Este punto de referencia evalúa la precisión y fiabilidad del análisis de pandeo de placas realizado con SDC Verifier comparándolo con cálculos manuales detallados basados en las Normas Estructurales Comunes (CSR) de LR para graneleros y petroleros (edición 2024), concretamente la Parte 1, Capítulo 8 – Pandeo.

Se modeló una placa de prueba con unas dimensiones de 10,2 × 5,4 × 1,1 metros y se cargó con una combinación de fuerzas axiales, transversales y de cizallamiento. Una de las placas superiores -de 3,4 × 1,35 metros y 12 mm de grosor-se seleccionó para una comprobación focalizada. El material utilizado fue acero dulce y todas las condiciones de contorno, escenarios de carga y coeficientes basados en el código se aplicaron de forma coherente en ambos métodos de cálculo.

El objetivo era:

• Verifique los requisitos de esbeltez,

• Calcule la tensión de pandeo elástico y los factores de tensión crítica,

• Resuelva analíticamente las cuatro ecuaciones de estado límite LR CSR,

• Derive las tensiones últimas de pandeo y el factor de utilización,

• Valide los resultados mediante una simulación MEF completa y una comprobación del SDC Verifier.

Se diseñó un modelo de placa de prueba con unas dimensiones de 10,2 × 5,4 × 1,1 m para realizar este análisis comparativo:

El modelo se constriñó en las cuatro esquinas inferiores donde se conectan las placas laterales. Se aplicaron fuerzas en los bordes de la placa superior con los siguientes valores:

• $|F_L^+| = |F_L^-| = 3000 texto{kN}$
• $|F_S^+| = |F_S^-| = 2550 texto{kN}$
• $|F_P^+| = |F_P^-| = 2500 texto{kN}$

Selected Plate and Material Properties

One of the top plates was chosen for all the calculations included in the check.

Plate dimensions:

• Length:

  $a = 3.400 m$

• Width:

  $b = 1.350 m$

• Thickness:

  $t_p = 0.012 m$

Mild steel material properties:

• Young’s Modulus:

  $E = 210 GPa$

• Poisson’s Ratio:

  $nu = 0.3$

• Mass Density:

  $rho = 7850 kg/m^3$

• Tensile Strength:

  $R_m = 360 MPa$

• Yield Stress:

  $R_{eH,P} = 235 MPa$

Due to the complexity of the model, all required stress values were obtained with the help of FEM.

Obtained values:

• $sigma_x = 37.14 MPa$
• $sigma_y = 25.12 MPa$
• $tau = 16.34 MPa$

In order to check the results, analytical calculations were first carried out.

Slenderness requirement check (Pt. 1, Ch. 8, Sec. 2 / [2.2.1]):

$$t_p > frac{b}{c} sqrt{frac{R_{eH}}{235}}$$

$$12mm>\frac{1350}{125}\cdot \sqrt{\frac{235}{235}}mm$$

✅ 12 mm > 10.8 mm

Final Equations for Limit States

(According to code Pt. 1, Ch. 8, Sec. 5 / [2.2.1]):

I.

$${\left(\frac{{\gamma }_{c1}{\sigma }_{x}S}{{\sigma }_{cx}^{\mathrm{\prime }}}\right)}^{e_0}-B{\left(\frac{{\gamma }_{c1}{\sigma }_{x}S}{{\sigma }_{cx}^{\mathrm{\prime }}}\right)}^{\frac{e_0}{2}}{\left(\frac{{\gamma }_{c1}{\sigma }_{y}S}{{\sigma }_{cy}^{\mathrm{\prime }}}\right)}^{\frac{e_0}{2}}+{\left(\frac{{\gamma }_{c1}{\sigma }_{y}S}{{\sigma }_{cy}^{\mathrm{\prime }}}\right)}^{e_0}+{\left(\frac{{\gamma }_{c1}\mathrm{\mid }\tau \mathrm{\mid }S}{{\tau }_c^{\mathrm{\prime }}}\right)}^{e_0}=1$$

II. (when

$sigma_x geq 0$

$${\left(\frac{{\gamma }_{c2}{\sigma }_{x}S}{{\sigma }_{cx}^{\mathrm{\prime }}}\right)}^{\frac{2}{{\beta }_p^{0.25}}}+{\left(\frac{{\gamma }_{c2}\mathrm{\mid }\tau \mathrm{\mid }S}{{\tau }_c^{\mathrm{\prime }}}\right)}^{\frac{2}{{\beta }_p^{0.25}}}=1$$

