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Beam Direct Stiff Lab CE 160 Lab – Beam Analysis by the Direct Stiffness Method Beam Element Stiffness Matrix in Local 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 2000 local 1 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 % ! Δ % # i # * k11 k12 k13 k14 -# i # M * - θ # i # k 21 k 22 k 23 k 24 # i # " & = * -" & V Δ # j # * k 31 k 32 k 33 k 34 -# j # # # * -# # M j θ j $ ' )* k 41 k 42 k 43 k 44 ,-$ ' where the sixteen terms in the 4x4 matrix are the stiffness influence coefficients which make up the element stiffness matrix in local coordinates. θj V x’ j M j y Δj y’ θi j V EI i M i Δi L i φ x Vukazich CE 160 Beam Direct Stiffness Lab 11 [L11] 1 We can find this relationship from the analysis of a fixed-fixed beam: ) 12EI 6EI 12EI 6EI , + 3 2 − 3 2 . + L L L L . ! V % ! Δ % # i # + 6EI 4EI 6EI 2EI .# i # + 2 − 2 . # Mi # L L L L # θi # " & = + ." & (1) Vj + 12EI 6EI 12EI 6EI . Δ j # # − − − # # + 3 2 3 2 . # M # L L L L # θ # $ j ' + .$ j ' + 6EI 2EI 6EI 4EI . 2 − 2 *+ L L L L -. we can write Eqn. (1) in shorthand form: � = � ∆ where the 4x4 beam element stiffness matrix in local coordinates is # 12EI 6EI 12EI 6EI & % 3 2 − 3 2 ( % L L L L ( % 6EI 4EI 6EI 2EI ( % 2 − 2 ( L L L L [k!] = % ( (2) % 12EI 6EI 12EI 6EI ( − − − % L3 L2 L3 L2 ( % ( % 6EI 2EI 6EI 4EI ( 2 − 2 $% L L L L '( Note that, as was the case in the truss element, the [k ] matrix is symmetric (kpq = kqp 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 1 and 2 as shown: Vukazich CE 160 Beam Direct Stiffness Lab 11 [L11] 2 y, y’ P 1 3 5 2 4 6 x, x’ 1 1 2 2 3 L 1 L2 Note that the global degrees of freedom are labeled related to the joint numbering. For a general joint with number n; 2n - 1 2n n For this example, the beam has 3 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: � = � ∆ (3) where for this example; [K] = 6x6 global stiffness matrix; {F} = 6x1 vector of applied joint forces (and support reactions); {Δ} = 6x1 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 1 and 2 A connectivity table is constructed that maps each element DOF with its corresponding global DOF (shaded) Element DOF Vukazich CE 160 Beam Direct Stiffness Lab 11 [L11] 3 1 (Δi) 2 (θi) 3 (Δj) 4 (θj) Associated global DOF for element 1 1 2 3 4 Associated global DOF for element 2 3 4 5 6 The structure (global) stiffness matrix is assembled from the two element contributions 1 2 3 4 3 4 5 6 1 1 1 1 1 3 2 2 2 2 k11 k12 k13 k14 k11 k12 k13 k14 1 2 1 1 1 1 2 2 2 2 = 2 = 4 [k] k 21 k 22 k 23 k 24 [k] k 21 k 22 k 23 k 24 3 1 1 1 1 5 2 2 2 2 k 31 k 32 k 33 k 34 k 31 k 32 k 33 k 34 4 1 1 1 1 6 2 2 2 2 k 41 k 42 k 43 k 44 k 41 k 42 k 43 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 1 2 3 4 5 6 1 1 1 1 1 k11 k12 k13 k14 0 0 1 1 1 1 2 k21 k22 k23 k24 0 0 1 1 1 2 1 2 2 2 [K] = 3 k31 k32 k33 + k11 k34 + k12 k13 k14 1 1 1 2 1 2 2 2 4 k41 k42 k43 + k21 k44 + k22 k23 k24 2 2 2 2 5 0 0 k31 k32 k33 k34 2 2 2 2 6 0 0 k41 k42 k43 k44 (5) Beam Structure (Global) System of Equations Next the beam structure system of equations can be assembled. Note that global DOF 3, 4 and 6 are unrestrained (free) and DOF 1, 2 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 160 Beam Direct Stiffness Lab 11 [L11] 4 ) , " V & K11 K12 K13 K14 0 0 " 0 & $ 1 $ + .$ $ K K K K 0 0 $ M $ + 21 22 23 24 .$ 0 $ 2 + . $ $ K31 K32 K33 K34 K35 K36 $ Δ3 $ # −P ' = + .# ' + . $ 0 $ K41 K42 K43 K44 K45 K46 $ θ4 $ + . $ V $ 0 0 K K K K $ 0 $ $ 5 $ + 53 54 55 56 .$ $ 0 + . θ6 % ( 0 0 K 63 K64 K65 K 66 % ( * - (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 K33 K34 K36 Δ3 $ $ + .$ $ # 0 ' = + K43 K44 K46 .# θ4 ' + . $ 0 $ K K K $ $ % ( *+ 63 64 66 -.% θ6 ( or equivalently ! $( , ( , K33 K34 K36 Δ3 −P # &* * * * # K43 K44 K46 &) θ4 - = ) 0 - (7) # & K K K * * * 0 * "# 63 64 66 %&+ θ6 . + . once Δ3, 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 % ( K K 0 +! Δ % # 1 # * 13 14 -# 3 # " M 2 & = * K23 K24 0 -" θ4 & (8) * - # V # K K K # # $# 5 '# )* 53 54 56 ,-$ θ6 ' and Eqn. 1 can be used to find the shear and bending moments at the ends of the individual beam members. Vukazich CE 160 Beam Direct Stiffness Lab 11 [L11] 5 CE 160 Direct Stiffness Beam Analysis Lab Problem y, y’ 12 k 1 3 5 2 4 6 x, x’ 1 1 2 2 3 12 ft 6 ft For the beam shown the properties of the elements are: Member Section I E 1 W8x10 30.8 in4 29000 ksi 2 W8x10 30.8 in4 29000 ksi Using the coordinate system given in the figure: 1. Find the 4x4 element stiffness matrices (be guided by Eqn. 1) and write the values in the spaces below. Use force units of kips and length units of inches for all calculations. 1 2 3 4 1 2 1 [k] = 3 4 Vukazich CE 160 Beam Direct Stiffness Lab 11 [L11] 6 3 4 5 6 3 4 2 [k] = 5 6 2. Assemble the 6x6 structure stiffness matrix (be guided by Eqn. 5) from the element contributions found in Step 1 and write the values in the spaces below. 1 2 3 4 5 6 1 2 3 [K] = 4 5 6 Vukazich CE 160 Beam Direct Stiffness Lab 11 [L11] 7 3. From the structure system of equations write the 3x3 system of equations (Eqn. 7) for the unrestrained DOF in the space below. Δ 3 θ 4 θ 6 4. Verify that the solution to the 3x3 system of equations from Step 3 is: ∆!= −1.238155 �� �! = 0.0051590 ��� �! = 0.0232154 ��� 5. Find the support reactions (V1, M2, and V5) 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 160 Beam Direct Stiffness Lab 11 [L11] 8 .
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