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If the original length of a specimen is L0 = 10"" and new length of the specimen after applied load is L = 12.5"". The value of true strain is: a) 0.5 b) 0.25 c) 0.223 d) 0.4

Respuesta :

InesWalston

Answer:

The correct answer is option 'b':0.25

Explanation:

By definition of strain we have

[tex]\epsilon =\frac{L_f-L_o}{L_o}[/tex]

where

[tex]\epsilon [/tex] is the strain

[tex]L_o[/tex] is the original length of specimen

[tex]L_f[/tex] is the elongated length of specimen

Applying the given values we get

[tex]\epsilon =\frac{12.5''-10''}{10''}=0.25[/tex]