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DIRECT STIFFNESS METHOD FOR 2D FRAMES CE 131 — Theory of Structures Henri Gavin Fall, 2006 1. Identify Degrees of Freedom. Number all the global degrees of freedom in your frame. In a planar frame, every joint has three degrees of freedom: one in the global X-direction, one in the global Y-direction, and one rotation about the global Z-axis (counter-clockwise). Using the method in this handout, every joint gets three degrees of freedom. 2. Number all of the elements. 3. Joint Coordinates. Write the coordinates of each joint using units consistent with E and I. In other words, if E and I are given in kN/cm2 and cm4 , write the coordinates, (x, y), in terms of centimeters. 4. Deﬁne each element. Draw each element of your frame individually and draw the local coordinates in the global directions. For example if element number N is a diagonal beam element, and the global directions are X: horizontal and Y: vertical, draw element number N like this: 5 2 6 4 Bar Direction θ 2 1 1 Local Joint Number 1 1 Local Degree of Freedom 3 Local Coordinate System where 1,2,3,4,5,6 are the LOCAL coordinates of the beam element in the GLOBAL di- rections. The local coordinates are always numbered 1,2,3,4 with 1 and 4 pointing in the global X direction (to the right), with 2 and 5 pointing in the global Y direction (up), and with 3 and 6 rotating about the global Z-axis (counter-clockwise). All of these six coordinates will line up with with the global degrees of freedom that you identiﬁed in step 1., above. 5. Element Stiﬀness Matrices in Global Coordinates, K. For each element, ﬁnd it’s (6 × 6) element stiﬀness matrix, by evaluating the equations 2 CE 131 – Theory of Structures – Duke University – Fall 2006 – H.P. Gavin EA 2 EA L c L cs − EA c2 − EA cs L L + 12EI s2 − 12EI cs − 6EI s − 12EI s2 + 12EI cs − 6EI s L3 L3 L 2 L 3 L 3 L 2 EA 2 L s − EA cs − EA s2 L L 12EI 2 6EI + 12EI cs − 12EI c2 6EI + L3 c L2 c L3 L3 L2 c 4EI 6EI L L2 s − 6EI c L2 2EI L K= EA 2 EA c cs L L 12EI 2 12EI 6EI + L3 s − L3 cs L2 s sym EA 2 s L 12EI 2 6EI + L3 c − L2 c 4EI L where x2 − x 1 c = cos θ = L y2 − y 1 s = sin θ = L You should understand where these equations come from, why this matrix is symmetric, why the diagonal terms are all positive, and what the oﬀ-diagonal terms mean. 6. Structural Stiﬀness Matrix, Ks . The structural stiﬀness matrix is a square, symmetric, matrix with dimension equal to the number of degrees of freedom. In this step we will ﬁll up the structural stiﬀness matrix using terms from the element stiﬀness matrices in global coordinates (from step 5.) This procedure is called matrix assembly. Recall from step 4. how the LOCAL element degrees of freedom (1,2,3,4,5,6) line up with the GLOBAL degrees of freedom in your problem. For example, local coordinates (1,2,3,4,5,6) might line up with degrees of freedom (13,14,15,7,8,9) of the frame. In that case: K(1,1) is added to Ks(13,13), K(1,2) is added to Ks(13,14), ... K(2,6) is added to Ks(14,9), ... K(5,6) is added to Ks(8,9), Stiffness Method for 2D Frames 3 K(6,6) is added to Ks(9,9), Add each element into the structural stiﬀness matrix in this way to build up K s . 7. Reactions. Ks must be modiﬁed to include the eﬀects of the reactions, which have been ignored up until now. Set every element of each row and column corresponding to a restrained degree of freedom (reaction) equal to zero. Set every diagonal element that is zero equal to 1. 8. External Loads, p. Create the load vector p, by ﬁnding the ﬁxed-end forces and moments of each member, and their components in the directions of the global degrees of freedom. Add the ﬁxed end forces and moments to any point loads applied directly to the joints. Create the force vector by placing these force components into the force vector at the proper coordinates. 