Slope Deflection Method:

Slides from Leet et al. 4th Ed.

Chapter 12

1
12.2 Illustration of the Slope-Deflection
Method

Continuous beam with applied loads

(deflected shape shown by dashed line)

Figure 12.1
2
12.2 Illustration of the Slope-Deflection
Method

Free bodies of joints and beams (sign convention:

Clockwise moment on the end of a member is positive)

Figure 12.1 (continued)

3
12.3 Derivation of the Slope-Deflection
Equation

Continuous beam whose supports settle under load

Figure 12.2
4
12.3 Derivation of the Slope-Deflection
Equation

Deformations of member
AB plotted to an
exaggerated vertical
scale

Figure 12.2 (continued)

5
12.3 Derivation of the Slope-Deflection
Equation

Figure 12.3 Simple beam moment curve

produced by a uniform load
6
12.3 Derivation of the Slope-Deflection
Equation

Figure 12.4
7
12.3 Derivation of the Slope-Deflection
Equation

Figure 12.5 Fixed-end moments

8
12.3 Derivation of the Slope-Deflection
Equation

Figure 12.5 Fixed-end moments (continued)

9
12.3 Derivation of the Slope-Deflection
Equation

Figure 12.5 Fixed-end moments (continued)

10
12.3 Derivation of the Slope-Deflection
Equation

Figure 12.5 Fixed-end moments (continued)

11
 Fixed end moment equation

A=L
B=R

2EI
MLR = F EMLR + (2L + R 2 LR )
L 12
 Example 12.1
Using Equations 12.12 and 12.13, compute the fixed-end moments
produced by a concentrated load P at midspan of the fixed-ended beam
in Figure 12.6a using the unit load method. We know that EI is
constant.

13
12.4 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 equal zero translate; chord rotations equal zero

Figure 12.7
14
12.4 Analysis of Structures by the Slope-
Deflection Method

Unbraced frames with chord rotations

Figure 12.7 (continued)

15
 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.

16
Example 12.2 Solution
Since joint A is fixed against rotation, A
= 0; therefore, the only unknown
displacement is B. Using the slope-
deflection equation

The member end moments are:

To determine B, write the equation of

moment equilibrium at joint B

17
 Example 12.2 Solution (continued)
Substituting the value of MBA and solving for B 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,

18
 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 B, express all variables in units of inches and kips.

19
 Example 12.2 Solution (continued)
Expressing B in degrees

Shear and moment curves

20
 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.

21
 Example 12.3 Solution

Use the slope-deflection equation

The fixed-end moments produced by the uniform load on member AB

22
 Example 12.3 Solution (continued)

Joint D Joint B

Express the member end moments as

23
 Example 12.3 Solution (continued)

Joint D Joint B

To solve for the unknown joint displacements B and D, write equilibrium

equations at joints D and B.

24
 Example 12.3 Solution (continued)
Express the moments in terms of displacements; write the equilibrium
equations as

Solving equations simultaneously gives

25
 Example 12.3 Solution (continued)
To establish the values of the member end moments, the values of B
and D are substituted

26
Example 12.3 Solution (continued)

Free bodies of members

and joints used to
compute shears and
reactions

27
Example 12.3 Solution (continued)

Free bodies of members

and joints used to
compute shears and
reactions

28
 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.

29
 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
B = and C = -,

30
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 produce

31
 Example 12.4 Solution (continued)
Substituting the value of given by Equation 5 into Equations 1, 2, and
3 gives

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 12.4 Solution (continued)

Free bodies of girder

BC and column AB
used to compute
shears; final shear
and moment curves
also shown

34
 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.

35
 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:

36
 Example 12.5 Solution (continued)

Free body of beam AB, joint B,

and column BD. Final shear and
moment diagrams for beam AB.
37
 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.

38
 Example 12.6 Solution
A = -0.009 rad. The settlement of
support B relative to support A
produces a clockwise chord
rotation

Angle B is the only unknown displacement. Expressing member end

moments with the slope-deflection equation

39
Example 12.6 Solution (continued)
Writing the equilibrium equation at
joint B yields

Substituting Equation 2 into

Equation 3 and solving for B yield

40
 Example 12.6 Solution (continued)
To evaluate MAB, substitute B into Equation 1:

Complete the analysis by using the equations of statics to compute the

reaction at B and the shear at A.

41
Example 12.6 Solution (continued)

Shear and moment curves

42
 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.

43
Example 12.7 Solution

The chord rotation BC of column

BC equals

Since the ends of girder AB are at

the same level, AB = 0. The
unknown displacements are B and
C

44
 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.

45
 Example 12.7 Solution (continued)
Writing equilibrium equations gives

Substituting and solving for B and C yield

Substituting C and B into Equations 1 to 3 produces

46
Example 12.7 Solution (continued)

47
 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

Figure 12.14
48
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)

49
 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.

50
Example 12.8 Solution
Identify the unknown displacements
B, C, and . Express the chord
rotations AB and CD in terms of :

Compute the relative bending

stiffness of all members.

51
 Example 12.8 Solution (continued)
Set 20E = K, then

Express member end moments in terms of displacements: MNF = (2EI/L)

(2N + F - 3NF) + FEMNF. Since no loads are applied to members
between joints, all FEMNF = 0.

Use Equations 1 to express AB in terms of CD, and use Equations 2 to

express all stiffness in terms of the parameter K.

52
 Example 12.8 Solution (continued)

The equilibrium equations are:

Substitute Equations 4 into Equations 5, 6, and 7 and combine terms.

53
 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.

54
Example 12.8 Solution (continued)

Reactions and moment diagrams

55
 Example 12.9
Analyze the frame in Figure 12.16a by the slope-deflection method.
Given: EI is constant for all members.

56
Example 12.9 Solution
Express member end moments
in terms of displacements with
Equation 12.16 (all units in kip-
feet).

57
 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 58

 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:

59
 Example 12.9 Solution (continued)
Solution of the equations

Establish the values of member end moments by substituting the

values of B, C, and AB into Equations 1.

60
Example 12.9 Solution (continued)

Reactions and shear

and moment curves

61
 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.

62
 Example 12.10 Solution

Express member end moments in terms

of displacements with the slope-
deflection equation.

63
 Example 12.10 Solution (continued)
To simplify slope-deflection expressions, set EI/15 = K.

64
 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.

65
 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 B, C, and into Equations 1,

66
 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), B = 5.86; substituting into

Equation 7,

67
Example 12.10 Solution (continued)

68
 12.6 Kinematic Indeterminacy

Indeterminate first degree, Indeterminate fourth degree

neglecting axial deformations

Figure 12.18 Evaluating degree of kinematic indeterminacy

69
 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)

70