A cylindrical specimen of an alloy (Fig. 1) 10mm in diameter and 100mm long that

Question

A cylindrical specimen of an alloy (Fig. 1) 10mm in diameter and 100mm long that is pulled in tension. A cylindrical specimen of an alloy (Fig. 1) 10mm in diameter and 100mm long that is pulled in tension. A cylindrical specimen of an alloy (Fig. 1) 10mm in diameter and 100mm long that is pulled in tension. A cylindrical specimen of an alloy (Fig. 1) 10mm in diameter and 100mm long that is pulled in tension. A cylindrical specimen of an alloy (Fig. 1) 10mm in diameter and 100mm long that is pulled in tension. A cylindrical specimen of an alloy (Fig. 1) 10mm in diameter and 100mm long that is pulled in tension. A. Determine the specimen length when a load of 78540 N is applied. B. What is the maximum load, in N, the specimen can sustain before failure? C. What is the material ductility?

Explanation / Answer

a) Load 78540 N

Diameter = 10 mm=10*10^-3 m

Specimen cross sectional area = 3.14/4*diameter^2=3.14/4*(10*10^-3)^2=7.85*10^-5 m2

Stress = Load/Cross sectional area = 78540/(7.85*10^-5)=10^9 Pa = 1000 MPa (Because 1 MPa=10^6 Pa)

Using curve, at stress=1000 Mpa, strain = .005

Length of specimen = Origninal length*(1+strain)=100*(1+.005)=100.5 mm

b) Material fails at stess value at last point of strain in stress-strain curve. In this curve, at last point of strain, stress = ~1800 MPa

Load at which material fails=Stress at which material fails*Cross sectional area = 1800*7.85*10^-5=0.14 MN=0.14*10^6 N

c) Material ductility = Maximum value of strain that material can hold without fail=Last point of strain in stress vs strain curve.

Here, last point of strain in curve = 0.08 = Material ductility

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