Problem 10.5. Local buckling of an I beam

The cross section of an I-beam is given in the Figure. The beam is subjected to an axial normal force, P. Give the maximal allowed value of P
a) based on the local buckling of the flange,
b) based on the local buckling of the web.
Assume hinged connection between the web and the flanges in both cases.
Walls are isotropic, their stiffness is: D = 2.4 kNm, ν = 0.3.

Solve Problem

Solve

Problem a)

Buckling load of the flange, Nx,cr[kN/m]=

Problem b)

Buckling load of the web, Nx,cr[kN/m]=

Critical load of the total cross section, Pcr [kN]=

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Steps

Step by step

Buckling loads of long plates are given in Table 10.4.

Problem a)

Step 1.  Determine the buckling load of the flange with one hinged and one free edges.

Show buckling load of the flange

8th row of Table 10.4.
Nx,cr=6DLy21ν=6×2.40.075210.3=1792 kN/m

where Ly in the formula is the half width of the flange as it is shown in the Figure.

Step 2.  Give the value of the load, P acting on the cross section which results the critical load in the flange.

Show critical load

Pcr=Nx,crtA=1792×1035150×5×2+220×5=931.8 kN

Problem b)

Step 1.  Determine the buckling load of the hinged web.

Show buckling load of the web

1st row of Table 10.4.

Nx,cr=4π2Ly2D=4π20.22522.4=1872 kN/m

where Ly in the formula is the height of the web as it is shown in the Figure.

Step 2.  Give the value of the load, P acting on the cross section which results the critical load in the flange.

Show critical load

Pcr=Nx,crtA=1871.6×1035150×5×2+220×5=973.2 kN

Step 3.  Choose the relevant buckling load.

Show critical load of the cross section

Pcr=min973.2, 931.8=931.8 kN

Flange loses its stability first, thus flange buckling is relevant.

Results

Worked out solution

Buckling loads of long plates are given in Table 10.4.

Problem a)

The buckling load of the flange with one hinged and one free edges is

8th row of Table 10.4.
Nx,cr=6DLy21ν=6×2.40.075210.3=1792 kN/m

where Ly in the formula is the half width of the flange as it is shown in the Figure.

The critical load, P acting on the cross section which causes the buckling of the flange is 

Pcr=min973.2, 931.8=931.8 kN

Problem b)

The buckling load of the hinged web is

1st row of Table 10.4.

Nx,cr=4π2Ly2D=4π20.22522.4=1872 kN/m

where Ly in the formula is the height of the web as it is shown in the Figure.

The critical value of the load, P acting on the cross section which cause web buckling is

Pcr=min973.2, 931.8=931.8 kN

Flange loses its stability first, thus flange buckling results the relevant critical load of the cross section.

Pcr=min973.2, 931.8=931.8 kN