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Strength of Materials
Structural Dynamics
Plates & Shells

Stress & Strain: Strain Energy, Resilience, Proof Resilience

Strain Energy When external forces are applied to a body the resisting force starts developing gradually till the deformation is fully developed. Work is done by the resisting force due to the deformation. This work is stored in the form of energy. We call this Strain Energy. Let's consider a bar of length L with cross-sectional area A under axial load P. Let Resistance developed be R. When deformation = 0, Resistance R = 0 When deformation = Δ, Resistance R = P Deformation = Δ = strain * original length = ε * L Work done by resisting force = average resistance * deformation = 0.5* (0+P) * Δ = P * ε * L/2 = p * A * ε * L/2 [Because, P = Force = Stress * Area = p * A ] = p * ε * A * L/2 = p * ε * V/2 = p * (p/E) * (V/2) [Because, ε = Strain = p/E ] = (p^2 * V) / 2E Resilience Strain Energy per unit Volume is defined as Resilience. Substituting V with 1, Work done by resisting f...

Double Integration Method: Useful Equations

Useful Equations Deflection = y Slope θ = dy/dx = y' Moment M = EI * y'' Shear Force F = - dM/dx = - EI * y''' Load Intensity q = dF/dx = - EI * y''''

GO : Inversing a Matrix

package main import "fmt" func main() { a := [][]float64{{1, 2, 3}, {0, 1, 4}, {5, 6, 0}} fmt.Println(a) ainv := invMatrix(a) fmt.Println(ainv) } func matRowScalarDiv(a [][]float64, I [][]float64, rowNumber int, divisor float64) [][]float64 { null := [][]float64{} if divisor != 0 { for i := 0; i < len(a); i++ { if i == rowNumber { for j := 0; j < len(a[rowNumber]); j++ { a[i][j] = a[i][j] / divisor I[i][j] = I[i][j] / divisor } } } return a } else { return null } } func neutralMatrixRow(a [][]float64, I [][]float64, neutralizingRow int) [][]float64 { for i := 0; i < len(a); i++ { if i != neutralizingRow { if a[neutralizingRow][neutralizingRow] != 0 || I[neutralizingRow][neutralizingRow] != 0 { neutralStar := a[i][neutralizingRow] / a[neutralizingRow][neutralizingRow] for j := 0; j < len(a[i]); j++ { a[i][j] = a[i][j] - a[neutralizingRow][j]*neutralStar I[i][j] = I[i][j] - I[ne...

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