Theory of Structures II Displacement Method of Analysis Beams Reference: Structural Analysis 10th Edition in SI Units by R.C. Hibbeler

1. CED 426
Structural Theory II
Lecture 13
Displacement Method of Analysis:
Analysis of Beams
Mary Joanne C. Aniñon
Instructor

2. Procedure for Analysis
• Degrees of Freedom
• Slope-Deflection Equations
• Equilibrium Equations

3. Step 1: Degrees of Freedom
1.a. Label all the supports and joints (nodes) in order to identify the spans of
the beam or frame between the nodes.
1.b. Draw the possible deflected shape of the structure.
1.c. Identify the number of degrees of freedom (angular displacement and
linear displacement)
1.d. Compatibility at the nodes is maintained provided the members that are
fixed connected to a node undergo the same displacement as the node
1.e. If these displacements are unknown, then for convenience assume they
act in the positive direction so as to cause clockwise rotation of the member
or joint.

4. Step 2: Slope-Deflection-Equations
2.a. Write the slope-deflection equations. The slope-deflection
equations relate the unknown internal moments at the nodes to the
displacements of the nodes for any span of the structure.
2.b. Calculate the FEM if a load exist on the span.
2.c. If the node has a linear displacement, calculate ψ = ∆/L for
adjacent spans

5. Step 2: Slope-Deflection-Equations
• Apply the slope deflection
equations (Eq. 10-8)
• However, if span at the end of a
continuous beam or frame is pin
supported apply Eq. 10-10 only to
the restrained end, thereby
generating one slope-deflection
equation for this span
• Note that Eq. 10-10 was derived
from Eq. 10-8 on the condition
that the end span of the beam or
frame is supported by pin or roller

6. Step 3: Equilibrium Equations
3.a. Write an equilibrium equation for each unknown degree of
freedom for the structure. Each of these equations should be
expressed in terms of the unknown internal moments as specified by
the slope-deflection equations.
3.b. For beams and frames, write the moment equation of equilibrium
at each support
3.c. For frames also write joint moment equations of equilibrium. If the
frame sidesways or deflects horizontally, column shears should be
related to the moments at the ends of column.

7. Step 3: Equilibrium Equations
3.d. Substitute the slope-deflection equations into the equilibrium
equations and solve for the unknown joint displacements.
3.e. The results are then substituted into the slope-deflection
equations to determine the internal moments at the ends of each
member
3.f. If any of the results are negative, they indicate counterclockwise
rotation; whereas positive moments and displacements create
clockwise rotation

8. Example 1
• Draw the shear and moment diagrams for the beam shown

9. Example 1
Step 1: Degrees of Freedom
1.a. Label all the supports and joints (nodes): Span AB and Span BC
1.b. Draw the possible deflected shape of the structure.
1.c. Identify the number of degrees of freedom: 𝜽𝑩

10. Example 1
Step 2: Slope-Deflection-Equation

11. Example 1
Step 2: Slope-Deflection-Equation
𝑀𝑁𝐹 = 2𝐸(
𝐼
𝐿
)(2𝜃𝑁 + 𝜃𝐹 − 3
∆
𝐿
+ (𝐹𝐸𝑀)𝑁𝐹

12. Example 1
Step 2: Slope-Deflection-Equation
𝑀𝑁𝐹 = 2𝐸(
𝐼
𝐿
)(2𝜃𝑁 + 𝜃𝐹 − 3
∆
𝐿
+ (𝐹𝐸𝑀)𝑁𝐹
2.a. Write the Slope Deflection Equation for Span AB:
Note:
• Since the support do not settle, ∆ = 0,
therefore AB = BA = 0
• (FEM)AB and (FEM)BA are zero because
there is no load on span AB.
2.b. Calculate the FEM for span AB

13. Example 1
Step 2: Slope-Deflection-Equation
𝑀𝑁𝐹 = 2𝐸(
𝐼
𝐿
)(2𝜃𝑁 + 𝜃𝐹 − 3
∆
𝐿
+ (𝐹𝐸𝑀)𝑁𝐹
2.a. Write the Slope Deflection Equation for Span BC:
Note:
• Since the support do not settle, ∆ = 0,
therefore AB = BA = 0
• (FEM)BC and (FEM)CB are computed from
the formulas for the FEMs given on the
inside back cover.
2.b. Calculate the FEM for span BC

14. Example 1
Step 2: Slope-Deflection-Equation
• The 𝐹𝐸𝑀𝐵𝐶 is negative since it acts counterclockwise on the beam at B.
• The Elastic curve for the beam is shown in Fig.10-10b. As indicated, there are four unknown
moments and an unknown slope at B. Since the supports do not settle ψ𝐴𝐵 𝑎𝑛𝑑 ψ𝐵𝐶
= -7.2 kN.m
10.8 kN.m=
From the formula:

15. Example 1
Step 3: Equilibrium Equations
3.a. Equilibrium equations for each unknowns.
• These four equations contain five unknowns. The necessary fifth equation comes from the
condition of moment equilibrium at support B.
𝑀𝐴𝐵 =
𝐸𝐼
4
𝜃𝐵
𝑀𝐵𝐴 =
𝐸𝐼
2
𝜃𝐵
𝑀𝐵𝐶 =
2𝐸𝐼
3
𝜃𝐵 − 7.2
𝑀𝐶𝐵 =
𝐸𝐼
3
𝜃𝐵 + 10.8

16. Example 1
Step 3: Equilibrium Equations
3.b. For beams and frames, write the moment equation of equilibrium at each support
• The free body diagram of segment of the beam at B is shown below
• Here, the moments act in the positive direction to be consistent
with the slope-deflection equations*.
• The beam shears contribute negligible moment about B since the
segment is of differential length

17. EQUAL AND OPPOSITE REACTIONS
+ SHEAR
+ MOMENTS

18. Example 1
Step 3: Equilibrium Equations
3.d. Substitute the slope-deflection equations into the equilibrium equations and solve for the
unknown joint displacements.
𝑀𝐴𝐵 =
𝐸𝐼
4
𝜃𝐵
𝑀𝐵𝐶 = 𝜃𝐵
𝑀𝐴𝐵 + 𝑀𝐵𝐶 = 0
𝐸𝐼
4
𝜃𝐵 + 𝜃𝐵 = 0 𝜃𝐵 =
6.17
𝐸𝐼

19. Example 1
Step 3: Equilibrium Equations
3.e. The results are then substituted into the slope-deflection equations to determine the internal
moments at the ends of each member
𝑀𝐴𝐵 =
𝐸𝐼
4
𝜃𝐵 =
𝐸𝐼
4
6.17
𝐸𝐼
= 1.54 𝑘𝑁. 𝑚
𝑀𝐵𝐴 =
𝐸𝐼
2
𝜃𝐵 =
𝐸𝐼
2
6.17
𝐸𝐼
= 3.09 𝑘𝑁. 𝑚
𝜃𝐵 =
6.17
𝐸𝐼
𝑀𝐵𝐶 =
2𝐸𝐼
3
𝜃𝐵 − 7.2 =
2𝐸𝐼
3
6.17
𝐸𝐼
− 7.2 = −3.09 𝑘𝑁. 𝑚
𝑀𝐶𝐵 =
𝐸𝐼
3
𝜃𝐵 + 10.8 =
𝐸𝐼
3
6.17
𝐸𝐼
− 10.8 = 12.86 𝑘𝑁. 𝑚
Note: The negative value for MBC indicates
that this moment acts counterclockwise on the
beam, NOT clockwise as shown in the figure
above.

20. Example 1
• Since we were asked to draw the shear and moment diagrams, the
next thing to do is to use these results to solve the shears at the end
spans.
• The free-body diagram of the entire beam are shown below.

21. Example 1
4.a. Solve the shear for span AB
𝑀𝐴𝐵 = 1.54 𝑘𝑁. 𝑚 𝑀𝐵𝐴 = 3.09 𝑘𝑁. 𝑚
A B
𝑀 = 0 at A
𝑀𝐴𝐵 + 𝑀𝐵𝐴 − 𝑅𝐵𝑌 8 = 0
𝑅𝐵𝑌 = 0.579 𝑘𝑁
𝐹𝑌 = 0
𝑅𝐵𝑌 − 𝑅𝐴𝑌 = 0
𝑅𝐴𝑌 = 0.579 𝑘𝑁
𝑅𝐴𝑌 𝑅𝐵𝑌

22. Example 1
4.b. Solve the shear for span BC
𝑀𝐵𝐶 = −3.09 𝑘𝑁. 𝑚 𝑀𝐶𝐵 = 12.86 𝑘𝑁. 𝑚
B C
𝑀 = 0 at A
𝑀𝐵𝐶 + 𝑀𝐶𝐵 − 𝑅𝐶𝑌 6 +
1
2
(6)(6)(
2
3
. 6) = 0
𝑅𝐶𝑌 = 13.63 𝑘𝑁
𝐹𝑌 = 0
𝑅𝐵𝑌 + 𝑅𝐶𝑌 −
1
2
(6)(6) = 0
𝑅𝐴𝑌 = 4.37 𝑘𝑁
𝑅𝐵𝑌 𝑅𝐶𝑌
6 𝑘𝑁/𝑚

23. Example 1
4.c. Draw the shear and moment diagrams.