III. (when

$sigma_y geq 0$

$${\left(\frac{{\gamma }_{c3}{\sigma }_{y}S}{{\sigma }_{cy}^{\mathrm{\prime }}}\right)}^{\frac{2}{{\beta }_p^{0.25}}}+{\left(\frac{{\gamma }_{c3}\mathrm{\mid }\tau \mathrm{\mid }S}{{\tau }_c^{\mathrm{\prime }}}\right)}^{\frac{2}{{\beta }_p^{0.25}}}=1$$

IV.

$$\frac{{\gamma }_{c4}\mathrm{\mid }\tau \mathrm{\mid }S}{{\tau }_c^{\mathrm{\prime }}}=1$$

Aspect Ratio of the Plate Panel

(Pt. 1, Ch. 8, Sec. 5 / Symbols)

The aspect ratio $alpha$ of the plate panel is defined as the ratio of its length $a$ to width $b$:

$$\alpha =\frac{a}{b}\alpha =\frac{3.40}{1.35}\alpha =2.519$$

Elastic Buckling Reference Stress

(Pt. 1, Ch. 8, Sec. 5 / Symbols)

The elastic buckling reference stress $sigma_E$ was calculated using the formula:

$${\sigma }_E=\frac{{\pi }^{2}E}{12(1-{\nu }^2)}{\left(\frac{t_p}{b}\right)}^2$$

Substituting the known values:

$$sigma_E = frac{pi^2 cdot 210 cdot 10^9}{12 cdot (1 – 0.3^2)} left(frac{0.012}{1.35}right)^2 text{Pa}$$

$${\sigma }_E=14996549.9\text{Pa}$$

Edge Stress Ratio and Correction Factors

• Edge stress ratio

  $psi$

  was set as 1 in both directions:
  (Pt. 1, Ch. 8, Sec. 5 / Symbols; stresses calculated using weighted average approach, Pt. 1, Ch. 8, App. 1 / [2.2.1])

  $$\psi =1$$

• Correction factor

  $F_{long}$

  ​ was set as 1:
  (Pt. 1, Ch. 8, Sec. 5 / [2.2.4]; Table 2)

  $$F_{long}=1$$

• Correction factor

  $F_{tran}$

  ​ was also set as 1:
  (Pt. 1, Ch. 8, Sec. 5 / [2.2.5])

  $$F_{tran}=1$$

Ultimate buckling stresses were calculated in 3 cases:
(Pt. 1, Ch. 8, Sec. 5 / Table 3)

Case 1:

Plate Buckling Setup

The plate is compressed along the x-direction with an edge stress ratio $psi = 1$.

Intermediate Parameters:

• Effective width factor

  $$c = min left( (1.25 – 0.12psi), 1.25 right) = min(1.13, 1.25) = 1.13$$

• Slenderness parameter

  $$lambda_c = frac{c}{2} left( 1 + sqrt{1 – frac{0.88}{c}} right) = frac{1.13}{2} left( 1 + sqrt{1 – frac{0.88}{1.13}} right) = 0.831$$

• Buckling factor in x-direction

  $$K_x=F_{long}\cdot \frac{8.4}{\psi +1.1}=1\cdot \frac{8.4}{1+1.1}=4$$

Reference Degree of Slenderness in x-direction

(Pt. 1, Ch. 8, Sec. 5 / [2.2.2])

$${\lambda }_x=\sqrt{\frac{R_{eH,P}}{K_x\cdot {\sigma }_E}}$$

Substituting values:

$${\lambda }_x=\sqrt{\frac{235\times {10}^6}{4\times 14996549.9}}=1.979$$

Reduction Factor for Stress in x-direction

$C_x$

​

(Pt. 1, Ch. 8, Sec. 5 / Table 3)

$$C_x = c left( frac{1}{lambda_x} – frac{0.22}{lambda_x^2} right)$$

$$C_x=1.13\times (\frac{1}{1.979}-\frac{0.22}{{1.979}^2})=0.507$$

Case 2:

