Usually, most of the industrial buildings have built-in overhead cranes for handling heavy equipment or goods. With the help of the overhead cranes, heavy equipment or goods can be lifted and moved from one point of work place to another. The cranes may be hand operated (generally they have a capacity of up to 5 tonnes) and electrically operated (EOT). A typical EOT crane system is shown in Figure 6.1.
The crane system consists of a crane bridge with two parallel girders spaced at a fixed distance apart known as a wheel base. The crane bridge which has a set of wheels at its ends moves on the rails attached to the gantry girders. A trolley or crab which carries a hook moves on the crane bridge. Any equipment or material to be handled is hung to the hook which may be moved horizontally in two perpendicular directions, viz., along the length and width of the working area. This is possible by the motion of the trolley/crab on the crane bridge and by the motion of the crane bridge on the rails of the gantry girders.
Over-head cranes can be made of different capacities and for a variety of applications. The probable range of data pertaining to over-head cranes up to 500 kN capacity is presented in the following:
Generally, gantry girders are supported on brackets connected to columns for lighter loads and are directly supported on columns for heavier loads (Figure 6.2). In the latter case, a stepped column is used. A suitable seat is fabricated to support and connect the bottom flange of the gantry girder. The gantry girder is connected to the column near the top flange to provide lateral support and torsional restraint by channel sections. In general, the compression flange of the gantry girder is unrestrained against the lateral torsional buckling. The compression flange is provided with wide plates or channel sections to minimize the lateral torsional buckling. The gantry girder may be assumed as simply supported and fully restrained against torsion at the ends with warping not restrained. If a walk way is provided along the gantry girder, it may provide continuous lateral restraint to the compression flange. Some of the types of sections provided for gantry girder are shown in Figure 6.3.
Gantry girders are subjected to vertical, lateral and longitudinal loads. Both lateral and longitudinal loads act at the level of the rail. It is considered that lateral and longitudinal loads do not act simultaneously. Hence, for the design of gantry girder, the following load combinations should be considered.
- Vertical and lateral loads,
- Vertical and longitudinal loads.
6.2.1 Vertical Loads
The vertical loads consist of the self weight of the crane, the maximum weight lifted by the crane (the crane capacity) and the weight of the crab. As these loads may act suddenly on the gantry girder, the additional static load is considered. As per IS 875 (Part 2):1987, for EOT cranes, the maximum static wheel loads should be increased by 25% to account for the impact effect.
Due to the sudden starting or stopping of the crab on the crane bridge, loads act on the gantry girder in the lateral direction. As per IS 875 (Part 2):1987, for EOT cranes, the total lateral load should be 10% of the weight of the crab and the maximum weight lifted by the crane. This load should be equally distributed amongst all the wheels of the crane bridge.
6.2.3 Longitudinal Loads
Due to the sudden starting or stopping of the crane bridge, loads act on the gantry girders in the longitudinal direction. As per IS 875 (Part 2):1987, the longitudinal force on each gantry girder should be 5% of the maximum static wheel loads depending upon the number of wheels which may pass within the span of a gantry girder.
As the gantry girder is subjected to concentrate wheel loads, it is necessary to check the web for buckling and crippling as explained in Sec. 5.5 of Chapter 5.
For the constant stress range, the actual normal stress range ƒ and shear stress range τ at a point of the structure subjected to Nsc cycles in life should satisfy
where ƒƒd and τƒd are the design normal fatigue strength range and the design shear fatigue strength range, respectively.
6.5.1 Low Fatigue
For the constant stress range, the fatigue assessment of a member or connection or detail is not required if the actual normal stress range ƒ ≤ 27/γmft and the actual shear stress range τ ≤ 27/γmft or if the number of stress cycles, Nsc < 5 × 106 (27 / γmft / γmƒ)3
where γmft = partial safety factor for fatigue strength (given in Table 6.1)
γm = partial safety factor for load (usually 1.0).
- fail-safe, if the local failure of one component due to fatigue crack does not result in the failure of the structure due to the availability of an alternative load path (the redundant system),
- non-fail-safe, if the local failure of one component leads rapidly to the failure of the structure due to the non-availability of an alternative load path (the non-redundant system).
6.5.2 Fatigue Strength Range
The design normal fatigue strength range ƒƒd for Nsc life cycles is given by
where ƒf = normal fatigue strength range
in which ƒfn is the normal fatigue strength for 5 × 106 cycles given under the relevant ‘detail category’ of Table 26 of IS800:2007.
In a similar manner, the design shear fatigue strength range τƒd for Nsc life cycles is given by
where τf = shear fatigue strength range
In general, the factor for thickness correction, μf = 1.0. When plates greater than 25 mm in thickness are joined together by a transverse fillet or butt welds, μf = (25/tp)0.25. No thickness correction is necessary when full penetration butt weld reinforcements are machined flush and proved free of defect through non-destructive testing (i.e. μf = 1.0).
6.5.3 Stress and Stress Range Limitations
The maximum (absolute) value of the normal and shear stresses should not exceed the elastic limit (ƒy, τy) for the material under cyclic loading. The maximum stress range should not exceed 1.5ƒy for normal stresses and for shear stresses.
