Springer Tracts in Civil Engineering
Isabella Giorgia Colombo ·
Matteo Colombo · Marco di Prisco ·
Anna Magri · Paolo Martinelli ·
Letizia Mazzoleni · Giulio Zani
Structural
Analysis
of Plane
Frames
Solved Examples with Force and
Displacement Methods
Springer Tracts in Civil Engineering
Series Editors
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Wuhan University, Wuhan, China
Marco di Prisco, Politecnico di Milano, Milano, Italy
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Athens, Athens, Greece
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Isabella Giorgia Colombo · Matteo Colombo ·
Marco di Prisco · Anna Magri · Paolo Martinelli ·
Letizia Mazzoleni · Giulio Zani
Structural Analysis of Plane
Frames
Solved Examples with Force
and Displacement Methods
Isabella Giorgia Colombo
DSC-Erba s.r.l.
Erba, Italy
Marco di Prisco
DSC-Erba s.r.l.
Erba, Italy
Department of Civil and Environmental
Engineering
Politecnico di Milano
Milan, Italy
Paolo Martinelli
Department of Civil and Environmental
Engineering
Politecnico di Milano
Milan, Italy
Matteo Colombo
Department of Civil and Environmental
Engineering
Politecnico di Milano
Milan, Italy
Anna Magri
Leviat s.r.l.
Bergamo, Italy
Letizia Mazzoleni
Cremonesi Workshop s.r.l.
Brescia, Italy
Giulio Zani
Department of Civil and Environmental
Engineering
Politecnico di Milano
Milan, Italy
ISSN 2366-259X
ISSN 2366-2603 (electronic)
Springer Tracts in Civil Engineering
ISBN 978-3-031-35266-9
ISBN 978-3-031-35267-6 (eBook)
https://doi.org/10.1007/978-3-031-35267-6
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature
Switzerland AG 2023
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Preface
This book contains a wide range of worked examples concerning the solution of statically indeterminate frame structures. The book is aimed at providing a useful tool for
anyone wishing to learn more about the application of practical methods for the evaluation of internal forces/moments in plane frames. It is intended to be both a valuable
reference guide for structural engineering professionals and an ideal learning resource
for students of civil engineering, building engineering and architecture programs.
This book has been written in the cultural tradition of the Civil Engineering School
and Building Engineering/Architecture School of the Politecnico di Milano and is
dedicated to Structural Design courses. The course of Structural Design is placed as
logical development after the course of Structural Mechanics. It is therefore assumed
that the student has taken university courses in Statics and/or Structural Mechanics.
This textbook presents applied examples of the main methods of structural analysis of statically indeterminate frame structures. The textbook begins with a brief
description of kinematic analysis for plane frames (Chap. 1). The force method, the
displacement method and the mixed method are applied for the solution of statically
indeterminate plane structures. The textbook first deals with the solution of simple
reference cases in which recurring structural situations such as inclined rods, extensional and rotational springs, thermal variations, symmetry and anti-symmetry (just
to name a few) are treated individually (Chap. 2). The textbook then presents the
complete solution of complex plane frames, in which the common structural situations individually analyzed in the previous chapter are combined (Chap. 3). The
tabular coefficients used for the force and displacement methods are collected in
Appendix A, while Appendix B summarizes the simple reference cases covered in
each complex frame example. The book, due to its peculiar nature, is suitable for
both sequential and subject-specific reading.
v
vi
Preface
We believe that most of the literature devoted to the structural analysis of statically
indeterminate frame structures covers mainly theoretical aspects. We decided to write
this book with the primary aim of providing practical and relevant worked examples
for real structures, with a focus on static schemes that are readily understandable to
students and designers alike.
Milan, Italy
Isabella Giorgia Colombo
Matteo Colombo
Marco di Prisco
Anna Magri
Paolo Martinelli
Letizia Mazzoleni
Giulio Zani
Acknowledgement
The authors acknowledge the assistance of students Andrea Baldi, Matteo Gianoli
and Giulia Lopez in preparing some of the figures in the textbook.
vii
Contents
1 Short Notes on Kinematic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1 Redundancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Centre of Instantaneous Rotation (CIR) . . . . . . . . . . . . . . . . . . . . . . .
1
1
1
2 Solution of Simple Reference Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 Problem A(I): Inclined Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Problem A(II): Inclined Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Problem B(I): Extensional Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Problem B(II): Extensional Spring . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Problem C(I): Rotational Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Problem C(II): Rotational Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7 Problem D(I): Thermal Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8 Problem D(II): Thermal Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9 Problem E(I): Symmetry and Anti-symmetry . . . . . . . . . . . . . . . . . .
2.10 Problem E(II): Symmetry and Anti-symmetry . . . . . . . . . . . . . . . . .
2.11 Problem E(III): Symmetry and Anti-symmetry . . . . . . . . . . . . . . . .
2.12 Problem F(I): Imposed Displacement/Rotation . . . . . . . . . . . . . . . . .
2.13 Problem F(II): Imposed Displacement/Rotation . . . . . . . . . . . . . . . .
2.14 Problem G(I): Infinitely Rigid Bending Rod . . . . . . . . . . . . . . . . . . .
2.15 Problem G(II): Infinitely Rigid Bending Rod . . . . . . . . . . . . . . . . . .
2.16 Problem H: Concentrated Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.17 Problem I: Statically Determined Portion . . . . . . . . . . . . . . . . . . . . .
2.18 Problem J: Unusual Restraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.19 Problem K(I): Distributed Loads/Moments . . . . . . . . . . . . . . . . . . . .
2.20 Problem K(II): Distributed Loads/Moments . . . . . . . . . . . . . . . . . . .
2.21 Problem L: Rod with Finite Axial Stiffness . . . . . . . . . . . . . . . . . . . .
7
7
19
27
35
42
53
59
65
71
75
82
92
98
103
106
111
116
122
127
132
137
3 Solution of Complex Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1 Worked Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Worked Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Worked Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Worked Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
143
143
158
164
172
ix
x
Contents
3.5
3.6
3.7
3.8
3.9
3.10
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.22
3.23
3.24
3.25
3.26
3.27
3.28
3.29
3.30
Worked Example 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Worked Example 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
179
185
191
197
203
210
216
222
228
233
238
245
253
259
265
272
279
285
290
295
299
307
313
319
325
331
Appendix A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
Appendix B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
Chapter 1
Short Notes on Kinematic Analysis
1.1 Redundancy
The force and displacement methods are used to solve statically indeterminate structures. Therefore, the first step to be performed is to count the degrees of freedom
(DOF) and the number of restraints (NOR). Table 1.1 provides the number of restraint
reactions for the most common restraint types.
1.2 Centre of Instantaneous Rotation (CIR)
If the structure turns out to be statically determined it is solved by the equilibrium
equations alone. If DOFs < NORs the structure is statically indeterminate.
The kinematic analysis allows understanding which restraints are overabundant
and have to be replaced by unknown reactions (statically indeterminate problem).
A frame with displaceable joints (i.e. sway frame) is defined as a frame in which
the joints, in addition to rotating, can also translate. If axial deformability is not
neglected, every frame always results as sway frame.
