Match the diagram conditions, then follow the example inputs and results. Change numerical values in the calculator.
1. Match the diagram to your question
細長い柱を軸方向に押すモデルです。端部の拘束は係数Kで表します。
Lは部材長さです。曲がる方向のI・断面積A・材料Eを確認します。
柱の長さ L
4,000 mm
有効長さ係数 K
1
ヤング率 E
200 GPa
検討する主軸まわり I
4,166,670 mm⁴
断面積 A
5,000 mm²
材料の降伏応力（比較用）
235 MPa
破線は座屈のイメージで、変形量の予測ではありません。Pcrは理論値で許容荷重ではありません。
Euler理論座屈荷重 Pcr
514.042 kN
理論座屈応力
102.808 MPa
細長比 KL/r
138.564
A slender column is compressed axially. K represents end restraint.
L is member length. Check I for the buckling direction, area A and material E.
Column length L
4,000 mm
Effective length factor K
1
Young’s modulus E
200 GPa
I about the checked principal axis
4,166,670 mm⁴
Area A
5,000 mm²
Yield stress for applicability check
235 MPa
The dashed buckle is illustrative, not a deflection prediction. Pcr is theoretical critical load, not allowable load.
Euler theoretical critical load
514.042 kN
Theoretical critical stress
102.808 MPa
Slenderness KL/r
138.564
Blue marks inputs; orange marks outputs. These diagrams explain the fixed example below; dimensions and deformation are not drawn to physical scale.
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2. Example inputs
Column length LWhat goes here?
Use the dimension marked in the diagram. Distinguish radius from diameter, bore from outside diameter, and overall length from support span.
4,000 mm
Effective length factor KWhat goes here?
Converts physical column length L to effective buckling length Le=KL, based on ideal rotational and lateral restraints.
These are theoretical factors. Check joint stiffness and sidesway, and use the applicable design method. A bolted end is not automatically fully fixed.
Choose values for actual contact, lubrication, temperature and other conditions. These initial values are examples.
1
Young’s modulus EWhat goes here?
E describes axial stiffness and G shear stiffness; neither is yield strength or allowable stress.
Use the actual material, temperature and direction. Table entries are comparisons, not material specifications. For isotropic elasticity, check consistency with G=E/[2(1+ν)].
Choose values for actual contact, lubrication, temperature and other conditions. These initial values are examples.
200 GPa
I about the checked principal axisWhat goes here?
A section geometry property relevant to bending and buckling. Obtain it about the bending axis from section properties.
4,166,670 mm⁴
Area AWhat goes here?
Cross-sectional or surface area. Check whether it excludes holes or represents an effective area.
5,000 mm²
Yield stress for applicability checkWhat goes here?
This is the stress comparison criterion. Yield stress differs from allowable stress; match material condition, use and safety factor.
235 MPa
These are fixed teaching inputs. Coefficients, properties and comparison limits are not recommendations. Enter your conditions in the calculator.
3. Results for this example
Euler theoretical critical load
514.042 kN
Theoretical critical stress
102.808 MPa
Slenderness KL/r
138.564
The first value is Euler theoretical critical load, not allowable load. This 4,000 mm example keeps theoretical buckling stress below the entered yield stress.
4. Which formula is used?
Pcr = π² E I / (K L)², r = √(I/A)
Table units are not always the units required in the formula. The calculator converts selected units to its internal base units before computing.
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Assumptions and exclusions
Ideal straight, uniform slender column in elastic buckling. Choose K for the actual restraints (ideal pinned/pinned 1, fixed/free 2, fixed/fixed 0.5, fixed/pinned about 0.699). Check the weakest principal axis and restraint in both directions. The elastic formula is inapplicable at or above yield stress. Imperfections, local or inelastic buckling and allowable loads are excluded.
