Alpha College of Engineering and Technology Civil-A Subject- Structure Analysis- 2 Group no-18 Group members- 1) Chaudhary Babubhai (130510106009) 2) Chaudhary Arvind (130510106010) 3) Chaudhary Madrup (130510106011) 4) Jain Mehul (130510106020) Guided By :- Prof. Vivek Patel Prof. Umesh Dobariya 1

SINKING FOR SLOPE DEFLECTION METHOD

SINKING ( SETTLEMENT): The settlement will be taken as positive, if it rotates the beam as a ᵟ= whole in the clockwise + ve direction. The settlement will be taken as negative, if it rotates the beam as a whole in the ᵟ= anticlockwise direction. + ve

In this problem A ϴ =0, ϴB≠0, ϴc≠ 0, δ =15mm Example (1): Analyse the continuous beam ABCD shown in figure by slope deflection method. The support B sinks by 15mm. Solution: In this problem A ϴ =0, ϴB≠0, ϴc≠ 0, δ =15mm

SINKING ( SETTLEMENT): FEM: FAB = - wab ˄2 / l˄2 = - 44.44 kn.m FAB = + wa˄2b / l˄2 = + 88.89 kn.m FBC = - wl˄2 / 8 = - 41.66 kn.m FCB = + wl˄2 / 8 = + 41.66 kn.m FEM due to yield of support B

SINKING ( SETTLEMENT): Slope deflection equation: MAB = FAB + 2EI / l ( 2ϴA+ ϴB - 3 δ/l ) = -44.44 +1/3 EI ϴB- 6 = -50.44 + 1/3 EI ϴB ..........(1) MBA = FBA + 2EI / l ( 2ϴA+ ϴB - 3 δ/l ) = + 88.89 + 2/3 EI ϴB- 6 = + 82.89 + 2/3 EI ϴB ............(2)

SINKING ( SETTLEMENT): MBC = FBC + 2EI / l ( 2ϴB+ ϴC - 3 δ/l ) = -44.44 + 4/5 EIϴB + 2/5 EI ϴC - 6 = -33.03 + 4/5 EIϴB + 2/5 EI ϴC .......... (3) MCB = FCB + 2EI / l ( 2ϴC+ ϴB - 3 δ/l ) = +41.67 + 4/5 EI ϴC + 2/5 EI ϴB + 8.64 = -50.31 + 4/5 EI ϴC + 2/5 EI ϴB .......... (4) MCB = - 30 KN.M .......... (5)

SINKING ( SETTLEMENT): There are only two unknown rotations ϴBand ϴC. Accordingly the boundary conditions are: MBA + MBC = 0 MCB + MCD = 0 Now, MBA + MBC = 49.66 + 22/15 EIϴB + 2/5 EIϴC = 0 MCB + MCD = 20.31 + 2/5 EIϴB + 4/5 EIϴC = 0 Solving the equation we get, ϴB = -31.95 / EI ϴC = -9.71 / EI

SINKING ( SETTLEMENT): Final Moments: MAB= -50.44 + 1/3 (-31.35) = -60.89 KN.M MBA= + 82.89 + 2/3 (-31.35) = + 61.99 KN.M MBC= -33.03 + 4/5 (-31.35) + 2/5 (-9.71) = - 61.99 KN.M MCB= -50.31 + 4/5 (-9.71) + 2/5 (-31.35) = + 30.00 KN.M MCD = - 30 KN.M

SINKING ( SETTLEMENT): Consider the free body diagram of continuous beam for finding reactions

SINKING ( SETTLEMENT):

SINKING ( SETTLEMENT): Reactions: Span AB: RB × 6 = 100 x 4 + 61.99 – 60.89 RB = 66.85 RA = 100 – RB = 33.15 KN Span BC: RB × 5 = 20 x 5 x25 + 61.99 – 30 RB = 56.40 KN RC = 20 x 5 - RB = 43.60 KN

SINKING ( SETTLEMENT): B.M. diagram: Span AB, M = Wab / l = (100×4×2) / 6 = 133.33 kn.m Span BC, M = Wl˄2 / 8 = ( 20×25) / 8 = 62.5 kn.m

Example (2):Determine the internal moments at the supports of the beam shown in Fig. The support at B is displaced (settles) 12 mm.

SINKING ( SETTLEMENT): Solution Two spans must be considered. FEMs are determined using

SINKING ( SETTLEMENT): 2. SLOPE DEFLECTION METHOD:

SINKING ( SETTLEMENT): 3. Equilibrium condition:

SINKING ( SETTLEMENT): In order to obtained the rotations equations (n) & (p) may then be solved simultaneously, it may be noted that since A is fixed support. Thus, Substituting these values into equations (i to l) yields