Design of a Trailer Adapted for Accommodation and Transport of Beehives

Abstract

There is relevant interest concerning beehives, taking into account climate change and its influence on bees’ behavior. A part of the industrial engineering sector is focusing on beekeeping applications. More specifically, the present study aims to develop a trailer for the transport of beehives adapted to be placed or fixed to a tractor or a vehicle trailer, with the objective of transporting the beehives safely and stably during transhumance. The proposed novel design relates to a trailer that incorporates a device for housing a rectangular section of the beehives, which can be adapted for fixing or housing in a vehicle or in a vehicle trailer. The device comprises a lower support structure, adapted to support a plurality of rectangular sections of beehives stacked horizontally on the lower structure, an upper frame adapted to house the beehives inside, and two or more connecting elements between the lower structure and the upper frame. The connection of the trailer with the device facilitates the loading and unloading of the beehives by mechanical means. The different parts have been designed as individual pieces and then assembly is carried out to achieve the complete design. This method of implementation is because the simulation of individual components is simpler and easier, since if it is carried out through assembly, the type of joint, such as welding, and the length of the weld would have to be indicated at each point of contact between components, along with its thickness and all the necessary parameters. Therefore, in those welding points, fixed fastenings are indicated and so will simplify it. In accordance with the individual creation of each part, its own load simulation has been carried out. Static analyses are performed taking into account structural elements of this proposed design, with restrictions and loads being established. The analysis, including upper bars and supports, has been completed with several situations. Based on stress values, deformations have been determined and calculations evaluated. The trays have been manufactured using flat steel bars and angled bars for the legs and support of the hives.

1. Introduction

Although traditionally beekeeping was developed statically, that is, the beekeeper placed his or her beehives in a single place in order to form a colony whose production would come from the flowering of the nearby environment, today transhumance in beekeeping is especially interesting [1,2,3,4,5]. Furthermore, the influence of climate change on bees’ behavior is recognized [6,7,8]. This means moving the hives from place to place looking for blooms or vegetation according to the seasons of the year, and according to the desired composition of the final product, etc., improving, in many cases, the quality and quantity of production of useful and valuable products, such as honey, and pollination.

The present study is part of the industrial engineering sector in beekeeping applications. More specifically, this study aims to develop a design of a trailer for transport adapted to be placed or fixed to a tractor or a vehicle trailer (Figure 1) with the aim of transporting the hives safely and stably during transhumance.

The types of beehives most frequently used in these transhumance beekeeping tasks are those called Langstroth and Layens (Figure 2), the first being of a vertical type and provided with frames and mobile supers, while the second is of a horizontal type (also of mobile frames), widely used in transhumance in several countries, such as Spain and France. These configurations favor the transportation of the hives, but are insufficient when it comes to transporting large quantities of them, as is currently the case on the most extensive farms.

To solve this problem, there are solutions which aimed at promoting and facilitating the transportation of beehives during transhumance. The literature on this subject is very scarce and, concerning designs, there are a lack of precedents. Patents are mainly known. Regarding the transport of beehives in trailers adapted to be coupled to a tractor or vehicle, patent application US4033620A [9] discloses a plurality of horizontal and vertical mobile frames that are sequentially coupled directly to a trailer and on which the hives are placed for transportation. This placement is done in rows, each with a height of two hives. In addition, some of the frames mentioned incorporate wheels that facilitate movement. The configuration of the hives inside the trailer allows it to be located in the field so that when the exit openings of the hives are opened, pollination can occur. This allows, when phytosanitary treatments are carried out or transported to another area, the opening of the hive can be closed, thus facilitating the safety of people and the survival of the bees.

On the other hand, the Utility Model ES1053438U [10] presents a trailer and a coupled structure that allows the transport and permanent accommodation of beehives during transhumance. The coupled structure can be used autonomously in the field. The trailer has a perforated bottom to facilitate ventilation of the beehives. As for the coupled structure, it allows the expansion of the different arms of the structure. The handling mechanism can be activated manually or hydraulically and consists of the sliding of trays carrying rows of hives. It is a three-dimensional mesh with a rectangular plan with two floors, which can be adapted according to the type of hive.

However, the use of the devices above described entails, in many cases, enormous physical effort on the part of the beekeeper, in addition to the fact that the load may shift during transport, with the risks that may occur for both the beekeeper and the driver of the vehicle, as for bees and production.

In the same way Salvachúa and Robles [11] present several devices for the placement on the ground and transport of hives, ranging from benches with folding legs where the hives are placed in the field, to bench and clamp systems for mechanized loading and even a container that facilitates the loading of hives in a transport vehicle. On this point, Salazar [12] reports in his final degree thesis that an Australian company uses a forklift with palletised load for loading and unloading trucks.

It is very interesting to mention the previous research reviewed by Wakgari and Yigezu [13] concerning the honeybee keeping constraints and future prospects. These authors indicated that the world market demand for honey (and other hive products) has tremendously increased in recent decades, since it is important for a wide variety of uses and applications. For this purpose, the protection of the bees and the environment is emphasized as well. Therefore, a design to develop adequate systems for the transport of beehives safely and stably during transhumance is a very important goal.

The present study proposes a design of a novel device, not described until now, which solves the aforementioned problem, allowing resistance, tensions and displacements of the structural material to be minimized. In this way, a minimum value is also taken into account in terms of the safety factor that must exist for safe and correct use of the adapted trailer. The analyses are completed with several situations. Based on stress values, deformations are determined and calculations are evaluated.

2. Description of Modelling

To resolve the limitations of the proposed designs described above, the purpose of the present study is to provide a novel device designed for transporting beehives, adapted to be easily placed in a vehicle. SOLIDWORKS SYSTEM, a software program using images, has been used as a tool for the proposed design. This program is very useful for machine designs. The device consists of a part that will be purchased directly from auxiliary companies and other components that will be designed. The latter are (1) body frame and floor; (2) impeller support; (3) spearhead; (4) side, front and back walls and (5) beehive tray.

The above parts will be made as individual pieces, and later an assembly will be made to see the complete design. This method of implementation is adopted because the simulation of individual components is simpler and easier, since if it is done through an assembly, the type of joint, such as welding, and the length of the weld would have to be indicated at each point of contact between components, with its thickness and all the necessary parameters. Therefore, in those welding points, it will indicate fixed fastenings and it will simplify it. Due to the individual creation of each part, its own load simulation will be carried out.

2.1. Frame and Body Floor

A frame measuring 6.300 × 2.000 mm² in size (length × width) has been created. This frame has an external structure formed by hollow structural rectangular tubes measuring 80 × 50 × 3 (mm³) (side × side × thickness) made of S-355 steel, with a total of 4 bars. The interior structure is made up of two sets of structural elements.

The first is where the ball impeller will be located, resting directly on the square formed by the four 80 mm UPE profiles, two perpendicular (those of shorter length) and two transversal (those of longer length). In addition, four square structural tubes of 30 × 50 × 3 (mm³) have been welded to better support the weight of the load in each interior corner of the square formed by the structural profiles mentioned above.

In the two lateral openings that are formed, structural profiles, IPN 80, have been installed, two in each opening. A tube of the same dimensions will be installed in the interior hole.

The second interior part will be the support for the rear leaf springs, to which the rear axle will later be attached. It will be mainly formed by two hollow structural rectangular tubes measuring 180 × 100 × 5 mm³, which will be placed transversally on the frame. In addition, three IPN 100 profiles will be added. They will be placed perpendicular to the two tubes mentioned above, and thus we will avoid permanent deformations in the sheet. To support the rear leaf springs we have placed four cuts of square tubes measuring 90 × 90 × 12 mm³, oriented to the vertical axis. There will be two elements in each cross tube.

The placement of these two large rectangular tubes and the four square tube cuts is due to the fact that the wheels are located below the frame, so that there is no contact between the tire and the structure when driving. It is also wants to prevent the doors on the side walls from having full travel when they are opened. The design of this element is found in Figure 3a–d.

A 4 mm thick S-355 steel sheet has been placed on the frame, supported by all the structural tubes. The size of the sheet will be equal to the dimensions of the exterior structure (Figure 3b,c).

All these structures will form the frame and will support the entire load, which we will demonstrate later (see later Section 3.5.1).

2.2. Ball Impeller Base

The purpose of this structure is to support the ball impeller that is attached to the frame and the coupling with the front axle, using the leaf springs. This base is made up of three types of structural elements (Figure 3d).

The first is the one that will be attached to the ball impeller. For this, a hollow rectangular structural tube measuring 140 × 50 × 8 mm³ has been used, forming a square where the lower ring of the ball impeller will rest. Four holes must be made, since this connection will be using bolts and not welding.

The second is four bars of hollow rectangular tubes measuring 50 × 30 × 2 (mm³) that will be placed in the corners, serving as support for the ball impeller ring.

