Answered

Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

A cylindrical bar of metal having a diameter of 19.9 mm and a length of 186 mm is deformed elastically in tension with a force of 42600 N. Given that the elastic modulus and Poisson's ratio of the metal are 67.1 GPa and 0.34, respectively, determine the following: (a) The amount by which this specimen will elongate (in mm) in the direction of the applied stress. (b) The change in diameter of the specimen (in mm). Indicate an increase in diameter with a positive number and a decrease with a negative number.

Sagot :

batolisis

Answer:

a) 0.347 mm

b)  - 0.0126 mm

Explanation:

Diameter of metal bar = 19.9 mm

length = 186 mm = 0.186 m

Tension force = 42600 N

elastic modulus = 67.1 GPa

Poisson's ratio = 0.34

a) calculate the amount by which the specimen will elongate

first calculate the area of metal bar

Area = πd^2 / 4 = π/4 ( 19.9 )^2  = 3.11 * 10^-4 m^2

Elongation ( E ) = б / e = P/A * L / ΔL

and ΔL = PL / AE

hence the elongation ( ΔL) = [ (42600 * 0.186 ) / ( 3.4*10^-4 *  67.1 * 10^9 ) ]

                                            = 3.47 * 10^-4 m ≈ 0.347 mm

b) Calculate the change in diameter of specimen

μ = -( Δd / d) / ( ΔL/L )

0.34 = - (Δd / 19.9 ) / ( 0.347 / 186 )

∴ Δd = 0.34 * 0.00186 * 19.9  = - 0.0126 mm

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.