In penalty stiffness method, it is assumed that a high-stiffness spring is connected between the contacting points. In this case, the elastic energy can be written as: U=Ub1(u1)+Ub2(u2)+Uc, where Ub1(u1) and Ub2(u2) are the strain energy of the first and the second beam respectively and Uc is the strain energy due to contact (stored in the spring), which can be split into normal, tangential, rotational (constraint between tangents to the beams), as well as other contributions (such as out-of-plane or torsional rotations which are ignored here): Uc=UNc+UTc+URc+…, where the strain energy due to normal contact, UNc, and tangential contact, UTc, have already been discussed in , , for beams with circular cross-sections. As a result, to formulate the finite element equations, we just need to find δURc and ΔδURc in terms of the nodal displacements. We have the following equation for URc: URc=12εRgR2 where gR is the rotational gap and εR is the rotational penalty stiffness. This equation leads to the following contributions of the rotational constraint: δURc=εRgRδgRΔδURc=εRΔgRδgR+εRgRΔδgR

We want to solve the Eq. () subjected to the constraint gR=0. Using Lagrange multiplier method, we can enforce this constraint by solving the following equation instead: δ(Π+λRgR)=δΠ+δ(λRgR)=0, where λR is the lagrange multiplier. The second term in this equation can be calculated as: δ(λRgR)=δλRgR+λRδgR To solve Eq. () using Newton–Raphson method, Δδ(λRgR) needs to be calculated as follows: Δδ(λRgR)=δλRΔgR+ΔλRδgR+λRΔδgR

Note that while with the first (consistent approach) formulation presented in Sect. (), both the Lagrange multiplier and penalty stiffness methods can be used, the second formulation (non-consistent approach) is derived only for the penalty stiffness method. Using either the penalty stiffness or Lagrange multiplier method, we need to find gR, δgR, ΔgR, and ΔδgR in terms of the nodal displacements of the beam elements in contact.

In this approach, in addition to the assumptions mentioned above, we assume that the beam elements have linear shape functions, both for their geometry and displacements (this is the same assumption as in ), to simplify the derivation. As a result, the unit tangent vectors to the beams (at the contact point) can be written as follows: t1=xb-xa‖(xb-xa)‖=xb-xal1t2=xd-xc‖(xd-xc)‖=xd-xcl2 where, as shown in Fig. , t1 and t2 are the tangents to Beams 1 and 2, xa and xb are the current positions of the starting and end nodes of Beam 1, xc and xd are the current positions of the starting and end nodes of Beam 2, and l1 and l2 are the current lengths of Beams 1 and 2, respectively. To find the variation of the rotational gap, δgR, (similarly for ΔgR), we have δgR=δ(θ-θ0)=δθ-δθ0=δθ-0=δθ We know that the inner product of two unit vectors gives the cosine of the angle between them: cos⁡θ=t1⋅t2 Using this equation, we can find the variation of the angle between the two tangent vectors as follows: δ(cos⁡θ)=δ(t1⋅t2)=δt1⋅t2+t1⋅δt2δ(cos⁡θ)=-sin⁡θδθ⇒δθ=δgR=-δt1⋅t2+t1⋅δt2sin⁡θ=-δt1⋅t2+t1⋅δt21-(t1⋅t2)2 As shown in Appendix , the variation of the unit tangent vectors t1 and t2 can be calculated by δt1=1l1(I-t1⊗t1)⋅(δub-δua)δt2=1l2(I-t2⊗t2)⋅(δud-δuc) Substituting Eqs. () into (), we can find δgR: δgR=-11-α2(1l1(t2-αt1)⋅(δub-δua)+1l2(t1-αt2)⋅(δud-δuc)) where the notation α=cos⁡θ=t1⋅t2 is used to shorten the equation. Using matrix notation and defining u=u1u2 with u1=uaub and u2=ucud, we can rewrite Eq. (): δgR=11-α2δuT1l1(t2-αt1)-1l1(t2-αt1)1l2(t1-αt2)-1l2(t1-αt2)=δuTGT Similarly, ΔgR can be written as ΔgR=GΔu To find ΔδgR, we use Eq. () and the following change of variables:

δgR=-ABA=δt1⋅t2+t1⋅δt2B=1-(t1⋅t2)2 As a result, we can write ΔδgR as ΔδgR=-(ΔA)B-A(ΔB)B2=-1B(ΔA+δgRΔB) Now we need to calculate ΔA and ΔB. ΔA=Δδt1⋅t2+δt1⋅Δt2+Δδt2⋅t1+δt2⋅Δt1 Using the results obtained in Appendices , , and the notation tij=ti⊗tj (i,j=1,2) to shorten the equation, we can write ΔA as ΔA=-1L12(δub-δua)⋅(αI+t12-2αt11)⋅(Δub-Δua)+1L1L2(δub-δua)⋅(I-t11-t22+αt12)⋅(Δud-Δuc)+-1L22(δud-δuc)⋅(αI+t21-2αt22)⋅(Δud-Δuc)+1L1L2(δud-δuc)⋅(I-t11-t22+αt21)⋅(Δub-Δua) Written in matrix notation, we have the following equation: ΔA=δuTP1-P1P3-P3-P1P1-P3P3P4-P4P2-P2-P4P4-P2P2Δu=δuTQ1Δu where P1=-1L12(αI+t12-2αt11)P2=-1L22(αI+t21-2αt22)P3=1L1L2(I-t11-t22+αt12)P4=1L1L2(I-t11-t22+αt21) To calculate ΔB, we can use B=sin⁡θ, obtaining ΔB=Δ(sin⁡θ)=cos⁡θΔθ=αΔgR Substituting Eqs. () and () into Eq. (), we get ΔδgR=-11-α2δuTQ1Δu-α1-α2δgRΔgR Using Eqs. () and (), we get δgRΔgR as follows: δgRΔgR=δuTGTGΔu=11-α2δuTP5-P5P7-P7-P5P5-P7P7P8-P8P6-P6-P8P8-P6P6Δu where P5, P6, P7, and P8 are given by P5=1L12(t22-αt12-αt21+α2t11)P6=1L22(t11-αt12-αt21+α2t22)P7=1L1L2(t21-αt11-αt22+α2t12)P8=1L1L2(t12-αt11-αt22+α2t21) Finally, substituting Eq. () into Eq. (), we get ΔδgR=δuT(-11-α2Q1+-α1-α2GTG)Δu=δuTQΔu

Using the results obtained in Eqs. (), (), and (), and substituting into Eq. (), we get the following tangent stiffness matrix for the rotational constraint, using the Lagrange multiplier method: KR=λRQGTG0

To derive the tangent stiffness matrix of the rotational constraint using a penalty stiffness method, we substitute Eqs. () and () into Eq. () to get the following equation: KR=εR(GTG+gRQ).

In this approach, in contrast to the consistent formulation presented above, we do not assume a particular form of the shape functions for the beam elements. Instead we assume that the contact normal and contact location are constant (independent of the deformations) in a single iteration. However, the contact normal and location can (and most often do) vary from one iteration to the next. For this reason we consistently update them between iterations. Using the assumption of independence of contact normal and location from the deformations (in a single iteration), the contribution from the change of contact normal to the linearization of the contact gap function vanishes. This simplifies the formulation significantly. Also, we will use a penalty stiffness method to enforce the rotational constraint. Using these assumptions we can write the incremental change of the rotational gap, ΔgR, as ΔgR=Δθ2N-Δθ1N where Δθ1N and Δθ2N are shown in Fig. . In this figure, the superscripts i-1 and i are iteration numbers and the subscript N refers to the projection onto the plane of the current contact normal. tj is the unit tangent vector to Beam j and θj is the rotation vector of Beam j at the contact point (j=1,2). We can observe that Δθ1N and Δθ2N are in fact projections of the changes in the rotation vectors of Beams 1 and 2 at the contact point, onto the plane of current contact normal (the plane spanned by the tangents to the beams at the contact point), respectively: Δθ1N=(θ1Ni-θ1Ni-1)=(θ1i-θ1i-1)⋅n=Δθ1⋅nΔθ2N=(θ1Ni-θ2Ni-1)=(θ2i-θ2i-1)⋅n=Δθ2⋅n where Δθ1 and Δθ2 are the incremental changes in the rotation vectors of Beams 1 and 2 evaluated at the contact point and n is the current contact normal which is updated at each iteration. Δθ1 and Δθ2 can be related to the nodal incremental displacements of Beams 1 and 2, Δu1 and Δu2, through the rotational shape functions of the beam elements, HR1 and HR2: Δθ1=HR1Δu1Δθ2=HR2Δu2 The present derivation is independent of the beam formulation, however, as examples, the rotational shape functions are derived in Appendices  and  for a linear and an Euler–Bernoulli beam element, respectively. Substituting Eqs. () and () into Eq. (), we obtain ΔgR=-nTnTHRΔu where HR and Δu are given by HR=HR100HR2Δu=Δu1Δu2 Similarly, δgR=δuHRT-nn Based on the assumption made for this approach, that the contact normal and location are independent of the nodal displacements in an iteration, we obtain ΔδgR=0 Using Eqs. (), (), (), and (), we can write the tangent stiffness matrix for the rotational constraint as KR=εRHRTNHR where N is N=n⊗n-n⊗n-n⊗nn⊗n=-nn-nTnT Using Eq. (), the internal force vector of this rotational constraint element can be written as FRint=εRgRHRT-nn

