1 Consider a cylindrical specimen of steel alloy Figure 110

1. Consider a cylindrical specimen of steel alloy (Figure 1)10 mm in diameter and 75 mm long that is pulled in tension. Determine its elongation when a load of 20,000 N is applied.

2. Calculate the modulus of elasticity for Figure 1.

3. What is the proportional limit for Figure 1?

4. What is the tensile strength for Figure 1?

ECO 500 400 300 200 ICO 0.00 0.04 500 n 400 300 H 200 100 0000 0002 0,004 0 C06 Stran 0,16 Strain Figure 1 0.20

Solution

Given, diameter of cylindrical specimen = 10mm,length = 75mm

1) Load = 20000 N

stress = load/crossectional area = 20000/(pi*(10)²/4) = 254.65 Mpa

from figure 1 for stress of 254.65 Mpa,corresponding strain = 0.0013

strain = (final length - initial length)/initial length

0.0013 = (final lenth-75)/75 so,final length = 75.0975 mm

2) for figure 1 , Modulus of elasticity is slope of stress vs strain graph in the linear proportional region (i,e up to strain of 0.0017)

from graph for stress of 100 Mpa strain is 0.0005 and for stress of 200 Mpa strain is 0.001

Modulus of elasticity = (200-100)/(0.001-0.0005) = 200000 Mpa = 200 Gpa

3) up to proportional limit stress and strain graph should be straight line which is up to stress of 330 Mpa

4) Tensile strength is maximum stress which a material can take before its failure.from figure 1 maximum stress is 500 Mpa.so, tensile strength = 500 Mpa