Problem 7.2. Load bearing capacity of columns

Determine the load bearing capacity of the 3 m high column, the cross section of which is given in the Figure. Both ends of the column are built-in. Material is glued laminated timber, grade of material is Gl36h, its compressive strength is: f = 31 MPa, the Euler slenderness is: λ1 = 61.6. Buckling reduction factor (χc) can be calculated in the function of the relative slenderness, λ by the following formula based on Eurocode (χc ≤ 1):
χc=1χ+χ2λ2, 

See Figure 7.14.

Solve Problem

Solve

Load bearing capacity, Nb [kN]=

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Steps

Step by step

Step 1.  Calculate the section properties.

Check section properties

Eq.(7-19)
A=50×200×2+100×50=25000 mm2yc=200×50×25+200×50×150+100×50×275A=125 mmIx=200×50312+200×50×1002+50×200312+50×200×252+100×50312+100×50×1502=2.552×108 mm4ix=IxA=2.55×10825000=101.0 mmIy=2003×5012+200×50312+1003×5012=3.958×107 mm4iy=IyA=3.96×10725000=39.8 mm

Step 2.  Give the relevant slenderness.

Check slenderness

Since the support conditions are the same in both directions, buckling occurs first about the axis of the smaller radius of gyration: iy . The relevant slenderness is

Eq.(7-18)
λmax=λy=l0iy=νliy=0.5×300039.8=37.7

Step 3.  Determine the buckling reduction factor.

Check buckling reduction factor

See Figure 7.14.
λ=λyλ1=37.761.6=0.612χ=0.51+0.2λ0.3+λ2=0.51+0.20.6120.3+0.6122=0.718χc=1χ+χ2λ2=10.718+0.71820.6122=0.913

Step 4.  Calculate the load bearing capacity of the column.

Check capacity

Eq.(7-20)
Nb=Aχf=25000×0.913×31=707.9×103 N=707.9 kN

Results

Worked out solution

First the section properties are calculated:

Eq.(7-19)
A=50×200×2+100×50=25000 mm2yc=200×50×25+200×50×150+100×50×275A=125 mmIx=200×50312+200×50×1002+50×200312+50×200×252+100×50312+100×50×1502=2.552×108 mm4ix=IxA=2.55×10825000=101.0 mmIy=2003×5012+200×50312+1003×5012=3.958×107 mm4iy=IyA=3.96×10725000=39.8 mm

Since the support conditions are the same in both directions, buckling occurs first about the axis of the smaller radius of gyration: iy . The relevant slenderness is

Eq.(7-18)
λmax=λy=l0iy=νliy=0.5×300039.8=37.7

Load bearing capacity of the column must be reduced by the buckling reduction factor:

See Figure 7.14.
λ=λyλ1=37.761.6=0.612χ=0.51+0.2λ0.3+λ2=0.51+0.20.6120.3+0.6122=0.718χc=1χ+χ2λ2=10.718+0.71820.6122=0.913

The capacity of the column becomes

Eq.(7-20)
Nb=Aχf=25000×0.913×31=707.9×103 N=707.9 kN