To examine if a frame is fixed-joint or sway, one must assess the degree of restraint
of the structure by inserting a hinge at each inner and outer joint where rotation
(absolute or relative) appears to be restrained, either it is perfectly restrained (rigid
restraint) or only partially restrained (elastic restraint).
The untied structure is called “associated truss structure”. If the associated truss
structure is statically determinate or statically indeterminate, the frame has fixed
joints. On the contrary, if the associated truss structure is hypostatic and allows
for the possibility of rigid body motions, the frame is sway. The number of simple
restraints to be introduced should be enough to lock any rigid body motion of the
associated truss structure. To figure out which restraints to add, one must look for
the centers of rotation of each member.
It is necessary to downgrade all the flexural restraints, as shown in Table 1.2
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
I. G. Colombo et al., Structural Analysis of Plane Frames, Springer Tracts in Civil
Engineering, https://doi.org/10.1007/978-3-031-35267-6_1
1
2
1 Short Notes on Kinematic Analysis
Table 1.1 Restraint numbering
Some examples are given below.
Example 1.1 Let’s consider the following structure:
After counting the number of restraints, it is concluded that the structure is statically indeterminate with redundancy grade equal to one. The downgrading of the
restraints can be performed.
1.2 Centre of Instantaneous Rotation (CIR)
3
Table 1.2 Downgrading of restraints
The structure is free to move according to the kinematics shown in the figure
below.
4
1 Short Notes on Kinematic Analysis
Rod number 3 is free to move horizontally, so the kinematic motion remains
unchanged even considering only rods 1 and 2.
The CIR of the entire structure can be found at the point of intersection of the
action lines (dotted grey lines).
To prevent the kinematic motion, it is possible to block the vertical displacement
of point B or, alternatively, the horizontal displacement of point D.
Example 1.2 Let’s consider the following structure:
The rod BC has an infinite flexural stiffness, so it transmits shear force and bending
moment directly onto the slider without deforming. Therefore, the structure can be
simplified as shown in the following figure.
1.2 Centre of Instantaneous Rotation (CIR)
5
Downgrading of the restraints is carried out. The structure turns out to be
hypostatic and the rod rotates around the hinge A.
A roller is then introduced to prevent the vertical translation of point B.
Chapter 2
Solution of Simple Reference Cases
2.1 Problem A(I): Inclined Rod
• E I = constant.
• EA →
/(∞(
l 2
• l∗ =
+ l2 =
2
√
5
l
2
Solution
The structure is statically indeterminate, having 6 restraints and 3 degrees of freedom
(NOR = 6 > DOF = 3).
The kinematic analysis of the structure is then carried out; to do this, it is necessary
to release the bending restraints. The member AB can be replaced with a roller with
vertical axis, while member CD can be replaced with a roller with axis parallel to the
member CD itself. In this way the member BC has a center of instantaneous rotation
(CIR) indicated in the figure as CIR.
Nodes B and C rotate with respect to the CIR and consequently the structure is a
sway frame.
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
I. G. Colombo et al., Structural Analysis of Plane Frames, Springer Tracts in Civil
Engineering, https://doi.org/10.1007/978-3-031-35267-6_2
7
8
2 Solution of Simple Reference Cases
The displacement method is used to solve the structure, adopting as hyperstatic
unknowns the rotations ϕ1 and ϕ2 and the horizontal translation δ3 .
An additional rotational restraint is introduced at nodes B and C (rigid block =
ground spring with infinite rotational stiffness) and at point B an additional translational restraint is also added (roller with horizontal axis) such as to prevent the
activation of the kinematics identified by the kinematic analysis.
For each of the unknowns it is possible to write an equilibrium equation. In total,
there will therefore be 2 equilibrium equations to the rotation and an equilibrium
equation to the horizontal translation:
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
2.1 Problem A(I): Inclined Rod
Kinematics
• ϕ1 = 1
9
10
2 Solution of Simple Reference Cases
4E I
6E I
4E I
+
=
l
2l
l
EI
2E I
=
m 21 =
2l
l
21E I
r31 = −
4l2
m 11 =
The reaction r31 is calculated by writing a rotational equilibrium equation of the whole
structure with respect to the CIR. The member CD has no flexural deformation and
consequently the shear force at the base D and the bending moment in D are zero.
The axial forces in A and D have a null lever arm with respect to the CIR. In writing
the rotational equilibrium, all the restraints reactions are considered, including those
on the rigid blocks (moments), which represent an on-ground restraint.
4E I
4E I
2E I
6E I
2E I
−
−
−
=0
5l + r31 4l −
2
l
l
l
2l
2l
21E I
r31 = −
4l2
• ϕ2 = 1
EI
l
(
√ )
√
4E I
10 + 8 5 E I
4E I
8 5 EI
+ ∗ = 2+
=
m 22 =
2l
l
5
l
5
l
√
15 − 48 5 E I
r32 =
20
l2
m 12 =
2.1 Problem A(I): Inclined Rod
11
• δ3 = 1
As a result of the global kinematics, node C undergoes a translation equal to δ.
Using the trigonometric relationship:
cos α =
l
√
5
l
2
2
=√
5
we have that:
√
1
5
δH
= 2 =
δ=
√
cos α
2
5
Node C undergoes a vertical translation equal to δV . Considering that
tan α =
1
l/2
=
l
2
the following relationship is obtained:
1
δH =
2
6E I
=
δV
4l2
6E I
=
δV
4l2
δV =
m 13
m 23
1
2
6E I
3E I
6E I
=
− 2 =
2
2
l
4l
l
√
6E I
3 EI
12 5
− ∗2 δ =
−
l
4 l2
5
−
−21 E I
4 l2
√
EI
15 − 48 5 E I
=
l2
20
l2
For the calculation of the roller restraint reaction (r33 ), as before, the global rotational
equilibrium equation of the whole structure is written:
12
2 Solution of Simple Reference Cases
12E I
12E I
6E I
6E I
6E I
6E I
5l + r33 4l − ∗3 δ5l∗ + 2 + ∗2 δ + 2 + ∗2 δ+
l3
l
l
l
l
l
6E I
6E I
− 2 δV −
δV = 0
4l
4l2
(
√ )
495 + 192 5 E I
r33 =
40
l3
−
It is worth noting that the displacements adopted are those referred to the
orthogonal direction with respect to the direction of the member considered.
• External load
1
m 10 = − pl2
3
1 2
m 20 = pl
3
1
r30 = pl
2
Calculation of unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
⎡
⎤ ⎡ ⎤ ⎡
⎤
6E I
21 E I
EI
1
− 4√l2
pl2
ϕ1
l
l
√
3
⎢ EI
10+8 5 E I
15−48 5 E I ⎥ ⎣
· ϕ2 ⎦ = ⎣ − 13 pl2 ⎦
⎣ l
5 √ l
20 √ l2 ⎦
E I 15−48 5 E I 495+192 5 E I
δ3
− 21 pl
− 21
4 l2
20
l2
40
l3
⎧
pl3
⎪
⎨ ϕ1 = 0.0448 E I
3
⇒ ϕ2 = −0.0926 pl
E I4
⎪
⎩
δ3 = −0.0300 pl
EI
2.1 Problem A(I): Inclined Rod
13
The stiffness matrix of the solving system is symmetrical. This is due to the fact
that we are using a pure solving method (the displacement method). Even in the case
of a solution using the force method (also pure), a compatibility matrix characterized
by symmetry should be obtained. On the contrary, in the case of the solution with the
mixed method, the stiffness/compatibility matrix should be symmetrical in terms of
absolute values, but antisymmetric if the signs are considered.
