BV NR615 (2023) Plate Buckling — Verified Benchmark (Hand Calcs vs SDC Verifier)

Hand Calculations

A test plate model with 10.2 × 5.4 × 1.1 m dimensions was designed for the purpose of this benchmark:

The model was constrained at four bottom corners where side plates are connected. Forces were applied on the edges of the top plate with the following values:

\[ \left| F_{L}^{+} \right| = \left| F_{L}^{-} \right| = 3000\,\mathrm{kN} \] \[ \left| F_{S}^{+} \right| = \left| F_{S}^{-} \right| = 2550\,\mathrm{kN} \] \[ \left| F_{P}^{+} \right| = \left| F_{P}^{-} \right| = 2500\,\mathrm{kN} \]

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: 𝜈 = 0.3
• Mass Density: 𝜌 = 7850 𝑘𝑔/𝑚³
• 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:

• 𝜎_𝑥 = 37.14 𝑀𝑃𝑎
• 𝜎_𝑦=25.12 𝑀𝑃𝑎
• 𝜏=16.34 𝑀𝑃𝑎

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

Slenderness requirement check (Sec. 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 Sec. 5 / [2.2.1]):

I.

\[
\left(\frac{\gamma_{c1}\,\sigma_x\,S}{\sigma_{cx}’}\right)^{e_{0}}
+ \left(\frac{\gamma_{c1}\,\sigma_y\,S}{\sigma_{cy}’}\right)^{e_{0}}
+ \left(\frac{\gamma_{c1}\,\left|\tau\right|\,S}{\tau_{c}’}\right)^{e_{0}}
– \Omega = 1
\] \[
\Omega
= B\,
\left(\frac{\gamma_{c1}\,\sigma_x\,S}{\sigma_{cx}’}\right)^{e_{0}/2}
\left(\frac{\gamma_{c1}\,\sigma_y\,S}{\sigma_{cy}’}\right)^{e_{0}/2}
\]

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

${\sigma }_y\ge 0$ $sigma_y geq 0$ $${\left(\frac{{\gamma }_{c3}{\sigma }_{y}S}{{\sigma }_{cy}^{\mathrm{\prime }}}\right)}^{\frac{2}{{\beta }_p^{0.25}}}+{\left(\frac{{\gamma }_{c3}\mathrm{\mid }\tau \mathrm{\mid }S}{{\tau }_c^{\mathrm{\prime }}}\right)}^{\frac{2}{{\beta }_p^{0.25}}}=1$$

IV.

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

Aspect Ratio of the Plate Panel

(Sec. 5 / Symbols)

The aspect ratio a 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

(Sec. 5 / Symbols)

The elastic buckling reference stress 𝜎_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: (Sec. 5 / Symbols; stresses calculated using weighted average approach, App. 1 / [2.2.1])

$$\psi =1$$

• Correction factor

  $F_{long}$​ was set as 1: (Sec. 5 / [2.2.4]; Table 3)

  $$F_{long}=1$$

• Correction factor

  $F_{tran}$​ was also set as 1: (Sec. 5 / [2.2.5])

  $$F_{tran}=1$$

Ultimate buckling stresses were calculated in 3 cases: (Sec. 5 / Table 4)

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

(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

$C_x$

(Sec. 5 / Table 4)

$$C_x = c left( frac{1}{lambda_x} – frac{0.22}{lambda_x^2} right)$$ $$C_x=1.13\times (\frac{1}{1.979}-\frac{0.22}{{1.979}^2})=0.507$$

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

(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

$c_1$

(Sec. 5 / Table 2)

The coefficient 𝑐₁ was calculated appropriately with the chosen 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

$C_y$

(Sec. 5 / Table 4)

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

• 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 𝜆_𝜏

$lambda_tau$

(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_𝜏

(Sec. 5 / Table 4)

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

• Substituted values:

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

Ultimate buckling stresses

(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
  (Sec. 5 / Table 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

$B$

(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

$e_0$

​

(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

(Sec. 5 / [2.2.1]) — Transformed to calculate stress multiplier factors acting on loads 

$gamma$

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

(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

${\gamma }_c$

$gamma_c$

– was found:

$${\gamma }_c=1.763$$

The utilization factor 𝜂_𝑎𝑐𝑡 was calculated  (Sec. 1 / [2.2.2]:

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

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

Top Plate Properties (T = 12 mm)

Properties Summary

Calculated for the CSys “0..Basic Rectangular”

FEM Loads

This paragraph contains information about applied loads to model.

1. Long edges

2. Short edges

3. Long edges parallel

Constraints

This paragraph contains information about constrained parts of the model.

Results

Context for the figure Output from SDC Verifier → BV NR615 Plate Buckling (2023) for Load Set 1 (element-averaged check, component 1..Long2). The table lists the checked plate sections, their geometry (L, W, t), FE stresses (σx, σy, τ), and the NR615 utilizations for Limit States 1–4 and Overall. The Slenderness Requirement column confirms the NR615 slenderness criterion (here = 1.00). Utilization < 1.0 = pass; the governing value in this set is Overall 0.562 (Section Z12).

Intermediate Results of σ′_cx, σ′_cy and τ′_c from Calculation Details of the Check

Comparison of Hand Calculations and SDC Verifier Results

Ultimate Buckling Stresses [MPa]

Inverse of Stress Multiplier Factors Acting on Loads

Utilization Factor

Note: Results from SDC Verifier are the same as those obtained with hand calculations.

Conclusion

The benchmark confirms a high level of agreement between hand calculations and the automated SDC Verifier check:

• The slenderness requirement was satisfied.

• Differences in ultimate buckling stresses and stress multiplier factors were within <0.2%, indicating precise consistency.

• The utilization factor obtained was nearly identical:
  → Hand calculations: η_act = 0.567
  → SDC Verifier: η_act = 0.562

This validation demonstrates that SDC Verifier accurately implements the BV NR615 (2023) buckling assessment procedures, making it a dependable tool for structural integrity checks in maritime and offshore applications.