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Definition NextReturn I. Beams 1. Definition A beam is a long, slender structural member generally subjected to transverse loading that produces significant bending effects (transverse displacement rotation).

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PrevNextReturn Simple beam theory 1 2. Simple beam theory. Assumptions: (A)member cross section is constant. (B)cross section dimension 10) (C) linear variation of stress and strain.(Small deformation theory) Equilibrium equation of a differential element of the beam. Beam under distributed load Differential beam element

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PrevNextReturn Simple beam theory 2 For constant EI and only nodal forces and moments, equation becomes Solution of displacement v(x) is of cubic polynomial function of x Curvature of the beam for small slope is given by : radius of deflected curve. E: modulus of elasticity. I: principal moment of inertia about Z-axis

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PrevNextReturn Beam element stiffness formulation 1 3. Beam element stiffness formulation. (Direct equilibrium approach) (1) Beam element (No axial effects are considered.) Sign convection of the beam element: Simple beam theory sign convection for positive shear forces and moments.

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PrevNextReturn Beam element stiffness formulation 2 (2) Assume displacement function (Without distributing loading, w(x)=0) Express (x) as a function of nodal displacements as follows.

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PrevNextReturn Beam element stiffness formulation 3 In matrix form, we have

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PrevNextReturn Beam element stiffness formulation 4 (3)Element stiffness Matrix and Equations Relationships between moment, force and displacement from elementary beam theory are Using the nodal and beam theory sign convections for shear forces and bending moments, we obtain Beam elementBeam theory

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PrevNextReturn Beam element stiffness formulation 5 Hence, the beam element equation relating nodal forces and nodal displacements is given as Use the following equations into above equation. Where [k] is the element stiffness matrix for a beam element with neglected axial effects.

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PrevNextReturn Example 1 4. Example: By direct stiffness method,the system eqn. For the beam is obtained as Constant P

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PrevNextReturn Example 1 The final set of equations is The transverse displacement at node 1 and rotations at node1 and 2 are where the minus sign indicates that displacement at node 1 and the positive signs indicate counterclockwise rotations at node1 and 2. By substituting the known global nodal displacements and rotations into the system equation, we can determine the global nodal forces. The resulting equations are The global nodal forces and moments are

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PrevNextReturn Example 2 Local nodal force for each element (used for stress analysis of the entire structure) Free body diagrams for element 1 and 2 are shown as follows. According the results of the global nodal forces and moments, the free body diagram for the whole beam is given as shown.

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PrevNextReturn Example 3 Shear force diagram for the beam Bending moment diagram for the beam By using the beam theory sign conventions, the shear force V and bending moment M diagrams are shown in the following figures.

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PrevNextReturn Distributed loading 5. Distributed loading Equivalent force system: Replace the distributed load by concentrated nodal forces and moments tending to have the same effect on the beam as the actual distributing load based on the concept of fixed-end reactions from structural analysis theory. Ex.Distributed load.equivalent force system Fixed-end reactions are those reactions at the ends of an element if the ends of the element are assumed to be fixed. Beam subjected to a distributed loadFixed-end reactions for the beam

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PrevNextReturn Dis EX 1 Example: Consider the cantilever beam subjected to the uniform load w. Find the right-end vertical displacement and rotation, and nodal forces. equivalent nodal force system for uniform load w cantilever beam subjected to the uniform load w Solution: One element is used to represent the whole beam. Based on the fixed-end reactions concept, the equivalent nodal force system is given as the following figure. The system equation for the beam is

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PrevNextReturn Dis EX 1 Applying the nodal forces and the boundary conditions, we obtain Solving the above equation for the displacements, we have Therefore, the reaction forces F 1e and M 1e for the equivalent nodal force system are

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PrevNextReturn Dis EX 2 The equivalent nodal forces are Hence, the effective global nodal forces are Thus, the correct global nodal forces By comparing the two equivalent system given in the previous page, we have relationships among the correct nodal forces, the effective nodal forces and the equivalent nodal forces.

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PrevNextReturn Bar Element 1 Description Uniaxial element with tension, compression, torsion, and bending capabilities. The more general Beam element is often used instead of this element. The figure, at the end of this section, defines both element types. For some analysis programs, MSC/N4W translates both types to the same element type. Application Used to model general beam/frame structures. Shape Line, connecting two nodes. A third node can be specified to orient the element Y axis. Element Coordinate System The element X axis goes from the first node to the second.The element Y axis is perpendicular to the element X axis. It points from the first node toward the orientation (or third) node. If you use an orientation vector, the Y axis points from the first node in the direction of the orientation vector. The element Z axis is determined from the cross product of the element X and Y axes. Properties Area, Moments of Inertia (I1, I2, I12), Torsional Constant, Shear Areas (Y, Z), Nonstructural Mass/Length, Stress Recovery Locations. Additional Notes Refer to the Beam element for further descriptions regarding Releases, Offsets and Stress Recovery Locations. Bar Element