Summary of Bending Moments
Below is a summary of the bending moments acting on the structure, written with
a positive or negative sign based on the convention defined in the figure. The
superimposition of the effects is exploited for the writing of the internal forces.
Summary of Shear Forces
Below is a summary of the shear forces acting on the structure, written with a positive
or negative sign based on the convention defined in the figure.
14
2 Solution of Simple Reference Cases
Calculation of Shear Forces and Bending Moments
• AB
2E I
6E I
ϕ1 − 2 δ3 = 0.269 pl2 (internal stretched fibers)
l
l
4E I
6E I
AB
ϕ1 + 2 δ3 = −0.359 pl2 (internal stretched fibers)
MB = −
l
l
6E I
12E I
AB
V A = − 2 ϕ1 −
δ3 = −0.628 pl
l
l3
6E I
12E I
VBAB = 2 ϕ1 −
δ3 = −0.628 pl
l
l3
M AAB =
• BC
2.1 Problem A(I): Inclined Rod
15
4E I
2E I
6E I
4 pl2
ϕ1 +
ϕ2 +
= −0.359 pl2 (external stretched fibers)
δV δ3 −
2
2l
2l
12
2l
2E I
4E I
6E I
4 pl2
ϕ1 −
ϕ2 −
= −0.170 pl2 (external stretched fibers)
δV δ3 −
MCBC = −
2
2l
2l
12
4l
6E I
6E I
12E I
2 pl
ϕ2 −
δV δ3 +
V BBC = − 2 ϕ1 −
= 1.094 pl
2
2l
2l2
2l3
6E I
6E I
12E I
2 pl
= 0.906 pl
ϕ2 −
δV δ3 −
VCBC = − 2 ϕ1 −
2
3
2
2l
2l
2l
M BBC =
The abscissa x of maximum bending moment can be identified by equating to zero
the derivative of the bending moment equation in the member BC :
M(x) = M BBC + VBBC · x − p · x ·
x
px 2
= −0.359 pl2 + 1.094 plx −
2
2
d M(x)
= 1.094 pl − px = 0 → x = 1.094l
dx
The abscissa x can also be determined by writing a proportion of similar triangles
on the shear force diagram.
Being V (x) = d M(x)
, the maximum of the function M(x) is obtained for x, such
dx
as V (x) = 0.
Moreover, the following proportion applies:
(|VB | + |VC |) : l = |VB | : x
so it is possible to calculate x:
x=
|VB | · 2l
1.094 pl · 2l
=
= 1.094l
(|VB | + |VC |)
(1.094 + 0.906) pl
16
2 Solution of Simple Reference Cases
For x = 1.094h, the value of bending moment is equal to:
M(1.094l) = −0.359 pl2 + 1.094 pl(1.094l) −
p(1.094l)2
= 0.240 pl2
2
• CD
4E I
6E I
ϕ2 − ∗2 δ · δ3 = −0.170 pl2
l∗
l
2E I
6E I
CD
M D = − ∗ ϕ2 + ∗2 δ · δ3 = 0.005 pl2
l
l
6E
I
12E
I
VCC D = − ∗2 ϕ2 + ∗3 δ · δ3 = 0.157 pl
l
l
6E I
12E I
VDC D = − ∗2 ϕ2 + ∗3 δ · δ3 = 0.157 pl
l
l
MCC D =
(external stretched fibers)
(internal stretched fibers)
The axial force is calculated with translational equilibrium equations at the node.
The shear forces in the rods are to be considered positive if they induce a clockwise
rotation of the section considered. They are therefore positive if a configuration like
the one reported in the figure is obtained:
2.1 Problem A(I): Inclined Rod
17
For the case under consideration, the following shear forces act on the rods:
The forces acting on the nodes have the opposite direction to those acting on the
rods:
N AB = −VBBC = −1.094 pl
N BC =
−VBAB
2VCBC
= −0.628 pl
NC D = √ +
5
(compression)
(compression)
2 · 0.906 pl (−0.628 pl)
N BC
=
= −1.091 pl (compression)
+
√
2
2
5
18
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
2 Solution of Simple Reference Cases
2.2 Problem A(II): Inclined Rod
19
It is worth noting that the zero points of the bending moment correspond to the
points of inflection (change of concavity) of the deformed shape.
2.2 Problem A(II): Inclined Rod
• E I = constant
• EA → ∞
• l∗∗ = 23 l
/( (
l 2
• l∗ =
+ (2l)2 =
2
√
17
l
2
Solution
The structure is statically indeterminate. The kinematic analysis is carried out,
releasing the flexural constraints and tracing the structure back to a rod with two
rollers at the ends. The analysis shows that the structure is a sway frame. The displacement method is used to solve the structure by adopting the rotations ϕ1 , ϕ2 and the
horizontal translation δ3 as kinematic unknowns.
The center of instantaneous rotation is placed at infinity, so horizontal translation
is allowed.
20
2 Solution of Simple Reference Cases
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
2.2 Problem A(II): Inclined Rod
m 11
m 21
21
√
8E I
24 + 8 17 E I
EI
= √
+
=
= 4.607
√
17
3l
l
l
3 17
l
2
2E I
4E I
EI
= √
=√
= 0.970
17
l
17l
l
2
4E I
r31 = −
6E I
24E I
=−
∗∗2
l
9l2
The roller restraint reaction (r31 ) is obtained from the equilibrium to the global
horizontal translation of the whole structure. It is worth noting that the rod CD has
no flexural deformation, so shear force and moment in D are zero.
• ϕ2 = 1
m 12 =
2E I
m 22 =
4E I
√
17
l
2
√
r32 = −
17
l
2
6E I
l2
= 0.970
+
EI
l
EI
4E I
= 5.940
l
l
22
2 Solution of Simple Reference Cases
• δ3 = 1
24E I
9l2
6E I
m 23 = − 2
l
96E I
12E I
EI
r33 =
+
= 15.56 3
3
3
27l
l
l
m 13 = −
• External load
The resultant P of the distributed load p can be decomposed into the two
components Q and Q' . Adopting the following trigonometric relations:
2l
cos α =
√
17
l
2
sin α =
√
17
l
2
4
=√
17
l/
2 = √1
17
2.2 Problem A(II): Inclined Rod
23
we have that:
• Q = P cos α =
• Q ' = P sin α =
√4 P
17
√1 P
17
The actions q and q' are derived by distributing the concentrated loads Q and Q'
over the length l∗ :
√4 P
16
8 p · 2l
17
√
=
p
=
17
17 l
17
l
2
√1 P
4
Q'
2 p · 2l
=
p
= √17 =
17
l∗
17
l
17
l
2
Q
q= ∗ =
l
q' =
From the equilibrium to horizontal translation of the entire structure it results that
r30 = 0, being the load p vertical and columns without bending deformation and,
therefore, no shear.
m 10 = −
m 20 =
∗2
ql
=−
12
16
17
p·
(√
)2
17
l
2
12
=−
4 2
1
pl = − pl2
12
3
1 2
pl
3
r30 = 0
The moments m10 and m20 can finally be derived through the relations:
24
2 Solution of Simple Reference Cases
p · (2l)2
1
4
= − pl2 = − pl2
12
12
3
4 pl2
1 2
=
= pl
12
3
m 10 = −
m 20
This procedure, in which the projection of the length in the direction perpendicular
to the load is used as the span of the beam, is valid only for the determination of
the bending moments; the shear forces acting at the ends of the rod (essential for the
calculation of the internal forces and moments) will have to be sought through the
writing of equilibrium equations.
