Learning Forum

[Rameez_ul_Haq]

This may be a simple concept to some, but is a little confusing for me. For example, I have a cantilevered beam and I apply a transverse force on one end and apply fixed support on the other. The beam will not undergo a rigid body motion because there will be reaction force and moment at the support. The nodes of the elements in fixed support cannot undergo any rotation as well as translation.nNow, if I only restrict the translational DOFs on the supported end, then this would automatically mean that the rotation at that end is restricted since rotation requires a tranlational motion of the nodes on the supported end. Will there be a rigid body motion in this case? Why won't I still see a moment reaction at the supported end? What is the basic difference in restricting the translational DOFs and rotational DOFs?n

[Rameez_ul_Haq]

Assume the beam is made of solid elements.n

[Rameez_ul_Haq]

Need help on this one please?n

[peteroznewman]

nThe nodes on solid elements only have translational DOF. There is no rotational DOF on a node connected to a solid element.nWhen you use Fixed Support on the face of a solid body, the nodes on that face are only getting XYZ translation set to 0. This is all that is needed.nIf you selected one straight edge on a solid body and applied Fixed Support, a Statics solution would fail, because there are no rotational DOF on the solid elements and holding just one straight edge creates a mechanism I would call a hinge, and that has no static solution.nShell elements have 6 DOF on their nodes, so if you mesh a surface and apply a fixed support to one straight edge of the surface, there is a static solution.n

[Rameez_ul_Haq]

Array, does rotational DOFs mean that the nodes can rotate about its center? Or does it mean they can rotate about any other axis or point?nWhy do the nodes on solid (rather it be on the surface of it or inside) have only 3 DOFs while the shell elements have 6 DOFs? Whats the difference between these two?nThere is no rotational DOF on a node connected to a solid element, you mentioned this. What does this mean? Does this mean that the solid element nodes cannot rotate at all? About which axis or location?nIf you selected one straight edge on a solid body and applied Fixed Support, a Statics solution would fail, because there are no rotational DOF on the solid elements and holding just one straight edge creates a mechanism I would call a hinge, and that has no static solution, Well, if the force is passing in such a manner that there is no net moment about the fixed support edge, then there is a rigid body motion right? Otherwise even if it is hinge, and no moment is occuring, then there will exist a static solution. Am I correct Sir?.n

[Rameez_ul_Haq]

Array, what does rotational DOFs mean? Does it mean that the node can rotate about its center (which requires no translation) or does it mean the node can rotate about any other point or axis (which actually requires a translational motion of the node).nWhy does the nodes of a solid element have only 3 DOFs while shell elements have 6 DOFs? What does this mean? DOF is an acronym for Degree Of Freedom, so if the rotational DOFs for a solid element are not there, then this means that the nodes of solid element cannot rotate at all? Am I correct? Or does it mean that the nodes of solid element are able to rotate but cannot prevent their rotations by any means (even if fixed support is applied to them)?nIf you selected one straight edge on a solid body and applied Fixed Support, a Statics solution would fail, because there are no rotational DOF on the solid elements and holding just one straight edge creates a mechanism I would call a hinge, and that has no static solution, WELL, only if the there is a moment creation about the fixed support line right? I mean if there is no moment creation by the applied force about the fixed support line then I think the static solution should work.

[peteroznewman]

nYes, a rotational DOF is the angular rotation about the node.nDegree of Freedom (DOF) also means unknown quantities that will be determined by the solver. Some nodal DOF are removed from the solution by being specified by Boundary Conditions where those DOF are specified, such as X = 0.nSolid elements have mathematical equations built into them that use nodal displacements. There are no equations for nodal rotations. This is because that is not necessary. If the element as a whole needs to rotate, that is done by translating the nodes.nShell elements represent 3D objects with zero thickness geometry. In order to represent bending, there must be equations that include nodal rotation, because you can bend a single linear shell element without any nodal displacement by just rotating the nodes at each end. nYou can?t ?bend? a single linear solid element. If you have many elements through the thickness and along the length, you can represent bending by translation of the nodes of many elements to make a curved shape.nA solid cube with a fixed support on one edge will fail to solve, even if there is no moment about that edge. n *** ERROR *** CP = 1.125 TIME= 17:15:41n There is at least 1 small equation solver pivot term n(e.g., at the UX degree of freedom of node 26). nPlease check for an insufficiently constrained model. nThe mathematical steps the solver uses to compute the deformation of the nodes requires a matrix inversion. If the matrix is singular, which happens if there are not at least six constraints to ground, that causes an error much like dividing by zero. This is done before the forces acting on the structure are used in the solution. nANSYS has a feature called ?Weak Springs? that will put six weak springs to ground for every body in the model. If you turn that feature on, then instead of dividing by zero, the solver gets to divide by a small number. In the case of a single straight edge support, no force will go through the weak springs, but their stiffness will allow the matrix to be inverted.n

