Modeling Prestressed Bolt Connection
Technical Article
When modeling surface models, such as a frame joint or similar structures, there is always the question of how to model a prestressed bolt connection. In this case, it is always necessary to find a compromise between the practicable and detailed solution. The following article describes the modeling procedure of such a connection based on the joint diagram calculation method.
Basics of Joint Diagram
The joint diagram is a graphical representation of forces in a prestressed bolt connection. In this case, the compression forces arising in the components to be connected and the accompanying deformations are contrasted with the forces and deformations in the bolt. Figure 01 shows such a diagram.
Figure 01  Simplified Joint Diagram
The blue line (characteristic line) represents a graph of the bolt, the yellow one a graph of the structural components. Typically, the bolt stiffness is smaller than the stiffness of the structural components. However, there are also various exceptions such as in the case of nuts. The intersection of both lines represents the preloading force in the connection without any external load applied. The end point of the bolt line is the maximum resistance force in the thread.
In addition to the bolt line and the component line, there is another important characteristic line of the external tensile force (also preload). This line is shown in gray in Figure 01. It originates from the characteristic line of the components on the y‑axis of the desired residual clamping force. The residual clamping force is the force that still holds the components together. For example, if there is a horizontal force to be absorbed by the connection (without shear strain of the bolt, so only by the component friction) in addition to the tensile component in the case of the existing working load, then the residual clamping force must be selected in a way that there is sufficient resistance.
In addition to these characteristic lines, there are other lines that can be used for a more detailed representation. However, since these lines have no influence on the basic procedure, they will not be explained further in this article, and the presented simplified joint diagram will only be used. For example, the additional characteristic lines would be due to the compression set or the eccentric stress and load.
Formulas of Simplified Joint Diagram
In order to create the joint diagram, it is necessary to calculate the corresponding stiffnesses, deformations, and forces first. In general, the spring stiffnesses can be calculated according to the Hooke's law as follows:
$$\mathrm c\;=\;\frac{\mathrm F}{\mathrm f}\;(1.1)$$where
In the case of a tension member with isotropic material, the spring constant can be calculated directly using the elastic modulus (modulus of elasticity):
$$\mathrm c\;=\;\frac{\mathrm E\;\cdot\;\mathrm A}{\mathrm l}\;(1.2)$$where
The bolt stiffness is simplified and the bolt shaft is only applied. Other possibilities are to apply the bolt head, thread, nut, different shaft diameters, etc. In that case, the elements with their reciprocal value are added to the total stiffness. The spring stiffness of the bolt is calculated using the following formula (Index S):