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
⎡
⎤ ⎡ 1 2 ⎤
EI
EI
24 E I ⎤ ⎡
4.607 l 0.970 l − 9 l2
pl
ϕ1
3
⎣ 0.970 E I 5.940 E I − 6E2 I ⎦ · ⎣ ϕ2 ⎦ = ⎣ − 1 pl2 ⎦
l
l
l
3
6E I
EI
EI
δ
0
− 24
−
15.56
3
2
2
3
l
l
⎧9 l
3
pl
⎪
⎨ ϕ1 = 0.07905 E I
3
⇒ ϕ2 = −0.09065 pl
E I4
⎪
⎩
δ3 = −0.02141 pl
EI
Summary of Bending Moments
2.2 Problem A(II): Inclined Rod
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
25
26
Qualitative Deformed Shape
2 Solution of Simple Reference Cases
2.3 Problem B(I): Extensional Spring
27
2.3 Problem B(I): Extensional Spring
• E I = constant
• EA → ∞
EI
• k = 10l
3
Solution
In order to control the shortening/elongation of the translational spring placed at node
D, it is necessary to introduce at that point a roller with a vertical reaction parallel
to the direction of the spring.
Therefore, the kinematic analysis of the structure is carried out considering the
presence of such a roller.
28
2 Solution of Simple Reference Cases
The structure turns out to be a sway frame. The allowable horizontal displacement
is prevented by the introduction of the roller, with horizontal reaction applied at node
C.
The displacement method is used to solve the structure, adopting as kinematic
unknowns the rotations ϕ 1 and ϕ 2 and the translations δ3 and δ4 (δ4 corresponds to
the vertical translation of the node D constrained to the ground by the translational
spring).
⎧
m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 14 δ4 + m 10 = 0
⎪
⎪
⎨
m 21 ϕ1 + m 22 ϕ2 + m 23 δ3 + m 24 δ4 + m 20 = 0
⎪
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r34 δ4 + r30 = 0
⎪
⎩
r41 ϕ1 + r42 ϕ2 + r43 δ3 + r44 δ4 + r40 = 0
2.3 Problem B(I): Extensional Spring
29
Kinematics
• ϕ1 = 1
2E I
6E I
4E I
+
=
l
l
l
EI
m 21 =
l
6E I
r31 = 2
l
3E I
r41 = − 2
2l
m 11 =
The reaction r31 is obtained by writing the horizontal translational equilibrium of
the whole structure knowing the base shear force at node A and knowing that the
base shear force at node D is zero, since member CD has no flexural deformation.
The reaction r41 is obtained by the rotational equilibrium of the whole structure
with respect to node A. In writing this equilibrium all the restraining reactions must
be considered, including those arising on the additional restraints (moments on the
rigid blocks and forces on the rollers):
4E I
4E I
2E I
2E I
+
+
+
− r31 l + r41 2l = 0
l
l
2l
2l
6E I
9E I
− 2 l + r41 2l = 0
l
l
3E I
r41 = − 2
2l
30
2 Solution of Simple Reference Cases
• ϕ2 = 1
EI
l
3E I
5E I
2E I
+
=
m 22 =
l
l
l
3E I
r32 = 2
l
3E I
r42 = − 2
2l
m 12 =
• δ3 = 1
2.3 Problem B(I): Extensional Spring
31
6E I
l2
3E I
m 23 = 2
l
15E I
r33 =
l3
r43 = 0
m 13 =
• δ4 = 1
3E I
2l2
3E I
m 24 = − 2
2l
r34 = 0
m 14 = −
For problems falling in the elasticity field, it is permissible to apply the principle
of superposition to subdivide the reaction r44 into the two contributions due to the
imposed settlement δ4 (r44 ' ) and to the presence of a spring with stiffness k (r44 '' ).
32
2 Solution of Simple Reference Cases
Exploiting the relationship of the elastic reaction F = k·x, with x the displacement
vector, for δ4 = 1 we have that:
FM = k · x = k · 1 = k
The compressed spring tends to move the roller upward, giving rise to a restraining
reaction with direction concordant to the displacement δ4 . In general terms, the
restraining reaction is assumed to be positive if it has concordant direction to the one
chosen for the related displacement.
3E I
2l3
=k
'
r44
=
''
r44
'
''
+ r44
=
r44 = r44
3E I
3E I
EI
8E I
+k =
+
=
2l3
2l3
10l3
5l3
• External load
pl2
3
pl2
m 20 =
3
r30 = 0
m 10 = −
r40 = − pl
2.3 Problem B(I): Extensional Spring
Calculation of Unknowns
⎧
m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 14 δ4 + m 10 = 0
⎪
⎪
⎨
m 21 ϕ1 + m 22 ϕ2 + m 23 δ3 + m 24 δ4 + m 20 = 0
⎪
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r34 δ4 + r30 = 0
⎪
⎩
r41 ϕ1 + r42 ϕ2 + r43 δ3 + r44 δ4 + r40 = 0
⎤ ⎡ ⎤
⎤
⎡
⎡
ϕ1
m 10
m 11 m 12 m 13 m 14
⎢ m 20 ⎥
⎢ m 21 m 22 m 23 m 24 ⎥ ⎢ ϕ2 ⎥
⎥ ⎢ ⎥
⎥
⎢
⎢
⎣ r31 r32 r33 r34 ⎦ · ⎣ δ3 ⎦ = −⎣ r30 ⎦
r41 r42 r43 r44
δ4
r40
⎡
⎤
⎡ 6E I
⎡
⎤
⎤
pl2
EI
6E I
3E I
−
−
ϕ
2
2
1
l
l
l
2l
⎢ pl32 ⎥
5E I
3E I
I ⎥ ⎢
⎢ EI
− 3E
ϕ ⎥
⎥
l
l2
2l2 ⎥ · ⎢ 2 ⎥ = −⎢
⎢ l
⎢ 3 ⎥
⎣ 6E2 I 3E2 I 15E3 I 0 ⎦ ⎣ δ3 ⎦
⎣
0 ⎦
l
l
l
3E I
3E I
8E I
δ4
− 2l2 − 2l2 0
− pl
5l3
⎧
pl3
⎪
ϕ1 = 1.19749 E I
⎪
⎪
3
⎨
ϕ2 = 0.84431 pl
EI 4
⇒
pl
⎪
⎪
⎪ δ3 = −0.64786plE4 I
⎩
δ4 = 2.53919 E I
Summary of Bending Moments
33
34
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
2 Solution of Simple Reference Cases
2.4 Problem B(II): Extensional Spring
Qualitative Deformed Shape
2.4 Problem B(II): Extensional Spring
35
36
2 Solution of Simple Reference Cases
• E I = constant
• EA → ∞
EI
• k = 10l
3
Solution
The structure is a sway frame.
