## 6

## Gantry Girder

#### 6.1 Introduction

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.

#### 6.2 Loads on Gantry Girder

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.

#### 6.2.2 Lateral Loads

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.

#### 6.3 Web Buckling and Web Crippling

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.

#### 6.4 Deflection Check

The calculated maximum deflection of the gantry girder should not exceed the limits given in Table 1.6 of Chapter 1. The limit is span / 750 for EOT cranes of up to 500 kN capacity.

#### 6.5 Design for Fatigue

For the constant stress range, the actual normal stress range ƒ and shear stress range τ at a point of the structure subjected to *N _{sc}* cycles in life should satisfy

*and τ ≤ τ*

_{ƒd}*(6.1)*

_{ƒd}

where ƒ* _{ƒd}* and τ

*are the design normal fatigue strength range and the design shear fatigue strength range, respectively.*

_{ƒd}#### 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/γ* _{mft}* and the actual shear stress range τ ≤ 27/γ

*or if the number of stress cycles,*

_{mft}*N*< 5 × 10

_{sc}^{6}(27 / γ

*/ γ*

_{mft}*)*

_{m}ƒ^{3}

where γ* _{mft}* = partial safety factor for fatigue strength (given in Table 6.1)

γ* _{m}* = partial safety factor for load (usually 1.0).

Based on the consequences of fatigue failure, a component or a detail is classified as:

*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 ƒ* _{ƒd}* for

*N*life cycles is given by

_{sc}

where ƒ* _{f}* = normal fatigue strength range

in which ƒ* _{fn}* is the normal fatigue strength for 5 × 10

^{6}cycles given under the relevant ‘detail category’ of Table 26 of IS800:2007.

In a similar manner, the design shear fatigue strength range τ* _{ƒd}* for

*N*life cycles is given by

_{sc}

where τ* _{f}* = shear fatigue strength range

in which τ* _{fn}* is the shear fatigue strength for 5 × 10

^{6}cycles given under the relevant ‘detail category’ of Table 26 of IS800: 2007.

In general, the factor for thickness correction, μ* _{f}* = 1.0. When plates greater than 25 mm in thickness are joined together by a transverse fillet or butt welds, μ

*= (25/*

_{f}*t*)

_{p}^{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. μ

*= 1.0).*

_{f}#### 6.5.3 Stress and Stress Range Limitations

The maximum (absolute) value of the normal and shear stresses should not exceed the elastic limit (ƒ* _{y}*, τ

*) for the material under cyclic loading. The maximum stress range should not exceed 1.5ƒ*

_{y}*for normal stresses and for shear stresses.*

_{y}**Example 6.1**

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 × 10^{6}

Grade of the steel = E250

*Maximum wheel load*

The maximum weight that the crane can lift (the crane capacity) and the weight of the crab act together as shown in Figure 6.2(a). Taking moments about point B in Figure 6.2(a)

or

*R*= 336 kN

_{A}

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 k*N*/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 *V*_{z} = 1.5 × 5.5 = 8.25 kN (when the lateral force acts on the support).

*Longitudinal force*

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 ƒ* _{bd}* = 100 N/mm

^{2}

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, *β** _{b}* = 1.0.

The location of the plastic neutral axis *z*:

*y*= 200 × 20 + (500 −

*y*) × 12

or

*y*= 116.7mm

The plastic section modulus about *z*-axis

*Z _{pz}* = 300 × 24 × (116.7 + 12) + 116.7 × 12 × (116.7/2) + 383.3 × 12 × (383.3/2) + 200 × 20 × (383.3 + 10) = 3,463 cm

^{3}

The plastic section modulus about *y*-axis

*Z _{py}* = 2[(150 × 24 × 150/2) + (100 × 20 × 100/2) + (500 × 6 × 6/2)] = 758 cm

^{3}

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*, *V*_{1} is the shear force in the upper flange of the section and *V*_{2} 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 / I _{y}* where

*V = P*

Since the shear flow varies parabolically in the flanges, *V*_{2} = 2/3 × 200× 0.00148 P = 0.197 P

Taking moments about a point on the line of the action of *V*_{1}

*Pe = V*

_{2}

*h*= 0.197

*P*(12 + 500 + 10)

or

*e*= 103 mm

*Elastic critical moment, M _{cr}*

The elastic critical moment *M _{cr}* is calculated using the expression given in Annexure E of IS800:2007 for sections symmetric about the minor axis.

where

*c*_{1}, *c*_{2}, *c*_{3} = factors which depend on the type of loading and supports

*c*_{1} = 1.365, *c*_{2} = 0.553, *c*_{3} = 1.78 from Table 42 of IS 800:2007 for a simply supported beam with a concentrated load at the centre.

