Method of Consistent Deformation, Summaries of Theory of Structures

Download Method of Consistent Deformation and more Summaries Theory of Structures in PDF only on Docsity! Method of Consistent Deformation Theory of Structures-II M Shahid Mehmood Department of Civil Engineering Swedish College of Engineering and Technology, Wah Cantt Structural Analysis By R. C. Hibbeler FRAMES • Method of consistent deformation is very useful for solving problems involving statically indeterminate frames for single story and unusual geometry. • Problems involving multistory frames, or with high indeterminacy are best solved using the slope deflection or moment distribution or the stiffness methods. 2 Solution Principle of Superposition • We will choose the horizontal reaction at support B as the redundant. • The pin at B is replaced by the roller, since a roller will not constraint B in the horizontal direction. University of Engineering & Technology, Taxila 5 A B 8 kN/m Primary Structure ΔB Solution Principle of Superposition University of Engineering & Technology, Taxila 6 A B 8 kN/m Primary Structure ΔB 5 m 4 m A B 8 kN/m Actual Frame = + A B redundant Bx applied Bx fBB Bx Solution Compatibility Equation University of Engineering & Technology, Taxila 7 A B 8 kN/m Primary Structure ΔB 5 m 4 m A B 8 kN/m Actual Frame = + A B redundant Bx applied Bx fBB Bx BBxB fBΔ        0 Solution University of Engineering & Technology, Taxila 10 A B 8 kN/m x1 20 kN 20 kN x2 2 11 1 420 xxM  02 M A B x1 0.8 kN 0.8 kN x2 11 8.0 xm  22 1xm  1 kN 1 kN      EIEI dx EI dxx EI dxxxxdx EI MmΔ L B 7.16607.166 108.04205 0 4 0 2211 2 11 0       Solution Compatibility Equation • For fBB we require application of real unit load acting at B and a virtual unit load acting at B University of Engineering & Technology, Taxila 11 A B x1 0.8 kN 0.8 kN x2 11 8.0 xm  22 1xm  1 kN 1 kN Solution University of Engineering & Technology, Taxila 12 A B x1 0.8 kN 0.8 kN x2 11 8.0 xm  22 1xm  1 kN 1 kN     EIEIEI dx EI dxx EI dxxdx EI mmf L BB 0.483.217.26 108.05 0 4 0 2 2 21 2 1 0     Solution Equilibrium Condition University of Engineering & Technology, Taxila 15 2.5 m 4 m A B 40 kN 2.5 m By 3.47 kN Ay Ax kN.A.AF xxx 473 0473 ;0         kN. B.-B.; M yyA 82204473552400  kN. A.; AF yyy 2170822400   Example 7 Determine the moment at fixed support A for the frame shown. EI is constant. University of Engineering & Technology, Taxila 16 8 ft A B Actual Frame C 4 ft 3 ft 5 f t10 0 l b/ ft Solution Principle of Superposition • By inspection the frame is indeterminate to the first degree. University of Engineering & Technology, Taxila 17 8 ft A B Actual Frame C 4 ft 3 ft 5 f t10 0 l b/ ft Solution Compatibility Equation • The terms θA and αAA will be computed using the method of virtual work. • The frame ’ s x coordinates and internal moments are shown in figure. University of Engineering & Technology, Taxila 20   (1) 0 AAAA M   Solution Compatibility Equation Reference to point A University of Engineering & Technology, Taxila 21 500 lb x1 11 17.29 xM  29.17 lb 300 lb 370.8 lb 222.5 lb 296.7 lb 2 222 507.296 xxM  3 4 5 x2 x1 11 0833.01 xm  0.0833 lb 0 0.0833 lb 0.05 lb 0.0667 lb 22 0667.0 xm  x2 1 lb.ft Solution For θA we require application of real loads and a virtual unit couple moment at A University of Engineering & Technology, Taxila 22 500 lb x1 11 17.29 xM  29.17 lb 300 lb 370.8 lb 222.5 lb 296.7 lb 2 222 507.296 xxM  3 4 5 x2 x1 11 0833.01 xm  0.0833 lb 0 0.0833 lb 0.05 lb 0.0667 lb 22 0667.0 xm  x2 1 lb.ft University of Engineering & Technology, Taxila 25 x1 11 0833.01 xm  0.0833 lb 0 0.0833 lb 0.05 lb 0.0667 lb 22 0667.0 xm  x2 1 lb.ft     EIEIEI EI dxx EI dxxdx EI mmL AA 04.4185.085.3 0067.00833.018 0 5 0 2 2 21 2 1 0        Solution Substituting these results into Eq. (1), and solving yields The negative sign indicates MA acts in opposite direction to that shown in figure. University of Engineering & Technology, Taxila 26 ANS . 