2D Frame III

Calculate deformation of node 4 of the isostatic system below, using the Virtual Work Method manually.

Hint for solution

According to the method; Integral should be considered separately for bars between 1-2 and 2-4 nodes.

Detailed solution

E= Modulus of elasticity A= Cross-sectional area I= Cross-section moment of inertia M= Moment on the element due to external loading M'= Moment on the element due to unit loading Δ= Deformation

Step 1: Finding Moment Function Based on Unit Load

1: Unit loading D ₁ , D ₂ , D ₃ : Support responses M ₁₂ : 1-2 bars 2 DN moment value
M ₂₄ : 2-4 bars 2 DN moment values ​​L ₁₂ : 1-2 bars length L ₂₄ : 2- Length of 4 bars
M ₁₂ (x): 1-2 bars moment function M ₂₄ (x): 2-4 bars moment function ΣM _1,dn : Total moment about 1 DN ΣM4,dn: Total moment about 4 DN ΣD: Total equilibrium

Moment Function : ,

Step 2: Finding the Moment Function Based on the Given Load

D ₄ , D ₅ , D ₆ : Support responses M'₁₂ : 1-2 bars 2 DN moment value M'₂₄ : 2-4 bars 2 DN moment values
​​L ₁₂ : 1-2 bars length L ₂₄ : 2-4 bars length M'₁₂ (x): 1-2 bars moment function
M'₂₄ (x): 2-4 bars moment function ΣM_1,dn : 1 Total moment relative to DN
ΣM4,dn: 4 Total moment relative to DN ΣD: Total equilibrium

Moment Function : ,

Step 3: Multiplying the M and M' Moment Functions in the Previous Steps and Getting the Integral

Integral for 1-2 Bar

Integral for 2-4 Rod

Deformation of node 4 = 0.91 m