The third structural element is the base that will be attached to the suspension leaf springs. It will be built by four cuts of hollow square tubes 90 × 90 × 10 (mm³), located in each corner of the square, and coinciding with the supports of the crossbow.

2.3. Upper Structure

This part is located on the frame, supported and welded on the steel plate that forms the floor of the body. This part of the design is divided into different parts (Figure 4):

• Main structure
• Front wall
• Back wall
• Side wall (in this case, it has two equal pairs)
• Side top bars
• Front top bars

These six elements have been designed according to the standards established for their subsequent homologation. Although each part is named individually, the design has been made in a single file, and in this way a single static study will be carried out, assuming the full forces of internal support, load, wind, etc.

Once this has been clarified, each wall will be described individually, in order to have a better visualization.

Figure 4. Details of a frame with bottom plate according to the present study: (1) main structure, (2) frontal wall, (3) rear wall, (4) side wall, (5) upper bars and (6) frontal upper bars.

Figure 4. Details of a frame with bottom plate according to the present study: (1) main structure, (2) frontal wall, (3) rear wall, (4) side wall, (5) upper bars and (6) frontal upper bars.

2.3.1. Main Structure

The main structure supporting all the walls and bars of the trailer is based on four rectangular tubes of 70 × 40 × 7 (mm³), located at each corner of the base of the frame, and two rectangular tubes of 80 × 40 × 7 (mm³), at the midpoint of the two sides of the trailer. These tubes will support the side and rear walls, as well as the side and front top rails. In the case of the front wall, it will be welded directly to the tubes, without allowing any movement.

As a large part of the side of the trailer was left uncovered, 3 mm-thick supports were installed in which to place rectangular bars, which will provide greater security to the trailer in the event of the load tipping over (Figure 5a). They have been installed in three different arrangements:

• On the front tubes of the main structure there will be a total of six supports, three on each perpendicular face, as can be seen in the picture.
• In the middle tubes of the main structure there will also be a tower of six supports, although this time in opposite ways.
• Only three supports will be installed on the rear tubes, just in front of the middle tubes.

Figure 5. (a–g): Details of a frame with top plate, according to the present study.

Figure 5. (a–g): Details of a frame with top plate, according to the present study.

As a safety element, vertical plates have also been installed to support the movable walls, thus avoiding forcing and damaging the hinges, as far as possible. These plates will have a thickness of 3 mm (Figure 5b).

2.3.2. Side Wall

The side wall will be the largest of all the walls installed on the trailer. It will be possible to move and dismantle it because three hinges will be installed at the bottom. These will be welded or bolted to the square tube of the wall structure and to the outer rectangular tube of the frame. Making these walls movable will facilitate the work of loading and unloading (Figure 5c). This part of the trailer has been designed with different elements.

• The main structure is based on a rectangle made of 40 × 40 × 5 (mm³) hollow square tubes. In addition, three 35 × 35 × 4 mm³ hollow square tubes have been installed in the inner rectangle, to give the structure a higher load resistance (Figure 5d).
• At the top, a railing has been installed with tubing of the same dimensions as the door structure, except for the thickness. This will give more height to the wall and will be a further reinforcement. It will consist of a total of five vertical tubes and a single horizontal tube that will rest on the previous ones and will be welded to it. The two vertical tubes and the upper horizontal tube, which forms the upper main structure, will have a thickness of 4 mm, while the remaining five vertical tubes will have dimensions of 30 × 30 × 1.5 mm³.
• On the outside, a 2 mm-thick folded metal sheet will be installed. This will be welded to the inner sides of the main structure. As it is a folded sheet, the stress resistance will be increased, giving more reliability to the trailer. In addition, the hitches that will serve as lashing points will be installed on this sheet.
• On the inside, a 3 mm-thick flat plate has been installed. This will be supported and welded to the inner sides of the main structure. It will also be supported on the folded part of the outer sheet and the inner tubes (Figure 5e).

2.3.3. Front Wall

The front wall will be closest to the vehicle. This wall will be directly welded to the base and the vertical tubes of the main structure (Figure 5f). This will be fixed, as this is where the drawbar end of the trailer is installed and, therefore, it will act as a stop if it is able to be unfolded, making it almost unusable. For this reason, it will be welded and will provide greater resistance to the force of the load in the event of overturning or braking.

The wall has been designed in the same way as the side wall described above, with the following small differences:

• The length will be shorter than the side wall.
• It will consist of two horizontal tubes in the inner structure as it will be directly welded to the main structure and will require fewer elements (Figure 5g).
• Only five couplings will be installed, respecting the spacing of each coupling and their dimensions.
• The arrangement of the top rail will have the crosspieces at a smaller equidistance and with a total number of three elements.

2.3.4. Rear Wall

The last wall will differ only in the elimination of the railing formed by the square tubes at the top, and those extended from the main square structure. The dimensions of the tubes and sheet metal used will be identical, varying only in length (Figure 6a).

2.3.5. Upper Side Bars

These hollow rectangular bars will be placed directly on the supports welded to the tubes of the main structure. They will be made of 200 × 40 × 3 mm³ aluminum tubes, with plastic stops installed at each end of the bar. We will have a total number of 12 side bars, 6 on each side (Figure 6b). In addition, a 2 mm S-355 steel U-bar is installed to reinforce the aluminum element against loads.

To protect the integrity of the bar, two buffers are fitted along both rubber edges, to prevent damage to the hives.

2.3.6. Front Top Bars

These bars only differ from the side bars in the length, which will be shorter. We will have a total of three units.

2.4. Spearhead

The function of the drawbar end is to join the trailer assembly made up of the frame, the base of the ball bearing, and the walls with the tractor. This drawbar end is made up of the visible elements (Figure 6c):

• Drawbar
• Lugs
• Jointing set

The connection to the ball impeller is made by means of lugs welded to the four square tube cut-outs installed at the bottom. These lugs have a hole on each side. These same holes (of equal diameter) will also be made at each end of the structure tube. By concentrically positioning the holes of the lugs and the tubes, a bolt will be passed through them in order to connect the components. The set of elements described is shown in Figure 7a.

2.4.1. Drawbar

The main structure of the drawbar has been designed with a S-355 structural steel tube, as has the frame (Figure 7b). The drawbar has been divided into two groups, where it will use two types of tubes of different thicknesses. The first group is formed by the tubes that connect to the trailer, whose tube dimensions are 50 × 50 × 6 mm³, consisting of a total of four tubes of variable length. The second group will form the inner tee of the drawbar, with tube dimensions of 50 × 50 × 3 mm³, with a total of two units. At the end of the drawbar, the system of connection with the trailer will be located.

2.4.2. Lugs

The lugs have been designed to be able to support the towing load of the trailer with its load, with the dimensions shown in Figure 7c and a thickness of 5 mm. It will have a total of two, for each end of the drawbar end.

2.4.3. Joint Assembly

In order to join the lug and the drawbar structure, a set consisting of a nut, a hexagonal screw and two washers will be placed between the inner face of the lug and the outer face of the tube that forms the drawbar (Figure 7d).

An overall dimension of M24 mm (inside diameter) has been chosen, the same as the holes drilled in the other components. This diameter has been chosen in relation to the calculations made in Part 4 of the detailed description of the invention. The bolts will be 90 mm long without full thread, the washers will be single washers with thickness 4.6 mm and outside diameter 44 mm and a heavy nut with a washer face.

As with the lugs, we will need two units of each element.

2.5. Hive Tray

This element (Figure 8) has been designed to facilitate the movement of hives, both for loading and unloading them from the trailer, as it will be possible to move in quantities of five units at a time. In addition, this element further secures the hives once loaded onto the trailer.

It has been manufactured using 60 × 3 (mm²) flat steel bars, consisting of a total of 11 units of various sizes. In addition, angled bars with dimensions of 40 × 3 (mm²) have been used for the legs and support of the hives. In this case, we have a total of eight units of varying sizes.

3. Calculations Concerning the Proposed Design

3.1. Load Calculation

3.1.1. Load on the Frame and Sheet

To carry out the static analysis, we must calculate the load value that is going to be exerted on the sheet metal floor. First of all, the load is defined as the maximum force it will have when fully loaded, taking into account the weight of the walls, trays and hives.

