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