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B 4/2 L By MCX) = -0.5Px1 El d' = dx² -0.5PX+G ΣM=OP - By L + 1.5 PL=0 = By 1.SPK K Σ= 0 A = 1.5 P -0.5PX3 2 E₁₁ = -0- SPX³ +4x+cz Ay L + PL = 0 2 Ay= -0.5 P A₁ = 0·5P+ Boundary Conditions @x=0 v=0 0 = -0.5P (0) + ((0) + Cz. C₂ =0 О @x₁ = L √=0 6=-0.5√(2)³ + GL 0.5P22 6 Elv₁ = -0.5Px,³ 2 + 1. PL² 6 -X, Elastic equatio M₂ (x) = - Px₂ Elds = -P² + Cs 2 0= - P(42)³ 6 + C½) + Ca EIV₂ = - Px Cate + Ca C₂L + C₁ = PL 6 48 @X=L, X₂ = 4/2 -0.5PX 0.522 2 + ما 0.5PL : dvi dv₂ 1 dx, dx = X2+ C3 2 -OSPL², OSR² = - P² + C3 052-PL + 2 ما C3= - PL² 24 8 A X1 L- B 22 Р x2

B 4/2 L By MCX) = -0.5Px1 El d' = dx² -0.5PX+G ΣM=OP - By L + 1.5 PL=0 = By 1.SPK K Σ= 0 A = 1.5 P -0.5PX3 2 E₁₁ = -0- SPX³ +4x+cz Ay L + PL = 0 2 Ay= -0.5 P A₁ = 0·5P+ Boundary Conditions @x=0 v=0 0 = -0.5P (0) + ((0) + Cz. C₂ =0 О @x₁ = L √=0 6=-0.5√(2)³ + GL 0.5P22 6 Elv₁ = -0.5Px,³ 2 + 1. PL² 6 -X, Elastic equatio M₂ (x) = - Px₂ Elds = -P² + Cs 2 0= - P(42)³ 6 + C½) + Ca EIV₂ = - Px Cate + Ca C₂L + C₁ = PL 6 48 @X=L, X₂ = 4/2 -0.5PX 0.522 2 + ما 0.5PL : dvi dv₂ 1 dx, dx = X2+ C3 2 -OSPL², OSR² = - P² + C3 052-PL + 2 ما C3= - PL² 24 8 A X1 L- B 22 Р x2

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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I asked this question from my book before "Determine the equations of the elastic curve for the beam using the X1 and X2 coordinates. Specify the beam's maximum deflection. EI is constant. Use the double integration method."

But I don't understand why X2 is L insteand of L/2.

B 4/2 L By MCX) = -0.5Px1 El d' = dx² -0.5PX+G ΣM=OP - By L + 1.5 PL=0 = By 1.SPK K Σ= 0 A = 1.5 P -0.5PX3 2 E₁₁ = -0- SPX³ +4x+cz Ay L + PL = 0 2 Ay= -0.5 P A₁ = 0·5P+ Boundary Conditions @x=0 v=0 0 = -0.5P (0) + ((0) + Cz. C₂ =0 О @x₁ = L √=0 6=-0.5√(2)³ + GL 0.5P22 6 Elv₁ = -0.5Px,³ 2 + 1. PL² 6 -X, Elastic equatio M₂ (x) = - Px₂ Elds = -P² + Cs 2 0= - P(42)³ 6 + C½) + Ca EIV₂ = - Px Cate + Ca C₂L + C₁ = PL 6 48 @X=L, X₂ = 4/2 -0.5PX 0.522 2 + ما 0.5PL : dvi dv₂ 1 dx, dx = X2+ C3 2 -OSPL², OSR² = - P² + C3 052-PL + 2 ما C3= - PL² 24 8
Transcribed Image Text:B 4/2 L By MCX) = -0.5Px1 El d' = dx² -0.5PX+G ΣM=OP - By L + 1.5 PL=0 = By 1.SPK K Σ= 0 A = 1.5 P -0.5PX3 2 E₁₁ = -0- SPX³ +4x+cz Ay L + PL = 0 2 Ay= -0.5 P A₁ = 0·5P+ Boundary Conditions @x=0 v=0 0 = -0.5P (0) + ((0) + Cz. C₂ =0 О @x₁ = L √=0 6=-0.5√(2)³ + GL 0.5P22 6 Elv₁ = -0.5Px,³ 2 + 1. PL² 6 -X, Elastic equatio M₂ (x) = - Px₂ Elds = -P² + Cs 2 0= - P(42)³ 6 + C½) + Ca EIV₂ = - Px Cate + Ca C₂L + C₁ = PL 6 48 @X=L, X₂ = 4/2 -0.5PX 0.522 2 + ما 0.5PL : dvi dv₂ 1 dx, dx = X2+ C3 2 -OSPL², OSR² = - P² + C3 052-PL + 2 ما C3= - PL² 24 8
A X1 L- B 22 Р x2
Transcribed Image Text:A X1 L- B 22 Р x2
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