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With the line of working I underlined, please explain where that formula comes from, particularly how they are using change of length in place of delta v? Thanks

With the line of working I underlined, please explain where that formula comes from, particularly how they are using change of length in place of delta v? Thanks

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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With the line of working I underlined, please explain where that formula comes from, particularly how they are using change of length in place of delta v? Thanks 

2.5.5 SI A rigid cylinder, inside diameter 15 mm, contains a column of water 500 mm long. What will the column length be if a force of 2 kN is applied to its end by a frictionless plunger? Assume no leakage. P₁ = 0; P₂ - - Force Area 2 kN T(0.0075 m)² 11320 kN/m² = 11.32 MPa E = 320,000 psi(6895) = 2.21 x 10° Pa = 2210 MPa Since the tube is rigid, using Eq. 2.3b: -(P₂-P₁) U2 - U1 U₁ 42-4 4₁ E so -0.5 = -(0.5 m) 11.32 - 0 = Please explain!! = -0.00256 m 2210 L₂ = 0.5-0.00256 0.497 m or 497 mm ;) L₁ 500 mm - Water 2 kN Rigid Water Figure X2.5.5
Transcribed Image Text:2.5.5 SI A rigid cylinder, inside diameter 15 mm, contains a column of water 500 mm long. What will the column length be if a force of 2 kN is applied to its end by a frictionless plunger? Assume no leakage. P₁ = 0; P₂ - - Force Area 2 kN T(0.0075 m)² 11320 kN/m² = 11.32 MPa E = 320,000 psi(6895) = 2.21 x 10° Pa = 2210 MPa Since the tube is rigid, using Eq. 2.3b: -(P₂-P₁) U2 - U1 U₁ 42-4 4₁ E so -0.5 = -(0.5 m) 11.32 - 0 = Please explain!! = -0.00256 m 2210 L₂ = 0.5-0.00256 0.497 m or 497 mm ;) L₁ 500 mm - Water 2 kN Rigid Water Figure X2.5.5
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