204 04.48.8210 ftlbM EI M EI A A        A C 10 0 l b/ ft MA =204 lb.ft Example 8 Determine the reactions and draw the shear and bending moment diagrams. EI is constant. University of Engineering & Technology, Taxila 27 30 ft 15 ft B D 2 k/ft Actual Frame A C 10 k Solution Principle of Superposition • Primary structure is obtained by removing the hinged support at point D. University of Engineering & Technology, Taxila 30 B D 2 k/ft Primary Structure A C 10 k Solution Principle of Superposition • Primary structure is subjected separately to the external loading and redundants Dx and Dy as shown. University of Engineering & Technology, Taxila 31 B D 2 k/ft Primary Structure A C 10 k ΔDx ΔDy Solution Principle of Superposition • Primary structure is subjected separately to the external loading and redundants Dx and Dy as shown. University of Engineering & Technology, Taxila 32 B D Redundant Dx applied A C Δ’DxDx=DxfDxDx Δ’DyDx=DxfDyDx Dx Solution Compatibility Equation University of Engineering & Technology, Taxila 35 B D 2 k/ft A C 10 k 60 k 10 k x1 M1 x2 M2 x3 M3 Member Origin Limits M (k-ft) AB A 0-15 -1050+10x1 CB C 0-30 -x2 2 DC D 0-15 0 1050 k-ft Solution Compatibility Equation University of Engineering & Technology, Taxila 36 B DA C 1 k x1 mDx1 x2 mDx2 x3 mDx3 Member Origin Limits M (k-ft) mDx (k-ft/k) AB A 0-15 -1050+10x1 -x1 CB C 0-30 -x2 2 -15 DC D 0-15 0 -x3 1 k Solution Compatibility Equation University of Engineering & Technology, Taxila 37 B DA C x1 mDy1 x2 mDy2 x3 mDy3 Member Origin Limits M (k-ft) mDx (k-ft/k) mDy (k-ft/k) AB A 0-15 -1050+10x1 -x1 30 CB C 0-30 -x2 2 -15 x2 DC D 0-15 0 -x3 0 1 k1 k 30 k-ft Solution University of Engineering & Technology, Taxila 40       2 30 0 2 1 15 0 1 0 1530 dx EI xdx EI xdx EI mm ff L DyDx DyDxDxDy      310125 ft EI ff DyDxDxDy  Solution University of Engineering & Technology, Taxila 41 3241875 ftk EI ΔDx  3641250 ftk EI ΔDy  39000 ft EI fDxDx  322500 ft EI fDyDy  310125 ft EI ff DyDxDxDy  Solution Now put these values in the Equations (1) and (2) By solving (1) and (2) simultaneously we get University of Engineering & Technology, Taxila 42 (1) 0125190002418750 yx DD  (2) 22500101256412500 yx DD   503.10 kDx  226.33 kDy Solution • Moment diagram University of Engineering & Technology, Taxila 45 B DA C 53.22 60.765 157.545 118.447 157.545 60.765 TRUSSES • The degree of indeterminacy of a truss can be find using Equation b+r > 2j. where b = unknown bar forces, r = support reactions, 2j = equations of equilibrium • This method is quite suitable for analyzing trusses that are statically indeterminate to the first or second degree. University of Engineering & Technology, Taxila 46 Example 9 Determine the force in member AC of the truss shown. AE is same for all members. University of Engineering & Technology, Taxila 47 8 ft 6 ft A B 400 lb Actual Truss D C Solution University of Engineering & Technology, Taxila 50 A B 400 lb Actual Truss D C = A B 400 lb D C ΔAC Primary Structure + A B D C Redundant FAC applied FAC FAC FACfACAC Solution Compatibility Equation • With reference to member AC, we require the relative displacement Δ AC, which occurs at the ends of cut member AC due to the 400-lb load, plus the relative displacement FACfACAC caused by the redundant force acting alone, be equal to zero, that is University of Engineering & Technology, Taxila 51A B 400 lb D C ΔAC Primary Structure A B D C Redundant FAC applied FAC FAC FACfACAC ACACACAC fF0 Solution Compatibility Equation • Here the flexibility coefficient fACAC represents the relative displacement of the cut ends of member AC caused by a real unit load acting at the cut ends of member AC. University of Engineering & Technology, Taxila 52A B 400 lb D C ΔAC Primary Structure A B D C Redundant