Question

on application of 300 kn force on a bar dia 3cm the elongation of bar of length 2m is 0.00005 cm what is the value of stress and strain is SI unit

Answer 1

$here \: is \: your \: answer$

Data: d=30 mm, L=200 mm, P =60 kN, SL=D0.09 mm, ôd = 0.0039 mm Calculate: u and E

A=Trd24=TTX30243706.858mm2

E-PLASL=60x103x20070 6.858x0.09=188628.08N/ mm2 E=1.89N/mm2

HHu= Lateral Strain Linear Strain =(6dd) (SLL)=(0.003930)(0.09200)=0.29=0.29

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Answer 2

Answer:

Data: d=30 mm

L=200 mm

P =60 kN, δL=0.09 mm

δd = 0.0039 mm

Calculate: μ and E

A=πd24=π×3024=706.858mm2

E=PLAδL=60×103×200706.

858×0.09=188628.08N/mm2

E=1.89N/mm2

μμμ= Lateral Strain Linear Strain

=(δdd)(δLL)=(0.003930)(0.09200)

=0.29=0.29

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