$${\mathrm c}_\mathrm S\;=\;\frac{{\mathrm E}_\mathrm S\;\cdot\;{\mathrm A}_\mathrm S}{{\mathrm l}_\mathrm K}\;(2.1)$$where
The thread flank diameter d_{3} is used for the cross‑section area in the bolt thread range. Thus, the total formula results:
$${\mathrm c}_\mathrm S\;=\;\frac{{\mathrm E}_\mathrm S\;\cdot\;\mathrm\pi}4\;\cdot\;\frac{{\mathrm d}_3²}{{\mathrm l}_\mathrm K}\;(2.2)$$The component stiffness is calculated in a similar way. Since there is one or more plates, the index P is used:
$${\mathrm c}_\mathrm P\;=\;\frac{{\mathrm E}_\mathrm P\;\cdot\;{\mathrm A}_\mathrm P}{{\mathrm l}_\mathrm K}\;(3.1)$$
where
The cross‑section area A_{P} depends on the thickness, in contrast to the bolt. It is assumed that the load is extended at an angle of approximately 60°. There are three various cases, as shown in Figure 02.
Figure 02  Load Extensions in Various Plate Dimensions
In Case 1, the components between the bolt and the nut are like a sleeve and this sleeve diameter is maximally equal to the bearing surface diameter of the bolt or nut.
Case 2 covers the range where this sleeve diameter is minimally equal to the bearing surface diameter of the nut or bolt, and maximally equal to the diameter of the load extension cone (marked red in Figure 02). This stretches symmetrically from both sides, and the diameter is the largest in the middle of the clamping length.
Case 3 covers the range from the maximum load extension cone to the infinite plate extension. For this reason, it is necessary to calculate the replacement area A_{ers}. A_{ers} corresponds to the cross‑section area of a replacement cylinder with a constant load extension.
For the following example, Case 3 is sufficient. A_{ers} is calculated using the following formula (see VDI 2230, edition 1986 [1]):
$${\mathrm A}_\mathrm{ers}\;=\;\frac{\mathrm\pi}4\;\cdot\;({\mathrm d}_\mathrm W²\;\;{\mathrm d}_\mathrm h²)\;+\;\frac{\mathrm\pi}8\;\cdot\;{\mathrm d}_\mathrm W\;\cdot\;{\mathrm l}_\mathrm K\;\cdot\;\left(\left(\sqrt[3]{\frac{{\mathrm l}_\mathrm K\;\cdot\;{\mathrm d}_\mathrm W}{({\mathrm l}_\mathrm K\;+\;{\mathrm d}_\mathrm W)²}}\;+\;1\right)^2\;\;1\right)\;(3.2)$$where
The bearing surface diameter can be applied in a simplified way as 90% of the width across flats:
$${\mathrm d}_\mathrm W\;=\;0.9\;\cdot\;\mathrm s\;\;\;(3.3)$$where
Since the load application point in a surface model is not necessarily at the top of the component (the plate), but always in the middle of the surface, the plate stiffness must be determined on this load application point. For this, the load application factor n is introduced, which reduces the clamping length accordingly. This problem is illustrated in Figure 03.
Figure 03  Conversion of Plate Solid Model to Plate Surface Model
The actual components, that is two plates in this case, are reduced to the middle of the surfaces. In the case of two plates, n is always 0.5 as there is always a half of each plate used. The new plate stiffness c_{Pn} is then calculated as follows:
$$\begin{array}{l}{\mathrm c}_\mathrm{Pn}\;=\;{\mathrm c}_\mathrm S\;\cdot\;\frac{1\;\;\mathrm n\;\cdot\;{\mathrm\Phi}_\mathrm K}{\mathrm n\;\cdot\;{\mathrm\Phi}_\mathrm K}\;(3.4)\\{\mathrm\Phi}_\mathrm K\;=\;\frac{{\mathrm c}_\mathrm S}{{\mathrm c}_\mathrm S\;+\;{\mathrm c}_\mathrm P}\;(3.5)\end{array}$$where