The displacement method is used to solve the problem by adopting as kinematic
unknowns the rotation ϕ1 of node B, the rotation ϕ2 of node C, the horizontal translations δ3 of node C, and the horizontal translations δ4 of node D. The horizontal
translation δ4 is introduced in order to control the deformations of the extensional
spring.
The solving system, consisting of 4 equilibrium equations, is the following:
⎧
m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 14 δ4 + m 10 = 0
⎪
⎪
⎨
m 21 ϕ1 + m 22 ϕ2 + m 23 δ3 + m 24 δ4 + m 20 = 0
⎪
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r34 δ4 + r30 = 0
⎪
⎩
r41 ϕ1 + r42 ϕ2 + r43 δ3 + r44 δ4 + r40 = 0
2.4 Problem B(II): Extensional Spring
37
Kinematics
• ϕ1 = 1
2E I
6E I
4E I
+
=
l
l
l
EI
m 21 =
l
6E I
r31 = 2
l
r41 = 0
m 11 =
The last equation (r41 = 0) is obtained from local equilibrium at node D, not
having shear force on member CD.
• ϕ2 = 1
38
2 Solution of Simple Reference Cases
EI
l
2E I
3E I
5E I
m 22 =
+
=
l
l
l
3E I
r32 = 2
l
3E I
r42 = 2
l
m 12 =
• δ3 = 1
6E I
l2
3E I
m 23 = 2
l
15E I
r33 =
l3
3E I
r43 = 3
l
m 13 =
2.4 Problem B(II): Extensional Spring
39
• δ4 = 1
m 14 = 0
3E I
m 24 = 2
l
3E I
r34 = 3
l
3E I
3E I
EI
31E I
r44 = 3 + k = 3 +
=
3
l
l
10l
10l3
• External load
pl2
3
pl2
m 20 =
3
r30 = 0
m 10 = −
r40 = 0
40
2 Solution of Simple Reference Cases
Calculation of Unknowns
⎧
m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 14 δ4 + m 10 = 0
⎪
⎪
⎨
m 21 ϕ1 + m 22 ϕ2 + m 23 δ3 + m 24 δ4 + m 20 = 0
⎪
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r34 δ4 + r30 = 0
⎪
⎩
r41 ϕ1 + r42 ϕ2 + r43 δ3 + r44 δ4 + r40 = 0
⎤ ⎡ ⎤
⎤
⎡
⎡
ϕ1
m 10
m 11 m 12 m 13 m 14
⎢ m 20 ⎥
⎢ m 21 m 22 m 23 m 24 ⎥ ⎢ ϕ2 ⎥
⎥ ⎢ ⎥
⎥
⎢
⎢
⎣ r31 r32 r33 r34 ⎦ · ⎣ δ3 ⎦ = −⎣ r30 ⎦
r41 r42 r43 r44
δ4
r40
⎡
⎤
⎡ 6E I E I 6E I
⎤ ⎡ ⎤
pl2
0
−
ϕ
2
1
l
l
l
⎢ pl32 ⎥
⎢ E I 5E I 3E2 I 3E2 I ⎥ ⎢ ϕ2 ⎥
⎥
l
⎢ l l l
⎥ · ⎢ ⎥ = −⎢
⎢ 3 ⎥
⎣ 6E2 I 3E2 I 15E3 I 3E3 I ⎦ ⎣ δ3 ⎦
⎣
0 ⎦
l
l
l
l
3E I 3E I 31E I
δ4
0 l2 l3 10l3
0
⎧
pl3
⎪
ϕ1 = 0.1870 E I
⎪
⎪
3
⎨
ϕ2 = −0.2440 pl
E I4
⇒
pl
⎪
⎪
⎪ δ3 = −0.0908plE4 I
⎩
δ4 = 0.3240 E I
Summary of Bending Moments
2.4 Problem B(II): Extensional Spring
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
41
42
Qualitative Deformed Shape
2.5 Problem C(I): Rotational Spring
• E I = constant
• EA → ∞
• k = E2lI
2 Solution of Simple Reference Cases
2.5 Problem C(I): Rotational Spring
43
Solution
The structure turns out to be a sway frame.
The displacement method is used to solve the structure, adopting as kinematic
unknowns the rotations ϕ1 of node B, ϕ2 of node C, ϕ3 of node C and the horizontal
translation δ4 of joint B. The two unknowns ϕ2 and ϕ3 at the ends of the spring
are introduced to know uniquely, in each of the cases considered, the rotation of
the spring and, consequently, the moment exchanged between the spring and the
structure.
⎧
m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0
⎪
⎪
⎨
m 21 ϕ1 + m 22 ϕ2 + m 23 ϕ3 + m 24 δ4 + m 20 = 0
⎪
m ϕ + m 32 ϕ2 + m 33 ϕ3 + m 34 δ4 + m 30 = 0
⎪
⎩ 31 1
r41 ϕ1 + r42 ϕ2 + r43 ϕ3 + r44 δ4 + r40 = 0
44
2 Solution of Simple Reference Cases
Kinematics
• ϕ1 = 1
2E I
6E I
4E I
+
=
l
l
l
EI
m 21 =
l
m 31 = 0
6E I
r41 = − 2
l
m 11 =
• ϕ2 = 1
2.5 Problem C(I): Rotational Spring
45
EI
l
5E I
2E I
+k =
m 22 =
l
2l
EI
m 32 = −k = −
2l
r42 = 0
m 12 =
The term k represents the moment exerted by the spring as a result of unit rotation;
for rotational type springs we have that M = k · ϕ, where ϕ represents the change
in angular opening. Clockwise (positive) unit rotation of rigid block 2 gives rise to
a clockwise moment of amplitude k within the spring, of direction concordant with
rotation ϕ2 . In compliance with the internal equilibrium equations, the spring will
exert a counterclockwise, hence negative, moment of amplitude k on block 3.