*K* = effective length factor = 0.75 (It is assumed that the ends are partially restrained against the rotation about the minor axis)

*K _{w}* = warping restraint factor = 1.0

*y _{g}* = 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,

*I _{fc}* = 24 × 300

^{3}/12 = 5,400 × 10

^{4}mm

^{4}

The moment of inertia of the tension flange about y-axis,

*I _{ft}* = 20 × 200

^{3}/12 = 1,333 × 10

^{4}mm

^{4}

*h _{y}* = distance between the shear centres of the two flanges

= 500 + 12 + 10 = 522 mm

*y _{j}* = 0.8 (2β

*–1)*

_{f}*h*/ 2 = 0.8 (2 × 0.8 –1) × 522/2 = 125 mm

_{y}*I _{t}* = torsion constant

*I _{w}* = warping constant = (1–

*β*

*)*

_{f}*β*

_{f}I_{y}h_{y}^{2}

= (1− 0.8) × 0.8 × 6,740.5 × 10^{4} × 522^{2} = 2.94 × 10^{12} mm^{6}

*L _{LT}* = effective length for the lateral torsional buckling (from Table 5.1)

= 1.2 × Span = 1.2 × 4,000 = 4,800 mm

Therefore,

*Design bending compressive strength*

Elastic critical bending stress,

*z*-axis,

*M*=

_{dz}*β*= 1.0 × 3,463 × 10

_{b}Z_{pz}f_{bd}^{3}× 194

*y*-axis,

*M*=

_{dy}*β*γ

_{b}Z_{py}f_{y/}_{m0}

^{3}× 250/1.1

*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 (*M _{t}*) 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.

*M*= 8.25 × 10

_{t}^{3}× (103 + 12 + 105) = 1.8 kNm

When the lateral force acts at mid span, the twisting moment *M _{t}* also acts at mid-span. Hence, a twisting moment of

*M*/ 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 (

_{t}*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 = M _{t}* /2

or *F = M _{t}* /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,

Since the gantry girder is subjected to a combined bending and twisting, the following interaction condition should be satisfied.

where *M _{dyf}* = 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) / 10^{3} = 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 *N _{d}* is the design strength in compression due to yielding.

*N*=

_{d}*A*

_{g}*f*/

_{y}*γ*

_{m0}= 17,200 × 250/1.1 = 3,909 kN

*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*,*f*_{cd}= 107MPaLet the bearing length,

*b*_{1}= 150mm*n*_{1}= 270mmUsing Eqn. 5.12,

*F*= (150 + 270) × 12 × 107 = 539kN > 324 kN_{w}*n*_{2}= 20 × 2.5 = 50mm*b*_{1}+*n*_{2}= 150 + 50 = 200mmUsing Eqn.5.14,

*F*_{x}= 200 × 12 × 250/1.1 = 545 kN > 324 kN*At wheel load**b*_{1}= 02*n*_{1}= 2 (105 + 24 + 250) = 758mmOK

2

_{n2}= 2 × 2.5 (105 + 24) = 645mm*F*= 645 × 12 × 250/1.1 = 1,759kN > 1.5 × 210kN_{x}OK

*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 *I _{z′}* = 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

The minimum 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* = *f*_{max} − *f*_{min} = 81 − 2.3 = 79 MPa

This is greater than 27 / *γ** _{mft}* = 27/1.35 = 20 MPa

Also, expected number of stress cycles 2 × 10^{6} 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 × 10^{6} cycles, *f _{fn}* = 92 (detail category 92).

The design normal fatigue strength range for *N _{sc}* = 2 × 10

^{6}cycles is

*f*=

_{fd}*μ*

_{r}*f*/

_{f}*γ*

_{mft}where

*μ* = 1.0

*γ _{mft}* = 1.35

∴ *f _{fd}* = 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

V_{min} = 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* = *τ** _{max}* −

*τ*

*= 46.4 − 1.3 = 45 MPa*

_{min}This is greater than 27/*γ** _{mft}* = 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 × 10^{6} cycles, *τ** _{fn}* = 67 (detail category 67).

The design shear fatigue strength range for *N _{sc}* = 2 × 10

^{6}cycles is

*τ*

_{fd}=

*μ*

_{r}*τ*/

_{f}*γ*

_{mft}

where

This is greater than the actual shear stress range of 45 MPa. So, OK.

#### Problems

*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 × 10
^{6}. - 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.