[peteroznewman]

Correction:you can bend a single linear shell element without any nodal displacement by just rotating the nodes at each end is not correct.nYou can bend a single linearbeamelement without any nodal displacement by just rotating the nodes at each end. The shape function in a beam element is quadratic. The shape function in a linear shell element is linear so you get no internal displacement in the element from applied rotation.nYou can bend a singlequadraticshell element by just rotating the nodes at each end because the element has a quadratic shape function. There is also a midside node which would translate.n

[Rameez_ul_Haq]

If the element as a whole needs to rotate, that is done by translating the nodes'', you mentioned this fact for solid elements. Now imagine a situation where there is a solid in reality and it is fixed in such a position that its only one edge is grounded (and not behaving as a hinge but actually restricting the rotations of the nodes on that edge as well), so in the ANSYS model you are trying to say that if I make that edge as fixed, it won't solve since it is unstable. What should I do then? Cause in reality only one of the edges of the solid is grounded, You said we can use WEAK SPRINGS option to make the mathematical equations work. Can I input constraint equations to keep the rotations of the nodes on that edge to be equal to zero? But you also mentioned at the same time that the rotational equations for a solid elements do not exist at all in the ANSYS solver.nAnd also, what would this mean if a pick a an edge of a solid model and try to apply a moment there? This cannot be done since the nodes of the solid elements cannot rotate at all? What do you think Array?nAlso, the thing you mentioned about fixing the edge of a solid and regarding it as like a hinge connection, so that does mean the nodes on that edge of solid elements can rotate about each of its center, but cannot resist the rotation? And that is why the the matrix is singular and rigid body motion is caused? Am I right? I mean the rotations of the nodes about each of its center is kind of possible though in the solid by making an analogy to a hinge connection.nAbout the WEAK SPRINGS, how correct and dependable will be my answers? All the forces and moments applied to the model will only travel to the boundary condition I inputted and none will go to the weak springs, is that what you mentioned Array? n''which happens if there are not at least six constraints to ground'', I think you meant to say that atleast six constraints to the ground IN THE CORRECT POSITION. Cause even if there are more than six constraints, there can still be a matrix singularity and hence a rigid body motion. Am I correct?n''You can’t "bend" a single linear solid element'', what about single quadratic solid element. By rotating the nodes on the ends, the mid side node can also translate for a quadratic solid element, as like you said for the quadratic shell element.

[Rameez_ul_Haq]

,I heavily appreciate your guidance on these topics. Because it is usually not possible for students or fresh engineers to talk in detail about these kind of topics either in schools or at work. nSince we have brought up the topic for a shell element, i want to ask a side question as well. For a shell element, along the thickness there is only one element. Then why do we see a difference in stress or displacement results along the thickness of shell elements in ANSYS?nAnd also why do shell elements take less time to solve than the solid elements even though the nodes on shell elements have 6 DOFs while for a solid element has 3 DOFs only?n

[peteroznewman]

