Concept of fixed end moments
Obtained using unit load method
1
Derivation of the Slope-Deflection
Equation
Figure 12.5 Fixed-end moments
2
Derivation of the Slope-Deflection
Equation
Figure 12.5 Fixed-end moments (continued)
3
Derivation of the Slope-Deflection
Equation
Figure 12.5 Fixed-end moments (continued)
4
Derivation of the Slope-Deflection
Equation
Figure 12.5 Fixed-end moments (continued)
5
Derivation of the Slope-Deflection
Equation
Continuous beam whose supports settle under load
Figure 12.2
6
§12.3 Derivation of the Slope-Deflection
Equation
Deformations of member
AB plotted to an
exaggerated vertical
scale
7
Derivation of the Slope-Deflection
Equation
Figure 12.4
8
Illustration of the Slope-Deflection
Method
Continuous beam with applied loads
(deflected shape shown by dashed line)
Figure 12.1
9
§12.3 Derivation of the Slope-Deflection
Equation
Deformations of member
AB plotted to an
exaggerated vertical
scale
10
§12.3 Derivation of the Slope-Deflection
Equation
Deformations of member
AB plotted to an
exaggerated vertical
scale
x
N F
11
Illustration of the Slope-Deflection
Method
Free bodies of joints and beams (sign convention:
Clockwise moment on the end of a member is
positive)
12
Analysis of Structures by the Slope-
Deflection Method
All joints restrained against Due to symmetry of structure and
displacement; all chord loading, joints free to rotate but not
rotations ψ equal zero translate; chord rotations equal zero
Figure 12.7
13
Analysis of Structures by the Slope-
Deflection Method
Unbraced frames with chord rotations
Figure 12.7 (continued)
14
Example 12.2
Using the slope-deflection method, determine the member end
moments in the indeterminate beam shown in Figure 12.8a. The beam,
which behaves elastically, carries a concentrated load at midspan. After
the end moments are determined, draw the shear and moment curves.
If I = 240 in4 and E = 30,000 kips/in2, compute the magnitude of the
slope at joint B.
15
Example 12.2 Solution
• Since joint A is fixed against rotation, θA
= 0; therefore, the only unknown
displacement is θB. Using the slope-
deflection equation
• The member end moments are:
• To determine θB, write the equation of
moment equilibrium at joint B
16
Example 12.2 Solution (continued)
• Substituting the value of MBA and solving for θB give
where the minus sign indicates both that the B end of member
AB and joint B rotate in the counterclockwise direction
• To determine the member end moments,
17
Example 12.2 Solution (continued)
• To complete the analysis, apply
the equations of statics to a free
body of member AB
Free body used
to compute end
shears
• To evaluate θB, express all variables in units of inches and kips.
18
Example 12.2 Solution (continued)
• Expressing θB in degrees
Shear and moment curves
19
Example 12.3
Using the slope-deflection method, determine the member end
moments in the braced frame shown in Figure 12.9a. Also compute the
reactions at support D, and draw the shear and moment curves for
members AB and BD.
20
Example 12.3 Solution
• Use the slope-deflection equation
• The fixed-end moments produced by the uniform load on member AB
21
Example 12.3 Solution (continued)
Joint D Joint B
• Express the member end moments as
22
Example 12.3 Solution (continued)
Joint D Joint B
• To solve for the unknown joint displacements θB and θD, write equilibrium
equations at joints D and B.
23
Example 12.3 Solution (continued)
• Express the moments in terms of displacements; write the equilibrium
equations as
• Solving equations simultaneously gives
24
Example 12.3 Solution (continued)
• To establish the values of the member end moments, the values of θB
and θD are substituted
25
Example 12.3 Solution (continued)
Free bodies of members
and joints used to
compute shears and
reactions
26
Example 12.3 Solution (continued)
Free bodies of members
and joints used to
compute shears and
reactions
27
Example 12.4
Use of Symmetry to Simplify the Analysis of a Symmetric Structure
with a Symmetric Load
Determine the reactions and draw the shear and moment curves for the
columns and girder of the rigid frame shown in Figure 12.10a. Given: IAB
= ICD = 120 in4, IBC = 360 in4, and E is constant for all members.