The weights of the different elements are known:

Walls = 1000 kg

Trays = 48 kg/Unit

Hives = 50 kg/Unit

It has been established that the maximum capacity of the trailer will be 24 trays, with 5 hives per tray. Therefore, the total load will be

Qtotal = 1000 kg + (48 kg/unit × 24 units) + (50 kg/unit × 120 units) = 8200 kg = 80,442 N

To avoid possible failures in more unfavorable situations, the following total load will be placed:

Q = 81,000 N

3.1.2. Pressure on Ball Impeller Base

As in the previous section, the maximum continuous pressure that the base of the designed ball impeller will withstand must be calculated and it must be checked whether they comply. Initially, the load exerted on the ball impeller must be obtained, which is where the front axle is located. To do this, through a static study, the total weight of the structure can be deducted:

Q = 81,000 N

Frame = 1000 kg = 9818 N

$$Q_2=\mathrm{81,000}+9818=\mathrm{90,818} \mathrm{N} \cong \mathrm{90,900} \mathrm{N}$$

The free body study is carried out (Figure 9a):

It is known that the center of mass will be located at the coordinates x and z = (−50 0 0), placing the center of coordinates in the center of the frame. Thus, the position of the point charge M is calculated. The forces can be added, as follows:

$$\sum F_X=0 ;0=0$$

$$\sum F_Y=0 ; R_{ya}+R_{yb}=W ; R_{ya}+R_{yb}=\mathrm{90,900} N$$

The sum of moments is as follows:

$$\sum M=0 ; R_{ya}\times 950+R_{yb}\times 5300-M\times 3000=0 ;$$

$$R_{ya}\times 950+R_{yb}\times 5500=\mathrm{90,900}\times 3000$$

Using a system of two equations with two unknowns the desired values can be calculated:

$$\left.\begin{array}{c}{R}_{ya}+R_{yb}=\mathrm{90,900} \mathrm{N}\\ R_{ya}\times 950+R_{yb}\times 5300=\mathrm{90,900}\times 3000\end{array}\right\}$$

$$R_{ya}=\mathrm{48,062} \mathrm{N} ; R_{yb}=\mathrm{42,837} \mathrm{N}$$

As it can be seen, the values obtained are almost identical, due to the similarity in position between the front axle and the rear axle. It thus takes a force of 48,062 N for this case, and it will be used it to carry out the subsequent simulation.

3.2. Unfavorable Situations

In this section, the most unfavorable situations that the trailer can be affected by during its journey will be studied.

Bump

This is one of the unfavorable situations that it is going to encounter during cargo transportation: the passage of the trailer over a rut or pothole (Figure 9b). It will be a very common situation, since you will be traveling through complicated terrain.

It will be taken as involving a pothole measuring 20 cm deep. It will consider the data of the fully loaded trailer, where it will obtain the most unfavorable results possible. With the pothole, only vertical forces are present, with the drop of 0.2 m (y) in height of the pothole.

In order to carry out the calculation, a series of initial values must be assumed. Zero is taken as an initial velocity (V₀), since the speed of the trailer is horizontal. The acceleration will be 9.81 (m/s²), taking into account the action of gravity. The time it would take to fall is calculated, using the equation of uniformly accelerated rectilinear motion (MRUA).

$$y=y_0+V_0\times t+{\displaystyle \frac{1}{2}}\times a\times t^2 \left(\mathrm{m}\right);$$

$$t=\sqrt{{\displaystyle \frac{2\times \left(y-y_0-V_0\times t\right)}{a}}}=\sqrt{{\displaystyle \frac{2\times \left(0.2(\mathrm{m})\right)}{9.81 (\mathrm{m}/{\mathrm{s}}^2)}}}=0.202 \mathrm{s}$$

The final velocity before hitting the ground, using the principle of conservation of energy, is now obtained:

$$W=∆E_c$$

$$W=-∆E_p$$

$$∆E_c=-∆E_p$$

$$E_{c0}+∆E_{p0}=E_{c1}+E_{p1}$$

$$0={\displaystyle \frac{1}{2}}\times m\times v^2+m\times \mathrm{g}\times y$$

$$V_1=\sqrt{2\times \mathrm{g}\times y}=\sqrt{2\times 9.81 (\mathrm{m}/{\mathrm{s}}^2)\times 0.2 \left(\mathrm{m}\right)}=1.98 \mathrm{m}/{\mathrm{s}}^2$$

The potential energy is now obtained, in order to later calculate the vertical force. It will use the greatest force that it will have on one of the two axles, in this case the rear one, Ray, with Q = 48,062 N = 4899 kg. The potential energy equation used previously is

$$E_p=m\times \mathrm{g}\times y (\mathrm{N}\mathrm{m})$$

$$E_p=4899\left(\mathrm{k}\mathrm{g}\right)\times 9.81\left({\displaystyle \frac{\mathrm{m}}{{\mathrm{s}}^2}}\right)\times 0.2\left(\mathrm{m}\right)=9611 (\mathrm{N}\mathrm{m})$$

Finally, the vertical force (V_y) can be obtained through the following equation:

$$V_{y total}={\displaystyle \frac{E_p}{y}}={\displaystyle \frac{9611 (\mathrm{N}\mathrm{m})}{0.2 (\mathrm{m})}}=\mathrm{48,059} \mathrm{N}$$

This value will be used in the static analysis for this situation on each axis, and divided if necessary by the supports it has with the frame.

If the four supports of the two leaf springs of the rear axle are considered, we have

$$V_y={\displaystyle \frac{V_{y total}}{4}}={\displaystyle \frac{\mathrm{48,059} (\mathrm{N})}{4}}=\mathrm{12,014} N/Support$$

3.3. Trailer Stability

3.3.1. Maximum Lateral Inclination

As it was mentioned previously, the trailer is of an agricultural type, so it will circulate on terrain with large unevenness, both frontal, as was calculated previously, and lateral. Calculating the maximum lateral inclination will help to ensure that the trailer does not overturn. Through the complete assembly in SolidWorks, the coordinates of the center of mass of the fully loaded trailer are obtained (Figure 10). It is taken into account that the coordinate origin would be located in the front left corner of the frame.

The specific coordinates of the center of mass are

X, Y, Z = (3080, 994, −998)

Knowing this information, the free body diagram can be made (Figure 11a), which will be produced during circulation on laterally inclined terrain.

Where

P: trailer weight (N), P_x: horizontal component of P (N), P_y: vertical component of P (N), N: normal force to the ground (N), Fy: friction force (N), and β: maximum inclination angle (ᵒ).

A value of p = 10,200 kg is applied, slightly increasing the actual weight, to obtain a larger dimensioning to avoid failures due to unexpected unfavorable situations. Observing the force diagram (Figure 11a), the following calculations can be performed to obtain the maximum inclination angle.

A sum of moments is considered:

$$\sum M_0=0; P_x\times 1916-P_y\times 920=0;$$

where

$$P_x=P\times sen \left(\beta \right)=\mathrm{100,000}\times sen (\beta )$$

$$P_y=P\times cos \left(\beta \right)=\mathrm{100,000}\times cos (\beta )$$

This leaves

$$\mathrm{100,000}\times sen \left(\beta \right)\times 1916-\mathrm{100,000}\times \cos \left(\beta \right)\times 920=0;$$

$${\displaystyle \frac{sen \left(\beta \right)}{\cos \left(\beta \right)}}={\displaystyle \frac{920}{1916}}; tg \left(\beta \right)=0.48;$$

β = 25.6°

In this way, a maximum inclination angle of 25.6° has been obtained. If the trailer exceeded this value, it would cause it to overturn, with all the dangers and material damage that this entails. It should be noted that this angle has been obtained assuming that the load is perfectly tied, without risk of detachment, since this would cause the trailer to overturn at a smaller angle.

The percentage of inclination can be obtained if we want to see the value obtained in a more representative way. Using trigonometry, and with the angle obtained, and taking the base (adjacent leg) as 100%, the percentage of inclination (opposite leg) will be obtained, as represented in Figure 11b:

tg (25.6) = d/100; d = tg (25.6) × 100;

D = 47.9%

It will be necessary to ensure that the terrain on which the trailer can circulate may have a maximum gradient of 47.9%.

3.3.2. Minimum Allowable Turning Radius

In this section, a particular study on the stability of the trailer in the event of a sharp curve or turn is considered. These are common situations that can occur due to the terrain that it will be traveling through and all the obstacles which will have to be overcome. There is a minimum radius, which will depend on factors such as speed, since the higher the speed, the easier it will be for the trailer to overturn.

We are going to make free-body diagrams to study the situation during the turn. Two diagrams will be prepared, one with a top view (Figure 12a) and another with a rear view (Figure 12b). By completing these two diagrams, it will be possible to see all the forces that act on the trailer during the journey, where

• P: Trailer weight (N)
• N: Normal force to the ground (N)
• Fr: Friction force (N)
• Fc: Centripetal force (N)
• V: Vehicle speed (m/s²)
• R: Turning radius (m)

Figure 12. Showing the free body diagram: minimum upper swing (a) and minimum rear swing (b).

Figure 12. Showing the free body diagram: minimum upper swing (a) and minimum rear swing (b).

The data to study will be R, which will depend on the variable V, as has been previously described. For this reason, we are going to analyze four different situations, giving a value to V and thus obtaining its corresponding R.

It is known that the trailer will be able to circulate, according to regulations, at a maximum speed of 25 km/h, but considering the most unfavorable case, where the tractor can circulate at a maximum speed of 40 km/h, the same value will be given to the trailer. Therefore, the situations to study will be the following: V = 10 km/h, V = 30 km/h, V = 20 km/h and V = 40 km/h.