9. Deﬂections, d. Find the deﬂections by inverting the stiﬀness matrix and multiplying it by the load vector. You can do this easily in matlab: d = Ks \ p 10. Internal beam forces, q. Again, recall how the global degrees of freedom line up with each element’s coordinates (1,2,3,4,5,6). For example, in element number ”N” from step 6., the local element de- ﬂections v1,v2,v3,v4,v5,v6 line up with the global deﬂections d13,d14,d15,d7,d8,d9. The internal bar forces can then be computed from: q = k T v − {F EF } 4 CE 131 – Theory of Structures – Duke University – Fall 2006 – H.P. Gavin Notation u = Element deﬂection vector in the Local coordinate system q = Element force vector in the Local coordinate system k = Element stiﬀness matrix in the Local coordinate system ... q = k u T = Coordinate Transformation Matrix ... T−1 = TT v = Element deﬂection vector in the Global coordinate system ... u = T v f = Element force vector in the Global coordinate system ... q = T f K = Element stiﬀness matrix in the Global coordinate system ... K = TT k T d = Structural deﬂection vector in the Global coordinate system p = Structural load vector in the Global coordinate system Ks = Structural stiﬀness matrix in the Global coordinate system ... p = Ks d {F EF } = the vector of ﬁxed end forces from internal applied loads. Local Global Element Deﬂection u v Element Force q f Element Stiﬀness k K Structural Deﬂection - d Structural Loads - p Structural Stiﬀness - Ks Stiffness Method for 2D Frames 5 Example 1 hudson8% matlab >> E = 30000; % modulus of elasticity >> I1 = 1000; I2 = 500; I3 = 250; % moments of inertia >> L1 = 150; L2 = 120; L3 = 100; % element lengths >> Ks = [ (4*E*I1/L1 + 4*E*I2/L2) 2*E*I2/L2 ; \ % stiffness > 2*E*I2/L2 4*E*I2/L2 + 4*E*I3/L3 ] % matrix Ks = 1300000 250000 250000 800000 >> w = 0.1; % uniform distributed load on member number 1 >> p = [ w*L1^2/12 0 ]’ % fixed end forces from member 1 applied to DoF 1 p = % Force vector 187.50000 0.00000 >> d = inv(Ks)*p d = % Displacement Vector 1.5345e-04 -4.7954e-05 >> M1 = 4*E*I1/L1 * d(1) M1 = 122.76 % Moment in member 1 at DoF 1 due to rotation D1 >> M2 = 4*E*I2/L2 * d(1) + 2*E*I2/L2 * d(2) M2 = 64.738 % Moment in member 2 at DoF 2 due to rotation D2 >> M1 - 187.5 % Moment in member 1 at DoF 1 MINUS the ans = -64.738 % fixed end moment equals the Moment M2 6 CE 131 – Theory of Structures – Duke University – Fall 2006 – H.P. Gavin Example 2 hudson10% matlab >> E=30e3; % modulus of elasticity >> I1=2000; I2=3000; I3=4000; % moments of inertia >> L1 = 120; L2 = 150; L3 = 120; % member lengths >> Ks = [ 4*E*I1/L1 + 4*E*I2/L2 2*E*I2/L2 6*E*I1/L1^2 ; > 2*E*I2/L2 4*E*I2/L2+4*E*I3/L3 6*E*I3/L3^2 ; > 6*E*I1/L1^2 6*E*I3/L3^2 12*E*I1/L1^3 + 12*E*I3/L3^3 ] Ks = % structural stiffness matrix 4400000 1200000 25000 1200000 6400000 50000 25000 50000 1250 >> w = 2; % uniform distributed load on element 2 >> P = 25; % lateral point load at joint 1 >> p = [ -w*L2^2/12 w*L2^2/12 P ]’ p = % force vector ( fixed end forces ) -3750 3750 25 >> d = inv(Ks)*p d = % structural displacements -1.1244e-03 6.7611e-04 1.5444e-02 % Now compute the Member Forces. % Moment in member 1 at joint 1 due to total deflections of member 1: >> Ma = 4*E*I1/L1 * d(1) + 6*E*I1/L1^2 * d(3) Ma = -1862.7 Stiffness Method for 2D Frames 7 % Moment in member 2 at joint 1 due to total deflections of member 2: >> Mb = 4*E*I2/L2 * d(1) + 2*E*I2/L2 * d(2) Mb = -1887.3 % Ma and Mb are not equal, although they should be! % The difference is the fixed end force at joint 1. >> M1 = Mb - p(1) M1 = 1862.7 % This is the actual moment at joint 1. % ... Likewise for joint 2 ... % Moment in member 3 at joint 2 due to total deflections of member 3: >> Ma = 4*E*I3/L3 * d(2) + 6*E*I3/L3^2 * d(3) Ma = 3476.6 % Moment in member 2 at joint 2 due to total deflections of member 2: >> Mb = 4*E*I2/L2 * d(2) + 2*E*I2/L2 * d(1) Mb = 273.36 % Ma and Mb are not equal, although they should be! % The difference is the fixed end force at joint 2. >> M2 = Mb - F(2) M2 = -3476.7 % This is the actual moment at joint 2. % What about the ractions? >> V = (M1+M2)/L2 % shear in member 2 due to moments V = -10.760 >> Vl = w*L2/2 + V % vertial reaction on the left Vl = 139.24 >> Vr = w*L2/2 - V % vertical reaction on the right Vr = 160.76 8 CE 131 – Theory of Structures – Duke University – Fall 2006 – H.P. Gavin % horizontal reaction on the right >> Hr = 12*E*I3/L3^3 * d(3) + 6*E*I3/L3^2 * d(2) Hr = 46.675 % horizontal reaction on the left >> Hl = 12*E*I1/L1^3 * d(3) + 6*E*I1/L1^2 * d(1) Hl = -21.675 >> Hl + Hr % horizontal equilibrium check. (P = 25) ans = 25.000 % moment reaction on the left >> Ml = 6*E*I1/L1^2 * d(3) + 2*E*I1/L1 * d(1) Ml = -738.32 % moment reaction on the right >> Mr = 6*E*I3/L3^2 * d(3) + 2*E*I3/L3 * d(2) Mr = 2124.4 % check global equilibrium >> -P*L1 - w*L2^2/2 + Mr + Ml + Vr*L2 ans = 0.098131 % a pretty small number, as compared to 2000

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