''You can’t "bend" a single linear solid element'', what about single quadratic solid element. By rotating the nodes on the ends, the mid side node can also translate for a quadratic solid element, as like you said for the quadratic shell element.nSolid elements are based on equations that use nodal translation. There are no equations that use nodal rotation, so the element has no knowledge of the nodal rotation.nwhy do we see a difference in stress or displacement results along the thickness of shell elements in ANSYS?nShell elements have equations that use the assigned thickness to calculate the stress through the thickness. That is why you can plot the stress on the top of the shell or the bottom of the shell. You can also render the thickness in Mechanical by turning on Thick Shells and Beams.nAnd also why do shell elements take less time to solve than the solid elements even though the nodes on shell elements have 6 DOFs while for a solid element has 3 DOFs only?nShell models take less time to solve because there are fewer equations in the model. Take an example of a 10 x 100 x 1 mm plate meshed with 1 mm linear shell elements. There are 6 x 1,000 = 6,000 equations in the model. If that solid is meshed with linear solid elements, you cannot represent the bending through the 1 mm thickness with only 1 solid element. For linear elements, a minimum of 4 solid elements are required through the thickness, which means there are 5 nodes through the thickness. That means the number of equations is 3 x 5,000 = 15,000 equations. So the shell model is less than 1/2 the size of the solid model.n''which happens if there are not at least six constraints to ground'', I think you meant to say that at least six constraints to the ground IN THE CORRECT POSITION. nYes, when you turn on Weak Springs, ANSYS puts springs to ground on each body that prevent rigid body motion. If you do that yourself, you must do it correctly.nAbout the WEAK SPRINGS, how correct and dependable will be my answers?nI usually turn off Weak Springs, but if you use them and expect there to be almost no force in the weak spring, you can probe the reaction force in the weak springs to verify that the forces are very small and having an insignificant effect on your results.nAll the forces and moments applied to the model will only travel to the boundary condition I inputted and none will go to the weak springsnNo. If the forces in your model are unbalanced and some force needs to be present in a weak spring for the solver to find equilibrium, then it will do that and you plot the reaction force in the weak spring to find that out.nthe nodes on that edge of solid elements can rotate about each of its center, but cannot resist the rotation?nYes, exactly, there is no stiffness in a solid element to resist rotation of a node. There is only stiffness to resist nodal translation. If there is a mechanism, like a hinge, in the model, the solver cannot solve (without weak springs).nif I pick a an edge of a solid model and try to apply a moment there, what happens?nNothing happens, a moment on a node connected to a solid element does nothing.nCan I input constraint equations to keep the rotations of the nodes on that edge to be equal to zero?nNo. The solid elements connected to the nodes do not care about nodal rotation. Constraint equations will only help if you pick nodes off the straight edge and choose nodes on the faces adjacent to the edge so that the collection of nodes defines an area.nthere is a solid in reality and it is fixed in such a position that its only one edge is groundednIn reality, a solid that is to be fixed must have a face with some area to be fixed. There is no way to hold a sharp edge of a solid, except in a pinned type of connection to another object. Imaging a cube rotated by 45 degrees and the edge with a 90 degree angle is sitting in a Vee shaped channel with a 120 degree angle. The edge of that cube has line contact with another body. It is free to rotate about that edge. If you want to prevent rotation about that edge, you need two lines that cover an area. Like the flat face of the cube on the flat face of another body. Put a drop of glue between them and you have a fixed support.n

[Rameez_ul_Haq]

Thank you.n''Constraint equations will only help if you pick nodes off the straight edge and choose nodes on the faces adjacent to the edge so that the collection of nodes defines an area'', even then I cannot constrain the rotational motion of the solid elements' nodes right. Meaning that rotational constraint equations for solid elements are basically useless?n

[peteroznewman]

nYes, rotational constraint equations are basically useless on nodes connected to only solid elements.n

[Rameez_ul_Haq]

,so what does making a fixed joint connection between a node of line body and a face on a solid mean? Because the face on the solid has nodes whose rotational motion cannot be restrained, while line body has nodes whose rotational motion can be restrained.n

[peteroznewman]

nSuppose you have a line body that ends at a node, which we can call node 99.nSuppose that node 99 is near one face of a solid body and that face has nodes spread around the YZ plane.nThe Fixed Joint creates Constraint Equations (or spider elements), which you can see in red after solving, go from node 99 to all the nodes on the face.nSince solid elements have stiffness, they prevent the nodes on the face from translating very much, which includes translating along the X axis.nSo if node 99 is rigidly connected to a bunch of nodes in the YZ plane that can't move much in the X direction, that prevents rotation of node 99, even though none of the nodes in the YZ plane are prevented from rotating themselves.n

[Rameez_ul_Haq]

,what if the connection is not rigid, but deformable?nAnd if we replace the same face of solid with a surface body, and if we rotate the vertex of the node99 of line body, will it cause the nodes on that surface body to rotate or translate? Because for a face of solid body, they should only translate.n

[peteroznewman]

nIf the connection is deformable, then instead of all the nodes on that face moving in a rigid manner with no relative motion of nodes within the surface, the forces and moments at node 99 will be distributed to all the nodes on the surface and the nodes on the surface can deform relative to one another.nIf you apply a fixed joint to the face of a surface, everything I said above is the same and when you rotate node 99, the face of the surface will move through space in a rigid manner so the nodes on that surface will be translating and rotating, but with no relative motion of nodes within the face. If the joint is deformable, then the nodes on the surface can translate and rotate relative to each other by the amount that the equilibrium with the shell elements that provide stiffness against relative translation and rotation allow.n