CE 6 ab Beam Analysis by the Direct Stiffness Method Beam Element Stiffness Matrix in ocal Coordinates Consider an inclined bending member of moment of inertia I and modulus of elasticity E subjected shear force and bending moment at its ends We will consider only bending and not include axial force for this lab Define a local coordinate system x y where the x axis is attached to the long dimension of the member and runs from the i initial) end of the member to the j terminal) end of the member Note that the SAP local axis is analogous to the x axis We seek a relationship between the shear and bending moment at the member ends and the transverse displacement and rotation at the ends of the form:! " V i M i V j M j * * = * * * )* k k k k4 k k k k 4 k k k k 4 k 4 k 4 k 4 k 44! - - -" - -,- Δ i θ i Δ j θ j where the sixteen terms in the 4x4 matrix are the stiffness influence coefficients which make up the element stiffness matrix in local coordinates θ j V j x y y θ i Δ j j M j V i EI M i Δ i i φ x Vukazich CE 6 Beam Direct Stiffness ab []
We can find this relationship from the analysis of a fixed-fixed beam:! " V i M i V j M j ) = * EI EI 4EI EI EI EI EI 4EI,! " - Δ i θ i Δ j θ j ) we can write Eqn ) in shorthand form: Q = k where the 4x4 beam element stiffness matrix in local coordinates is [ k! ] = EI EI 4EI EI EI EI EI 4EI ) Note that, as was the case in the truss element, the [k ] matrix is symmetric k pq = k qp for p q) Note that all stiffness matrices are symmetric Example of Direct Stiffness Assembly of the Beam Structure System of Equations In order to illustrate the concept of the assembly of a system of equations for the entire beam structure, consider the beam that is made of two elements labeled and as shown: Vukazich CE 6 Beam Direct Stiffness ab []
y, y P 5 4 6 x, x Note that the global degrees of freedom are labeled related to the joint numbering For a general joint with number n; n - n n For this example, the beam has joints and so there are a total of 6 global DOF for the structure and so, similar to the truss problem, the structure system of equations will be of the form: F = K ) where for this example; [K] = 6x6 global stiffness matrix; {F} = 6x vector of applied joint forces and support reactions); {Δ} = 6x vector of joint displacements Assembly of the Truss Structure Stiffness Matrix We will assemble the 6x6 structure stiffness matrix from the 4x4 element stiffness matrices Eqn 7) for elements and A connectivity table is constructed that maps each element DOF with its corresponding global DOF shaded) Element DOF Vukazich CE 6 Beam Direct Stiffness ab []
Δ i ) θ i ) Δ j ) 4 θ j ) Associated global DOF for element 4 Associated global DOF for element 4 5 6 The structure global) stiffness matrix is assembled from the two element contributions [ k] = 4 k k k k4 k k k k 4 k k k k 4 4 k 4 k 4 k 4 k 44 [ k] = 4 5 6 k k k k4 4 k k k k 4 5 k k k k 4 6 k 4 k 4 k 4 k 44 4) Note that the rows and columns of the element stiffness matrices are labeled with their corresponding global DOF in order to aid the assembly of the structure system of equations which yields the 6x6 structure global) stiffness matrix [ K] = k k k 4 k 4 4 5 6 k k k k 4 k k k k k 4 k 5 k 6 k 4 k 4 k 4 k 4 k k 44 k k k 4 k k k k 4 k 4 k 4 k 4 k 44 5) Beam Structure Global) System of Equations Next the beam structure system of equations can be assembled Note that global DOF, 4 and 6 are unrestrained free) and DOF, and 5 are restrained supported) We can partition the structure system of equations by restrained and unrestrained DOF With the unrestrained partitions shaded below Vukazich CE 6 Beam Direct Stiffness ab [] 4
" V M P V 5 ) = * K K K K 4 K K K K 4 K K K K 4 K 5 K 6 K 4 K 4 K 4 K 44 K 45 K 46 K 5 K 54 K 55 K 56 K 6 K 64 K 65 K 66," - Δ 6) Note at the unrestrained DOF we know the forces or moments applied to the joints but the joint displacements and rotations are unknown At the restrained DOF we know that the displacements or rotations) are equal to zero at the supports but we do not know the support reactions We can solve the following system of equations for the unknown displacements at the unrestrained DOF " P ) = * K K 4 K 4 K 6 K 44 K 64 K 6 K 46 K 66," - Δ or equivalently! " K K 4 K 4 K 6 K 44 K 64 K 6 K 46 K 66 * ) * Δ, * * - = ) * * P, * - * 7) once Δ, θ4, and θ6 are found by solving the system of equations shown in Eqn 7, the support reactions can be found by performing the matrix multiplication! " V M V 5 * = * * )* K K 4 K K 4 K 5 K 54 K 56! - -" -,- Δ 8) and Eqn can be used to find the shear and bending moments at the ends of the individual beam members Vukazich CE 6 Beam Direct Stiffness ab [] 5
CE 6 Direct Stiffness Beam Analysis ab Problem y, y k 5 4 6 x, x ft 6 ft For the beam shown the properties of the elements are: Member Section I E W8x 8 in 4 9 ksi W8x 8 in 4 9 ksi Using the coordinate system given in the figure: Find the 4x4 element stiffness matrices be guided by Eqn ) and write the values in the spaces below Use force units of kips and length units of inches for all calculations 4 [k] = 4 Vukazich CE 6 Beam Direct Stiffness ab [] 6
4 5 6 [k] = 4 5 6 Assemble the 6x6 structure stiffness matrix be guided by Eqn 5) from the element contributions found in Step and write the values in the spaces below 4 5 6 [K] = 4 5 6 Vukazich CE 6 Beam Direct Stiffness ab [] 7
From the structure system of equations write the x system of equations Eqn 7) for the unrestrained DOF in the space below Δ 4 Verify that the solution to the x system of equations from Step is:! = 855 in θ! = 559 rad θ! = 54 rad 5 Find the support reactions V, M, and V 5 ) using Eqn 8 and your results from Step 4 6 Using statics, verify that the results from Step 5 satisfy equilibrium of the beam Vukazich CE 6 Beam Direct Stiffness ab [] 8