Beam_Example - CES 4141 - Stress Analysis Direct Stiffness...

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CES 4141 - Stress Analysis 1 of 7 Direct Stiffness - Beam Example Direct Stiffness - Beam Example Equations we will use from the beam notes (1) (4) , (12) k AE L ------- 00 L ---------- 0 12EI L 3 ----------- 6EI L 2 -------- 0 12EI L 3 L 2 0 L 2 4EI L 0 L 2 2EI L L L 0 12EI L 3 L 2 –0 12EI L 3 L 2 0 L 2 L 0 L 2 L = a Lx Ly 00 00 Ly Lx 100 0 0 0 0 0 000 1 = Ke L L 2 x 12EI L 3 L 2 y + L 12EI L 3   LxLy L 2 ------------------ L L 2 x 12EI L 3 L 2 y + L ------ 12EI L 3 L 2 L 12EI L 3 L L 2 y 12EI L 3 L 2 x + L 2 --------------- L 12EI L 3 L L 2 y 12EI L 3 L 2 x + L 2 L 2 L 2 L L 2 L 2 L L L 2 x 12EI L 3 L 2 y + L 12EI L 3 L 2 L L 2 x 12EI L 3 L 2 y + L 12EI L 3 L 2 L 12EI L 3 L L 2 y 12EI L 3 L 2 x + L 2 L 12EI L 3 L L 2 y 12EI L 3 L 2 x + 6 –E I L 2 L 2 L 2 L L 2 6 I L 2 L =
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CES 4141 - Stress Analysis 2 of 7 Direct Stiffness - Beam Example Example : Analyze the following frame structure for displacements, reactions and member forces. Stiffness matrix for elements: Note - All units converted to kips and inches as we go. .. Element #1: Finding Lx and Ly and applying Eq. . : Lx = change in x / L = (20 - 0)/25 = 0.8 Ly = change in y / L = (15 - 0)/25 = 0.6 near end D.O.F. (N = 4, 5, 6) far end D.O.F. (F = 1, 2, 3) Element #2: Finding Lx and Ly and applying Eq. . : Lx = change in x / L = 1 Ly = change in x / L = 0 near end D.O.F. (N = 1, 2, 3) far end D.O.F. (F = 7, 8, 9) r1 r2 20 ft 1 2 r3 r7 r9 r8 r4 r6 r5 20 ft 15 ft 3 k/ft I = 600 in 4 A = 12 in 2 E = 29000 ksi r4 r6 r5 r1 r2 r3 1 N F Ke 1 745.18 553.09 696 745.18 –5 5 3 . 0 9 –6 9 6 422.55 928 553.09 –4 2 2 . 5 5 –9 2 8 232 e 3 696 928 –1 1 6 e 3 745.18 553.09 696 422.55 928 232 e 3 = r4 r5 r6 r1 r4 r5 r6 r1 r2 r3 r2 r3 symmetric r7 r9 r8 r1 r2 r3 F N 1 1450 0 0 1450 –0 0 15.10 1812.50 0 15.10 1812.50 290 e 3 0 1812.50 4 5 e 3 1450 0 0 15.10 1812.50 290 e 3 = r1 r2 r3 r7 r1 r2 r3 r7 r8 r9 r8 r9 symmetric
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CES 4141 - Stress Analysis 3 of 7 Direct Stiffness - Beam Example Assembly of Global Stiffness matrix: After assembly we have Define known loads: Using the fixed end moment table and some statics to reduce the uniform load to equivalent nodal loads (E.N.L.) KG Ke 1 2 + = 2195.18 553.09 696 745.18 553.09 696 1450 –0 0 553.09 437.65 884.5 553.09 422.55 –9 2 8 0 15.10 1812.50 696 884.50 522 e 36 9 6 2 8 1 1
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This note was uploaded on 06/10/2011 for the course CES 4141 taught by Professor Lybas during the Spring '08 term at University of Florida.

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Beam_Example - CES 4141 - Stress Analysis Direct Stiffness...

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