To create the characteristic lines, various forces are still required, in addition to stiffnesses. The residual clamping load F_{KR,} the working load F_{A,} and the tightening factor α_{A} (anglecontrolled tightening) must be specified. On the other hand, the resulting minimum and maximum assembly preload F_{Mmin} and F_{Mmax,} must be calculated. Below is the formula for the assembly preloads at an anglecontrolled tightening:
$$\begin{array}{l}{\mathrm F}_\mathrm{Mmin}\;=\;{\mathrm F}_\mathrm{Kmin}\;+\;{\mathrm F}_\mathrm{PA}\;\;\;(3.6)\\{\mathrm F}_\mathrm{Mmax}\;=\;{\mathrm\alpha}_\mathrm A\;+\;{\mathrm F}_\mathrm{Mmin}\;\;\;(3.7)\end{array}$$where
The additional plate load F_{PA} is the force arising when applying the working load. It is calculated according to the formula:
$${\mathrm F}_\mathrm{PA}\;=\;(1\;\;\mathrm n\;\cdot\;{\mathrm\Phi}_\mathrm K)\;\cdot\;{\mathrm F}_\mathrm A\;\;\;(3.8)$$where
In the case of simplification without considering embedding, the preload F_{V} corresponds to the minimum preload F_{Mmin}. To consider the working load line, the maximum bolt force F_{Smax} is missing, which arises in the bolt when concerning the working load:
$${\mathrm F}_\mathrm{Smax}\;=\;{\mathrm F}_\mathrm{Mmax}\;+\;{\mathrm F}_\mathrm{SA}\;\;\;(3.9)$$where
The additional bolt force F_{SA}, is again calculated similarly to Formula 3.8:
$${\mathrm F}_\mathrm{SA}\;=\;\mathrm n\;\cdot\;{\mathrm\Phi}_\mathrm K\;\cdot\;{\mathrm F}_\mathrm A\;\;\;(3.10)$$The maximum loading capacity of the bolt (F_{0,2}) as the last missing force must be determined using the crosssection area of the bolt in the thread. This is calculated using the crosssection area diameter d_{s}, which results from the mean value of the core diameter d_{k} (d_{3}) and the flank diameter d_{fl} (d_{2}):
$${\mathrm F}_{0,2}\;=\;{\mathrm A}_\mathrm S\;\cdot\;{\mathrm f}_\mathrm{ub}\;=\;\frac{\mathrm\pi}4\;\cdot\;\left(\frac{{\mathrm d}_2\;+\;{\mathrm d}_3}2\right)^2\;\cdot\;{\mathrm f}_\mathrm{ub}\;(3.11)$$where
In addition to the forces, the deformations must be determined as the corresponding values so the characteristic lines can be entered in the joint diagram. For this, Formula 1.1 converted according to f is used. Below are the formulas for the deformations f to the respective forces F:
$$\begin{array}{l}{\mathrm F}_{0,2}\;\rightarrow\;{\mathrm f}_{0,2}\;=\;\frac{{\mathrm F}_{0,2}}{{\mathrm c}_\mathrm S}\;(4.1)\\{\mathrm F}_\mathrm{Mmax}\;\rightarrow\;{\mathrm f}_\mathrm{SMmax}\;=\;\frac{\mathrm{FMmax}}{{\mathrm c}_\mathrm S}\;(4.2)\\{\mathrm F}_\mathrm{Mmax}\;\rightarrow\;{\mathrm f}_\mathrm{Mmax}\;=\;{\mathrm F}_\mathrm{Mmax}\;\cdot\;\left(\frac1{{\mathrm c}_\mathrm{Pn}}\;+\;\frac1{{\mathrm c}_\mathrm S}\right)\;(4.3)\\{\mathrm F}_\mathrm{SA}\;\rightarrow\;{\mathrm f}_\mathrm{SA}\;=\;{\mathrm f}_\mathrm{PA}\;=\;\frac{{\mathrm F}_\mathrm{SA}}{{\mathrm c}_\mathrm S}\;(4.4)\\{\mathrm F}_\mathrm{Smax}\;\rightarrow\;{\mathrm f}_\mathrm{Smax}\;=\;\frac{{\mathrm F}_\mathrm{Smax}}{{\mathrm c}_\mathrm S}\;(4.5)\end{array}$$This results in the following line points/values for the joint diagram:
Line  Deformation  Force 

