Code_Aster Study Ax¶

Daniel Weschke

October 22, 2018

List of Symbols¶

\(A\)	mm²	Cross-section
\(E\)	MPa	Young’s modulus
\(F\)	N	Force
\(h\)	mm	Height
\(L\)	mm	Length
\(t\)	mm	Thickness
\(u\)	mm	Displacement

\(\varepsilon_{x}\)		Normal strain
\(\sigma_{xx}\)	MPa	Normal stress

1 Objective¶

Displacement \(u(x)\)
Stresses \(\sigma_{xx}(x)\)

2 Geometry & Mesh¶

Geometry.

\(L = \units[1000]{mm}\)
\(t = \units[10]{mm}\)
\(h(x) = (h(L)-h(0))/L \, x + h(0)\)
- \(h(0) = \units[100]{mm}\)
- \(h(L) = \units[10]{mm}\)
- \(h(x) = (-0.09x + 100) \units{mm}\)
\(A = h \, t\)
- \(A(0) = \units[1000]{mm^2}\)
- \(A(L) = \units[100]{mm^2}\)
- \(A(x) = (-0.9x + 1000) \units{mm^2}\)

3 Material¶

\(E = \units[3\times10^4]{MPa}\)

4 Boundary Conditions and Loads¶

Boundary conditions.

\(F = \units[2\times10^4]{N}\)

5 Results¶

5.1 Analytic¶

Displacement

\[u(x) = \int \dif u\]

Kinematics

\[\dif u = \varepsilon(x) \dif x\]

Constitutive equation

\[\varepsilon(x) = \frac{\sigma(x)}{E}\]

Mechanical equilibrium

\[\sigma(x) = \frac{F}{A(x)}\]

Displacement

\[u(x) = \int_0^L \dif u = \int_0^L \frac{F}{E\,A(x)} \dif x = \frac{F}{E} \int_0^L \frac{1}{A(x)} \dif x\]

\[u(x) = \frac{20000}{30000} \int_0^L \frac{1}{-0.9x + 1000} \dif x \cdot \units{mm}\]

With the integration rule

\[\int \frac{f'(x)}{f(x)} \dif x = \ln|f(x)| + C\]

we can integrate by re-writing the function as (because the derivative of the denominator is -0.9)

\[u(x) = \frac{2}{3} \cdot \frac{1}{-0.9} \int_0^L \frac{-0.9}{-0.9x + 1000} \dif x \cdot \units{mm} = -\frac{20}{27} \int_0^L \frac{-0.9}{-0.9x + 1000} \dif x \cdot \units{mm}\]

\[u(x) = -\frac{20}{27} \ln(-0.9x+1000) \cdot \units{mm} + C\]

For the constant of integration we look the known value \(u(0)=0\)

\[u(0) = -\frac{20}{27} \ln(1000) \cdot \units{mm} + C = 0\]

\[C = \frac{20}{27} \ln(1000) \cdot \units{mm}\]

Displacement

\[u(x) = -\frac{20}{27} [ \ln(-0.9x+1000) - \ln(1000) ] \cdot \units{mm}\]