<p> Y 30 kN Stiffness Method (Frame Example) 3 Analyze the frame shown using Stiffness method. 3 m 2 E is constant for all members. 20 kN Section properties: 2 member Area Moment of Inertia* 1 5 m 1 A I 1 2 1.5A 3.5I 1 X</p><p>2 4 * A=0.04 m , I=0.0004/3 m  A=300I 4 m</p><p>(I) Member stiffness Matrices in Global coordinates :</p><p>(1) Member 1: i-node=1, j-node=2, L= 5 m, 5 j= 2 4 x j  xi 0  0 y j  yi 5  0 6 c  cos    0, s  sin    1 x L 5 x L 5 1 EA 12EI 6EI 1 a   0.2EA  60EI, d  3  0.096EI, e  2  0.24EI, L L L 2 4EI 2EI f   0.8EI , g   0.4EI. i= 1 1 L L 3</p><p>2 2 K11  a c  d s  0.096EI, K12  (a  d)c s  0, K13  e s  0.24EI 2 2 K 22  a s  d c  60EI, K 23  ec  0</p><p>K 33  f  0.8EI  Coord. # 1 2 3 4 5 6  0.096 0  0.24  0.096  0  0.241  2  0 60 0  0  60 0    0.24 0 0.8 0.24  0 0.4 3 [K]  EI 1  456 (See Stiffness Handout)  0.096 0 0.24 .096 0 0.24   0  60 0 0 60 0 6     0.24 0 0.4 0.24 0 0.8  (2) Member 2: i-node=2, j-node=3, L= 5 m, </p><p> x  x 4  0 y  y 8  5 c  cos  j i   0.8, s  sin  j i   0.3 x L 5 x L 5 E(1.5A) 12E(3.5I) 6E(3.5I) a   0.3EA  90EI, d   0.336EI, e   0.84EI, L L3 L2 4E(3.5I) 2E(3.5I) f   2.8EI , g   1.4EI. 8 L L 7 9 j= 3 2 5 1 Coord. # i= 2 4 6 4 5 6 7 8 9 2 2 K11  a c  d s  57.7210EI   57.7210 43.0387 - 0.5040 - 57.721 - 43.0387 - 0.5040  4   K  (a  d)c s  43.0387  43.0387 32.6150 0.6720 43.0387 - 32.6150 0.6720   12    5   K13  e s  0.5040EI  - 0.5040 0.6720 2.8000 0.5040 - 0.6720 1.4000    [K]2  EI   6 2 2 - 57.7210 - 43.0387 0.5040 57.7210 43.0387 0.5040 K 22  a s  d c  32.6150EI      K  ec  0.6720  - 43.0387 - 32.6150 - 0.6720 43.0387 32.6150 - 0.6720 7  23       - 0.5040 0.6720 1.4000 0.5040 - 0.6720 2.8000 8 K33  f  2.8EI   </p><p>(II) Structure Stiffness Matrix (Global Stiffness Matrix):</p><p>[K]99  [K]1  [K]2</p><p>1 2 3 4 5 6 7 8 9 0.096 0 -0.24 -0.096 0 -0.24 1 0 60 0 0 -60 0 2 -0.24 0 0.8 0.24 0 0.4 3 -0.096 0 0.24 0.096+ 0+ 0.24+ 4 57.721 43.0387 -0.504 -57.721 -43.0387 -0.504 =EI  0 -60 0 0+ 60+ 0+ 5 43.0387 32.615 0.672 -43.0387 -32.615 0.672 -0.24 0 0.4 0.24+ 0+ 0.8+ 6 -0.504 0.672 2.8 0.504 -0.672 1.4 -57.721 -43.0387 0.504 57.721 43.0387 0.504 7 -43.0387 -32.615 -0.672 43.0387 32.615 -0.672 8 -0.504 0.672 1.4 0.504 -0.672 2.8 9 or</p><p>0.0960 0 - 0.24 - 0.0960 0 - 0.2400 0 0 0     0 60.0 0 0 - 60.0 0 0 0 0   - 0.240 0 0.8000 0.2400 0 0.4000 0 0 0     - 0.096 0 0.2400 57.8170 43.0387 - 0.264 - 57.721 - 43.0387 - 0.504  [K]  EI  0 - 60.0 0 43.0387 92.6150 0.672 - 43.0387 - 32.6150 0.672     - 0.24 0 0.4000 - 0.2640 0.6720 3.600 0.504 - 0.6720 1.400   0 0 0 - 57.7210 - 43.0387 0.504 57.7210 43.0387 0.504     0 0 0 - 43.0387 - 32.6150 - 0.672 43.0387 32.6150 - 0.672     0 0 0 - 0.5040 0.6720 1.400 0.5040 - 0.6720 2.8000 </p><p>(III) Force Vector { F }</p><p>{F}91 ={F0}-{FE} where </p><p>{F0} = vector of nodal forces , and {FE} = vector of equivalent end forces ( fixed end forces and moments due to span loadings. )</p><p>For our problem;</p><p>T T {F0}={F1, F2, …,F9} = {Fx1, Fy1, Mz1, 20, 