28
Example 12.4 Solution
Moments
acting on
joint B
• Expressing member end moments with Equation 12.16, reading the value
of fixed-end moment for member BC from Figure 12.5d, and substituting
θB = θ and θC = -θ,
29
Example 12.4 Solution (continued)
• Writing the equilibrium equation at joint B yields
Moments
acting on • Substituting Equations 2 and 3 into Equation 4
joint B and solving for θ produce
30
Example 12.4 Solution (continued)
• Substituting the value of θ given by Equation 5 into Equations 1, 2, and
3 gives
31
Example 12.4 Solution (continued)
Free bodies of girder
BC and column AB
used to compute
shears; final shear
and moment curves
also shown
32
Example 12.4 Solution (continued)
Free bodies of girder
BC and column AB
used to compute
shears; final shear
and moment curves
also shown
33
Example 12.5
Using symmetry to simplify the slope-deflection analysis of the frame in
Figure 12.11a, determine the reactions at supports A and D. EI is
constant for all members.
34
Example 12.5 Solution
• Since all joint and chord rotations are zero, the member end moments
at each end of beams AB and BC are equal to the fixed-end moments
PL/8 given by Figure 12.5a:
35
Example 12.5 Solution (continued)
Free body of beam AB, joint B,
and column BD. Final shear and
moment diagrams for beam AB.
36
Example 12.6
Determine the reactions and draw the shear and moment curves for the
beam in Figure 12.12. The support at A has been accidentally
constructed with a slope that makes an angle of 0.009 rad with the
vertical y-axis through support A, and B has been constructed 1.2 in
below its intended position. Given: EI is constant, I = 360 in4, and E =
29,000 kips/in2.
37
Example 12.6 Solution
• θA = -0.009 rad. The settlement of
support B relative to support A
produces a clockwise chord
rotation
• Angle θB is the only unknown displacement. Expressing member end
moments with the slope-deflection equation
38
Example 12.6 Solution (continued)
• Writing the equilibrium equation at
joint B yields
• Substituting Equation 2 into
Equation 3 and solving for θB yield
39
Example 12.6 Solution (continued)
• To evaluate MAB, substitute θB into Equation 1:
• Complete the analysis by using the equations of statics to compute the
reaction at B and the shear at A.
40
Example 12.6 Solution (continued)
Shear and moment curves
41
Example 12.7
Although the supports are constructed in their correct position, girder AB
of the frame shown in Figure 12.13 is fabricated 1.2 in too long.
Determine the reactions created when the frame is connected into the
supports. Given: EI is a constant for all members, I = 240 in4, and E =
29,000 kips/in2.
42
Example 12.7 Solution
• The chord rotation ψBC of column
BC equals
• Since the ends of girder AB are at
the same level, ψAB = 0. The
unknown displacements are θB and
θC
43
Example 12.7 Solution (continued)
• Using the slope-deflection equation (Equation 12.16), express member
end moments in terms of the unknown displacements. Because no loads
are applied to the members, all fixed-end moments equal zero.
44
Example 12.7 Solution (continued)
• Writing equilibrium equations gives
• Substituting and solving for θB and θC yield
• Substituting θC and θB into Equations 1 to 3 produces
45
Example 12.7 Solution (continued)
46
§12.5 Analysis of Structures That Are
Free to Sidesway
Unbraced frame, deflected shape shown to an exaggerated scale by
dashed lines, column chords rotate through a clockwise angle ψ
Figure 12.14
47
§12.5 Analysis of Structures That Are
Free to Sidesway
Free-body diagrams of
columns and girders;
unknown moments shown
in the positive sense, that
is, clockwise on ends of
members
Figure 12.14 (continued)
48
Example 12.8
Analyze the frame in Figure 12.15a by the slope-deflection method. E is
constant for all members; IAB = 240 in4, IBC = 600 in4, and ICD = 360 in4.
