engi6705-classnoteshandout6
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7/28/2019 ENGI6705-ClassNotesHandout6
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6. SLOPE DEFLECTION METHOD
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6.1 SLOPE DEFLECTION METHOD - AN OVERVIEW
• 6.1 SLOPE DEFLECTION METHOD - AN OVERVIEW
• 6.2 INTRODUCTION
• 6.3 DETAILS OF SLOPE DEFLECTION METHOD
• 6.4 SOLUTION OF PROBLEMS
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6.2 INTRODUCTION
As mentioned earlier, it is a displacement based analysis forindeterminate structures - Unknown displacements are first
written in terms of the loads by using load-displacement
relationships; then these equations are solved for the
displacements. Once the displacements are obtained, unknown
loads are determined from the compatibility equations usingload-displacement relationships.
- Nodes: Specified points on the structure that undergo displacements
(and rotations)
- Degrees of Freedom: These displacements (and rotations) are
referred to as degrees of freedom
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6.2 Introduction
To clarify these concepts we will consider some examples, beginning with the
beam in Fig. 1(a). Here any load P applied to the beam will cause node A onlyto rotate (neglecting axial deformation), while node B is completely restricted
from moving. Hence the beam has only one degree of freedom, A. The
beam in Fig. 1(b) has node at A, B, and C, and so has four degrees of
freedom, designed by the rotational displacements A, B, C, and the
vertical displacement C. Consider now the frame in Fig. 1(c). Again, if weneglect axial deformation of the members, an arbitrary loading P applied to the
frame can cause nodes B and C to rotate nodes can be displaced horizontally by an
equal amount. The frame therefore has three degrees of freedom, A, B, B.
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6.2 INTRODUCTION (Cont’d)
Fig. 1 (a)
Fig. 1 (b)
Fig. 1 (c)
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6.3 Slope Deflection Method
Consider portion AB of a continuous beam, shown below, subjected to a
distributed load w(x) per unit length and a support settlement of at B;
EI of the beam is constant.
A B
B
A
= rigid body motion
= /L
=
FEMAB FEMBA
MBA=(2EIA)/L MAB=(4EIA)/L
A
(i) Due to externallyapplied loads
(ii) Due to rotation A
at support A
+
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MBA=(4EIB)/L
B
MAB=(2EIB)/L
+
A B(iii) Due to rotation
B at support B
L
MAB=(-6EI)/L2
A B
L
MBA=(-6EI)/L2
+
(iv) Due to differential
settlement of (between A
and B)
BA(FEM))
L
3Δ
B2θ
A(θ
L
2EI2L
6EIΔ
L
B4EIθ
L
A2EIθ
BA(FEM)
BAM
BAM
BAM
BA(FEM)
BAM
AB(FEM))
L
3Δ
Bθ
A(2θ
L
2EI
2L
6EIΔ
L
B2EIθ
L
A4EIθ
AB(FEM)
ABM
ABM
ABM
AB(FEM)
ABM
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A
B
=
(FEM)AB
(FEM)BA
(FEM)BA/2
(FEM)BA
+
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