0 , 0, Fx3, Fy3, Mz3}</p><p>Fixed End Moments and Forces:</p><p>Span 2 is loaded with concentrated load acting at its middle. The fixed end forces in member lcs are calculated as shown below:</p><p> f F 9  x1    f =-15 kN.m f =9 kN     e6 e4 f y1 12     30 kN 24 kN M 1   15  18 kN fe2       (in member local coordinates)  f x2   9  f =12 kN  f   12  e5  y2    M  15  2    f =15 kN.m e3</p><p> f =9 kN e1 f =12 kN e2 Member 2</p><p>Coord. # Fixed End Forces In global coordinates:</p><p>4 0.8  0.6 0 0 0 0 9   0       5 0.6 0.8 0 0 0 0 12   15  6 T  0 0 1 0 0 0 15   15  FE2  T 2  fe2        7  0 0 0 0.8  0.6 0 9   0   0 0 0 0.6 0.8 0 12   15  8      9  0 0 0 0 0 115 15</p><p>The Global load vector becomes:</p><p>Unkown forces and moents (Reactions) F1  Fx1   0   Fx1  1   0  F  F   0   F     0   2   y1     y1   2     ⋮  M 1   0   M 1   ⋮   0              Unkown d.o.f.    20   0   20     x2               ⋮    0    15    15  The global disp. vector   ⋮    y2     0   15   15                     Fx 3   0   Fx3     0              ⋮ F 15 F 15 ⋮ 0    y3     y3      F9  M 3  15 M 3 15  9   0 </p><p>The global stiffness equation: [K]{}={F}</p><p>1 2 3 4 5 6 7 8 9</p><p>1 0.0960 0 - 0.24 - 0.0960 0 - 0.2400 0 0 0  0   Fx1   0 60.0 0 0 - 60.0 0 0 0 0  0   F  2     y1  3  - 0.240 0 0.8000 0.2400 0 0.4000 0 0 0  0   M 1       4  - 0.096 0 0.2400 57.8170 43.0387 - 0.264 - 57.721 - 43.0387 - 0.504  x2   20       5 EI 0 - 60.0 0 43.0387 92.6150 0.672 - 43.0387 - 32.6150 0.672  y2    15    - 0.24 0 0.4000 - 0.2640 0.6720 3.600 0.504 - 0.6720 1.400     15  6   2      7 0 0 0 - 57.7210 - 43.0387 0.504 57.7210 43.0387 0.504  0   Fx3       0 0 0 - 43.0387 - 32.6150 - 0.672 43.0387 32.6150 - 0.672 0 F 15 8     y3       9  0 0 0 - 0.5040 0.6720 1.400 0.5040 - 0.6720 2.8000  0  M 3 15</p><p>(Eq. I) Solution of Stiffness Equations</p><p>After applying the boundary conditions, the stiffness matrix equation in free coordinates is</p><p>57.817 43.0387  0.2640 x2   20      EI  92.615 0.672    15   y2         3.60   2  15 which have the solution</p><p> x2   0.6517    1    y2    0.4355   EI     2   4.0376</p><p>Support Reactions are found by back substituting in equation in (Eq. I) above:</p><p>Fx1   0.9065  F   26.13   y1    M 1  1.459      Fx3   20.91 F   3.87   y3    M 3   21.27</p><p>Local effect (end shear forces and Moments in each member) can be obtained by substituting the displacement results in the member stiffness equations.</p><p>Typical Computer Output:</p><p>NODAL DISPLACEMENTS node x-disp. y-disp. z-rotation 1 000 000 000 2 6.517e-001 -4.355e-001 -4.038e+000 3 000 000 000</p><p>SUPPORT REACTIONS supp. x-reac y-reac z-reac 1 9.065e-001 2.613e+001 -1.459e+000 3 -2.091e+001 3.870e+000 -2.127e+001</p><p>MEMBER FORCES: MEM. END SHEAR FORCES END MOMENTS END AXIAL FORCES I-NODE J-NODE I-NODE J-NODE I-NODE J-NODE 1 -9.065e-001 9.065e-001 -1.459e+000 -3.074e+000 2.613e+001 -2.613e+001 2 8.360e+000 1.564e+001 3.074e+000 -2.127e+001 3.240e+001 -1.440e+001</p>