• ϕ3 = 1
m 13 = 0
m 23 = −k = −
9E I
4E I
+k =
l
2l
6E I
=− 2
l
m 33 =
r43
EI
2l
46
2 Solution of Simple Reference Cases
• δ4 = 1
m 14 = −
m 24 = 0
6E I
l2
6E I
l2
24E I
=
l3
m 34 = −
r44
• External load
pl2
3
pl2
m 20 =
3
m 30 = 0
r40 = 0
m 10 = −
2.5 Problem C(I): Rotational Spring
47
Calculation of Unknowns
⎧
m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0
⎪
⎪
⎨
m 21 ϕ1 + m 22 ϕ2 + m 23 ϕ3 + m 24 δ4 + m 20 = 0
⎪
m ϕ + m 32 ϕ2 + m 33 ϕ3 + m 34 δ4 + m 30 = 0
⎪
⎩ 31 1
r41 ϕ1 + r42 ϕ2 + r43 ϕ3 + r44 δ4 + r40 = 0
⎤ ⎡ ⎤
⎤
⎡
⎡
ϕ1
m 10
m 11 m 12 m 13 m 14
⎢ m 20 ⎥
⎢ m 21 m 22 m 23 m 24 ⎥ ⎢ ϕ2 ⎥
⎥ ⎢ ⎥
⎥
⎢
⎢
⎣ m 31 m 32 m 33 m 34 ⎦ · ⎣ ϕ3 ⎦ = −⎣ m 30 ⎦
r41 r42 r43 r44
δ4
r40
⎡
2
⎡ 6E I E I
⎡
⎤
⎤
6E I
0
−
− pl3
ϕ
2
1
l
l
l
⎢ pl2
⎢ E I 5E I − E I
⎢ ⎥
0 ⎥
2l
2l
⎢ l
⎥ · ⎢ ϕ2 ⎥ = −⎢
⎢ 3
E
I
6E
I
9E
I
⎣ 0 −
− l2 ⎦ ⎣ ϕ3 ⎦
⎣ 0
2l
2l
6E I
6E I 24E I
δ4
− l2
0 − l2
0
l3
⎧
pl3
⎪
ϕ1 = 0.1242 E I
⎪
⎪
3
⎨
ϕ2 = −0.1765 pl
E3 I
⇒
pl
⎪
⎪
E I4
⎪ ϕ3 = 0.0327 pl
⎩
δ4 = 0.0392 E I
Summary of Bending Moments
⎤
⎥
⎥
⎥
⎦
48
2 Solution of Simple Reference Cases
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
Remarks:
i. The bending moment at point C has a value included between zero (Case a) and
the moment MC ’ that would occur in the case of a continuous frame (Case B).
The bending continuity restraint can in fact be seen as a special case of rotational
spring with k → ∞, while the hinge as a special case of rotational spring with
k → 0.
ii. It is possible to verify the correctness of the calculated moment at point C by
computing the moment in the spring (Mspring ):
Mspring = k(ϕ3 − ϕ2 ) =
EI
pl3
· 0.2092
= 0.105 pl2
2l
EI
2.5 Problem C(I): Rotational Spring
49
Qualitative Deformed Shape
Mixed Method
The mixed method allows the number of unknowns in the problem to be reduced to
3. The structure can be solved by adopting as kinematic unknowns the rotation ϕ1 of
node B, the horizontal translation δ2 of node B and as static unknown the moment
X3 of node C. In this case, the unknown X3 directly represents the moment exerted
on the spring and, consequently, exchanged with the structure.
50
2 Solution of Simple Reference Cases
In this case, the solving system is represented by the following equations (an
equilibrium equation at rotation referring to node B, an equilibrium equation at horizontal translation, again referring to node B, and a compatibility equation, referring
to node C):
⎧
⎨ m 11 ϕ1 + m 12 δ2 + m 13 X 3 + m 10 = 0
r ϕ + r22 δ2 + r23 X 3 + r20 = 0
⎩ 21 1
ϕ31 ϕ1 + ϕ32 δ2 + ϕ33 X 3 + ϕ30 = 0
Kinematics
• ϕ1 = 1
3E I
11E I
4E I
+
=
l
2l
2l
6E I
r21 = − 2
l
)
(
1
1
= (positive rotation if clockwise)
ϕ31 = ϕright − ϕleft = 0 − −
2
2
m 11 =
2.5 Problem C(I): Rotational Spring
• δ2 = 1
6E I
l2
12E I
3E I
15E I
r22 =
+ 3 =
3
l
l
l3
3
3
ϕ32 = ϕright − ϕleft =
−0=
2l
2l
m 12 = −
• X3 = 1
51
52
2 Solution of Simple Reference Cases
1
2
3
=−
2l
m 13 = −
r23
ϕ33 = ϕright − ϕleft
(
)
2l
3l
2l
11l
1
l
− −
+
=
+ =
=
4E I
4E I
k
4E I
EI
4E I
• External load
m 10 = −
1
4 pl2
= − pl2
8
2
r20 = 0
pl3
6E I
ϕ30 =
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 δ2 + m 13 X 3 + m 10 = 0
r ϕ + r22 δ2 + r23 X 3 + r20 = 0
⎩ 21 1
ϕ31 ϕ1 + ϕ32 δ2 + ϕ33 X 3 + ϕ30 = 0
⎤ ⎡ ⎤
⎡
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ r21 r22 r23 ⎦ · ⎣ δ2 ⎦ = −⎣ r20 ⎦
⎡
ϕ31 ϕ32 ϕ33
11E I
2l
⎣ − 6E2 I
l
1
2
I
− 6E
l2
15E I
l3
3
2l
− 21
3
− 2l
11l
4E I
X3
ϕ30
⎤⎡ ⎤ ⎡ 1 2 ⎤
pl
ϕ1
2
⎥
⎦⎣ δ2 ⎦ = ⎢
⎣ 0 ⎦
3
pl
X3
− 6E
I
2.6 Problem C(II): Rotational Spring
53
Note that the stiffness/flexibility matrix of the solving system is antisymmetric:
terms arising from the displacement method have opposite signs to those arising
from the force method.
⎧
pl3
⎪
⎨ ϕ1 = 0.1242 E I
4
⇒
δ2 = 0.0392 pl
EI
⎪
⎩ X = −0.1046 pl2
3
2.6 Problem C(II): Rotational Spring
• E I = constant
• EA → ∞
• k = E2lI
Solution
The structure is statically indeterminate and results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotations ϕ1 of node B, ϕ2 of node C, ϕ3 of node D and the horizontal
translation δ4 of node B. The rotation unknown ϕ3 allows to uniquely control the
rotation of the spring and consequently to uniquely identify the moment that the
spring exchanges with the structure.
54
2 Solution of Simple Reference Cases
⎧
m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0
⎪
⎪
⎨
m 21 ϕ1 + m 22 ϕ2 + m 23 ϕ3 + m 24 δ4 + m 20 = 0
⎪
m ϕ + m 32 ϕ2 + m 33 ϕ3 + m 34 δ4 + m 30 = 0
⎪
⎩ 31 1
r41 ϕ1 + r42 ϕ2 + r43 ϕ3 + r44 δ4 + r40 = 0
Kinematics
• ϕ1 = 1
2E I
6E I
4E I
+
=
l
l
l
EI
m 21 =
l
m 31 = 0
6E I
r41 = − 2 (global equilibrium to horizontal translation)
l
m 11 =
2.6 Problem C(II): Rotational Spring
55
• ϕ2 = 1
EI
l
4E I
6E I
2E I
+
=
m 22 =
l
l
l
2E I
m 32 =
l
6E I
r42 = − 2
l
m 12 =
• ϕ3 = 1
m 13 = 0
2E I
m 23 =
l
9E I
4E I
+k =
m 33 =
l
2l
6E I
r43 = − 2
l
In the calculation of m33 , the term k represents the moment exerted by the spring
as a result of unit rotation; for rotational type springs we have that M = k · ϑ, where
56
2 Solution of Simple Reference Cases
ϑ represents the change in angular opening. The clockwise (positive) unit rotation
of block 3 gives rise to a clockwise moment of amplitude k within the spring, of
direction concordant with rotation ϕ3 .
