Consider the brass alloy the stress-strain behavior of which is shown in the Ani

Question

Consider the brass alloy the stress-strain behavior of which is shown in the Animated Figure 7.12. A cylindrical specimen of this material 16.3 mm in diameter and 141.4 mm long is pulled in tension after which the tensile load is released; after the load is released the total length has still increased to 141.7 mm. Calculate the magnitude of the load (in N) necessary to cause this elongation.

References - le Chrome Secure https://edugen ileyplus.com/edugen/player references index uni?mode-help80linkobiect-callister978 111806 1602c07 anim-sec-0001&itemid;=nopoli References Tensile strength 450 MPa (65,000 psi) Strain= 0 Stress = 0 MPa Stress 0 psi 500 70 60 Strain=0 400 Stress = 0 MPa Stress = 0 psi 50 MPa 40 Yield strength 250 MPa (36,000 psi) 300 40 200 30 200 30 3 20 100 10 100 0 10 0.005 0 0.40 0.10 0.20 0.30 Strain

Explanation / Answer

deltaL = 141.7 - 141.4 = 0.3 mm Or 3 x 10^-4 m

strain = deltaL / L

= (3 x 10^-4) / (141.4 x 10^-3)

= 2.12 x 10^-3

for this strain, Y = 250 M Pa = 250 x 10^6 Pa ( from Curve)

Applying,

Stress = Y (strain)

F / A = (250 x 10^6) (2.12 x 10^-3)

F / (pi (16.3 x 10^-3 / 2)^2) = 0.53 x 10^6

F = 110.7 N ..........Ans

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