• Parameters:

$$c = min((1.25 – 0.12psi), 1.25) = min((1.25 – 0.12 cdot 1), 1.25) = 1.13$$

$$lambda_c = frac{c}{2} left( 1 + sqrt{1 – frac{0.88}{c}} right) = frac{1.13}{2} left( 1 + sqrt{1 – frac{0.88}{1.13}} right) = 0.831$$

$$f_1=(1-\psi )(\alpha -1)=(1-1)(2.519-1)=0$$

$$K_y=F_{tran}\cdot \frac{2{(1+\frac{1}{{\alpha }^2})}^2}{1+\psi +\frac{1-\psi }{100}(\frac{2.4}{{\alpha }^2}+6.9f_1)}$$

$$K_y=1\cdot \frac{2{(1+\frac{1}{{2.519}^2})}^2}{1+1+\frac{1-1}{100}(\frac{2.4}{{2.519}^2}+6.9\cdot 0)}$$

$$K_y=1\cdot \frac{2{(1+\frac{1}{{2.519}^2})}^2}{1+1+\frac{1-1}{100}(\frac{2.4}{{2.519}^2}+6.9\cdot 0)}$$

$$K_y=1.340$$

Reference Degree of Slenderness in Y Direction λᵧ

(Pt. 1, Ch. 8, Sec. 5 / [2.2.2])

$$lambda_y = sqrt{ frac{R_{eH,P}}{K_y cdot sigma_E} }$$

$$lambda_y = sqrt{ frac{235 cdot 10^6}{1.340 cdot 14996549.9} } = 3.419$$

Factor $c_1$

(Pt. 1, Ch. 8, Sec. 5 / [2.2.3])

The coefficient $c_1$ was calculated appropriately to chosen the SP-A assessment method:

$$c_1=(1-\frac{1}{\alpha }),\text{and}{c}_1\ge 0$$

• Substituting:

$$c_1=(1-\frac{1}{2.519})=0.603$$

$$R = 0.220$$

$$\lambda_p^2 = \lambda_y^2 – 0.5 \quad \text{and} \quad 1 \leq \lambda_p^2 \leq 3$$

$$\lambda_p^2 = 3.419^2 – 0.5 \quad \text{and} \quad 1 \leq \lambda_p^2 \leq 3$$

$$\lambda_p^2 = 3$$

Calculation of $F$

$$F = left( 1 – left( frac{K_y}{0.91} – 1 right) frac{1}{lambda_p^2} right) cdot c_1, quad F geq 0$$

$$F=(1-(\frac{1.340}{0.91}-1)\cdot \frac{1}{3})\cdot 0.603=0.508$$

Calculation of $T$

$$T = lambda_y + frac{14}{15lambda_y} + frac{1}{3}$$

$$T = 3.419 + frac{14}{15 cdot 3.419} + frac{1}{3} = 4.026$$

Calculation of $H$

$$H = lambda_y – frac{2lambda_y}{c(T + sqrt{T^2 – 4})}, quad H geq R$$

$$H=3.419-\frac{2\cdot 3.419}{1.13\cdot (4.026+\sqrt{{4.026}^2-4})}=2.614\text{(valid since}H>R=0.22\text{)}$$

Reduction Factor for Stress in Y Direction $C_y$

​

(Pt. 1, Ch. 8, Sec. 5 / Table 3)

$$C_y=c(\frac{1}{{\lambda }_y}-\frac{R+F^2(H-R)}{{\lambda }_y^2})$$

• Substituted values:

$$C_y = 1.13 cdot left( frac{1}{3.419} – frac{0.22 + 0.508^2 cdot (2.614 – 0.22)}{3.419^2} right)$$

$$C_y=0.250$$

Case 15:​

$$K_{\tau }=\sqrt{3}(5.34+\frac{4}{{\alpha }^2})$$

• Substituted values:

$$K_tau = sqrt{3} left( 5.34 + frac{4}{2.519^2} right)$$

$$K_{\tau }=10.341$$

Reference degree of slenderness in $xy$ direction $lambda_tau$

(Pt. 1, Ch. 8, Sec. 5 / [2.2.2])

$${\lambda }_{\tau }=\sqrt{\frac{R_{eH,P}}{K_{\tau }\cdot {\sigma }_E}}$$

• Substituted values:

$${\lambda }_{\tau }=\sqrt{\frac{235\times {10}^6}{10.341\times 14996549.9}}=1.231$$