Design a suitable section for a simply supported gantry girder for the following data:
Spacing of columns = 4 m
Crane capacity = 160 kN
Weight of the crane excluding the crab = 250 kN
Weight of the crab = 60 kN
Minimum clearance of cross travel = 0.8 m
Wheel base = 5.3 m
Centre to centre distance between gantry girders = 20 m
Height of the rail = 105 mm
Expected number of stress cycles = 2 × 106
Grade of the steel = E250
Maximum wheel load
Maximum static wheel load = 336/2 = 168 kN
Maximum wheel load including the impact effect @ 25% =168 + (25 /100) × 168 = 210 kN
Maximum bending moment and shear force due to vertical forces
The self weight of the gantry girder including rail @ 3 kN/m = 4.0 × 3.0 = 12 kN. As the wheel base is more than the span of the gantry girder, only one wheel load can act on the gantry girder. For the maximum bending moment, its position is shown in Figure 6.2(b), whereas for the maximum shear force, its position is shown in Figure 6.2(c).
The design bending moment in the gantry girder due to vertical forces
The design shear force in the gantry girder due to vertical forces
Maximum bending moment and shear for due to lateral force
Total lateral force @10% of 220 kN = 10/100 × 220 = 22 kN
Lateral force acting at each wheel on the gantry girder = 22/4 = 5.5 kN
Factored lateral force acting at each wheel on the gantry girder = 1.5 × 5.5 = 8.25 kN
Design bending moment in the gantry girder due to the lateral force
The design shear force in the gantry girder in the horizontal plane due to the lateral force Vz = 1.5 × 5.5 = 8.25 kN (when the lateral force acts on the support).
The longitudinal force on the gantry girder @ 5 % of 168 kN = (5/100) × 168 = 8.4 kN
The design longitudinal force = 1.5 × 8.4 = 12.6 kN
Design of cross section for gantry girder
The gantry girder is laterally unsupported and hence the design bending stress is assumed initially.
Let ƒbd = 100 N/mm2
The plastic section modulus to be provided
The welded section shown in Figure 6.5 may be tried.
∴ From Table 1.7, the section is plastic and hence, βb = 1.0.
The location of the plastic neutral axis z:
The plastic section modulus about z-axis
Zpz = 300 × 24 × (116.7 + 12) + 116.7 × 12 × (116.7/2) + 383.3 × 12 × (383.3/2) + 200 × 20 × (383.3 + 10) = 3,463 cm3
The plastic section modulus about y-axis
Zpy = 2[(150 × 24 × 150/2) + (100 × 20 × 100/2) + (500 × 6 × 6/2)] = 758 cm3
The momemt of inertia about y-axis,
Determination of shear centre (SC) of entire section
Let P be the force acting at the shear centre of the section which is at a distance e from the centroid of the upper flange as shown in Figure 6.6.
Due to force P, V1 is the shear force in the upper flange of the section and V2 is the shear force in the lower flange of the section as shown in Figure 6.6.
The maximum shear flow in the lower flange of the section = VA / Iy where V = P
Since the shear flow varies parabolically in the flanges, V2 = 2/3 × 200× 0.00148 P = 0.197 P
Taking moments about a point on the line of the action of V1
Elastic critical moment, Mcr
The elastic critical moment Mcr is calculated using the expression given in Annexure E of IS800:2007 for sections symmetric about the minor axis.
c1, c2, c3 = factors which depend on the type of loading and supports
c1 = 1.365, c2 = 0.553, c3 = 1.78 from Table 42 of IS 800:2007 for a simply supported beam with a concentrated load at the centre.
Kw = warping restraint factor = 1.0
yg = distance between the point of the application of the load and the shear centre of the section = 12 + 103 = 115 mm
The moment of inertia of the compression flange about y-axis,
Ifc = 24 × 3003/12 = 5,400 × 104 mm4
The moment of inertia of the tension flange about y-axis,
Ift = 20 × 2003/12 = 1,333 × 104 mm4
hy = distance between the shear centres of the two flanges
= 500 + 12 + 10 = 522 mm
yj = 0.8 (2βf –1) hy / 2 = 0.8 (2 × 0.8 –1) × 522/2 = 125 mm
It = torsion constant
Iw = warping constant = (1– βf) βf Iy hy2
= (1− 0.8) × 0.8 × 6,740.5 × 104 × 5222 = 2.94 × 1012 mm6
LLT = effective length for the lateral torsional buckling (from Table 5.1)
= 1.2 × Span = 1.2 × 4,000 = 4,800 mm
Design bending compressive strength
Elastic critical bending stress,
Checking of cross-section for vertical and lateral forces
The gantry girder is subjected to twisting in addition to bending about z- and y-axes since the lateral force acts at the top of the rail (Figure 6.7). The design twisting moment (Mt) is obtained by multiplying the factored lateral force with the distance between the shear centre of the entire section to the top of the rail.
