this problem

Hello Friends,

Watch Complete Video Series of Subject Strength of Materials only on Ekeeda Application.

Use Coupon Code "NEWUSER" and access any one Subject free for 3 days!

Download Ekeeda Application and take advantage of this offer.

Android:- https://play.google.com/store/apps/details?id=student.ekeeda.com.ekeeda_student&hl=en

iOS:- https://itunes.apple.com/tt/app/ekeeda/id1442131224

Sir but how can you say that both will have same strain

Doesn't it depends on the young's modulus? Please explain sir