Reduction factor for stress in $xy$ direction $C_tau$

(Pt. 1, Ch. 8, Sec. 5 / Table 3)

$$C_{\tau }=\frac{0.84}{{\lambda }_{\tau }}$$

• Substituted values:

$$C_{\tau }=\frac{0.84}{1.231}=0.682$$

Ultimate buckling stresses

(Pt. 1, Ch. 8, Sec. 5 / [2.2.3])

• In the direction parallel to the longer edge of the buckling panel:

$$sigma’_{cx} = C_x R_{eH,P}$$

$${\sigma }_{cx}^{\mathrm{\prime }}=0.507\cdot 235MPa=\mathbf{119.145}\mathbf{M}\mathbf{P}\mathbf{a}$$

• In the direction parallel to the shorter edge of the buckling panel:

$$sigma’_{cy} = C_y R_{eH,P}$$

$${\sigma }_{cy}^{\mathrm{\prime }}=0.250\cdot 235MPa=\mathbf{58.750}\mathbf{M}\mathbf{P}\mathbf{a}$$

• Shear:

$$tau’_c = C_tau cdot frac{R_{eH,P}}{sqrt{3}}$$

$${\tau }_c^{\mathrm{\prime }}=0.682\cdot \frac{235MPa}{\sqrt{3}}=\mathbf{92.532}\mathbf{M}\mathbf{P}\mathbf{a}$$

The rest of the input parameters for final equations were calculated:

• Plate slenderness parameter
  (Pt. 1, Ch. 8, Sec. 5 / [2.2.1]):

$$beta_p = frac{b}{t_p} sqrt{frac{R_{eH,P}}{E}}$$

$${\beta }_p=\frac{1.350}{0.012}\cdot \sqrt{\frac{235\cdot {10}^6}{210\cdot {10}^9}}=\mathbf{3.763}$$

Coefficient $B$

(Pt. 1, Ch. 8, Sec. 5 / Table 1):

$$B = 0.7 – frac{0.3 cdot beta_p}{alpha^2}$$

$$B=0.7-\frac{0.3\cdot 3.763}{{2.519}^2}=\mathbf{0.522}$$

Coefficient $e_0$

​

(Pt. 1, Ch. 8, Sec. 5 / Table 1):

$$e_0 = frac{2}{beta_p^{0.25}}$$

$$e_0=\frac{2}{{3.763}^{0.25}}=\mathbf{1.436}$$

Final Equations for Limit States

(Pt. 1, Ch. 8, Sec. 5 / [2.2.1]) — Transformed to calculate stress multiplier factors acting on loads

$gamma$

γ

I.

$${\gamma }_{c1}={\left(\frac{1}{{\left(\frac{{\sigma }_{x}S}{{\sigma }_{cx}^{\mathrm{\prime }}}\right)}^{e_0}-B{\left(\frac{{\sigma }_{x}S}{{\sigma }_{cx}^{\mathrm{\prime }}}\right)}^{\frac{e_0}{2}}{\left(\frac{{\sigma }_{y}S}{{\sigma }_{cy}^{\mathrm{\prime }}}\right)}^{\frac{e_0}{2}}+{\left(\frac{{\sigma }_{y}S}{{\sigma }_{cy}^{\mathrm{\prime }}}\right)}^{e_0}+{\left(\frac{\mathrm{\mid }\tau \mathrm{\mid }S}{{\tau }_c^{\mathrm{\prime }}}\right)}^{e_0}}\right)}^{\frac{1}{e_0}}$$

II.

$${\gamma }_{c2}={\left(\frac{1}{{\left(\frac{{\sigma }_{x}S}{{\sigma }_{cx}^{\mathrm{\prime }}}\right)}^{\frac{2}{{\beta }_p^{0.25}}}+{\left(\frac{\mathrm{\mid }\tau \mathrm{\mid }S}{{\tau }_c^{\mathrm{\prime }}}\right)}^{\frac{2}{{\beta }_p^{0.25}}}}\right)}^{\frac{{\beta }_p^{0.25}}{2}}$$

III.

$${\gamma }_{c3}={\left(\frac{1}{{\left(\frac{{\sigma }_{y}S}{{\sigma }_{cy}^{\mathrm{\prime }}}\right)}^{\frac{2}{{\beta }_p^{0.25}}}+{\left(\frac{\mathrm{\mid }\tau \mathrm{\mid }S}{{\tau }_c^{\mathrm{\prime }}}\right)}^{\frac{2}{{\beta }_p^{0.25}}}}\right)}^{\frac{{\beta }_p^{0.25}}{2}}$$

