In the design and construction of technical products, the Mechanics of Materials has a defining role in ensuring their safe operation.

Mechanics of Materials known also as Strength of Materials, highlights the behaviour of solid deformable bodies under the action of various types of loading. The main objective of mechanics of materials is to determine the stresses, strains and displacements in structures and components due to the loads acting on them. These determinations define the mechanical behaviour of that structures and components.

A mechanics of materials analysis must be done in the context of a continuous relationship between the loading scheme and the mechanical characteristics of the loaded material (elasticity and strength). Also, such an analysis is based on three essential elements, whose accuracy depends on the results obtained. First of all, the loading scheme is essential in solving a mechanics of materials problem. The distribution of loads on the analysed structure or component is the condition for which such analysis is needed. The precision of the loading scheme determines the accuracy of the analysis results.

Secondly, identifying the loading modes is essential in establishing the computational theories needed to solve the problem. The loading scheme can cause simple loading modes (tensile/compression, bending, shear, torsion) or composed loading modes.

Last but not least, knowing the mechanical behaviour of the material is also essential in conducting a mechanics of materials analysis. The load bearing capacity of the material determines, on the one hand, the maximum load that the analysed structure or component can support, and on the other hand it determines the overall dimensions of the structure or component.

To illustrate the above, let us consider a hand-operated crane used in construction, which has to raise a maximum weight of 150 kg, fig. 1. For this case, the lengths of the crane arm and the column are known and a dimensional calculation is required for the components of the crane.

Crane1

Fig. 1. Hand-operated crane

First, it is necessary to determine the maximum load corresponding to the weight of 150 kg. Since the crane driving is manual, meaning a low lifting speed of the weight, the maximum force can be estimated based on the gravitational acceleration:

eq 1.png

The crane uses a steel wire rope for lifting the weight, which passes over two pulleys and is run on a drum.

The action of the maximum load cause the steel wire rope to be loaded in tension by an internal force N = 1471.5 N. If the internal force is reported to the cross section area of the steel wire rope results the maximum stress on the cable:

eq 2.png

In this problem, we must determine the minimum steel wire rope diameter that can support the given weight. This means that a steel wire rope diameter has to be calculated which loaded with the given weight determines a tensile stress lower or at most equal to an admissible cable strength.

eq 3.png

The admissible cable strength, σa, is determined as the ratio between the nominal tensile strength and a safety factor:

eq 4.png

The safety coefficients are provided in standardized rules depending on the operating risk of the structure or component. In this case, a safety factor c = 2 can be considered.

According to the Reference [1], the steel wire ropes for lifting are of several types, made according to different standard norms. At the same, the nominal minimum tensile strength of these cables is σult,cable = 1320 MPa.

Based on the relationship (4), therefore the admissible cable strength is σa = 660 MPa.

From relations (2) and (3) it follows:

eq 5.png

This results in a minimum cable diameter of d = 1.684 mm.

Comparing this calculation with the steel wire ropes catalogs, it can be found that the minimum diameter of the manufactured cables is d = 6 mm.

Furthermore, a cable stiffness calculation is required, by which the elongation over 1 m length is calculated for the maximum applied force. This means:

eq 6.png

According to the Reference [2] the elasticity modulus of the steel core ropes can be considered E = 55 000 MPa.

Thus:

– For d = 1.684 mm  =>  Δl = 12 mm;

– For d = 6 mm  => Δl = 0.946 mm.

In the first case, even if the stress in the cable is less than the permissible strength, the 12 mm elongation per cable meter is very high compared to the second case. Therefore, it is much more justified to choose the diameter of 6 mm for the cable.

The second part of this analysis is the calculation of the crane structure consisting of the arm and the column. These are also loaded with the maximum force resulted from the lifting weight, according to the loading scheme in Figure 2.

Crane2.png

Fig. 2. The loading scheme of the crane structure

The maximum force determines in each of the two components of the structure a combined loading of compression and bending.

Thus:

– The crane arm is loaded with a force F2 = 1040.5 N and a bending moment that is maximum in the connection point with the column, M1 = 1040.5·750 = 780375 Nmm.

– The column is also loaded with a compression force F = 1471.5 N and a constant bending moment equal of M = 1471.5·530.3 = 780380.6 Nmm.

If the internal forces diagrams are drawn, the following situation is obtained:

crane3.png

Fig. 3. The internal forces diagrams

The combined effect of the internal forces gives a normal stress in the structure in the following form:

eg 7.png

where A is the cross section area of the structural elements, I is the second moment of inertia.

The problem consists in choosing and dimensioning the two elements of the structure. Therefore, we need to determine what type of profile will be used to create physical structure, following the calculation to determine the required size of the section profile.

Thus, we consider a tubular profile with a square section, defined by the size of the outer side and the thickness of the wall:

Crane4-1.png

Fig. 4. Hollow square section

For the square section, we can calculate:

eqs 8-9.png

Normally these profiles are made of S355 steel with the following mechanical properties:

– Yield strength: σy = 355 MPa

– Ultimate tensile strength: 600 MPa.

– Young’s modulus: E = 205000 MPa.

Based on the yield strength, the admissible strength corresponding to a safety factor of 2 is:

eq 10.png

The equation (7) is applied to calculate the normal stress at the points furthest from the x-axis. Thus, for the crane arm:

eqs 11-12.png

For the column:

eqs 13-14.png

The strength condition of the structure is fulfilled if for different values of the parameters a and t, the normal stress in the relations (11) – (14) is less than the admissible strength calculated in the relation (10).

The problem is also moved forward in checking the rigidity of the structure. For this, it is necessary to calculate the vertical displacement of the loading point using the principle of potential strain energy by means of one of the known methods: Mohr-Maxwell, Castigliano or Vereshchagin’s rule.

For this problem, the Vereshceagin’s rule is used for the following loading scheme:

Crane5.png

Fig. 5. The loading scheme for calculation of the vertical displacement of the loading point

After calculations, the vertical displacement of the loading point is given by the following equation:

eq 15.png

In our problem, l = 750 mm, m = 2000 mm and r = 530.33 mm. Also, in this calculation, only the effect of the bending moment, which is dominant for the analysed structure, was considered.

For a 70 mm square tubular section with a wall thickness of 4 mm, the normal stress in the two elements of the structure is considerably small compared to the admissible strength (σB = -37.48 MPa in the column), but the vertical displacement of the loading point is 5.7 mm.

If the angle increase to 65°, the vertical displacement of the loading point decrease to 2.32 mm.

Therefore, a 70×4 mm hollow square section and a change in the angle α from 45° to 65° can be recommended.

Also, the column being loaded in compression, with a much longer length relative to the section dimensions, a critical buckling load verification is required.

Thus, considering the Euler’s elastic buckling formula, it follows:

eq 16.png

where n is factor accounting for the end conditions, m is the column length.

Crane6.png

Fig. 6. The values of the n factor for different end conditions

The critical buckilng force is much higher than the maximum force, indicating that there is no risk of buckling the column.

 

References

[1] Steel wire ropes in elevators, Elevator products, Pfeifer Drako, 2015;

[2] Elevator Ropes, usha martin