To begin with, moments are added (Figure 12b):

$$\sum M_0=0; F_c\times 1916-P\times 0.92=0;$$

where the value of the centripetal force is obtained through the following equations:

$$F_c=m_t\times a_c$$

$$a_c={\displaystyle \frac{V^2}{R}}$$

This leaves, knowing that $m_t$ = 10,200 kg,

$$F_c=m_t\times {\displaystyle \frac{{\mathrm{V}}^2}{\mathrm{R}}}=\mathrm{10,200} \mathrm{k}\mathrm{g}\times {\displaystyle \frac{{\mathrm{V}}^2}{R}}$$

Substituting into the moment sum equation, and solving for R,

$$\mathrm{10,200} \mathrm{k}\mathrm{g}\times {\displaystyle \frac{{\mathrm{V}}^2}{\mathrm{R}}}\times 1.916-\mathrm{100,000}\times 0.92=0;$$

$$R={\displaystyle \frac{\mathrm{10,200}\times 1.916\times {\mathrm{V}}^2}{\mathrm{100,000}\times 0.92}}={\displaystyle \frac{\mathrm{19,543}\times {\mathrm{V}}^2}{\mathrm{92,000}}}$$

Then, the equation of R dependent on V is obtained.

Using the four cases described above, with the variable V, the corresponding R values are obtained:

$$R={\displaystyle \frac{\mathrm{19,543}\times {\mathrm{V}}^2}{\mathrm{92,000}}}$$

• V = 20 km/h = 5.55 m/s → R = 6.53 m
• V = 25 km/h = 6.94 m/s → R = 10.22 m
• V = 30 km/h = 8.33 m/s → R = 14.72 m
• V = 40 km/h = 11.11 m/s → R = 26.2 m

This has the maximum turning radii that it can make for the different speeds.

3.4. Spear Set

3.4.1. Lance

According to UNE-EN 1853:1999+A1:2010 standard [14], the trailer is dragged, not semi-supported, so the drawbar only transmits longitudinal forces. The vertical effort can be neglected. This section aims to calculate the maximum stress that the lance will have to withstand during use. To carry out this calculation in the design, the most unfavorable conditions must be considered. This would be during the climb of the highest allowable slope, with a full load.

These data would be

Load (P) = 10,200 kg; Slope (p) = 18%

The chosen load would be the sum of bodywork, cargo and additional elements plus a margin of error in the case where the maximum suggested load is exceeded. A slope with an 18% gradient is assumed, as established by UNE-EN 1853:2018+AC standard [15], where this data will be taken to carry out the design safety tests.

As described, there would be the following scenario:

The variables can be described as follows (Figure 13a):

FH: horizontal force; FR: friction force; Q: trailer weight; PY: vertical component of P; Px: horizontal component of P and α: angle of inclination

Figure 13. Shows the free-body diagram (maximum slope) (a) and the force diagram on the lance (b).

Figure 13. Shows the free-body diagram (maximum slope) (a) and the force diagram on the lance (b).

The calculation of the angle α can be the first one. Using trigonometry, in the right triangle (Figure 13a), this value is easily obtained:

α: arctg(α) = 18/100; α = 10.2°

The objective is to calculate the F_H value, since it is the force that the drawbar must withstand while driving on a slope. This calculation would come through the sum of forces:

$$\sum {\mathrm{R}}_{\mathrm{x}}=0 ; {\mathrm{F}}_{\mathrm{H}}-{\mathrm{F}}_{\mathrm{R}}-{\mathrm{P}}_{\mathrm{x}}=0 ; {\mathrm{F}}_{\mathrm{H}}={\mathrm{F}}_{\mathrm{R}}+{\mathrm{P}}_{\mathrm{x}}$$

We move on to calculate the values of the unknowns that are in the previous equation.

Starting with the components of weight (P):

$$\mathrm{P}=m\times g=\mathrm{10,200} \mathrm{k}\mathrm{g}\times 9.81 \mathrm{m}/{\mathrm{s}}^2=\mathrm{100,000} \mathrm{N}$$

$${\mathrm{P}}_{\mathrm{x}}=\mathrm{P}\times sen\left(\alpha \right)=\mathrm{100,000} (\mathrm{N})\times sen\left(10.2\right)=\mathrm{17,708} \mathrm{N}$$

$${\mathrm{P}}_{\mathrm{y}}=\mathrm{P}\times \cos \left(\mathsf{\alpha }\right)=\mathrm{100,000} (\mathrm{N})\times \cos \left(10.2\right)=\mathrm{98,419} \mathrm{N}$$

It can be continued with the calculation of F_R. This value would come through the following equation:

$${\mathrm{F}}_{\mathrm{R}}={\displaystyle \frac{\mathsf{\rho }}{\mathrm{R}}}\times \mathrm{N} (\mathrm{N})$$

The variables are as follows:

$\mathsf{\rho }$: rolling coefficient for rubber-earth, grass, and mud = 0.06 m; N: normal component of weight, P_y; A: wheel radius

$$F_R={\displaystyle \frac{\rho }{R}}\times N={\displaystyle \frac{0.06 (m)}{0.3945 (m)}}\times \mathrm{100,000} \left(N\right)=\mathrm{15,209} N$$

Having calculated the two unknowns, we can proceed to obtain the value of F_H:

$$F_H=F_R+P_x ; F_H=\mathrm{15,209}+\mathrm{17,708}=\mathrm{32,917} \mathrm{N}$$

In this way we obtain the value of the force that will be exerted on the drawbar, more specifically at the tip, where the coupling with the tractor will be located.

Looking at the design, it shows that the lance will be attached to the base of the impeller by means of two contacts. Therefore, the force obtained previously is divided in half in each of the two contacts (Figure 13b).

The spearhead design has been made with hollow square structural tubes. To verify that this structural member will have sufficient strength to withstand the load calculated above, a series of calculations are performed. The basic document SE-A is used [16].

The lance will be subjected to traction and compression forces, either when ascending or descending slopes. Given the situation of the slope, it will be faced with a case of tensile stress. It is ensured that our structure is safe an acceptable SDS is obtained, through the following:

$${\displaystyle \frac{\mathrm{A}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g} \mathrm{e}\mathrm{f}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{t}}{\mathrm{R}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t} \mathrm{e}\mathrm{f}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{t}}}={\displaystyle \frac{{\mathrm{N}}_{\mathrm{E}\mathrm{d}}}{{\mathrm{N}}_{\mathrm{t},\mathrm{R}\mathrm{d}}}}\le 1$$

with

$${\mathrm{N}}_{\mathrm{E}\mathrm{d}}=\mathrm{32,917} \mathrm{N}$$

The value of ${\mathrm{N}}_{\mathrm{t},\mathrm{R}\mathrm{d}}$ is obtained through the equation considered from the document mentioned above, knowing that bars are working in tension.

$${\mathrm{N}}_{\mathrm{t},\mathrm{R}\mathrm{d}}\le {\mathrm{N}}_{\mathrm{p}\mathrm{l},\mathrm{R}\mathrm{d}}=\mathrm{A}\times {\mathrm{f}}_{\mathrm{y}\mathrm{d}}$$

where, according to the safety document, there is an indication that the pure tensile strength of the bar (${\mathrm{N}}_{\mathrm{t},\mathrm{R}\mathrm{d}}$), will be the plastic resistance of the gross section (${\mathrm{N}}_{\mathrm{p}\mathrm{l},\mathrm{R}\mathrm{d}}$).

This leaves

$${\mathrm{N}}_{\mathrm{t},\mathrm{R}\mathrm{d}}={\mathrm{N}}_{\mathrm{p}\mathrm{l},\mathrm{R}\mathrm{d}}=\mathrm{A}\times {\mathrm{f}}_{\mathrm{y}\mathrm{d}}$$

where ${\mathrm{N}}_{\mathrm{p}\mathrm{l},\mathrm{R}\mathrm{d}}$ is the pure tensile strength of the bar (N); A is the section area $\left({\mathrm{m}}^2\right)$; and ${\mathrm{f}}_{\mathrm{y}\mathrm{d}}$ is the strength of steel calculation $(\mathrm{N}/{\mathrm{m}}^2)$.

We will take ${\mathrm{f}}_{\mathrm{y}\mathrm{d}}$ as

$${\mathrm{f}}_{\mathrm{y}\mathrm{d}}={\displaystyle \frac{{\mathrm{f}}_{\mathrm{y}}}{{\mathsf{\gamma }}_{\mathrm{M}1}}}$$

where ${\mathrm{f}}_{\mathrm{y}}$ is the elastic limit of the material and ${\mathsf{\gamma }}_{\mathrm{M}1}$ is the partial safety coefficient related to instability phenomena (${\mathsf{\gamma }}_{\mathrm{M}1}$ = 1.05).