Bolt  0  0 
f_{0,2}  F_{0,2}  
Plate  f_{SMmax}  F_{Mmax} 
f_{SMmax} + f_{PMmax} or f_{Mmax}  0  
Working load  f_{SMmax} + f_{SA}  F_{Mmax}  F_{PA} 
f_{SMmax} + f_{SA}  F_{Mmax} + F_{SA} = F_{Smax} 
Table 1  Line Points/Values for Joint Diagram
Modeling Prestressed Bolt Connection in RFEM
The model should be a good mix of accuracy and practicability. Therefore, the connection will consist of surfaces, members, and contact solids.
The following parameters are specified for the calculation example:
F_{A} = 25 kN
F_{K} = 10 kN
E_{S} = E_{P} = 210,000 N/mm²
t_{1} = t_{2} = 10 mm
l_{K} = t_{1} + t_{2} = 20 mm
d_{h} = 10 mm
D_{A} > d_{W} + l_{K}
n = 0.5
α_{A} = 1.0
Bolt:
M10 8.8
f_{ub} = 800 N/mm²
d_{2} = 9.03 mm
d3 = 8.16 mm
s = 17 mm
The model includes two superimposed square surfaces with a hole (diameter d_{h}) in the middle, which have the dimensions 60 x 60 mm (to fulfil D_{A} > d_{W} + l_{K}). Since t_{1} = t_{2}, this results in a plate spacing of 10 mm. The load acts directly in the middle of the plate (neutral fibre). Thus, the resulting n is 0.5. The model is supported by a fixed support at the bottom end of the bolt member. In order to achieve the total support force equal to zero, the load must be applied to both the upper and the bottom plate. The load is 6.95 N/mm² for 25 kN of the total force.
For a good load transfer between the bolt (beam) and plates, a rigid surface (ring) with the outer diameter d_{W} is modeled around the hole. The connection between the plates is generated using three contact solids. One solid is around the hole without the rigid surface part, two contact solids rest around the hole like two shells. The contact solids must have the same material as the plates to reflect the stiffness between the plates accurately. Furthermore, the contact fails when lifting, and it has a rigid friction in the horizontal direction with the factor of 0.1.
Figure 04  FEA Model of Bolt Connection
The structure is displayed in Figure 04. Number 1 shows the surfaces and members with the actual dimensions. Number 2 shows the upper surface with the beam (bolt) and the rigid members, which represent the connection between the bolt and the plate. The rigid surface (pink) has also a rigid member on the inner edge to be able to transfer any moments.
Another important point is the FE mesh. Because of small dimensions, the main FE mesh size for l_{FE} has been set to 2 mm. Moreover, the surface mesh refinement has been defined with l_{FE} 0.2 mm on the rigid surfaces.
Since the bolt diameter nor the working force on the bolt are known in practice, it is possible to model the structure without the hole and to use a rigid member instead of a beam for the first design of the model and for the determination of the bolt forces. This model for pre‑dimensioning is shown in Figure 05.
Figure 05  Simplified FEA Model for PreDimensioning
In order to be able to detect the residual prestress in the model, a result member was attached parallel to the bolt (distance 0.1 mm). This includes all internal forces of the contact solid.
Comparison of Analytical and Numerical Solution
To compare the solutions, it is necessary to create the joint diagram first. The required values are listed in Table 1. By substituting the values for the practical example (see above), the intermediate values and characteristic lines shown in Table 2 are obtained. Table 3 includes the summary of the most important values analogous to Table 1, and Figure 06 shows the complete joint diagram.
Symbol  Formula Number  Value 

