Example for determining the bearing load
Portal crane
The maximum load is to be determined in accordance with the following formulae. The determined loads are multiplied by the load factors before a bearing is selected. For the examples presented, the following applies:
Cargo operation: Load factor fstat. = 1.25
Grab operation: Load factor fstat. = 1.45
1 Lifting capacity at maximum loading
1.1) Maximum operational load including wind
Axial load, Fa = Q1 + A + O + G
Res. moment, Mk = Q1 · lmax + A · amax + W · r - O · o - G · g
1.2) Load incl. 25% increase in lifting capacity without wind
Axial load, Fa = 1.25 · Q1 + A + O + G
Res. moment, Mk = 1.25 · Q1 · lmax + A · amax - O · o - G · g
2 Lifting capacity at minimum loading
2.1) Maximum operational load including wind
Axial load, Fa = Q2 + A + O + G
Res. moment, Mk = Q2 · Imin + A · amin + W · r - O · o - G · g
2.2) Load incl. 25% increase in lifting capacity without wind
Axial load, Fa = 1.25 · Q2 + A + O + G
Res. moment, Mk = 1.25 · Q2 · Imin + A · amin - O · o - G · g
Crane for cargo operations
At maximum loading
Q = 220 kN lmax = 23 m
A = 75 kN amax = 11 m
O = 450 kN o = 0.75 m
G = 900 kN g = 3 m
W = 27 kN r = 6.5 m
1) Maximum operational load with wind
Fa = Q + A + O + G
Fa = 220 + 75 + 450 + 900
Fa = 1,645 kN
Mk = Q · lmax + A · amax + W · r - O · o - G · g
Mk = 220 · 23 + 75 · 11 + 27 · 6,5 - 450 · 0,75 - 900 · 3
Mk = 3,023.0 kNm
2) Loading condition incl. 25% increase in lifting capacity without wind
Fa = Q · 1.25 + A + O + G
Fa = 275 + 75 + 450 + 900
Fa = 1,700 kN
Mk = Q · 1.25 · l max + A · amax - O · o - G · g
Mk = 275 · 23 + 75 · 11 - 450 · 0,75 - 900 · 3
Mk = 4,112.5 kNm
3) Maximum operational load without wind
Fa = 1,645 kN
Mk = Q · lmax + A · amax - O · o - G · g
Mk = 220 · 23 + 75 · 11 - 450 · 0,75 - 900 · 3
Mk = 2,847.5 kNm
For the bearing selection, loading condition 2) is to be used for static dimensioning and loading condition 3) for service life. The static carrying capacity of the bearing is checked, taking into account the load factor fstat = 1.25 against “static limit load curve” with the reading load:
Load condition 2)
Fa' = 1,700 kN · 1.25 = 2,125 kN
Mk' = 4,112.5 kNm · 1.25 = 5,140.6 kNm
.
For a service life on 45,000 full load rotations, a load factor, fL = 1.15 is used.
Reference load:
Load condition 3)
Fa' = 1,645 kN · 1.15 = 1,891.7 kN
Mk' = 2,847.5 kNm · 1.15 = 3,274.6 kNm
Number of bolts and strength class are determined without a factor for maximum loads:
Load condition 2)
Fa = 1,700 kN
Mk = 4,112.5 kNm
Crane for grab operations
At maximum loading
Q = 180 kN lmax = 19 m
A = 110 kN amax = 9 m
O = 450 kN o = 0.75 m
G = 900 kN g = 3 m
W = 27 kN r = 6.5 m
1) Maximum operational load with wind
Fa = Q + A + O + G
Fa = 180 + 110 + 450 + 900
Fa = 1,640 kN
Mk = Q · lmax + A · amax + W · r - O · o - G · g
Mk = 180 · 19 + 110 · 9 + 27 · 6,5 - 450 · 0,75 - 900 · 3
Mk = 1,548.0 kNm
2) Loading condition incl. 25% increase in lifting capacity without wind
Fa = Q · 1.25 + A + O + G
Fa = 225 + 110 + 450 + 900
Fa = 1,685 kN
Mk = Q · 1.25 · l max + A · amax - O · o - G · g
Mk = 225 · 19 + 110 · 9 - 450 · 0,75 - 900 · 3
Mk = 2,227.5 kNm
3) Maximum operational load without wind
Fa = 1,640 kN
Mk = Q · lmax + A · amax - O · o - G · g
Mk = 180 · 19 + 110 · 9 - 450 · 0,75 - 900 · 3
Mk = 1,372.5 kNm
For the bearing selection, loading condition 2) is to be used for static dimensioning and loading condition 3) for service life. The static carrying capacity of the bearing is checked, taking into account the load factor fstat = 1.45 against “static limit load curve” with the reading load:
Load condition 2)
Fa' = 1,685 kN · 1.45 = 2,443.3 kN
Mk' = 2,227.5 kNm · 1.45 = 3,230.0 kNm
.
For a fixed service life of 150,000 full load rotations, a load factor of fL = 1.7 is used.
Reference load:
Load condition 3)
Fa' = 1,640 kN · 1.7 = 2,788 kN
Mk' = 1,372.5 kNm · 1.7 = 2,333.3 kNm
Number of bolts and strength class are determined without a factor for maximum loads:
Load condition 2)
Fa = 1,685 kN
Mk = 2,227.5 kNm