FAC applied FAC FAC FACfACAC ACACACAC fF0 Solution University of Engineering & Technology, Taxila 55 A B 400 lb D C A B D C 1 lb 1 lb 300 lb 400 lb 300 lb +400 +400 +300 0 0 -500 x -0.8 -0.6 -0.8 -0.6 +1 +1                     AE AEAEAEAEAE AE nNL AC 11200 100110500163006.0606.084008.02               Solution Compatibility Equation • For fACAC we require application of the real unit forces acting on the cut ends of member AC, and virtual unit forces acting on the cut ends of member AC University of Engineering & Technology, Taxila 56 A B D C 1 lb 1 lb x -0.8 -0.6 -0.8 -0.6 +1 +1 A B D C 1 lb 1 lb -0.8 -0.6 -0.8 -0.6 +1 +1 Solution University of Engineering & Technology, Taxila 57 A B D C 1 lb 1 lb x -0.8 -0.6 -0.8 -0.6 +1 +1           AE AEAEAE AE Lnf ACAC 56.34 101266.0288.02 222 2                        A B D C 1 lb 1 lb -0.8 -0.6 -0.8 -0.6 +1 +1 Example 11 Determine the reactions and the force in each member of the truss shown in Fig. shown. E = 29,000 ksi University of Engineering & Technology, Taxila 60 3 panels at 20 ft = 60 ft 15 ft A B Actual Truss D C E F 28 k (6 in.2 ) (6 in.2) (6 in.2) (6 in.2) (6 in.2) (6 in. 2)(4 in.2 ) (4 in .2 ) (4 in .2 ) 25 k 25 k Solution The truss is statically indeterminate to the first degree. b + r = 2j 9 + 4 = 2(6) 13 > 12 13 – 12 = 1st degree University of Engineering & Technology, Taxila 61 3 panels at 20 ft = 60 ft 15 ft A B Actual Truss D C E F 28 k (6 in.2 ) (6 in.2) (6 in.2) (6 in.2) (6 in.2) (6 in. 2)(4 in.2 ) (4 in .2 ) (4 in .2 ) 25 k 25 k Solution • Dx at hinged support D is selected as Redundant. • Primary structure is obtained by removing the effect of Dx and replacing hinge by roller support there. • Primary structure is subjected separately to external loading and redundant Force Dx. University of Engineering & Technology, Taxila 62 A B Actual Truss D C E F 28 k Ax Ay Dy Dx 25 k 25 k Solution University of Engineering & Technology, Taxila 65 A B Redundant Dx is applied DC E F A B DC E F 1 k 0 1 1 1 0 00 0 0 Dx fDD 1 k Δ’DD=DxfDD Solution Δ’DD is horizontal deflection at point ‘D’ due to redundant force Dx. University of Engineering & Technology, Taxila 66 A B Redundant Dx is applied DC E F Dx Δ’DD=DxfDD Solution fDD is horizontal deflection at point ‘D’ due to unit force. University of Engineering & Technology, Taxila 67 A B DC E F 1 k 0 1 1 1 0 00 0 0 fDD 1 k Solution • We will use virtual work method to find ΔD and fDD. • Deflection of truss is calculated by where n = axial force in truss members due to real unit load acting at joint and in the direction of ΔD n = axial force in truss members due to virtual unit load acting at joint and in the direction of ΔD University of Engineering & Technology, Taxila 70  AE LnfDD 2 Solution University of Engineering & Technology, Taxila 71 TABLE Member L (in.) A (in.2) N (k) n (k) nNL/A (k/in.) n2L/A F=N+nDx AB 240 6 52 1 2,080 40 6.22 BC 240 6 42.67 1 1,706.8 40 -3.11 CD 240 6 42.67 1 1,706.8 40 -3.11 EF 240 6 -24 0 0 0 -24 BF 180 4 18 0 0 0 18 CF 180 4 25 0 0 0 25 AE 300 6 -30 0 0 0 -30 BF 300 4 11.67 0 0 0 11.67 DF 300 6 -53.33 0 0 0 -53.33 ∑ 5,493.6 120E k/in. 6.493,5   AE nNL D E (1/in.) 2012  AE LnfDD Solution University of Engineering & Technology, Taxila 72  AE nNL D EEED 6 1220167.42 6 1220167.42 6 1220152       EEED 6 8.1706 6 8.1706 6 2080  ED k/in 6.5493  University of Engineering & Technology, Taxila 75 A B Primary Structure DC E F 28 k Ax =17.78 Ay =18 Dy = 32 25 k 25 k A B DC E F 28 k 25 k 25 k 30 6.22 3.11 3.11 24 53.3311.67 18 25 Dx = 45.78 Ay =18 Ax =17.78 Dy = 32 Dx = 45.78 Example 12 Determine the reactions and the force in each member of the truss shown in Fig. shown. EA = constant. E = 200 GPa., A = 4000 mm2 University of Engineering & Technology, Taxila 76 Actual Truss A B C D E F G H 70 kN 80 kN 80 kN 4 panels at 10m= 40m 10m Solution Principle of Superposition