• δ4 = 1
6E I
l2
6E I
m 24 = − 2
l
6E I
m 34 = − 2
l
24E I
r44 =
l3
m 14 = −
• External load
2.6 Problem C(II): Rotational Spring
57
pl2
3
pl2
m 20 =
3
m 30 = 0
r40 = 0
m 10 = −
Calculation of Unknowns
⎧
m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0
⎪
⎪
⎨
m 21 ϕ1 + m 22 ϕ2 + m 23 ϕ3 + m 24 δ4 + m 20 = 0
⎪
m ϕ + m 32 ϕ2 + m 33 ϕ3 + m 34 δ4 + m 30 = 0
⎪
⎩ 31 1
r41 ϕ1 + r42 ϕ2 + r43 ϕ3 + r44 δ4 + r40 = 0
⎡
⎤ ⎡ ⎤
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13 m 14
⎢ m 20 ⎥
⎢ m 21 m 22 m 23 m 24 ⎥ ⎢ ϕ2 ⎥
⎢
⎥ ⎢ ⎥
⎥
⎢
⎣ m 31 m 32 m 33 m 34 ⎦ · ⎣ ϕ3 ⎦ = −⎣ m 30 ⎦
r41 r42 r43 r44
δ4
r40
⎡ 2
⎡ 6E I
⎤
⎤
⎡
pl
6E I
EI
0
−
ϕ
2
1
l
l
l
⎢ 3pl2
6E I ⎥ ⎢
6E I
2E I
⎢ EI
⎥
− l2 ⎥ ⎢ ϕ2 ⎥ ⎢ − 3
l
l
⎢ l
=⎢
2E I
I ⎦·⎣
9E I
⎣ 0
− 6E
ϕ3 ⎦ ⎣ 0
l
2l
l2
I
I
I 24E I
δ4
− 6E
− 6E
− 6E
0
l2
l2
l2
l3
⎧
3
pl
⎪
ϕ = 0.084175 E I
⎪
⎪ 1
3
⎨
ϕ2 = −0.07071 pl
E 3I
⇒
⎪
ϕ3 = 0.053872 pl
⎪
E I4
⎪
⎩
δ4 = 0.016835 pl
EI
Summary of Bending Moments
⎤
⎥
⎥
⎥
⎦
58
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
2 Solution of Simple Reference Cases
2.7 Problem D(I): Thermal Variation
Qualitative deformed shape
2.7 Problem D(I): Thermal Variation
• E I = constant
• EA → ∞
pl2
• αΔT
= 5E
t
I
l
• t = 10
Solution
The structure is statically indeterminate and results to be a sway frame.
59
60
2 Solution of Simple Reference Cases
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3
of node B.
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
4E I
6E I
4E I
+
=
l
2l
l
EI
m 21 =
l
6E I
r31 = − 2
l
m 11 =
2.7 Problem D(I): Thermal Variation
61
• ϕ2 = 1
EI
l
4E I
6E I
4E I
+
=
m 22 =
2l
l
l
6E I
r32 = − 2
l
m 12 =
• δ3 = 1
6E I
l2
6E I
m 23 = − 2
l
24E I
r33 =
l3
m 13 = −
• Thermal variation
The trapezoidal thermal distribution can be decomposed into two contributions:
one constant and one linearly variable along the thickness.
62
2 Solution of Simple Reference Cases
Therefore, the principle of superposition can be used to study the contribution of
thermal variation by analyzing the structure subjected to each of the two contributions
separately.
m 10 = 0
6E I 3 pl4
9
·
= − pl2
l2 50E I
25
18
12E I 3 pl4
=
pl
=+ 3 ·
l
50E I
25
m 20 = −
r30
m 10 = E I ·
4 pl2
1
4 pl2
=
= pl2
20E I
20
5
1
m 20 = − pl2
5
r30 = 0
2.7 Problem D(I): Thermal Variation
63
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
⎧
⎡ 6E I
⎡
⎡ 1 2⎤
⎤
⎤
pl3
EI
I
⎪
− 6E
ϕ1
− 5 pl
⎨ ϕ1 = −0.076 E I
l
l
l2
3
I ⎦ ⎣
6E I
⎣ EI
· ϕ2 ⎦ = ⎣ 14
− 6E
pl2 ⎦ ⇒
ϕ2 = 0.076 pl
l
l
l2
25
EI 4
⎪
I
I 24E I
⎩
δ3
− 18
pl
− 6E
− 6E
δ3 = −0.030 pl
25
l2
l2
l3
EI
Summary of Bending Moments
Summary of Shear Forces
64
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
2 Solution of Simple Reference Cases
2.8 Problem D(II): Thermal Variation
65
Remark The displacement δ3 of node B is negative and, therefore, directed to the
left. Since the structure is symmetrically loaded, the displacement of node C is equal
to δ3 and directed to the right. The total elongation of rod BC, due to the effect of
constant thermal variation, results to be equal to δT . The latter value results to be
equal to two times δ3 .
2.8 Problem D(II): Thermal Variation
• E I = constant
• EA → ∞
pl2
• αΔT
= 5E
t
I
l
• t = 10
Solution
The structure is statically indeterminate and results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3
of node B.
66
2 Solution of Simple Reference Cases
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
4E I
6E I
4E I
+
=
l
2l
l
EI
m 21 =
l
6E I
r31 = − 2
l
m 11 =
2.8 Problem D(II): Thermal Variation
67
• ϕ2 = 1
EI
l
4E I
6E I
4E I
+
=
m 22 =
2l
l
l
6E I
r32 = − 2
l
m 12 =
• δ3 = 1
6E I
l2
6E I
m 23 = − 2
l
24E I
r33 =
l3
m 13 = −
68
2 Solution of Simple Reference Cases
• Thermal variation
3
3 pl4 6E I
9
6E I
m 10 = − αΔT l ·
=−
=−
·
pl2
2
2
4l
2 50E I 4l2
200
9
pl2
m 20 = −
200
r30 = 0
m 10 = −
m 20 = 0
r30 = 0
pl2
αΔT · E I
=−
t
5
2.8 Problem D(II): Thermal Variation
69
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
⎧
⎡ 6E I
⎡
⎡ 49 2 ⎤
⎤
⎤
pl3
EI
I
⎪
− 6E
pl
ϕ1
⎨ ϕ1 = 0.0562 E I
l
l
l2
200
3
I ⎦ ⎣
6E I
9
⎣ EI
· ϕ2 ⎦ = ⎣ 200
− 6E
pl2 ⎦ ⇒ ϕ2 = 0.0162 pl
l
l
l2
E I4
⎪
I
I 24E I
⎩
δ3
0
− 6E
− 6E
δ3 = 0.0181 pl
l2
l2
l3
EI
Summary of Bending Moments
Summary of Shear Forces
70
2 Solution of Simple Reference Cases
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
Note that, in rod AB, there is a bending (mechanical) moment stretching the
right fibers; the resulting mechanical curvature is compensated for by the thermal
curvature of that rod; globally, it results in a deformation of that rod as in the figure,
with the thermal curvature prevailing over the mechanical curvature.
