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13. A steel rod of 20 mm diameter passes centrally through a copper tube of external diameter 40 mm and internal diameter 20 mm: The tube is closed at each end with the help of rigid washers (of negligible thickness) which are screwed by the nuts The nuts are tightened until the compressive load on the copper tube is 50 kN. Determine the stresses in the rod and the tube, when the temperature of whole assembly falls by 50°C. Take Es = 200 GPa Ec = 100 GPa αs = 12 x 10^-6/°C and αc = 18 x 10^-6/°C. [Ans. 99.6 MPa; 19.8 MPa] 14 A bar of 2 m length; 20 mm breadth and 15 mm thickness is subjected to a tensile load of 30 kN. Find the final volume of the bar; if the Poisson's ratio is 0.25 and Young's modulus is 200 GN/m². [Ans. 600 150 mm³] 15 A bar of 12 mm diameter gets stretched by 3 mm under a steady load of 8 kN. What stress would be produced in the bar by a weight of 800 N, which falls through 80 mm before commencing the stretching of the rod, which is initially unstressed. Take E = 200 kN/mm² [Ans: 170.6 N/mm²]

          13. A steel rod of 20 mm diameter passes centrally through a copper tube of external diameter 40 mm and internal diameter 20 mm: The tube is closed at each end with the help of rigid washers (of negligible thickness) which are screwed by the nuts The nuts are tightened until the compressive load on the copper tube is 50 kN. Determine the stresses in the rod and the tube, when the temperature of whole assembly falls by 50°C. Take Es = 200 GPa Ec = 100 GPa αs = 12 x 10^-6/°C and αc = 18 x 10^-6/°C. [Ans. 99.6 MPa; 19.8 MPa]

14 A bar of 2 m length; 20 mm breadth and 15 mm thickness is subjected to a tensile load of 30 kN. Find the final volume of the bar; if the Poisson's ratio is 0.25 and Young's modulus is 200 GN/m². [Ans. 600 150 mm³]

15 A bar of 12 mm diameter gets stretched by 3 mm under a steady load of 8 kN. What stress would be produced in the bar by a weight of 800 N, which falls through 80 mm before commencing the stretching of the rod, which is initially unstressed. Take E = 200 kN/mm² [Ans: 170.6 N/mm²]

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Added by Daniela C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Supreeta N

09/06/2023

Step 1

ΔLs = αs * L * ΔT ΔLc = αc * L * ΔT Since the assembly is rigid, the change in length for both steel and copper should be equal. ΔLs = ΔLc Now, we can find the stress in the steel rod and copper tube due to the temperature drop. Stress in steel rod (σs) = Es Show more…

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13. A steel rod of 20 mm diameter passes centrally through a copper tube of external diameter 40 mm and internal diameter 20 mm: The tube is closed at each end with the help of rigid washers (of negligible thickness) which are screwed by the nuts The nuts are tightened until the compressive load on the copper tube is 50 kN. Determine the stresses in the rod and the tube, when the temperature of whole assembly falls by 50°C. Take Es = 200 GPa Ec = 100 GPa αs = 12 x 10^-6/°C and αc = 18 x 10^-6/°C. [Ans. 99.6 MPa; 19.8 MPa] 14 A bar of 2 m length; 20 mm breadth and 15 mm thickness is subjected to a tensile load of 30 kN. Find the final volume of the bar; if the Poisson's ratio is 0.25 and Young's modulus is 200 GN/m². [Ans. 600 150 mm³] 15 A bar of 12 mm diameter gets stretched by 3 mm under a steady load of 8 kN. What stress would be produced in the bar by a weight of 800 N, which falls through 80 mm before commencing the stretching of the rod, which is initially unstressed. Take E = 200 kN/mm² [Ans: 170.6 N/mm²]

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00:01 Hello students, in this question we have to determine the stresses in the rod and the tube.

00:06 So, these are the values which are given in the question.

00:10 We know that the area of the steel rod, area of steel rod that is as is equal to pi by 4 d square s which is equal to pi by 4 into 20 raised to 2 and that is equal to 100 pi millimeter square.

00:34 Now, similarly let us find the area of copper tube, area of the copper tube that is ac is equal to pi by 4 into edc square minus idc square.

00:54 Now, substituting the values we get ac is equal to pi by 4 into 40 square minus 20 square and this is equal to 300 pi millimeter square.

01:07 Now, es is given as 200 and that is equal to 200 giga newton per meter square which is equal to 200 into 10 raised to 9 newton per meter square or in terms of millimeter square we can write it as 200 into 10 raised to 3 newton millimeter square.

01:39 Similarly, we can find ec is equal to 100 giga pascal and that is equal to 100 giga newton per meter square which is equal to again in terms of millimeter square we can write it as 100 into 10 raised to 3 newton per millimeter square.

02:03 So, as we know that alpha c is more than that of alpha s.

02:12 That is the alpha for copper is more than, is more than that of steel.

02:26 And that of steel therefore the free contraction of copper is more than that of steel.

02:34 Therefore the free contraction of copper is more than, is more than that of steel.

02:49 Is more than that of steel.

02:55 So which means that the steel rod, steel rod is subjected to, is subjected to compressive stress, is subjected to compressive stress.

03:09 Okay and the copper tube, copper tube is subjected to tensile stress, tensile stress.

03:28 Therefore the tensile load on copper tube is equal to the compressive load, compressive load on steel rod.

03:50 That is we can write equation sigma cac is equal to sigma s as.

03:58 That is sigma c into 300 pi is equal to sigma s into 100 pi which means that sigma s is equal to 3 times sigma c and let this be 3 times sigma c.

04:11 Let this be equation number 1...

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