In this way, the final equation can be obtained:

$${\mathrm{N}}_{\mathrm{t},\mathrm{R}\mathrm{d}}={\mathrm{N}}_{\mathrm{p}\mathrm{l},\mathrm{R}\mathrm{d}}=\mathrm{A}\times {\displaystyle \frac{{\mathrm{f}}_{\mathrm{y}}}{{\mathsf{\gamma }}_{\mathrm{M}1}}} (\mathrm{N})$$

The lance is designed using 50 × 50 × 6 (mm³) hollow square structural tubes, with S-355 material.

The section area of the square tube can be obtained from a catalog:

$$\mathrm{A}=10\times {10}^{-4} ({\mathrm{m}}^2)$$

It is obtained like this:

$${\mathrm{N}}_{\mathrm{t},\mathrm{R}\mathrm{d}}={\mathrm{N}}_{\mathrm{p}\mathrm{l},\mathrm{R}\mathrm{d}}=10\times {10}^{-4} ({\mathrm{m}}^2)\times {\displaystyle \frac{355\times {10}^6(\mathrm{N}/{\mathrm{m}}^2)}{1.05}}=\mathrm{338,095} (\mathrm{N})$$

Finally, the previous relationship can be calculated:

$$(\mathrm{A}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g} \mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s})/(\mathrm{R}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g} \mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s})={\displaystyle \frac{{\mathrm{N}}_{\mathrm{E}\mathrm{d}}}{{\mathrm{N}}_{\mathrm{t},\mathrm{R}\mathrm{d}}}}={\displaystyle \frac{\mathrm{32,917} \left(N\right)}{\mathrm{338,095} \left(N\right)}}\le 1$$

0.097 ≤ 1

It can be visualized with the Security Factor (FDS), giving a result of

$${\displaystyle \frac{\mathrm{338,095} \left(\mathrm{N}\right)}{\mathrm{32,917} \left(\mathrm{N}\right)}}\ge 1;\mathrm{F}\mathrm{D}\mathrm{S}=10.2$$

It has been decided to use the tubes that form the T of the lance with those having equal dimensions, but changing the thickness, using 3 mm. The resistance to effort during use can be calculated in the same way.

With 50 × 50 × 3 mm tubes, it is

$$A=5.33\times {10}^{-4} ({\mathrm{m}}^2)$$

We obtain the following:

$${\mathrm{N}}_{\mathrm{t},\mathrm{R}\mathrm{d}}={\mathrm{N}}_{\mathrm{p}\mathrm{l},\mathrm{R}\mathrm{d}}=5.33\times {10}^{-4} ({\mathrm{m}}^2)\times {\displaystyle \frac{355\times {10}^6(\mathrm{N}/{\mathrm{m}}^2)}{1.05}}=\mathrm{180,204} (\mathrm{N})$$

Finally, the previous relationship is calculated:

$$(\mathrm{A}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g} \mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s})/(\mathrm{R}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g} \mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s})={\displaystyle \frac{{\mathrm{N}}_{\mathrm{E}\mathrm{d}}}{{\mathrm{N}}_{\mathrm{t},\mathrm{R}\mathrm{d}}}}={\displaystyle \frac{\mathrm{32,917} \left(\mathrm{N}\right)}{\mathrm{180,204} \left(\mathrm{N}\right)}}\le 1$$

0.18 ≤ 1

It can be visualized with the FDS, giving a result of

$${\displaystyle \frac{\mathrm{180,204} \left(N\right)}{\mathrm{32,917} \left(N\right)}}\ge 1;\mathrm{F}\mathrm{D}\mathrm{S}=5.4$$

Important consequences can be deduced, as follows. Analyzing the results obtained, it can be seen how the lance structure will more, rather than withstanding the maximum stresses to which it has theoretically been subjected.

The two types of tubes used in the design will be sufficient to ensure safety. There will be a slight oversizing in the structural part, with FDS = 10.2, where the union with the lugs will be made through the union assembly, since it is the area where it can have the most problems with respect to the effort.

For the part where the connection point with the tractor will be located, the thickness will be reduced, since it will continue to have a higher resistance, with FDS = 5.4.

It must be taken into account that the study has been carried out on firm surfaces, not to mention that during the work we may encounter stones and various obstacles that will increase the load that has been previously obtained. Therefore, the slight oversizing of the lance will ensure that it withstands more unfavorable conditions without problems.

3.4.2. Screw

As described, the lance is attached to the frame by means of two lugs. But, at the union between the structure of the lance and the lugs, they are joined by means of two bolts of ∅ = 24 mm (Figure 14).

Shear Resistance

In this section, we are going to calculate the shear stress resistance of the screws at the maximum load. This will avoid errors like those in Figure 15:

The trailer will be placed in the most unfavorable scenario. It will be the case of ascending a slope, as mentioned in the previous section, where it will have a load of 28.850 N. Distributing this between the two bolts, it will finally have a load of 14.425 N in each screw. This force will be divided between the faces of each element that is in contact with the screw. In the case of the lug attached to the screw by means of two walls, therefore,

$${F_1={\displaystyle \frac{F_H}{2}}=\mathrm{32,197} \mathrm{N} ; F}_2={\displaystyle \frac{F_H}{4}}=\mathrm{16,458} \mathrm{N}$$

In this way, we have the following force diagram (Figure 16a):

Looking at the diagram obtained (Figure 16a), it can be known that it is facing double shear failure, as it can be seen in Figure 16b.

It is known that the shear stress that the screw will have under the load will come from the following equation:

τ = F/2A (MPa)

where A is the area of the section of the chosen screw.

$$\mathrm{A}=\pi \times r^2={\displaystyle \frac{\pi \times D^2}{4}} \left({\mathrm{m}}^2\right);$$

$$\mathrm{A}={\displaystyle \frac{\pi \times {0.024}^2}{4}}=4.5\times {10}^{-4}\left({\mathrm{m}}^2\right)$$

In this way, we have

$$\tau ={\displaystyle \frac{\mathrm{32,917} (N)}{2\times 4.5\times {10}^{-4} ({\mathrm{m}}^2)}}=37 (\mathrm{M}\mathrm{P}\mathrm{a})$$

The material used, S-355 steel [17], has an elastic limit of 355 MPa, but with the given safety percentage, it is 284 MPa. Therefore,

37 (MPa) << 284 (MPa)

An FDS is obtained from the following:

284 (MPa)/(37 (MPa)); FDS = 7.6

Crush Resistance

In this subsection, we will calculated the resistance of the profile that holds the screw to the crushing that will be exerted on it. For this, the following formula will be used:

$$\mathsf{\sigma }={\displaystyle \frac{\mathrm{N}}{\mathrm{A}}} \left(\mathrm{M}\mathrm{P}\mathrm{a}\right)$$

The value of N is known to be N = 16,458 N, and A will be the area of the hole where the screw will be positioned. It is shown in Figure 17. Therefore, this area will be

$$\mathrm{A}=\mathrm{h}\times \mathrm{e}=0.024\times 0.006=1.4\times {10}^{-4 }{\mathrm{m}}^2$$

This leaves

$$\mathsf{\sigma }={\displaystyle \frac{\mathrm{16,458} (\mathrm{N})}{1.4\times {10}^{-4} ({\mathrm{m}}^2)}}=117 \left(\mathrm{M}\mathrm{P}\mathrm{a}\right)$$

$$117 \left(\mathrm{M}\mathrm{P}\mathrm{a}\right)<284 (\mathrm{M}\mathrm{P}\mathrm{a})$$

The FDS can be obtained:

$${\displaystyle \frac{284 \mathrm{M}\mathrm{P}\mathrm{a}}{117 \mathrm{M}\mathrm{P}\mathrm{a}}};\mathit{}\mathrm{F}\mathrm{D}\mathrm{S}=2.4$$

Several interesting consequences can be deduced from these results. As can be seen, there were favorable results for the entire lance assembly. The results obtained in the resistance to tensile stress of the lance structure and the results of the resistance to shear stress of the screw are very favorable, leading to the conclusion that the design is oversized. But, with the study of the crushing resistance of the lance material, it has been seen how the result is more adjusted, leaving us with an FDS of 2.4. This is higher than unity, but, taking into account that these calculations are theoretical and not practical, in real situations it may encounter more adverse conditions. This is rocky or muddy terrain.

Therefore, we can conclude that the spearhead design is optimal, and more than meets all the necessary requirements, being able to adapt to adverse conditions during its work use.

3.5. Static Analysis

3.5.1. Frame and Floor

We are going to observe the results obtained from the element called the frame and its complementary sheet that would form the floor of the trailer. For this static analysis, we have worked with structural elements, so defined joints and contacts between beams and solids have been defined. These solids are the floor plate and the supports, which will be the leaf springs and the upper ring of the ball impeller.

The external loads that it will be applied to the model will be the force exerted by the total load next to the upper structure. This value will be 81,000 N, calculated in Section 4, giving a margin greater than the load designed for the structure to be safe under some unforeseen event. The force of gravity, exerted by the weight of the frame, will also be placed.