Degree of Indeterminacy = 2 b + r > 2 j 14 + 4 > 2 × 8 18 > 16 University of Engineering & Technology, Taxila 77Actual Truss A B C D E F G H 70 kN 80 kN 80 kN Ax Ay Dy Ey Solution Principle of Superposition This determinate truss is subjected separately to actual loading, redundant ‘ Dy ’ and redundant force in the redundant member BG. University of Engineering & Technology, Taxila 80Primary structure subjected to actual loading A B C D E F G H 70 kN 80 kN 80 kN ΔD ΔBG Solution Principle of Superposition This determinate truss is subjected separately to actual loading, redundant ‘ Dy ’ and redundant force in the redundant member BG. University of Engineering & Technology, Taxila 81Redundant Dy applied A B C D E F G H Dy Δ’DD=DyfDD Δ’BG,D=DyfBG,D Solution Principle of Superposition This determinate truss is subjected separately to actual loading, redundant ‘ Dy ’ and redundant force in the redundant member BG. University of Engineering & Technology, Taxila 82Redundant FBG applied A B C D E F G H Δ’D,BG=FBGfD,BG Δ’BG,BG=FBGfBG,BG FBG FBG Solution University of Engineering & Technology, Taxila 85 Redundant FBG applied A B C D E F G H Δ’D,BG=FBGfD,BG Δ’BG,BG=FBGfBG,BG FBG FBG A B C D E F G H fD,BG fBG,BG 11 Unit force in member BG applied University of Engineering & Technology, Taxila 86 Redundant FBG applied A B C D E F G H Δ’D,BG=FBGfD,BG Δ’BG,BG=FBGfBG,BG FBG FBG Actual Truss A B C D E F G H 70 kN 80 kN 80 kN Ax Ay Dy Ey= A B C D E F G H 70 kN 80 kN 80 kN ΔD ΔBG Primary structure + A B C D E F G H Dy Δ’DD=DyfDD Δ’BG,D=DyfBG,D + Redundant Dy applied Compatibility Equation BGDBGDDyD fFfD ,0  BGBGBGDBGyBG fFfD ,,0  ΔD = vertical deflection at joint D of primary truss due to external loading ΔBG = relative displacement b/w cutting ends of member BG due to external loading fDD = vertical deflection at joint D due to a unit load at joint D fBG,D = relative displacement b/w cutting ends of member BG due to unit load at D fBG,BG = relative displacement b/w cutting ends of member BG due to unit force fD,BG = vertical deflection at joint D due to a unit force in member BG University of Engineering & Technology, Taxila 87 Compatibility Equation BGDBGDDyD fFfD ,0  BGBGBGDBGyBG fFfD ,,0  Solution N = member forces due to external loading University of Engineering & Technology, Taxila 90 A B C D E F G H 70 kN 80 kN 80 kN 152.5 152.5 77.5 77.5 85 85 80116.673 109.602109.602 3.536 0 0 77.582.5 70 Solution nD = member forces due to unit load at joint D University of Engineering & Technology, Taxila 91 A B C D E F G H 0.25 0.25 0.75 0.75 0.5 0.5 0 0.354 1.0610.354 0.354 0 1 0.750.25 0 1 kN Solution nBG = member forces due to unit force in member BG University of Engineering & Technology, Taxila 92 A B C D E F G H 0 0.707 0 0 0.707 0 0. 70 70 00 1 0 00 0 0. 70 71 1 University of Engineering & Technology, Taxila 95 Compatibility Equation Solving these equations simultaneously for Dy and FBG 0773.6736.48642.472,4  BGy FD 0284.48773.6819.992  BGy FD  507.96 kNDy kNFBG 1.34 University of Engineering & Technology, Taxila 96 The remaining reactions of the indeterminate truss can now be determined by superposition of reactions of primary truss due to the external loading and due to each of the redundants. The forces in the remaining members of the indeterminate truss can be determined by using the superposition relationship University of Engineering & Technology, Taxila 97 A B C D E F G H 70 kN 80 kN 80 kN 152.5 152.5 77.5 77.5 85 85 80 116.673 109.602109.6023.536 0 0 77.582.5 70 A B C D E F G H 0.25 0.25 0.75 0.75 0.5 0.5 00.354 1.0610.3540.354 0 1 0.750.25 0 1 kN A B C D E F G H 0 0.707 0 0 0.707 0 0. 70 70 00 1 0 00 0 0. 70 71 1 + + = Actual Truss Ax=70 A B C D E F G H 70 kN 80 kN 80 kNAy=58.373 Dy=96.507 Ey=5.12 82.51 128.373 104.265 5.12 5.12 60.855 36.747 7.208 55 .8 91 24 .1 09 96 .5 07 34.1 143.765 3.473