c_{S}  2.2  549 kN/mm 
A_{ers}  3.2  303 mm² 
c_{P}  3.1  3,182 kN/mm 
Φ_{K}  3.5  0.147 
c_{Pn}  3.4  6,921 kN/mm 
F_{SA}  3.10  1.8 kN 
f_{SA}  4.4  3 μm 
F_{PA}  3.8  23.2 kN 
F_{Mmax}  3.6, 3.7  33.2 kN 
f_{SMmax}  4.2  60 μm 
f_{Mmax}  4.3  65 μm 
F_{0,2}  3.11  46.2 kN 
f_{0,2}  4.1  84 μm 
Table 2  Intermediate Results and Results of Calculation Example
Characteristic Line  Deformation [μm]  Force [kN] 

Bolt  0  0.0 
84  46.2  
Plate  60  33.2 
65  33.2  
Working load  63  10.0 
63  35.0 
Table 3  Characteristic Line Points/Values of Calculation Example
Figure 06  Simplified Joint Diagram of Calculation Example
For the numerical solution, two load cases were initially created. The first load case (LC1 Prestress) includes the member load of the prestress, and the second case (LC2 Working load) includes the working load. Furthermore, the load combination of both load cases (factor 1.0) was generated (LC1: LC1 + LC2). The calculation is based on the linear static analysis with 15 load steps (better convergence in the case of contact solids with failure).
For the prestress, it is possible to apply the member load type initial prestress or end prestress to the member. The actual preload is the end prestress. Since the end prestress load requires a lot of computing time, it is recommended to use the member load of initial prestress. However, this has the disadvantage that this load does not include the reacting force through the plates. Therefore, the axial force in the member is too small after the calculation, since a part can be reduced by deformation of the plates. This difference can be reduced in two ways. On the one hand, this can be predicted by means of the plate deformation and converted into an additional force F_{Zus,v} (predicted) according to the following formula:
$${\mathrm F}_{\mathrm{Zus},\mathrm v}\;=\;{\mathrm f}_\mathrm{PMmax}\;\cdot\;{\mathrm c}_\mathrm S\;\;\;(5.1)$$On the other hand, this can also be determined iteratively. For this, the prestress load case must be calculated. The difference between the applied initial prestress and the resulting axial force in the member corresponds to the additional force F_{Zus,i} (iterative). The following formula can be used:
$${\mathrm F}_{\mathrm{Zus},\mathrm i}\;=\;{\mathrm F}_\mathrm{Mmax}\;\;{\mathrm N}_\mathrm S\;\;\;(5.2)$$where
The additional force F_{Zus,v} results from the values in Table 2 as follows:
$${\mathrm F}_{\mathrm{Zus},\mathrm v}\;=\;{\mathrm f}_\mathrm{PMmax}\;\cdot\;{\mathrm c}_\mathrm S\;=\;({\mathrm f}_\mathrm{Mmax}\;\;{\mathrm f}_\mathrm{SMmax})\;\cdot\;{\mathrm c}_\mathrm S\;=\;5\;\mathrm{μm}\;\cdot\;549\;\mathrm{kN}/\mathrm{mm}\;=\;2.8\;\mathrm{kN}$$The iterative additional force F_{Zus,i} can be obtained after calculating the load case on the member in Figure 07.
Figure 07  First Calculation of Prestress Load Case Without Preload Equilibrium
$${\mathrm F}_{\mathrm{Zus},\mathrm i}\;=\;33.2\;\mathrm{kN}\;\;30.4\;\mathrm{kN}\;=\;2.8\;\mathrm{kN}$$Thus, the resulting prestress is 36 kN in both cases. This allows for a recalculation of the load case. The result is shown in Figure 08.
Figure 08  Results of Prestress Load Case
The additional result member, which adds up the contact forces of all contact solids, has the result of 34.2 kN. This is about 1.2 kN more than the axial force of the bolt member that is 33 kN. The deformation of both surfaces (S1 and S27) shown in the diagram must be added in order to be able to compare it with f_{PMmax}. On average, this results as follows:
$${\mathrm u}_{\mathrm z,1}\;=\;\frac{0.00555\;+\;0.00552}2\;+\;\frac{0.00001+\;0.00003}2\;=\;5.6\;\mathrm{μm}\;>\;5.0\;\mathrm{μm}\;=\;{\mathrm f}_\mathrm{Mmax}$$The deformation is thus 0.6 μm greater than the calculated deformation f_{Mmax}.
The result of the calculations subjected to the working load in LC1 is shown in Figure 09.
Figure 09  Results of Load Combination (Preload and Working Load)
The bolt member has the result of 33.9 kN. This member force can be compared to the force F_{Smax} = 35 kN (see Table 1 and Table 3, Working load). The difference is 1.1 kN. The deviation of the differences is also important here. According to the analytical calculation, the difference should be equal to the force F_{SA} = 1.8 kN. However, the difference of the FE model is only half as great with 33.9 kN − 33 kN = 0.9 kN.
Similar deviations are obtained in the case of deformation (see the diagram in Figure 09). The value shown is the value reduced by the working load. Thus, the deformation must be calculated by the working load using the deformation by the prestress. The actual deformation is the difference between u_{z,1} and the mean deformation in the diagram. The analytical reference value is f_{SA}. This results in the deformation u_{z,2}:
$${\mathrm u}_{\mathrm z,2}\;=\;{\mathrm u}_{\mathrm z,1}\;\;\left(\frac{0.00385\;+\;0.00381}2\;+\;\frac{0.00001\;+\;0.00004}2\right)\;=\;5.5\;\mathrm{μm}\;\;3.9\;\mathrm{μm}\;=\;1.6\;\mathrm{μm}\;<\;3\;\mathrm{μm}\;=\;{\mathrm f}_\mathrm{SA}$$Thus, the deformation is about 1.4 μm smaller than the calculated deformation f_{Mmax}.
Finally, the results in the result member are compared. As you can see in Figure 09, the load for the result member is the compressive force of 10.6 kN. This value must be compared with the clamping load F_{K} = 10 kN. This results in a deviation of 0.6 kN. Table 4 includes a summary of all results.
Symbol  Analytical Value  FEM Calculation  Difference  