Once all the restrictions and loads have been established, the meshing is performed (Figure 18a):

• Stresses (Von Mises and STRMAX)
  • Solids and shells (Von Mises) (Figure 18b):
    • Maximum value: 53.9 (MPa)
    • Minimum value: 0.2 (MPa)
  • Structural elements (STRMAX) (Figure 18c):
    • In this section, after looking at the Von Mises stresses in the solids and casts, we are going to enter into the main structure of the frame, taking into account all the structural elements that have been designed. For this case, the axial and bending stress will be studied at the upper limit in Mpa: maximum value: 113 (MPa) and minimum value: 1.41 (MPa).
  • Displacements (URES) (Figure 18d)
    • Maximum value: 3.6 (mm)
    • Minimum value: 1 ×10⁻³⁰ (mm)
  • Safety Factor (FDS) (Figure 18e)
    • Minimum value: 3.1

Figure 18. Static analysis of frame and floor frame and floor: mesh (a), stresses (Von Mises and STRMAX) of solids and shells (b), stresses (Von Mises and STRMAX) of structural elements (c), displacements (d) and safety factor (FDS) (e). Note: e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

Figure 18. Static analysis of frame and floor frame and floor: mesh (a), stresses (Von Mises and STRMAX) of solids and shells (b), stresses (Von Mises and STRMAX) of structural elements (c), displacements (d) and safety factor (FDS) (e). Note: e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

Consequences deduced from the above: we will now carry out a study of the results obtained. Knowing that he material used, S-355 steel [15], has an elastic limit of 355 Mpa, it is ensured that the proposed design (Figure 18b), in the solids and casts, will perfectly withstand the load to be transported without suffering plastic deformations, since it does not exceed our elastic limit, with a maximum theoretical stress of 53.9 Mpa, a visibly lower value.

In the case of stresses in the structural elements (Figure 18c), the maximum stress will be higher than the previous one, reaching a value of 113 Mpa. Even so, our design will easily withstand the loads to which it will be subjected.

We see how the maximum displacement value will be 3.6 mm (Figure 18d), a very low value taking into account the dimensions of the element; therefore, it would be negligible.

Regarding the value of the safety factor obtained (Figure 18e), there is a value of 3.1, verifying that the minimum is greater than 1, with respect to unity, thus ensuring reliability of the element studied.

3.5.2. Ball Impeller Support

In this section, a static study will be carried out on the structure where the ball impeller installed in the frame will be supported. This structure will join the frame with the front axle, also leveling the position between both axles.

The loads that this element will support will be located in the lower ring of the ball impeller, since it will be the only element supported on it. The value of this will be 42.649 N, as we have calculated previously (Section 3.1). There is a force that would be located in the lower ring, at the base of the impeller. In addition, we will add the force of gravity exerted by the weight of the element itself. We will place the fixed supports on the four fastening elements of the two front leaf springs.

Before obtaining the results, we mesh the element to be studied with all the established restrictions and loads (Figure 19a):

• Voltages (STRMAX) (Figure 19b)
  • Maximum value: 44.7 (MPa)
  • Minimum value: 4.3 (MPa)

• Displacements (URES) (Figure 19c)
  • Maximum value: 0.27 (mm)
  • Minimum value: 1 × 10⁻³⁰ (mm)

• Safety Factor (FDS) (Figure 19d)
  • Minimum value: 7.8

Figure 19. Static analysis of ball impeller support: mesh (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g. 3,674 = 3.674).

Figure 19. Static analysis of ball impeller support: mesh (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g. 3,674 = 3.674).

Important consequences can be deduced. Looking at the results, very favorable values have been obtained to ensure the structural reliability of our design in the face of the loads to which it will be subjected. Regarding the stresses produced (Figure 19b), a result of 44.7 (MPa) has been obtained as a maximum value. The margin with respect to the elastic limit of the material used is very wide, being 355 MPa.

In the case of the displacements produced, a maximum value of 0.27 mm is obtained, a more-than-insignificant value (Figure 19c).

Finally, it has an FDS greater than unity in the solids and casts (Figure 19d), of 7.8, so reliability can be ensured under the maximum load in these elements.

Viewing these results, we may think that there is an oversizing, but it is here that the loads increase and the structural elements may suffer, so this oversizing is acceptable.

3.5.3. Pothole Situation

In Section 2.1 Pothole, a very common situation has been studied, which the trailer will be subjected to. This is when one of the wheels passes through a rut, which can cause serious structural damage to the vehicle. To check the safety of the trailer in this situation, a static study is carried out on both the front and rear axles, using the most unfavorable load in both cases.

A load of 11,985 N will be used in each of the leaf spring supports that will be welded onto the frame. Knowing that only one wheel will have to resist this situation, the two supports that are located on that leaf spring will be selected.

The other leaf spring supports are defined as fixed supports, with a total of six elements, without taking into account the function of the leaf springs. This places the trailer in an even more unfavorable scenario.

It will be taken into account the fact that the trailer is fully loaded, with a load of 81,000 N. That said, the two static analyses will be performed.

Pothole Situation on the Rear Wheel

In this case, the bump will be passed with one of the rear wheels. By placing the loads and fixings mentioned above, the model is meshed (Figure 20a):

• Voltages (STRMAX) (Figure 20b)
  • Maximum value: 264.8 (MPa)
  • Minimum value: 2.64 × 10⁻² (MPa)

• Displacements (URES) (Figure 20c)
  • Maximum value: 12 (mm)
  • Minimum value: 1 × 10⁻³⁰ (mm)

• Factor of Safety (FDS) (Figure 20d)
  • Minimum value: 1.34

Figure 20. Static analysis of pothole situation on the rear wheel: mesh (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

Figure 20. Static analysis of pothole situation on the rear wheel: mesh (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

As in preceding subsections, some consequences can be deduced. Once the analysis is completed, the results are displayed to check if the design meets its functionality. It is begun with the results obtained in the stresses of the structural elements (Figure 20b), where favorable values have been obtained, since the elastic limit established by the chosen material (355 MPa) has not been reached, this value being 264 MPa. The margin is sufficient to avoid permanent deformations.

Regarding the displacements obtained (Figure 20c), the maximum value was 12 mm. Taking into account the dimensions of the element studied, this result would be more than acceptable, since it is minimal in comparison.

Finally, and in accordance with the previous results, an FDS of 1.34 is obtained (Figure 20d), a value that provides the structure with sufficient security in the case of a more unfavorable situation than the one studied occurring.

Therefore, it can be concluded by ensuring that the structure would support the established load without problems and the design would be ideal.

Pothole Situation on the Front Wheel

In this case, the bump will be crossed with one of the front wheels. Placing the loads and fixings named at the beginning of the section (the load on one of the front-axle leaf spring supports), the model is meshed (Figure 21a):

• Voltages (STRMAX) (Figure 21b)
  • Maximum value: 260 (MPa)
  • Minimum value: 6.5 × 10⁻¹³ (MPa)

• Displacements (URES) (Figure 21c)
  • Maximum value: 11 (mm)
  • Minimum value: 1 × 10⁻³⁰ (mm)

• Factor of Safety (FDS) (Figure 21d)
  • Minimum value: 1.36

Figure 21. Static analysis of pothole situation on the front wheel: mesh (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 =3.674).

Figure 21. Static analysis of pothole situation on the front wheel: mesh (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 =3.674).

Once the analysis of the bump situation in the front wheel is finished, some consequences can be deduced. With respect to stresses (Figure 21b), a maximum value of 260 MPa has been obtained, with a wide margin up to the elastic limit of the material (355 MPa) and avoiding plastic deformation.

The results of the displacements have also been favorable (Figure 21c), obtaining a value of 11 mm. As in the previous case, it is a small value compared to the dimensions of the element studied.

Finally, with respect to the FDS (Figure 21d), the value obtained was 1.36.

As in the previous section, the design obtained has been correct, ensuring acceptable structural safety.

3.5.4. Upper Structure

This upper structure will have to differentiate several parts for analysis, since it has movable elements, such as the side doors, rear doors and upper bars.

Rear Wall Analysis

This analysis looks at the design of the rear door. This door can be folded down, thanks to a pair of hinges, and can be left fixed by means of two handles located on the sides. These will be the fixed points.

In the case of the loads to be supported, we will take the most unfavorable situation as the case in which the total load moves due to inappropriate acceleration or the ascent of a very steep slope. These situations are very unlikely, since the load must be well tied and secured, but it must provide the trailer with a structure that can withstand unfavorable conditions.

It will give a load value perpendicular to the interior wall of 70.700 N, a value that it has given to the maximum load.

