Topic3_Displacement Method of Analysis Frames Sideway.pptx

CED 426
Structural Theory II
Lecture 15
Displacement Method of Analysis:
Analysis of Frames: Sideway
Mary Joanne C. Aniñon
Instructor

Analysis of Frames:
Sidesways
• A frame will sidesway, or be
displaced to the side, when it or the
load acting on it is nonsymmetrical.
• For Fig. 10-18, the loading P causes
unequal moments 𝑀𝐵𝐶 and 𝑀𝐶𝐵

Analysis of Frames:
Sidesways
• 𝑀𝐵𝐶 tends to displace joint B to the
right whereas 𝑀𝐶𝐵 tends to
displace joint C to the left
• Since the two moments are not
equal, the resulting net
displacement is a sidesway of both
joint B and C to the right

Procedure for Analysis
• Degrees of Freedom
• Slope-Deflection Equations
• Equilibrium Equations
30 30

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.

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

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

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.

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

Analysis of Frames:
Sidesways
• Determine the moments at each
joint of the frame shown
Example 1
∆𝐵= ∆𝐶= ∆

Example 1
Step 1: Degrees of Freedom
1.a. Label all the supports and joints (nodes): Span AB, Span BC, and Span CD
1.b. Draw the possible deflected shape of the structure.
1.c. Identify the number of degrees of freedom: 𝜽𝑩 , 𝜽𝑪 and ∆
Note:
• Sidesway occurs since both the applied loading and the geometry of the frame are nonsymmetric.
• As shown in the figure, both joints B and C are assumed to be displaced an equal amount ∆.
𝜃𝐵 𝜃𝐶
∆𝐵 ∆𝐶

Example 1
Step 2: Slope-Deflection-Equation
𝑀𝑁𝐹 = 2𝐸(
𝐼
𝐿
)(2𝜃𝑁 + 𝜃𝐹 − 3
∆
𝐿
+ (𝐹𝐸𝑀)𝑁𝐹
Note:
• Since the ends A and D are fixed, the
equation below applies for all 3 spans of the
frame.

Example 1
𝐼
𝐿
)(2𝜃𝑁 + 𝜃𝐹 − 3
∆
𝐿
2.a. Write the Slope Deflection Equation for Span AB:
Note:
• A = D = 0 because it is fixed supports.
• (FEM)AB and (FEM)BA are zero because
there the load is applied directly to joint
B., and A and D are fixed supports
2.b. Calculate the FEM for span AB
𝜓𝐴𝐵 = 𝜓𝐵𝐴 =
Δ
4
𝜓𝐷𝐶 =
Δ
6
(𝜓𝐴𝐵)(4) = (𝜓𝐷𝐶)(6) (𝜓𝐴𝐵) = 𝜓𝐵𝐴 =
6
4
(𝜓𝐷𝐶)

Example 1
𝐼
𝐿
)(2𝜃𝑁 + 𝜃𝐹 − 3
∆
𝐿
2.a. Write the Slope Deflection Equation for Span BC:
Note:
• ΔB and ∆C are equal, therefore for span
BC, ∆ can be considered zero.
• (FEM)BC and (FEM)CB are zero because
2.b. Calculate the FEM for span BC

Example 1
𝐼
𝐿
)(2𝜃𝑁 + 𝜃𝐹 − 3
∆
𝐿
2.a. Write the Slope Deflection Equation for Span CD:
Note:
• (FEM)CD and (FEM)DC are zero because
2.b. Calculate the FEM for span CD
𝜓𝐶𝐷 =
Δ
6
𝜓𝐷𝐶 =
Δ
6
(𝜓𝐶𝐷)(6) = (𝜓𝐷𝐶)(6) (𝜓𝐶𝐷) = (𝜓𝐷𝐶)

Example 1
3.a. Equilibrium equations for each unknowns.
• These six equations contain nine unknowns.
𝑀𝐴𝐵 = 𝐸𝐼(0.5𝜃𝐵 − 2.25𝜓𝐷𝐶)
𝑀𝐵𝐴 = 𝐸𝐼(1𝜃𝐵 − 2.25𝜓𝐷𝐶
𝑀𝐵𝐶 = 𝐸𝐼(0.8𝜃𝐵 − 0.4𝜃𝐶
𝑀𝐶𝐵 = 𝐸𝐼(0.8 + 0.4𝜃𝐵
𝑀𝐶𝐷 = 𝐸𝐼(0.667𝜃𝐶 − 1𝜓𝐷𝐶
𝑀𝐷𝐶 = 𝐸𝐼(0.333𝜃𝐶 − 1𝜓𝐷𝐶

Example 1
• The first 2 equilibrium equations come from moment equilibrium
at joints B and C. The free body diagram of segment of the frame at
B and C is shown below
+M
+V
-M
+V
-M
+V
+M
+V

Example 1
• Since a horizontal displacement ∆ occurs, we will consider
summing forces on the entire frame in the x direction. This
yields (the 9th equation)
• The free body diagrams of span AB and CD are shown
VA
VB
-M
+V
+M
+V

EQUAL AND OPPOSITE REACTIONS
+ SHEAR
+ MOMENTS
LEFT SIDE RIGHT SIDE

Example 1
3.d. Substitute the slope-deflection equations into
the equilibrium equations and solve for the unknown
joint displacements.
(1)
(2)
(3)

Example 1
3.e. The results are then substituted into the slope-
deflection equations to determine the internal
moments at the ends of each member

Topic3_Displacement Method of Analysis Frames Sideway.pptx

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