1. Define the load cases that must be considered. For this purpose, use the British Standard BS EN 19913:2006 (retrieve it using the iCAT service of our University).
2. Find the position of the trolley for which the bending moment is maximized
3. Plot the bending moment diagram for the bridge girder
4. Find the position of the trolley for which the shearing force is maximized
5. Plot the shearing force diagram for the bridge girder
6. Find the position of the trolley for which the vertical deflection is maximized
7. Plot the deflection diagram for the bridge girder
8. For the defined load cases and the selected positions:
o Estimate the corresponding stresses
o Check adequacy with respect to the constraints.
Design Data Considered
An Overhead Traveling (OT) crane or rather commonly referred to as ‘bridge crane’ is normally configured by having parallel runways such that the gap between the rails is spanned by a horizontally traveling bridge. The degree of freedom is restricted only in the xy plane such that the bridge moves in the longitudinal direction. The crane has got a hoist which does the lifting of objects via electrically excited system. The trolley therefore moves horizontally on the beam bridge. Admittedly, these crane types have found wide applications in various industries such as shipping, machining and other industrial stations and whose major functionality is to lift and move heavy objects like wheels of bogies among others. For instance, in the rolling stock industry, the OT crane is used to facilitate replacement of old wheel and axles of bogies (Globalspec.com, 2018). The bridge is normally built using either plate welding or hot rolling of steel (Cranes, 2018). However, most of commercially available steel material comes from hot rolling which itself is an opportunity to provide sufficient structural integrity of the bridge. The strength of the beam must be adequate to carry the live and dead loads and endure the various stresses as the system is in operation (Directindustry.com, 2018). Therefore, hereinafter, a major design to size and select the dimensions of the bridge girder ensues. This is a preliminary design and analysis report whose aim is to: undertake the design calculation and illustrate NQM diagrams by applying the singularityfunctions methods and the equivalent stress techniques; then we base it in a reallife engineering application (Mathalino.com, 2018).
Notably, for the purpose of design constraints, we consider the following standards: BS EN 19913:2006 and BS EN 19936: 2007.
Consider the following given design data for the system:
Table 1: Given Design Data
PARAMETER 
VALUE 
Crane bridge span 
6m 
Lifting capacity 
3.2tn 
Trolley wheel base 
600mm 
Trolley Load distribution 
4060%, end beam approach at 100mm (on both sides) 
The system must operate within defined limits. Therefore, the following are the given constraints that define the operational limits:
 Maximum vertical displacement: 20mm or span/600 (whichever is smaller)
 Maximum developed stress: less than yield stress
 Partial Factor Of Safety (applied on yield stress): 1.15
 Fundamental frequency: greater than 1.2Hz
In the given system, the following are the assumptions that have been made to facilitate design of the crane system:
 The bridge girder is considered simply supported beam
 The loading is uniformly distributed but with varying positions along the rail , check figure 2 for the free body diagram of the load system (Brighthub Engineering, 2018)
 Lifting capacity is inclusive of the trolley weight and the hoist
 Cables and festoon pushbutton are of negligible weight
Defining the load cases being considered
For this purpose, we use the British Standard BS EN 19913:2006 (Shop.bsigroup.com, 2018). The provided standard designates the loading arrangement which is adopted in the system.
Finding the position of the trolleyfor which the bending moment is maximized
From the given data, we know that the rail shares the load in the ration of 40 to 60% hence correspondingly R1 and R2 are determined:
Design Calculation and Analysis
But Qr= total loading (equivalent to the lifting loading plus other weight components)= 3.2x9.81=31.392 (fix at 31.4kN)
R1= 0.4x31.4= 12.56kN
R2= 0.6x 31.4= 18.84kN
Next, we find the reactions at supports A and B shown in the FBD figure 2
Given the system is at static equilibrium such that:
Ray+Rby=R1+R2=Qr and with a little consideration from the FBD we see that the reactions are actually sharing the load equally hence: R1=R2= Qr/2= 31.4/2=15.7kN
We then find the Bending moment equation by considering moments about A:
Ma= 12.56x18.84(x+d1)+15.7x6
= 12.56x18.84x18.84d1)+94.2
Ma= 31.4x +82.896….(i)
For the position of maximum bending moment, we put equation (i) as being equal to 0 hence:
31.4(x) +82.896= 0; 31.4x=82.896, x= 2.64 (that is 2.64m from point A)
Plot of the bending moment diagram for the bridge girder
X 
BMx 
0 
82.896 
1 
51.496 
2 
20.096 
3 
11.304 
4 
42.704 
5 
74.104 
6 
105.504 
Figure 3: Bending moment diagram
Finding the position of the trolley for which the shearing force is maximized
For shear force, we set Vx +ve
Hence Va= 31.415.7= 15.7kN (shear force is constant all over the beam)
Plot of the shearing force diagram for the bridge girder
Shear force diagram (15.7kN) 

