d = 20 mm; E = 2.1011 Pa; F = 1,57.105 N
Tiết diện: \(S = \pi {R^2} = \pi {\left( {{d \over 2}} \right)^2} = {{\pi {d^2}} \over 4}\)
Ta có:
\(\eqalign{
& {F_{dh}} = k\left| {\Delta l} \right| = E{S \over {{l_0}}}\left| {\Delta l} \right|\cr& \Rightarrow {{\left| {\Delta l} \right|} \over {{l_0}}} = {{{F_{dh}}} \over {E.S}} = {{{F_{dh}}} \over {E.{{\pi {d^2}} \over 4}}} = {{4{F_{dh}}} \over {E.\pi {d^2}}} \cr
& \Rightarrow {{\left| {\Delta l} \right|} \over {{l_0}}} = {{{{4.1,57.10}^5}} \over {{{2.10}^{11}}.3,14.{{\left( {{{20.10}^{ - 3}}} \right)}^2}}} = {2,5.10^{ - 3}} \cr} \)