Stab  Plate  
F_{Mmax} [kN]  33.2  33.0  34.2  0.2 / 1.2 
f_{PMmax} [μm]  5.0    5.6  0.6 
f_{SA} [μm]  3.0    1.6  1.4 
F_{Mmax} + F_{SA} [kN]  35.0  33.9    1.1 
F_{Mmax}  F_{PA} [kN]  10.0    10.6  0.6 
Table 4  Comparison Values of Analytical Model/FEM Calculation
Evaluation
As shown in Table 4, there are partially large differences between the models. Generally, the largest matches are in the Prestress load case. Depending on the evaluation of the result member (plate) or the bolt member, the deviations from F_{Mmax} are 3.6% or 0.6% (referred to the analytical result).
The largest deviation is the result of the bolt member and the plate deformation after applying the working load. In this case, there is a deviation of 1.1 kN between the axial force on the member and the analytical solution. This deviation, referred to the analytical solution, is initially 3%. However, the difference is much greater when referred to the bolt additional force. The deviation of the FEA model is as follows:
$${\mathrm F}_{\mathrm{SA},\mathrm{FEA}}\;=\;({\mathrm F}_\mathrm{Mmax}\;+\;{\mathrm F}_\mathrm{SA})\;\;{\mathrm F}_\mathrm{Mmax}\;=\;33.9\;\mathrm{kN}\;\;33\;\mathrm{kN}\;=\;0.9\;\mathrm{kN}\;<\;1.8\;\mathrm{kN}\;=\;{\mathrm F}_\mathrm{SA}$$These deviations can be caused by the fact that both plates have no rigidity in the z‑direction and the contact solid has only one FE element in its thickness. Thus, there can be no load extension within the solid. The load transfer in the solid is done exclusively via the deformation of the plate by bending and shear force. It is obvious from the values that the FE model in the Prestress load case in combination of plates and contact solids has smaller stiffness than in the analytical model (see the smaller deformation). At this point, the higher stiffness of the bolt can be excluded as this is determined by the beam theory and the cross‑section.
On the other hand, there is a smaller deformation in the case of the load combination in the FEA model, or the bolt force has significantly smaller increment. This again indicates the higher stiffness in the plate. In summary, there is only one explanation for this: The composite of a plate and contact solids has a different load extension so the approach from Formula 3.2 in the FEA model is not valid in the form. It would be probably necessary to examine on a real example or on an extended FEA model to find out which one of both solutions is closer to reality.
However, it is important to note that the residual clamping force is almost identical in both variants. Thus, the prestress in the connection is modeled well and can be used for the joint analysis.
Summary
Modeling a prestressed bolt connection by using contact solids, surfaces, and beams is a combination of practical solution and real display. Practical means that the computing time is significantly smaller, compared to the calculation with FEA solids, which would probably represent the connection more accurately. However, it is necessary to improve the bolt design or to perform further analysis, which sets the results in relation with the reality.
Since the preloading forces and the residual clamping forces largely correspond to those from the analytical calculation, it can be assumed that this modeling type can be used for the connection analysis.
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Figure 01  Simplified Joint Diagram

Figure 02  Load Extensions in Various Plate Dimensions

Figure 03  Conversion of Plate Solid Model to Plate Surface Model

Figure 04  FEA Model of Bolt Connection

Figure 05  Simplified FEA Model for PreDimensioning

Figure 06  Simplified Joint Diagram of Calculation Example

Figure 07  First Calculation of Prestress Load Case Without Preload Equilibrium

Figure 08  Results of Prestress Load Case

Figure 09  Results of Load Combination (Preload and Working Load)