x 
Vx 

0 
15.7 

1 
15.7 

2 
15.7 

3 
15.7 

4 
15.7 

5 
15.7 

6 
15.7 
Figure 4: Shear force diagram
Finding the position of the trolley for which the vertical deflection is maximized
For Maximum Deflection, we use the double integration method as follows (Hsu, Weng and Yang, 2014):
From this expression: EId^{2}y/dx^{2} = M = 31.4x+82.896 we integrate it to obtain the slope:
Hence: EIdy/dx=31.4x^{2}/2+82.896x
EIdy/dx=15.7x^{2}+82.896x+C1
The boundary conditions: dy/dx=0, when x=0
Substituting in the above quadratic
0= 15.7(0)^{2}+82.896(0)+c1, C1= 0
Hence EIdy/dx= =15.7x^{2}+82.896x…. (ii)
Further integrating equation (ii) to obtain the deflection:
EIy =15.7x^{3}/3+82.896x^{2}/2+C2…. (ii)
EIy= 5.23x^{3}+41.448x^{2}+C2
Again invoking the boundary condition: at x=0, y=0
Substituting in (ii):
0=5.23(0)^{3}+41.448(0)^{2}+C2
0= +C2
Hence the slope equation: dy/dx =5.23x^{3}+41.448x^{2}… (iii)
For maximum deflection, we differentiate or rather we pick the slope equation (ii) and equate to zero hence:
EIdy==15.7x^{2}+82.896x=0
X(82.89615.7x)= 0
X=0, or x=82.896/15.7=5.22
We take the latter as it is more practical considering the system in figure 2 hence position is at 5.22m from point A
2.4.7 Plot of the deflection diagram for the bridge girder
x 
y 
0 
0 
1 
36.218 
2 
123.952 
3 
231.822 
4 
328.448 
5 
382.45 
6 
362.448 
Figure 5: Deflection equation diagram
For the defined load cases and the selected positions:
(a) Estimating the corresponding stresses
First we need to go back to the position at which maximum deflection occurs hence x=5.22 is the position and we substitute in the bending moment equation:
M= 31.4(5.22)+ 82.896= 81.012kNm
Now we need to size the beam hence from this expression, we can get the moment of Inertia I (Codecogs.com, 2018):
Selection of Girder Bridge Profile
M/I=E/R
I= MR/E
Taking E= 203GPa and R= 300/2=150mm (neutral axis)
I= 81.012 x 0.15/203x10^{9}= 5.986x10^{11}m^{4}
Fixing b at 0.3, d can be found since we know: I= bd^{3}/12 (assume rectangular section)
Hence d= {(12x 5.986x10^{11})/0.3}^{1/3}= 1.338m
Now, Ax= 6x 1.338= 8.028m^{2}
And Az= 0.3 x 6= 1.8m^{2}
Hence:
Local shear stress τ,Ed
= Qr/Ay= 31.4/(0.3x 1.338)= 78.226MPa
Longitudinal stress σx,Ed
= Qr/Ax= 31.4/1.8= 17.44MPa
Traverse stress σz,Ed
= Qr/Az= 31.4/8.028= 3.911MPa
Yield stress fy
=200MPa, fy/Ym= 200/1.15= 173.91MPa
(b) Checking adequacy with respect to the constraints.
Note: we check the Ultimate Limit State of the bridge girder using the following equation:
Where:
σx,Ed: design value of the local longitudinal stress at the point of consideration
σz,Ed: design value of the local transverse stress at the point of consideration
τEd: design value of the local shear stress at the point of consideration
fy: yield stress for the selected material
γMo: partial factor of safety
We substitute the respective values in the above expression of constraints:
Let us break into terms:
(ρx/f/γ)^{2}= (17.44/173.91)^{2}= 0.010056
(ρz/f/γ)^{2} = (3.911/173.91)^{2} = 5.057x 10^{4}
(ρz/f/γ) (ρx/f/γ)= 0.10028x0.02248= 4.4609
3(τ/f/γ) = 3(78.226/173.91) = 0.6069
Checking by substituting in the above expression of constraint:
=0.010056+ 5.057x 10^{4}4.4609+ 0.6069= 3.843<1
Hence the design is safe
Based on the design output and using a theoretical approach, we undertake a selection of a girder bridge profile. It must meet the given design specifications and be of minimum weight.
From Harrington hoists catalogue (Qu et al., 2014), we select the bridge girder: S15x42.9 with a capacity of 6.1m span and capacity of 3 to 5 tones
Conclusion
From the foregoing, we can conclude that the design output and the selected commercially available bridge girder match (bridges, 2018). In the calculation, a capacity of 3.2 ton and span of 6m was used to design the system by considering the structural integrity of the bridge girder. From checking of structural constraints, it has been established that the design is safe (Makeitfrom.com, 2018). However, it should be noted that the design did not consider various operational parameters like fracture toughness among others. The calculations only assumed static conditions. Therefore, in future design work of overhead travelling crane, dynamic conditions will have to be considered so that the loading system can aptly be established. Otherwise, as far as the given constraints are concerned, this design is safe and will operate sufficiently with minimum break downs.
References
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