1 Displacement Method (Stiffness) The Slope-Deflection approach entails writing expressions that relate the unknown deflections (displacements or rotations) at the ends of a member to the moments at the ends of a member. This method only considers flexural deformation!!!(i.e. it does not consider axial, shear, or torsion deformation) To illustrate how these equations can be used for structural analysis, consider the following two-span continuous beam under applied load P. Step 1: Draw the deflected shape and identify unknown deflections Based on this figure, we have three unknown deflections, !A, !B, !C. Note: in the slope-deflection method we are only concerned with the deflections at the ends of the members. P Moments at the ends of a flexural memberDeflections (displacements and rotations) at the ends of a flexural member Slope-Deflection equations P !C !B!A A C B A C B 2 Step 2: Write the slope-deflection equations for each member in terms of the unknown displacements We will derive these equations shortly, for now we may write them as: ) , , ( P f MB A AB! ! = ) , , ( P f MB A BA! ! = ) , (C B BCf M ! ! = ) , (C B CBf M ! ! = Step 3: Write equilibrium equations at each joint to supplement the slope-deflection equations written in Step 2. -Begin by Drawing the FBD of each joint and member: -Note the sign convention. -Positive moments are clockwise on the members -Do not confuse this convention (external moments), with the convention we use to plot moment diagrams (internal moments), i.e. -Positive shear forces are vertical upwards on members We have seven unknowns: -Rotations !A, !B, !C -Moments MAB, MBA, MBC, MCB Therefore, we need three additional equations! RA RB RC VAB MAB MABMBA MBAMBC MBCMCB MCB VBA VBCVCB (+) 3 -Writing the moment equilibrium equations for each joint: 0 =ABM BC BAM M = 0 =CBM Step 4: Solve for the unknown deflections and end moments -Substitute the expressions obtained in Step 2 for end moments (slope-deflection eqns) into the equations of equilibrium (Step 3) -Solve for deflections -Substitute the calculated deflections back into the slope-deflection equations and solve for the moments. -Calculate reactions, plot moment and shear diagrams, etc. Derivation of Slope-Deflection Equations We begin by defining a sign convention and terms. "AB = the rotation of the chord of member AB #AB = L("AB) (small angle theory) VA VB MAB MBA !B !A #AB P(x) A B The equilibrium equations supplement the slope-deflection eqns to give us 7 eqns and 7 unknowns. "AB L EI 4 Sign Convention Summary:-Moments are positive clockwise -Rotations are positive clockwise -Displacements are positive when they cause a clockwise rotation of " In general, the end moments are a function of end rotations, member displacement, and applied load (along with the member properties E, I, L) ) , , , ( P f MAB B A AB! = " " ) , , , ( P f MAB B A BA! = " " To simplify the derivation, we will solve for the moment due to each of the identified factors independently and then sum the solutions together using the principle of superposition. Step 1: Solve for MAB and MBA due to !A only (i.e. !B = #AB = P = 0) -Drawing the moment diagram: -Note that the sign conventions are different for external moments and internal moments. External moments are positive clockwise, whereas internal moments are positive when the beam is concave up. VA VB MAB MBA !A A L EI B MAB MBA M 5 -Assuming linear elastic behavior and constant EI, the curvature diagram becomes: Aside: Moment-Area Theorem (1) The change in slope between any two points on a continuous elastic beam is equal to the area under the $-diagram between the points. (2) The vertical distance between a point A on a continuous horizontal elastic beam to the tangent line drawn from point B is the moment of the $-diagram, between point A and B, about point A. Invoking the first portion of the moment area theorem and noting the difference in slope between points A and B is !A: ABA ABBA BABA ABAB ABM ML MEIMM ML MEIM! =""#$%%&'+"#$%&'"#$%&'(""#$%%&'+"#$%&'"#$%&'2121 ABA ABBA ABM MM MEIL! =""#$%%&'+("#$%&'2 22 ( )( )ABA ABBA AB BA ABM MM M M MEIL! =""#$%%&'++ ("#$%&'2 EIMAB $ EIMBA L ( )LM M MBA ABAB+ ( )LM M MBA ABBA+ 6 ( )A BA ABM MEIL! = "#$%&'(2(1) Invoking the second portion of the moment area theorem and noting the vertical displacement between point A and the tangent line drawn to point B is 0: 032213121=!!!!!"#$$$$$%&+!!"#$$%&!"#$%&+!!"#$$%&+!"#$%&!"#$%&'!!"#$$%&+!"#$%&!!"#$$%&+!"#$%&!"#$%&BA ABBA ABBA ABBA BABA ABABBA ABAB ABM ML M MM ML MEIMM ML MM ML MEIM03 2 6322232=!!"#$$%&'!!"#$$%&'!!"#$$%&BA AB BA ABMEILM MEILMEIL 0 2 33 2 3= ! !BA AB BA ABM M M M AB BAM M21=(2) Substituting Eqn. (2) into Eqn (1): LEIMAAB! 4= LEIMABA! 2= Step 2: Solve for MAB and MBA due to !B only (i.e. !A = #AB = P = 0) Relationship between end moments and rotation at end A VA VB MABMBA !B A L EI B 7 By inspection, this is the same problem we just solved. In addition, due to symmetry we can argue that the moment at A caused by a rotation at A should be equal to the moment at B caused by a rotation at B. LEIMBAB! 