INTRODUCTION: STRUCTURAL ANALYSIS ! Deflected Shape of Structures ! Method of Consistent Deformations ! Maxwell’s Theorem of Reciprocal Displacement 1 Deflection Diagrams and the Elastic Curve P fixed support ∆ = 0 θ = 0 -M P roller or rocker θ support ∆ = 0 inflection point +M -M 2 P θ pined support ∆ = 0 -M 3 inflection point • Fixed-connected joint P fixed-connected inflection point joint Moment diagram 4 • Pined-connected joint pined-connected P joint Moment diagram 5 P inflection point Moment diagram 6 P1 C A B D P2 M +M x -M inflection point 7 P1 P2 +M x -M inflection point 8 Method of Consistent Deformations Beam 1 DOF P MA Ax = 0 B A C = A RB y P B A C ∆´B + f BB x RB B A C ∆´ + f R = ∆ = 0 B BB B B 1 9 Beam 2 DOF Compatibility Equations. w ∆´1 + f11R1 + f12R2 = ∆1 = 0 3 ∆´2 + f21R1 + f22R2 = ∆2 = 0 4 1 2 5 = 0 w ∆´ f f R 1 11 12 1 ∆1 0 + = ∆´ f12 f ∆ 2 22 R2 2 ∆´1 ∆´2 + f11 f21 ×R1 1 + f12 f22 ×R2 1 10 Beam 3 DOF Compatibility Equations. P1 w P2 θ´ + f M + f R + f R = θ = 0 1 6 1 11 1 12 2 13 3 1 = ∆´2 + f21M1 + f22R2 + f23R3 = ∆2 = 0 4 2 3 5 ∆´ + f M + f R + f R = ∆ = 0 P1 w P2 3 31 1 32 2 33 3 3 0 θ´ f f f M 1 11 12 13 1 θ1 0 ∆´2 θ´ ∆´3 + 1 ∆´2 f21 f22 f23 R f f + 2 = ∆2 0 1 11 21 f31 ∆´3 f31 f32 f33 R3 ∆3 ×M1 + f f12 22 f32 ×R2 1 + f33 f13 f23 ×R3 1 11 Compatibility Equation for n span. Equilibrium Equations ∆´1 + f11R1 + f12R2 + f13R3 = ∆ 1 ∆´ + f R + f R + f R = ∆ 2 21 1 22 2 23 3 2 Fixed-end force matrix ∆´n + fn1R1 + fn2R2 + fnnRn = ∆n [Q] = [K][D] + [Qf] If ∆1 = ∆2=……= ∆n = 0 ; Stiffness matrix ∆ ´1 f11 f12 f13 R1 0 Solve for displacement [D]; ∆´2 f21 f22 f23 R + 2 = 0 ∆´ f n 1 f n 2 f -1 f n n n R n 0 [D] = [K] [Q] - [Q ] or [∆´] + [f][R] = 0 1 k = ij f [R] = - [f]-1[∆´] ij [fij] = Flexibility matrix dependent on 1 1 1 , , EI EA GJ 12 Maxwell’s Theorem of reciprocal displacements 1=i 2=j 1 EI A B f = f f11 = fii 21 ji dx f = m M ij ∫ i j EI Mi = mi dx = m m ∫ i j EI dx 1 f = m M A B ji ∫ j i EI f22 = fjj f = f dx 12 ij = m m ∫ j i EI M = m j j fij = f ji 13 Example 1 Determine the reaction at all supports and the displacement at C. 50 kN B A C 6 m 6 m 14 SOLUTION • Principle of superposition 50 kN MA B A C 6 m 6 m RA RB = 50 kN ∆´B + x R fBB B -----(1) 1 kN Compatibility equation : ∆'B + f BB RB = 0 15 • Use conjugate beam in obtaining ∆´B and fBB 50 kN A B 300 kN•m Real beam ∆´B 50 kN 6 m 6 m 9000/EI 6 + (2/3)6 = 10 m Conjugate beam 900/EI /EI 900/EI 300 ∆´B = M´B = -9000/EI , 12 kN•m fBB Real beam 1 kN 1 kN 72/EI 12 /EI (2/3)12 = 8 m 576/EI Conjugate beam f = M´´ = 576/EI, BB B 72/EI 16 • Substitute ∆´B and fBB in Eq. (1) 9000 576 + ↑: − + ( )R = 0 EI EI B RB = +15.63 kN, (same direction as 1 kN) 50 kN 300 kN•m ∆´B 50 kN + f 12 kN•m BB x RB = 15.63 kN 1 kN = 1 kN 50 kN B C 34.37 kN•m A 34.37 kN 15.63 kN 17 Use conjugate beam in obtaining the displacement 50 kN 113 kN•m 6 mC 6 m Real Beam ∆ B A C 15.6 kN 34.4 kN 93.6 M (kN•m) x (m) 3.28 m 6 m 12 m -113 93.6/EI Conjugate Beam -113/EI 223/(EI) 281/(EI) M´ 281 223 776 C M ' = (2) − (6) = − C EI EI EI 776 V´ ∆C = M 'C = − , ↓ C 2 m 4 m 223/(EI) EI 18 Example 2 Determine the reaction at all supports and the displacement at C. Take E = 200 GPa and I = 5(106) mm4 10 kN 3EI 2EI A B C 4 m 2 m 2 m 19 10 kN 3EI 2EI A B C 4 m 2 m 2 m = 10 kN ∆´B + fBB x RB 1 kN Compatibility equation: ∆´B + fBBRB = ∆B = 0 20 • Use conjugate beam in obtaining ∆´ B 10 kN 40 kN•m 3EI 2EI A B Real Beam C 4 m 2 m 2 m 10 kN 10 V (kN) 10 + x (m) M x (m) (kN•m) - 40 177.7/EI Conjugate Beam 40/3EI = 13.33/EI 26.66/EI ∆´ = M´ = 177.7/EI B B 21 • Use conjugate beam for fBB 3EI 2EI 8 kN•m A B Real Beam C 1 kN 1 kN 4 m 2 m 2 m V (kN) x (m) - -1 -1 8 4 + M x (m) (kN•m) 4/(3EI)=1.33EI 2.67 8 4/(2EI)=2EI = EI 3EI 60.44/EI 12/EI Conjugate Beam fBB = M´B = 60.44/EI 22 10 kN 3EI 2EI A B C 4 m 2 m 2 m = 10 kN ∆´B + fBB x RB 1 kN Compatibility. equation: ∆´B + fBBRB = ∆B = 0 −177.7 60.44 + ↑: − + ( )R = 0 EI EI B RB = +2.941 kN, (same direction as 1 kN) 23 • The quantitative shear and moment diagram and the qualitative deflected curve 10 kN 16.48 kN•m 3EI 2EI A B ∆ C C 7.06 kN 4 m 2 m 2 m 2.94 kN 7.06 + V (kN) - x (m) -2.94 -2.94 11.76 M 2.33 m (kN•m) + x (m) - 1.67 m -16.48 24 • Use the conjugate beam for find ∆C 10 kN 3EI 16.48 kN•m C 2EI A B Real beam ∆C 7.059 kN 4 m 2 m 2 m 2.941 kN 11.76 2EI 11.76 2.335 m 3EI Conjugate beam 1.665 m 16.48 3EI 3.263 EI 3.263 6.413 −18.85 M´ = (0.555) − (3.222) = ,↓ C EI EI EI 6.413 (1.665)/3=0.555 m EI −18.85 ∆C = M'C = = −18.85 mm,↓ 1.665+(2/3)(2.335) = 3.222 m (200×5) 25 Example 3 Draw the quantitative Shear and moment diagram and the qualitative deflected curve for the beam shown below.EI is constant. Neglect the effects of axial load. 5 kN/m AB 4 m 4 m 26 SOLUTION • Principle of superposition 5 kN/m AB θ θ A 4 m 4 m B = 5 kN/m θ´A θ´B 1 kN•m + × M A αAA αBA + 1 kN•m × M B α AB α Compatibility equations: BB θ A = 0 = θ 'A + f AAM A + f AB M B − − − (1) θ = 0 = θ ' + f M + f M − − − (2) B B BA A BB B 27 • Use formula provided in obtaining θ´A, θ´B, αAA, αBA, αBB, αAB 5 kN/m θ´A θ´B 4 m 4 m 3wL3 3(5)(8)3 60 θ ' = = = A 128EI 128EI EI 7wL3 7(5)(8)3 46.67 θ 'B = = = 1 kN•m 384EI 384EI EI 1 kN•m α α αAB AA BA α 8 m 8 m BB M o L 1(8) 2.667 M o L 1(8) 2.667 α = = = α BB = = = AA 3EI 3EI EI 3EI 3EI EI M L 1(8) 1.333 M L 1(8) 1.333 α = o = = α = o = = BA 6EI 6EI EI AB 6EI 6EI EI Note: Maxwell’s theorem of reciprocal displacement, αAB = αBA 28 • Use conjugate beam for αAA, αBA, αBB, αAB Real Beam 1 kN•m 1 kN•m Real Beam α α α AA BA AB α (1/8) BB (1/8) (1/8) (1/8) 8 m 8 m 4/EI 4/EI 1/EI 1/EI Conjugate Beam Conjugate Beam 2.67/EI 1.33/EI 1.33/EI 2.67/EI − 2.667 −1.333 α = V ' = α = V ' = AA A EI AB A EI 1.333 2.667 α = V ' = α = V ' = BA B EI BB B EI 29 5 kN/m ABCompatibility equation θ θ A 4 m 4 m B 60 2.667 1.333 + + ( )M A + ( )M B = 0 = EI EI EI 5 kN/m 46.67 1.333 2.667 + ( )M + ( )M = 0 + EI EI A EI B 60 46.67 θ ' = θ 'B = A EI EI Solve simultaneous equations, 1 kN•m + M = -18.33 kN•m, + × M A A 1.333 MB = -8.335 kN•m, + 2.667 α BA = V 'B = α AA = EI EI + 1 kN•m × M B 1.333 2.667 α AB = α = EI BB EI 30 MA = -18.33 kN•m, MB = -8.335 kN•m, 5 kN/m 18.33 kN•m AB8.335 kN•m 4 m 4 m RA RB 18.33 − 20(2) + R (8) −8.355 = 0 R = 3.753 kN, + ΣMA = 0: B B 3.753 R = 16.25 kN, + ΣFy = 0: RA + RB − 20 = 0 a 31 • Quantitative shear and bending diagram and qualitative deflected curve 5 kN/m 18.33 kN•m AB8.36 kN•m 16.25 kN 4 m 4 m 3.75 kN 16.25 V diagram 3.25 m -3.75 M 8.08 diagram 6.67 -8.36 -18.33 Deflected Curve 32 Example 4 Determine the reactions at the supports for the beam shown and draw the quantitative shear and moment diagram and the qualitative deflected curve.