4) A cylindrical specimen of an alloy with a diameter of 9 mm is subjected to te

Question

4) A cylindrical specimen of an alloy with a diameter of 9 mm is subjected to tensile load. a)Calculate the force that will produce an elastic reduction of 4 x 103 mm in the diameter. (Assume that the material is stressed elastically in tension). (25 Points) b)ls there any plastic deformation in the material after applying the force? Justify your answer. (5 Points) c)ls there any necking in the material after applying the force? Justify your answer. (5 Points) Given Data:(Poisson's ratio for this material is 0.32, Elasticity Modulus is 138 GPa, Yield strength is 250 MPa, Tensile Strength is 350 MPa.) 5) A cylindrical rod of brass originally 10 mm in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 380 MPa and a ductility of at least 15 %EL are desired. Furthermore, the final diameter must be 7.5 mm. Explain how this may be accomplished. (40 Points) 14070 po 00- 040 Steel 60 3 700 100 Brass 80 30 500 70 Copper o 1040 0 10 20 30 40 50 60 TO Percent cold work 0 10 20 30 40 50 60 70 Percent cold work Figure 1: For 1040 Steel, Brass and Copper, a) the increase in tensile strength c) the decrease in ductility %EL) with % cold work.

Explanation / Answer

SOLUTION:(4)

given:
original diametre = D = 9mm
v = poisson;s ratio = 0.32

lateral strain =?L = -(?L/D) = -(4.0x10-3 /9)

poisson's ratio = v = -(?L/?) where ? = tensile strain

0.32 = -(?L/?) so ? = 1.38x10-3

from Hooke's law,   ? = E? where, ? = tensile stress, E = modulus of elasticity

? = Force/area = F/(? 20.25 ) N/m2 =0.0157F N/m2

hence E = ?/? = 0.0157F/1.38x10-3

138Gpa=0.0157F/1.38x10-3

138x1000N/mm2=11.39F

F=138x1000/11.39

F=12115.89N=12.115KN

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