2= LEIMBBA! 4= Step 3: Solve for MAB and MBA due to #AB only (i.e. !A = !B = P = 0) -Due to symmetry, MAB = MBA.Drawing the moment diagram: -Assuming linear elastic behavior and constant EI, the curvature diagram becomes: Relationship between end moments and rotation at end B VA VB MAB MBA #AB A L EI B MAB MBA M 8 Invoking the second portion of the moment area theorem and noting the vertical displacement between point A and the tangent line drawn to point B is #AB: ABBA ABL LEIM L LEIM! ="#$%&'"#$%&'"#$%&'"#$%&'("#$%&'"#$%&'"#$%&'"#$%&'"#$%&'652 212 312 21 Noting that MAB = MBA ABAB ABEIL MEIL M! =""#$%%&'(""#$%%&'245242 2 26LEIMABAB!" = 26LEIMABBA!" = EIMAB $ EIMBA 2L 2L Relationship between end moments and differential displacement 9 Step 4: Solve for MAB and MBA due to P only (i.e. !A = !B = #AB = 0) The moments due to P only are commonly referred to as fixed-end moments (MFAB; MFBA) Common cases to memorize: VA VB MAB MBA A EI B L P(x) 22LPbaMBA =A EI B L P ba 22LPabMAB! =122wLMBA =AEIB L w (force/length) 122wLMAB! =10 Step 5: Combine all of the cases (principle of superposition) to obtain the slope-deflection equations. (Hint: memorize these equations) FABAB B AABMLEILEILEIM !" + =26 2 4 # # FBAAB B ABAMLEILEILEIM !" + =26 4 2 # # Example 1: Solve for the moment and shear diagrams of the following structure: Step 1: Draw the deflected shape and identify unknown deflections The only unknown we have is !A 202wLMBA =AEIB L w (force/length) 302wLMAB! =A EI B L P L/2L/2 PL M =A B P L/2 L/2 PL M =!A 11 Step 2: Write the slope-deflection equations for each member in terms of the unknown displacements Recognizing that !B = #AB = 0: FABAABMLEIM =! 4 FBAABAMLEIM =! 2 84 PLLEIMAAB! =" 82 PLLEIMABA+ =! We have three unknowns: -Rotation !A -Moments MAB and MBA Therefore, we need one additional equation. Step 3: Write equilibrium equations at each joint to supplement the slope-deflection equations written in Step 2. -Free body diagram: PL MAB = 2R MBA! = R1 VAB MAB MABMBA MBA R2 VBA R3 This equation introduces another unknown (R2) and thus doesnt help. 22LPba 22LPab!From the common cases discussed PL M =12 Step 4: Solve for the unknown deflections and end moments -Substitute the expressions obtained in Step 2 for end moments (slope-deflection Eqns) into the equations of equilibrium (Step 3) 84 PLLEIPLA! =" EIPLA3292= ! -Substitute the calculated deflections back into the slope-deflection equations and solve for the moments. 8 329 22PLEIPLLEIMBA+!!"#$$%&= 1611PLMBA = -Calculate reactions, plot moment and shear diagrams, etc. PL PL PL 1611PL 1611PL 1611PL P 163P 163P1619P 1619P A B PL P PL 1611PL!3213PL M 13 Example 2: Solve for the moment diagram of the following structure (only consider flexural deformation): Step 1: Draw the deflected shape and identify unknown deflections

The unknown displacements are !B and !D. 2L 2L LL L P P EI EI 2EI A BC D P P A BC D !B !D A B PL P V 1619P 1635P 1635P 14 Step 2: Write the slope-deflection equations for each member in terms of the unknown displacements Recognizing that !A = !C = #AB = #BC = #BD = MFAB= MFBA = MFBD = MFDB =0: For Member AB: LEIMBAB! 2= (1) LEIMBBA! 4= (2) For Member BC: ( )( )FABBBCMLEIM =22 4 ! ( )( )( )( )22222 2232 4LL L PLL LPLEIMBBC!"#$%&'"#$%&'! =( 3217 4 PLEIMBBC! =" (3) Similarly, 3211 2 PLEIMBCB+ =!(4) For Member BD: LEILEIMD BBD! ! 2 4+ = (5) LEILEIMD BBD! ! 4 2+ = (6) !"22LPab From the common cases discussed 15 We have eight unknowns: -Rotation !B and !D -Moments MAB, MBA, MBC, MCB, MBD, and MDB Therefore, we need two additional equations Step 3: Write equilibrium equations at each joint to supplement the slope-deflection equations written in Step 2. -Free body diagram (only showing moments): 3R MAB = 0 = + +BB BC BAM M M (7) 6R MCB! = 0 =DBM (8) Step 4: Solve for the unknown deflections and end moments (note, we now have 8 equations and 8 unknowns) -Substitute Eqn. (6) ! Eqn. (8) 04 2= +LEILEID B! ! B MDB A MAB MBA A R3 These equations introduce additional unknowns MBC A MCB A MBD A R6 A C D 16 2BD!! " = (9) -Substitute Eqn. (4), (5), (6) ! Eqn. (7) 02 43217 4 4= + + ! +LEILEI PLLEILEID B B B" " " "(10) -Substitute Eqn. (9) ! Eqn. (10) 022 43217 4 4= !"#$%&' + + ' +B B B BLEILEI PLLEILEI ( ( ( ( 3217 11 PLLEIB=! EIPLB352172= ! EIPLD704172! = " -Substitute back into the slope-deflection equations (i.e. Eqn. (1)-(6)) and solve for the end moments. 35234PLMAB =35268PLMBA = 352119PLMBC! =352155PLMCB = 35251PLMBD = 0 =DBM As a check, make sure these end moments satisfy the equilibrium equations: MBA+ MBC + MBD = 0 % MDB = 0 % 17 Complete free body diagram: Plotting the moment diagram on the compression side of the member: 0 35234PL 35268PL 352119PL 352155PL 35251PL PP 352102P 352102P 352422P 352282P 35251P 35251P P P 35234PL 35268PL 352119PL 352155PL35251PL 352127PL 35292PL M