When the lateral force acts at mid span, the twisting moment Mt also acts at mid-span. Hence, a twisting moment of Mt / 2 is produced in each half span of the gantry girder. It is assumed that this twisting moment is resisted only by the flanges. Therefore, horizontal shear forces (F) are developed in the flanges as shown in Figure 6.7 due to which each flange in the half span bends as a cantilever in the horizontal plane with a built-in end at the supports.
i.e. Fh = Mt /2
or F = Mt /2h
where h = The distance between the centroids of the flanges
= 500 + 12 + 10 = 522 mm
The design bending moment in each flange in the horizontal plane,
where Mdyf = the design bending strength of the lower flange about y-axis
Checking of cross-section for vertical and longitudinal forces
The distance of the centroid C of the cross section from the top (Figure 6.5),
The longitudinal force acts at the top rail and is equivalent to an axial compressive force of 12.6 kN acting at the centroid and a bending moment of 12.6 (105 + 225) / 103 = 4.2 kNm about the major axis through the centroid. As the gantry girder is subjected to a combined axial force and bending moment, the following interaction condition should be satisfied.
where N is the factored axial force and Nd is the design strength in compression due to yielding.
Check for shear
Web buckling and web crippling
- At support
The radius of the gyration of the web section,
From Table 4.3, for the buckling class c, fcd = 107MPa
Let the bearing length, b1 = 150mmn1 = 270mm
Using Eqn. 5.12,Fw = (150 + 270) × 12 × 107 = 539kN > 324 kNn2 = 20 × 2.5 = 50mmb1 + n2 = 150 + 50 = 200mm
Using Eqn.5.14, Fx = 200 × 12 × 250/1.1 = 545 kN > 324 kN
- At wheel load
b1 = 02n1 = 2 (105 + 24 + 250) = 758mm
2n2 = 2 × 2.5 (105 + 24) = 645mmFx = 645 × 12 × 250/1.1 = 1,759kN > 1.5 × 210kN
Connection between the flanges and web
Fillet welds are provided to connect the flanges to the web. The horizontal shear force per unit length for which the welds should be designed is given by
where Iz′ = moment of inertia of the section about z′ axis (Figure 5.5)
The maximum horizontal shear force per unit length on the welds connecting the upper flange to the web
The vertical shear force per unit length on the welds connecting the upper flange to the web when the wheel load is at support (Figure 6.9)
The resultant shear force on the welds connecting the upper flange to the web
If ‘s’ is the size of the weld,
or s = 4.9 mm
However, to connect 24 mm thick flange, the minimum size of the weld should be 6 mm. Hence, 6 mm size fillet welds as shown in Figure 6.5 may be provided throughout the length of the gantry girder to connect the flanges to the web. As the gantry girder is subjected to fatigue, intermittent welds should not be provided.
Assessment for fatigue
The maximum bending moment at the mid-span of the gantry girder under a service condition
(when wheel load is on support)
The maximum and minimum normal stresses at the mid-span in the bottom flange are given by
The normal stress range, f = fmax − fmin = 81 − 2.3 = 79 MPa
This is greater than 27 / γmft = 27/1.35 = 20 MPa
Also, expected number of stress cycles 2 × 106 are greater than 81,130 cycles.
So, a fatigue assessment for the normal stress range is needed.
From Table 26(b) of IS 800: 007, for illustration number 8, the normal fatigue strength of the detail for 5 × 106 cycles, ffn = 92 (detail category 92).
The design normal fatigue strength range for Nsc = 2 × 106 cycles is
μ = 1.0
γmft = 1.35
∴ ffd = 1.0 × 125/1.35 = 92.6 MPa
This is greater than the actual normal stress range of 79 MPa. So, OK.
The maximum shear force at the support of the gantry girder under the service condition
Vmin = 12/2 = 6kN (when no wheel load acts on gantry girder)
The maximum and minimum horizontal shear forces per unit length at the interface between the top flange and the web
The maximum and minimum shear stresses in the welds are given by
The shear stress range, τ = τmax − τmin = 46.4 − 1.3 = 45 MPa
This is greater than 27/γmft = 27/1.35 = 20 MPa
So, the fatigue assessment for the shear stress range is needed.
From Table 26(b) of IS 800: 2007, for illustration number 39, the shear fatigue strength of the detail for 5 × 106 cycles, τfn = 67 (detail category 67).
The design shear fatigue strength range for Nsc = 2 × 106 cycles is
This is greater than the actual shear stress range of 45 MPa. So, OK.
For the following problems, consider the grade of steel as E250
- Design a simply supported gantry girder of 6 m effective span to carry two cranes of the capacity of 100 kN each working in tandem. The weight of each crane excluding the crab is 150 kN and the weight of each crab is 20 kN. The weight of the rail is 300 N/m. The minimum approach of the crane hook is 1.0 m. The wheel base is 3.8 m. The height of the rail is 75 mm. Assume that the gantry girder is laterally unsupported. The expected number stress cycles = 2 × 106.
- Redo Example 6.1 assuming that the lateral support is provided to the compression flange throughout its length.
- Redesign Example 6.1 using the rolled I Section with a plate connected to the compression flange by bolts.
- Redesign problem 6.1 using the rolled I section with a plate connected to the compression flange by bolts.