The mesh can be created (Figure 22a):

• Stresses (Von Mises and STRMAX)
  • Solids and shells (Von Mises) (Figure 22b):
    • Maximum value: 284 (MPa)
    • Minimum value: 2 ×10⁻¹ (MPa)
  • Structural elements (STRMAX) (Figure 22c):
    • Maximum value: 341 (MPa)
    • Minimum value: 22.5 (MPa)
  • Displacements (URES) (Figure 22d)
    • Maximum value: 10 (mm)
    • Minimum value: 1 ×10⁻³⁰ (mm)
  • Factor of Safety (FDS) (Figure 22e)
    • Minimum value: 1.04

Figure 22. Static analysis of rear wall (upper structure): mesh (a), stresses (Von Mises and STRMAX) of solids and shells (b), stresses (Von Mises and STRMAX) of structural elements (c), displacements (d) and safety factor (FDS) (e). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 =3.674).

Figure 22. Static analysis of rear wall (upper structure): mesh (a), stresses (Von Mises and STRMAX) of solids and shells (b), stresses (Von Mises and STRMAX) of structural elements (c), displacements (d) and safety factor (FDS) (e). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 =3.674).

Consequences: with the most unfavorable condition possible, favorable results are achieved, at the limit of what is admissible but acceptable knowing that it is almost impossible for this situation to happen, but that for safety reasons we must take it into account.

It can be seen how, in the cases of voltage and FDS, they have limit values. In the case of the stress for the structural elements (Figure 22c), a maximum value of 341.2 MPa has been obtained with the elastic limit being 355 MPa. The maximum allowable tension according to the theory will not be exceeded.

In the case of solids and casts (Figure 22b), the value of the maximum stress is lower, being 284 MPa, giving a wider margin up to the elastic limit, compared to the case of structural elements.

Regarding the FDS, it has a value of 1.04 (Figure 22e). As had already been mentioned, for a structure to have minimum security, a value greater than or equal to unity must be obtained. Therefore, the value is acceptable.

The deformation will have a maximum value of 8 mm, an admissible value given the situation (see Figure 22d).

Therefore, the back door would have a correct design.

Front Wall Analysis

This wall would be the only one that we are going to find to be fixed, since having the spearhead and other elements would make it difficult to maneuver. On this wall we are going to have three upper mobile bars, but we will carry out the individual analysis (only of the fixed wall).

We will take as the most unfavorable situation that where during the transport of the load some emergency braking is carried out, and even though the trays are secured, they move and exert a perpendicular force on the wall. The established value of force exerted will be that of the entire load supported on that wall, approximately 70,700 N.

We begin by performing the meshing for subsequent analysis (Figure 23a):

• Stresses (Von Mises and STRMAX)
  • Solids and shells (Von Mises) (Figure 23b):
    • Maximum value: 261 (MPa)
    • Minimum value: 4192 (MPa)
  • Structural elements (STRMAX) (Figure 23c):
    • Maximum value: 252 (MPa)
    • Minimum value: 11.9 (MPa)
  • Displacements (URES) (Figure 23d)
    • Maximum value: 5.3 (mm)
    • Minimum value: 1 ×10⁻³⁰ (mm)
  • Factor of Safety (FDS) (Figure 23e)
    • Minimum value: 1.108

Figure 23. Static analysis of front wall (upper structure): mesh: (a), stresses (Von Mises and STRMAX) of solids and shells (b), stresses (Von Mises and STRMAX) of structural elements (c), displacements (d) and safety factor (FDS) (e). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

Figure 23. Static analysis of front wall (upper structure): mesh: (a), stresses (Von Mises and STRMAX) of solids and shells (b), stresses (Von Mises and STRMAX) of structural elements (c), displacements (d) and safety factor (FDS) (e). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

In this scenario, we encounter great difficulties for a correct design without exceeding the use of material. The structure is made of small tubes, so more are needed to be able to support such an amount of load. After analyzing the element, it has been managed to obtain favorable results.

In the case of stresses we have very similar values for both the solids and casts (Figure 23b) and for the structural elements (Figure 23c), 261 and 252, respectively), which do not cause the material to undergo plastic deformation.

The deformations are minimal, with a maximum value of 5.3 mm (Figure 23d). This displacement is negligible with respect to the measurements of the element and the load to which it is subjected.

The problem came in the case of the FDS, in which a final minimum value of 1.1 has been obtained(Figure 23e). It is true that, although it is a low value, it is greater than unity and, therefore, it has a safety margin.

It should be noted that the loads to which the element has been subjected would occur in almost impossible situations, since an accumulation of errors would have to occur. In any case, the present design would support that established load.

Side Wall Analysis

In this analysis we are going to study one of the two side walls that the trailer presents. Unlike the analyses of the other walls, this one it will not have a maximum force exerted of 70,700 N, since the load will be distributed to the two equal walls on each side. Therefore, without taking into account the fact that part of the load may rest on the main structure, it will take half of the load on each wall, so it will place a force perpendicular to the interior face of 35,850 N. This load would be due to a lateral overturn or a sudden turn where the load became loose from its moorings and fell to the side.

It Is true that these walls are larger than the others, so, although the force is lower, similar efforts can be produced as in the other cases. It should also be noted that, like the rear pair, these walls are foldable and removable.

The meshing is carried out (Figure 24a):

• Stresses (Von Mises and STRMAX)
  • Solids and shells (Von Mises) (Figure 24b):
    • Maximum value: 201.6 (MPa)
    • Minimum value: 1.43 ×10⁻² (MPa)
  • Structural elements (STRMAX) (Figure 24c):
    • Maximum value: 246.5 (MPa)
    • Minimum value: 5.43 (MPa)
  • Displacements (URES) (Figure 24d)
    • Maximum value: 11.7 (mm)
    • Minimum value: 1 ×10⁻³⁰ (mm)
  • Factor of Safety (FDS) (Figure 24e)
    • Minimum value: 1.44

Figure 24. Static analysis of side wall (upper structure): mesh: (a), stresses (Von Mises and STRMAX) of solids and shells (b), stresses (Von Mises and STRMAX) of structural elements (c), displacements (d) and safety factor (FDS) (e). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

Figure 24. Static analysis of side wall (upper structure): mesh: (a), stresses (Von Mises and STRMAX) of solids and shells (b), stresses (Von Mises and STRMAX) of structural elements (c), displacements (d) and safety factor (FDS) (e). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

After the analysis, acceptable results have been obtained, validating the proposed design. It can be seen how the main structure (Figure 24c) and the solids (Figure 24b) resist well the force applied to them, without reaching, under the maximum load, the value of the admissible elastic limit. The maximum stress value is obtained in the fixed clamping zone of 246.5 MPa in the structural elements, which has been established in a general way, since it will use a gripping system chosen from a catalogue adapted to high loads. This means that the wall is subjected to acceptable stresses, due to its good design.

The displacements, as expected, have been obtained mainly at the top of the railing (Figure 24d). This is due to the fact that the wall is very long and high stresses are exerted. In any case, the maximum value obtained is 11.7 mm, which is an acceptable displacement for a large element and the stresses it is subjected to.

As for the safety factor, it is found it to be greater than unity, with a minimum value of 1.44 (Figure 24e). It thus has a structure with acceptable safety in the event of failure.

3.5.5. Main Structure Analysis

This analysis will study which of the tubes of the structure will be the one that will withstand the worst conditions.

• Front tubes: these two tubes will be the safest, since in addition to being welded to the frame, they will have the front door attached, which is fixed and will be welded both to the two tubes of the structure and to the floor of the frame, so the fixed points will be larger than in any other case.
• Rear tubes: these two tubes will be responsible for holding the rear sliding door and the removable rear doors. The conditions of these elements will be more unfavorable than the previous ones, since they have fewer fixed supports, although the load they will support will be the same. It should be noted that this load will be in an almost impossible situation, since a series of failures must occur.
• Side tubes: these last two tubes will be the ones installed at the midpoint of the side of the trailer. Its function will be to hold the side doors on both sides.

Having studied the different situations, the front tubes were discarded and were left with the tubes located at the midpoint and at the rear.

The analysis is carried out by applying the forces to the side and rear doors, to study how the tubes act under such a load. Therefore, two different scenarios can be considered, as follows:

(1)

A first scenario with the load overturning on one of the two sides; therefore, the force applied will only be on the two removable side walls.

(2)

A second scenario will be when, due to heavy braking or ascent of a slope that is too steep, the load moves until it collides with the rear door.

The load that will impact the structure will not be completed, since to do so the trailer must lie down completely, so we will enter a load value of 60% of the total load, 4320 kg (42.379 N), a value that will be sufficient.

If we take into account the calculations carried out for the maximum angle of lateral inclination to which the trailer can be subjected without overturning, it would have a value of the component normal to the walls of 30,519 N. This value is less than the proposed percentage, but the results must be oversized to provide extra security in unfavorable situations.

Lateral Load on Main Structure

The first scenario will be shown in Figure 25a:

• Voltages (STRMAX) (Figure 25b)
  • Maximum value: 250 (MPa)

• Displacements (URES) (Figure 25c)
  • Maximum value: 54.8 (mm)

• Factor of Safety (FDS) (Figure 25d)
  • Minimum value: 1.1

Figure 25. Static analysis of lateral load on main structure: the first scenario (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 =3.674).