In this example we consider two crossing perpendicular beams. The translational degrees of freedom are constrained at the two ends of one beam, while the other beam is given a prescribed in-plane displacement at the two ends, which would cause a rigid body rotation of 15∘ around the mid-point of the beam if the rotational constraint were not present. The initial configuration and boundary conditions are shown in Fig. a. We have no out-of-plane deformation and we assume bonded constraint (inseparable contact) between the beams. A sample deformed configuration is shown in Fig. b. To crosscheck the results, we consider another case, where the beams are located in the same plane and have a common node instead of the contact point. Although these two problems are not exactly the same, as the beams inter-penetrate in the case of a shared node, the in-plane displacements should be similar and we will have a comprehension of the convergence rate, which we can use as a reference. In Fig. , the deformed shape for the case with the common node is compared with the deformed shape for the case with the rotational constraint element, when using a rotational penalty stiffness of 105. It should be noted that although the deformations of the two cases are similar, the configurations are different: in the case with a contact point, the beams are located at different z levels, whereas in the case with the common node, they are in the same plane. The effect of the size of the rotational penalty stiffness on the rotational gap is shown in Fig. . Note that the common node is similar to a contact point with an infinitely large penalty stiffness, which results in a zero gap. The change in the imbalance of energy during the analysis is shown in Fig.  for the case with the common node and the case with the contact point using both consistent and non-consistent formulations. As can be seen, including the rotational constraint has a small effect on the convergence of the solution. Due to the fact that the contact point and normal remain unchanged during the iterations of this analysis, the proposed consistent and non-consistent methods behave identically in solving this example.

In this example, we consider two crossing perpendicular beams which are fully constrained at one end. We apply an in-plane force of Fy= 50 units and an out-of-plane force of Fz=-50 units to the free end of the upper beam, as shown in Fig. a. We assume a bonded constraint (inseparable contact) between the beams and analyze the problem with and without the rotational constraint. The deformed configuration of the beams is shown in Fig. b and in Fig.  the deformed shapes of the beams, including and ignoring the rotational constraint, are compared.

The change in the energy imbalance during the solution process is shown in Fig.  when using consistent and non-consistent formulations, and when ignoring the rotational constraint. We can see that including or ignoring the rotational constraint has a small effect on the convergence. It should also be noted that the consistent and non-consistent formulations both lead to the same results for nodal deformations and forces. The consistent method has a better convergence if very narrow tolerances are considered and the non-consistent method has better stability with larger load steps.

In this example, we start with a structure consisting of 4 intertwined beams of length 10, as shown in Fig. a. The endpoints of the beams at the left side of the structure are fixed to avoid translation and a unit displacement Δ=1 is applied to the endpoints located at the right side of the structure, as shown in Fig. b. The stiffness of the structure is calculated as follows: k=-RΔ where R is the sum of the reaction forces in the x direction (direction of the prescribed displacement) collected from the constrained nodes on the left side, and Δ is the unit displacement applied on the right side. Without the rotational constraints the structure has zero stiffness and undergoes rigid body rotations.

We then modified our model, by adding beams of length 5 that passed through the intertwined beams in order to get the models shown in Fig. . To study the effect of the rotational penalty stiffness on the stiffness of the structure, we included the rotational constraint and changed the rotational penalty stiffness. The results are shown in Fig.  for the different models of Fig. ). As can be seen in this figure, if the rotational penalty stiffness is above a certain value (104 in this example) the results are not affected. Figure a shows the structure stiffness of the different models, with and without the rotational constraint. In Fig. b the ratio between the stiffness of the models when including and excluding the rotational constraint is plotted. As can be seen, with increasing the number of beams and bonds between them, this ratio decreases, however, its effect is considerable even in the last case (Model 4). In the next example, we will demonstrate that even with a random network consisting of hundreds of fibers, the rotational constraint has a big effect on the stiffness.

In this example, we generated a random fiber network of size 20×20 units, using a deposition technique . The network consists of 776 fibers meshed with 11 789 beam elements of equal length coming into contact at 1322 points. An initial inter-penetration which happened between certain beams was resolved by offsetting the contact distance by the initial value of penetration. The mean, min, max and standard deviations for the fiber lengths and cross-sectional radii are given in Table (). It should be noted that the width of the network is larger than the length of the longest fiber and as a result, no fibers are extending from one side of the network to the other. We have defined 20 different materials with equally spaced values of the elasticity modulus, in the range of 5×105 to 1.5×106, and randomly assigned one of these materials to each fiber.

All the contact points are assumed to be inseparable (bonded constraints) throughout the analysis. We assumed that no new contacts are established during the deformation. A total prescribed displacement of 3 units corresponding to 15 % of the network length is applied to the model, as shown in Fig. b. The 3-D configuration of the network is shown in Fig. a. To demonstrate how the rotational constraint affects the stiffness of the network, we performed a geometrically non-linear analysis with and without the rotational constraint. The sum of the reaction forces in the direction of the applied displacement is plotted versus the applied displacement for both cases in Fig. . In both case, the initial response was linear. However, the inclusion of the rotational constraint increases the stiffness by a factor of 3.5 approximately. This contribution from the rotational constraint is dependent on the geometry of network, for example with a more compact network (with more bonds between fibers and shorter distance between contact points along a fiber), the effect of the rotational constraint decreases. It is worth mentioning that the problem computed with rotational constraints required significantly fewer iterations, as the added constraints effectively prevented any rigid body rotations of the loosely constrained fibers.