49
Example 12.8 Solution
• Identify the unknown displacements
θB, θC, and ∆. Express the chord
rotations ψAB and ψCD in terms of ∆:
• Compute the relative bending
stiffness of all members.
50
Example 12.8 Solution (continued)
• Set 20E = K, then
• Express member end moments in terms of displacements: MNF = (2EI/L)
(2θN + θF - 3ψNF) + FEMNF. Since no loads are applied to members
between joints, all FEMNF = 0.
• Use Equations 1 to express ψAB in terms of ψCD, and use Equations 2 to
express all stiffness in terms of the parameter K.
51
Example 12.8 Solution (continued)
• The equilibrium equations are:
• Substitute Equations 4 into Equations 5, 6, and 7 and combine terms.
52
Example 12.8 Solution (continued)
• Solving the equations simultaneously gives
Also,
Since all angles are positive, all joint rotations
and the sidesway angles are clockwise.
• Substituting the values of displacement above into Equations 4, establish
the member end moments.
53
Example 12.8 Solution (continued)
Reactions and moment diagrams
54
Example 12.9
Analyze the frame in Figure 12.16a by the slope-deflection method.
Given: EI is constant for all members.
55
Example 12.9 Solution
• Express member end moments
in terms of displacements with
Equation 12.16 (all units in kip-
feet).
56
Example 12.9 Solution (continued)
• Write the joint equilibrium equations
Moments at B and C. Joint B:
acting on
joint B
• Joint C:
• Shear equation:
Free
body of
column
AB
•Solving for V1 gives
Moments acting on joint C 57
Example 12.9 Solution (continued)
Free body of girder
used to establish third
equilibrium equation
• Isolate the girder and consider • Express equilibrium equations in
equilibrium in the horizontal terms of displacements by
direction. substituting Equations 1 into
Equations 2, 3, and 4. Collecting
terms and simplifying,
• Substitute Equation 4a into
Equation 4b:
58
Example 12.9 Solution (continued)
• Solution of the equations
• Establish the values of member end moments by substituting the
values of θB, θC, and ψAB into Equations 1.
59
Example 12.9 Solution (continued)
Reactions and shear
and moment curves
60
Example 12.10
Analyze the frame in Figure 12.17a by the slope-deflection method.
Determine the reactions, draw the moment curves for the members, and
sketch the deflected shape. If I = 240 in4 and E = 30,000 kips/in2,
determine the horizontal displacement of joint B.
61
Example 12.10 Solution
• Express member end moments in terms
of displacements with the slope-
deflection equation.
62
Example 12.10 Solution (continued)
• To simplify slope-deflection expressions, set EI/15 = K.
63
Example 12.10 Solution (continued)
• The equilibrium equations are:
• Shear equation:
where
• Substituting V1 and V2 given by Equations 4b into 4a gives
Alternatively, set Q = 0 in Equation 12.21 to produce Equation 4.
64
Example 12.10 Solution (continued)
• Express equilibrium equations in terms of displacements by substituting
Equations 1 into Equations 2, 3, and 4. Combining terms and simplifying
give
• Solving the equations simultaneously,
• Substituting the values of the θB, θC, and ψ into Equations 1,
65
Example 12.10 Solution (continued)
• Compute the horizontal displacement of joint B. Use Equation 1 for MAB.
Express all variables in units of inches and kips.
• From the values in Equation 5 (p. 485), θB = 5.86ψ; substituting into
Equation 7,
66
Example 12.10 Solution (continued)
67
Determine all relations at points A and D in Figure shown. EI is constant.
74
§12.6 Kinematic Indeterminacy
Indeterminate first degree, Indeterminate fourth degree
neglecting axial deformations
Figure 12.18 Evaluating degree of kinematic indeterminacy
75
§12.6 Kinematic Indeterminacy
Indeterminate eighth degree, Indeterminate eleventh degree,
imaginary rollers added at points 1 imaginary rollers added at points
and 2 1, 2, and 3
Figure 12.18 Evaluating degree of kinematic indeterminacy (continued)
76
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