Figure 25. Static analysis of lateral load on main structure: the first scenario (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 =3.674).

Some consequences can be deduced. In this first case, where we have the load on the interior side, the results will be very fair, although it should be noted that the situation to which the trailer is subjected in the study is a very unfavorable and difficult case to occur.

In the case of tension (Figure 25b) there is a maximum value of 250 MPa, located very close to the elastic limit that has been used, giving an FDS (Figure 25d) of 1.1.

The displacement that it is going to have in this case will be 54.8 mm (Figure 25c). This is a large displacement, but we must take into account the fact that it is facing the most unfavorable situation and the dimensions of the structure. The important thing is that this displacement does not lead to permanent deformation.

Rear Load on Main Structure

The second scenario is shown in Figure 26a:

• Voltages (STRMAX) (Figure 26b)
  • Maximum value: 49.2 (MPa)

• Displacements (URES) (Figure 26c)
  • Maximum value: 2 (mm)

• Factor of Safety (FDS) (Figure 26d)
  • Minimum value: 5.59

Figure 26. Static analysis of rear load on main structure: the second scenario (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674) and ^2 should be superscript ².

Figure 26. Static analysis of rear load on main structure: the second scenario (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674) and ^2 should be superscript ².

It must be taken into account the fact that it will only study the FDS of the structural elements of the main structure, and not of the elements of the walls or frame, since these elements have already been analyzed.

As it can be seen after the analysis, the results obtained are very favorable. In this second most unfavorable case of the main structure, the maximum load to which it will be subjected will not be a cause of failure.

The stress (Figure 26b) is much lower than the elastic limit, being 49.2 MPa, and having an FDS (Figure 26d) of 5.59. This value is more than enough for us to say that our structure is safe without being oversized, since this is a theoretical calculation and not a practical one.

3.5.6. Analysis of Upper Bars and Supports

In the case of the upper bars and supports, we will only analyze the components that are on the side. This is because the side bars are longer, but of the same dimensions as the front ones. Thus, by analyzing and obtaining favorable values for the side bars, the front bars will perform in the same way, or with even greater reliability and safety.

These bars, as has been mentioned previously, are designed so that in the event of any of the hive trays overturning, they can support their weight without the risk of it falling out of the trailer. This is difficult since, thanks to the ties that will be made, the trays will be well secured, but the worst-case scenario must be analyzed.

Knowing that the side has six bars, three on each side of the hollow structural tube located in the middle, and that each row of eight trays has a total weight with hives of 2400 kg, it will be assumed that, in the event of overturning, the bars can drop the weight of two complete rows, which is 4800 kg. Dividing this weight by the six side bars, a final load of 800 kg = 7848 N can be calculated. This value will be approximated to 8000 N.

Knowing all the restrictions and loads, the mesh of the part can be created (Figure 27a):

• Stresses (Von Mises) (Figure 27b)
  • Maximum value: 152 (MPa)
  • Minimum value: 4.6 × 10⁻³ (MPa)

• Displacements (URES) (Figure 27c)
  • Maximum value: 11 (mm)
  • Minimum value: 1 × 10⁻³⁰ (mm)

• Factor of Safety (FDS) (Figure 27d)
  • Minimum value: 1.4

Figure 27. Static analysis of upper bars and supports: mesh (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

Figure 27. Static analysis of upper bars and supports: mesh (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

As can be seen, the bar has the capacity to withstand up to 800 kg of force in the event of any of the hive trays overturning, and even more.

In the case of tension (Figure 27b), the elastic limit of 215 MPa is not reached, obtaining a value of 152 MPa, avoiding permanent deformations. The margin obtained is sufficient, as will be seen later.

In the case of deformation (Figure 27c), a maximum value of 11 mm is obtained. It is considered to be t an acceptable displacement, taking into account the weight it will carry.

Once these values were obtained, it was assumed that the FDS was at the limit. A result of 1.4 has been obtained (Figure 27d).

Seeing that the resistance of this element is more than sufficient against the imposed load, and knowing that it is longer than the front one, it can be ensured that it will be able to work with a greater effort without problems.

3.5.7. Static Analysis of Lance

We are going to perform the static analysis of the lance. This element is essential for handling the trailer. In Section 4, a theoretical study of the structure of the lance and its components was carried out, but it can be re-checked with respect to the resistance to the maximum load it will support, which is 32,917 N (obtained in the same calculation).

The study will be carried out on the main structure, without taking into account screws or lugs, since more specific calculations have been made for these elements. It will be begun with the meshing of the element to be studied (Figure 28a):

• Voltages (STRMAX) (Figure 28b)
  • Maximum value: 62 (MPa)
  • Minimum value: 2.46 (MPa)

• Displacements (URES) (Figure 28c)
  • Maximum value: 0.21 (mm)
  • Minimum value: 1 × 10⁻³⁰ (mm)

• Factor of Safety (FDS) (Figure 28d)
  • Minimum value: 4.4

Figure 28. Static analysis of lance: mesh (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

Figure 28. Static analysis of lance: mesh (a), voltages (b), displacements (c) and safety factor (FDS) (d). Note: Min = minimum and Max = maximum, e = 10 (e.g., e-30 = 10⁻³⁰ and e+02 = 10²), plus commas = full stops (e.g., 3,674 = 3.674).

As it can be seen in this last part of the present study, the results of the analysis agree with the results of the previous calculations (Section 4). The structure of the lance will easily withstand the most unfavorable stress to which it will be subjected.

In the case of stresses (Figure 28b) a result of 47 MPa is obtained, with a large margin with respect to the elastic limit of the chosen material.

Based on these stress values, it was to be expected that the deformations were going to be minimal, with a maximum value of 0.18 mm.

Regarding the value of the FDS (Figure 28d) of 4.4, we obtained a result very similar to that calculated in Section 4, of 5.4. Furthermore, the area where this minimum value has been obtained is the same in both cases. Thus, our calculations are correct.

4. Summary and Conclusions

The present study aimed to develop a trailer for the transport of beehives adapted to be placed or fixed to a tractor or a vehicle trailer. The objective was the transport of the beehives safely and stably during transhumance. A novel design has been investigated and proposed, which is related to a trailer that incorporates a device for housing rectangular sections of beehives, which can be adapted for fixing or housing in a vehicle or in a vehicle trailer. The device comprises a lower support structure, adapted to support by itself a plurality of rectangular section of beehives stacked horizontally on the lower structure, an upper frame adapted to house the beehives inside, and two or more connecting elements between the lower structure and the upper frame. The connection of the trailer with the device facilitates the loading and unloading of the beehives by mechanical means.

Static analyses have been performed, taking into account structural elements of this proposed design, with restrictions and loads being established. The analysis, including upper bars and supports, was completed with several situations. Based on stress values, deformations were determined and calculations were evaluated. Then, the design of the trailer for transporting beehives was achieved.

The device comprised a part that can be purchased directly from auxiliary companies and other components that must be designed as presented in this study. These components were (a) body frame and floor; (b) impeller support; (c) spearhead; (d) side, front and back walls and (e) beehive tray

The different parts of the trailer were designed as individual pieces, and later an assembly must be carried out to achieve the complete design. This method of implementation is because the simulation of individual components was simpler and easier, since if it is carried out through an assembly, the type of joint, such as welding, and the length of the weld would have to be indicated at each point of contact between components, with its thickness and all the necessary parameters. Therefore, in those welding points, fixed fastenings are indicated, and so it will simplify it. Furthermore, due to the individual creation of each part, its own load simulation has been carried out.

Taking into account the design proposed in this study, the trailer can be loaded with trays specially designed for transporting hives. These trays have been designed to facilitate the movement of hives, both to load and unload them from the trailer, since it will be possible to move in quantities of five units at a time, with a pen. In addition, this element further secures the hives once loaded on the trailer.

The trays of this design have been manufactured using 60 × 3 (mm) flat steel bars, consisting of a total of 11 units of various sizes. Also, angled bars have been used for the legs and support of the hives with dimensions of 40 × 3 (mm). In this case, a quantity of eight units of variable sizes has been proposed. The lower structure of this tray is adapted for fixing or accommodation in a vehicle, or in a vehicle trailer. The lower structure, the upper frame and/or the connecting elements comprise one or more braced reinforcements and one or more tubes. These tubes have a substantially rectangular section. The trays have utility for the transport of rectangular-section hives, which can be carried out in beekeeping transhumance tasks.

Finally, the proposed design performance according to the present study allows us to conclude that the lower structure, and the upper frame and/or the connecting elements comprise the means for hooking or gripping a mechanical lifting system. The elements such as the lower structure, upper frame, connecting elements, longitudinal slats, bars, braced reinforcements, tubes, hooking or gripping means and support and/or fastening means were composed of alloy steel. The authors hope that this design of the trailer will be of great interest for the accommodation and transport of beehives safely and stably during transhumance.