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PrevNextReturn Bar Element 2 Third Node, or Orientation Vector Plane 1 (XY) Offset A Offset B Plane 2 (XY)

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PrevNextReturn Beam Element 1 Description Uniaxial element with tension, compression, torsion, and bending capabilities. This element can be tapered. You can specify different properties at each end of the beam. Application Used to model beam/frame structures. Shape Line, connecting two nodes. A third node can be specified to orient the element Y axis. Element Coordinate System The element X axis goes from the first node to the second. The element Y axis is perpendicular to the element X axis. It points from the first node toward the orientation (or third) node. If you use an orientation vector, the Y axis points from the first node in the direction of the orientation vector. The element Z axis is determined from the cross product of the element X and Y axes. Properties Area, Moments of Inertia (I1, I2, I12), Torsional Constant, Shear Areas (Y, Z), Nonstructural Mass/Length, Stress Recovery Locations, Neutral Axis Offsets (Nay, Naz, Nby and Nbz). If the beam is tapered, you can specify different properties at each end of the element. Additional Notes You can specify Releases which remove the connection between selected element degrees of freedom and the nodes. Offset vectors defined on the Element move the neutral axis and shear center from the nodes. Neutral Axis Offsets (Y,Z) defined on the Property card move the neutral axis away from the shear center. If there are no Neutral Axis Offsets, the neutral axis and shear center are coincident. Beam Element

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PrevNextReturn Beam Element 2 Third Node, or Orientation Vector Plane 1 (XY) Offset A Offset B Plane 2 (XY) If there are no offsets,both the neutral axis and shear center lie directly between the nodes. Stress Recovery Locations define positions in the elemental YZ plane (element cross-section) where you want the analysis program to calculate stresses. Specifying moments of inertia for Beam (and Bar) elements can sometimes be confusing. In MSC/N4W, I1 is the moment of inertia about the elemental Z axis. It resists bending in the outer Y fibers of the beam. It is the moment of inertia in plane 1. Similarly, I2 is the moment of inertia about the elemental Y axis. If you are familiar with one of the analysis program conventions, the following table may help you convert to MSC/N4W's convention. MSC/N4W I1 I2 MSC/pal & CDA/Sprint Iww Ivv NASTRAN Izz Iyy ANSYS IZ1 IY1 STARDYNE I3 I2 ALGOR, mTAB & SAP I3 I2 ABAQUS I22 I11 WECAN Izz Iyy COSMOS Izz Iyy STAAD IZ IY

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PrevNextReturn Element coordinate sys. Element coordinate system. ( orientation of a beam element) Interpretation of Element Output. ． Bar element internal forces and moments. (1) plant 1 (2) plant 2

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PrevNextReturn Element coordinate sys. II. Frames and Grids 1. Rigid plane frame : (1) Definition A frame consists of a series of beam elements rigidly connected to each other. (i) joint angles between elements remain unchanged after deformation (ii) moment continuity exists at the rigid joint

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PrevNextReturn Element coordinate sys. (iii) element centroids and applied loads lie in the pane of the structure (2) Two Dimensional Beam element

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PrevNextReturn Element coordinate sys. Vectors transform law in 2D For the beam element, we have (use the 2nd eq. of above relation)

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PrevNextReturn Element coordinate sys. Global stiffness matrix for 2D Beam element where Here, the global stiffness matrix for a beam element including shear and bending effect is given as.

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PrevNextReturn Element coordinate sys. (3) 2D Beam element including axial force effect Axial force effect Combined with shear force and bending moment effects

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PrevNextReturn Element coordinate sys. or and relate the local to the global displacement by

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PrevNextReturn Element coordinate sys. The global stiffness matrix for beam element including axial, shear and bending effects where The analysis of a rigid plane frame can be performed by using above stiffness matrix.

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PrevNextReturn Grid-1 2. Grid (1) Definition A Grid is a structure on which loads are applied perpendicular to the plane of the structure. (2) Grid Element (Shear + bending + torsion)

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PrevNextReturn Grid-2 Torsional bar element stiffness matrix. where G = shear modulus ; J =centroidal polar moment of inertia Fig. Nodal and element torque sign conventions The relationship between torque and twist is

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PrevNextReturn Grid-3 We assume the shear loading to go through the shear center of these open cross sections to prevent twisting of the cross section. By Combining the torsional effect with the shear and bending effects, the local stiffness matrix equation for a grid element is written as: Hence, the stiffness matrix for the torsional bar is

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Grid-4 PrevReturn The transformation matrix relating local to global D.O.F for a grid is Hence, the global stiffness matrix for a grid element is

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