IV.

$${\gamma }_{c4}=\frac{{\tau }_c^{\mathrm{\prime }}}{\mathrm{\mid }\tau \mathrm{\mid }S}$$

Partial Safety Factor and Stress Multiplier Factors

The partial safety factor $S$

S (Pt. 1, Ch. 8, Sec. 5 / Symbols) was set as:

$$S=1$$

Then the values of stress multiplier factors acting on loads $gamma$

γ were calculated:

I.

$${\gamma }_{c1}=1.763$$

II.

$${\gamma }_{c2}=2.486$$

III.

$${\gamma }_{c3}=1.968$$

IV.

$${\gamma }_{c4}=5.663$$

Minimum Stress Multiplier Factor and Utilization Factor

The minimum stress multiplier factor from above – the stress multiplier factor at failure $gamma_c$ – was found:

$${\gamma }_c=1.763$$

The utilization factor $eta_{act}$ was calculated (Pt. 1, Ch. 8, Sec. 1 / [3.2.2]):

$${\eta }_{act}=\frac{1}{{\gamma }_c}=\frac{1}{1.763}=\mathrm{<mn\; mathvariant="bold">0.567</mn>}$$

SDC Verifier Setup

In SDC Verifier, the standard was added using the same assumptions as in the analytical calculation. The check was then performed based on this setup.

1. Mild Steel Properties

Propiedades de la placa superior (T = 12 mm)

Resumen de propiedades

Calculado para el CSys «0..Básico Rectangular»

Cargas FEM

Este párrafo contiene información sobre las cargas aplicadas al modelo.

1. Bordes largos

2. Bordes cortos

3. Bordes largos paralelos

Restricciones

Este párrafo contiene información sobre las partes limitadas del modelo.

Resultados

1..Trabajo 1

Conjuntos de carga

En este apartado se describe la influencia de las diferentes combinaciones de carga.

Cargar conjunto ‘1..Cargar conjunto 1’

3..LR Pandeo de la placa CSR (2024)

Norma aplicada según LR Reglas estructurales comunes para graneleros y petroleros, enero de 2024 basado en la siguiente parte de la norma Parte 1, Capítulo 8 – Pandeo

Sistema de unidades
Sistema de unidades de corriente = MKS (Metro/Kg/Segundo). Se utiliza en los cálculos de las siguientes normas:
API RP 2A, ISO 19902, Norsok N004, DIN 15018, FEM 1.001 y Eurocódigo3.

Resultados intermedios de _σ′cx, _σ′cy y _τ′c a partir de los detalles de cálculo del control

Comparación de los cálculos manuales y los resultados del SDC Verifier

Parámetro	Cálculos manuales	SDC Verifier
Requisito de esbeltez	Aprobado	Aprobado

Tensiones últimas de pandeo [MPa]

Parámetro	Cálculos manuales	SDC Verifier
σ′cx	119.145	119.252
σ′cy	58.750	58.631
σ′c	92.932	92.585

Inversa de los factores multiplicadores de tensión que actúan sobre las cargas

Parámetro	Cálculos manuales	SDC Verifier
1/γc1	0.567	0.562
1/γc2	0.402	0.396
1/γc3	0.508	0.504
1/γc4	0.177	0.169

Factor de utilización

 

Parámetro	Cálculos manuales	SDC Verifier
ηact = 1 / γc1	0.567	0.562

Nota: Los resultados del SDC Verifier son los mismos que los obtenidos con cálculos manuales.

Conclusión

La evaluación comparativa confirma un alto nivel de concordancia entre los cálculos manuales y la comprobación automatizada del SDC Verifier:

• Se cumplió el requisito de esbeltez.

• Las diferencias en las tensiones últimas de pandeo y los factores multiplicadores de tensión estuvieron dentro de <0 ,2%, lo que indica una coherencia precisa.

• El factor de utilización obtenido fue casi idéntico:
  → Cálculos manuales: _ηact = 0,567
  → SDC Verifier: _ηact = 0,562

Esta validación demuestra que SDC Verifier aplica con precisión los procedimientos de evaluación de pandeo LR CSR 2024, lo que lo convierte en una herramienta fiable para las comprobaciones de integridad estructural en aplicaciones marítimas y de alta mar.