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 f_stat. = 1.25

Grab operation: Load factor f_stat. = 1.45

1 Lifting capacity at maximum loading

1.1) Maximum operational load including wind

Axial load, F_a = Q₁ + A + O + G

Res. moment, M_k = Q₁ · l_max + A · a_max + W · r - O · o - G · g

1.2) Load incl. 25% increase in lifting capacity without wind

Axial load, F_a = 1.25 · Q₁ + A + O + G

Res. moment, M_k = 1.25 · Q₁ · l_max + A · a_max - O · o - G · g

2 Lifting capacity at minimum loading

2.1) Maximum operational load including wind

Axial load, F_a = Q₂ + A + O + G

Res. moment, M_k = Q₂ · I_min + A · a_min + W · r - O · o - G · g

2.2) Load incl. 25% increase in lifting capacity without wind

Axial load, F_a = 1.25 · Q₂ + A + O + G

Res. moment, M_k = 1.25 · Q₂ · I_min + A · a_min - O · o - G · g

Crane for cargo operations

At maximum loading

Q = 220 kN l_max = 23 m

A = 75 kN a_max = 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

F_a = Q + A + O + G
F_a = 220 + 75 + 450 + 900
F_a = 1,645 kN

M_k = Q · l_max + A · a_max + W · r - O · o - G · g
M_k = 220 · 23 + 75 · 11 + 27 · 6,5 - 450 · 0,75 - 900 · 3
M_k = 3,023.0 kNm

2) Loading condition incl. 25% increase in lifting capacity without wind

F_a = Q · 1.25 + A + O + G
F_a = 275 + 75 + 450 + 900
F_a = 1,700 kN

M_k = Q · 1.25 · l max + A · amax - O · o - G · g
M_k = 275 · 23 + 75 · 11 - 450 · 0,75 - 900 · 3
M_k = 4,112.5 kNm

3) Maximum operational load without wind

F_a = 1,645 kN

M_k = Q · lmax + A · amax - O · o - G · g
M_k = 220 · 23 + 75 · 11 - 450 · 0,75 - 900 · 3
M_k = 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 f_stat = 1.25 against “static limit load curve” with the reading load:

Load condition 2)

F_a' = 1,700 kN · 1.25 = 2,125 kN

M_k' = 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)

F_a' = 1,645 kN · 1.15 = 1,891.7 kN

M_k' = 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)

F_a = 1,700 kN

M_k = 4,112.5 kNm

Crane for grab operations

At maximum loading

Q = 180 kN l_max = 19 m

A = 110 kN a_max = 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

F_a = Q + A + O + G
F_a = 180 + 110 + 450 + 900
F_a = 1,640 kN

M_k = Q · l_max + A · a_max + W · r - O · o - G · g
M_k = 180 · 19 + 110 · 9 + 27 · 6,5 - 450 · 0,75 - 900 · 3
M_k = 1,548.0 kNm

2) Loading condition incl. 25% increase in lifting capacity without wind

F_a = Q · 1.25 + A + O + G
F_a = 225 + 110 + 450 + 900
F_a = 1,685 kN

M_k = Q · 1.25 · l max + A · amax - O · o - G · g
M_k = 225 · 19 + 110 · 9 - 450 · 0,75 - 900 · 3
M_k = 2,227.5 kNm

3) Maximum operational load without wind

F_a = 1,640 kN

M_k = Q · l_max + A · a_max - O · o - G · g
M_k = 180 · 19 + 110 · 9 - 450 · 0,75 - 900 · 3
M_k = 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 f_stat = 1.45 against “static limit load curve” with the reading load:

Load condition 2)

F_a' = 1,685 kN · 1.45 = 2,443.3 kN

M_k' = 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 f_L = 1.7 is used.

Reference load:

Load condition 3)

F_a' = 1,640 kN · 1.7 = 2,788 kN

M_k' = 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)

F_a = 1,685 kN

M_k = 2,227.5 kNm