2.9 Problem E(I): Symmetry and Anti-symmetry
71
2.9 Problem E(I): Symmetry and Anti-symmetry
• E I = constant
• EA → ∞
Solution
The structure is statically indeterminate and results to be a sway frame.
Since the frame is symmetrically and symmetrically loaded, it is possible to
analyze half structure by introducing, at the axis of symmetry, an appropriate
constraint that takes into account the actual conditions of the overall structure. Such
a constraint is represented by a slider with a sliding plane parallel to the axis of
symmetry. The constraint should be applied at node C located on the axis of symmetry.
Since this node is located at the top of an axially inextensible rod, it is possible to
further simplify the structure by transforming the slider and the vertical connecting
rod in a fixed support (see image below). The frame is solved by the displacement
method by adopting the rotation ϕ1 as the only unknown.
72
2 Solution of Simple Reference Cases
m 11 ϕ1 + m 10 = 0
Kinematics
• ϕ1 = 1
2.9 Problem E(I): Symmetry and Anti-symmetry
m 11 =
73
4E I
8E I
4E I
+
=
l
l
l
• External load
m 10 = −
pl2
12
Calculation of Unknowns
m 11 ϕ1 + m 10 = 0
8E I
pl2
· ϕ1 −
=0
l
12
l
pl3
pl2
ϕ1 =
·
= 0.0104
12 8E I
EI
Summary of Bending Moments
74
2 Solution of Simple Reference Cases
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
The solution of the system allows to plot the diagrams of the internal forces and
moments and to plot the deformation of half structure. In order to represent the drawings of the entire structure, the following observations—valid for any symmetrically
and symmetrically loaded structure—can be drawn:
(a) the deformation of the structure results to be symmetrical;
(b) the moment and axial force diagrams are symmetrical;
(c) the shear force diagram is anti-symmetrical, meaning that a clockwise shear at
a generic point of the structure results to be counterclockwise at its symmetrical
point and vice-versa.
2.10 Problem E(II): Symmetry and Anti-symmetry
Qualitative Deformed Shape
2.10 Problem E(II): Symmetry and Anti-symmetry
75
76
2 Solution of Simple Reference Cases
• E I = constant
• EA → ∞
Solution
The frame is symmetrical and anti-symmetrically loaded. Due to the symmetry of the
structure and the anti-symmetry of the loads, it is possible to simplify the problem
by analyzing half of the structure and introducing, at the node located on the axis of
symmetry, a roller with an axis parallel to the axis of symmetry in order to keep the
loading condition of the whole structure in consideration. In this case, the introduction
of the roller is more formal than substantial, given the presence of the rod CF with
infinite axial stiffness.
It should be noted that, in the study of the simplified problem, the bending stiffness
of the rod CF must be halved, as the CF rod lies exactly on the axis of symmetry:
(E I )C F =
EI
2
The structure results to be a sway frame, so we add, at node C, a roller with
horizontal reaction and associated displacement unknown δ3 .
The displacement method is used to solve the structure, adopting as kinematic
unknowns the rotations ϕ1 of node B, ϕ2 of node C, and the horizontal translation δ3
of node C.
2.10 Problem E(II): Symmetry and Anti-symmetry
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
4E I
8E I
4E I
+
=
l
l
l
2E I
m 21 =
l
6E I
r31 = 2
l
m 11 =
77
78
2 Solution of Simple Reference Cases
• ϕ2 = 1
2E I
l
4E I
6E I
4E I
+
=
m 22 =
l
2l
l
3E I
r32 = 2
l
m 12 =
• δ3 = 1
2.10 Problem E(II): Symmetry and Anti-symmetry
6E I
l2
3E I
m 23 = 2
l
18E I
r33 =
l3
m 13 =
• External load
pl2
12
=0
m 10 =
m 20
r30 = pl −
pl
pl
=
(global equilibrium of horizontal translation)
2
2
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
⎧
⎡ 8E I 2E I 6E I ⎤ ⎡ ⎤ ⎡ pl2 ⎤
pl3
⎪
ϕ
−
⎨ ϕ1 = 0.011719 E I
2
1
l
l
l
12 ⎥
3
⎢
⎣ 2E I 6E I 3E2 I ⎦ · ⎣ ϕ2 ⎦ = ⎣ 0 ⎦ ⇒
ϕ2 = 0.013021 pl
l
l
l
EI 4
⎪
6E I 3E I 18E I
⎩
δ3
− pl
δ3 = −0.033854 pl
l2
l2
l3
2
EI
79
80
2 Solution of Simple Reference Cases
Summary of Bending Moments
Summary of Shear Forces
Solving the system of equilibrium equations makes it possible to evaluate internal
forces and moments and displacements in the analyzed half structure.
In order to represent the diagrams of internal forces and moments and deformation for the entire structure, the following considerations, valid for all symmetrical
structures loaded with anti-symmetrical loads, are taken into account:
(a) the deformation results to be anti-symmetrical;
(b) the shear force diagram results to be symmetrical (same sign on both sides of
the axis of symmetry);
(c) the bending moment and axial force diagrams result to be anti-symmetrical
(opposite signs at two symmetrical points).
Therefore, on the central column CF placed on the axis of symmetry we have:
(i) the moment of the right side that sums to that of the left side;
(ii) the shear of the right part that sums to that of the left part;
(iii) the axial force of the right side being compensated by that of the left side
(globally zero axial force).
2.10 Problem E(II): Symmetry and Anti-symmetry
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
81
82
2 Solution of Simple Reference Cases
2.11 Problem E(III): Symmetry and Anti-symmetry
• E I = constant
• EA → ∞
Solution
In the situation of symmetrical structure and generic loads, it is possible to decompose
the loads into a symmetrical part and an anti-symmetrical part. In this way, it is
possible to exploit the superposition of effects by analyzing the structure under
consideration as the sum of a symmetrical structure symmetrically loaded and a
symmetrical structure with anti-symmetrical loads.
2.11 Problem E(III): Symmetry and Anti-symmetry
83
Solution of Symmetrically Loaded Frame
The frame is symmetrically and symmetrically loaded; by introducing an appropriate restraint to consider this condition, only half frame can be analyzed. The
displacement method is used to solve the structure.
m 11 ϕ1 + m 10 = 0
84
2 Solution of Simple Reference Cases
Kinematics
• ϕ1 = 1
m 11 =
EI
5E I
4E I
+
=
l
l
l
• External load
m 10 =
pl2
15
p l2
5
·
−
= − pl2 = − pl2
4 12
3
48
16
Calculation of Unknowns
m 11 ϕ1 + m 10 = 0
5E I
15 2
· ϕ1 −
pl = 0
l
48
15 2
l
1 pl3
ϕ1 =
pl ·
=
·
48
5E I
16 E I
2.11 Problem E(III): Symmetry and Anti-symmetry
Summary of Bending Moments
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
85
86
Qualitative Deformed Shape
Solution of Anti-symmetrically Loaded Frame
2 Solution of Simple Reference Cases
2.11 Problem E(III): Symmetry and Anti-symmetry
87
The frame is symmetrically and anti-symmetrically loaded; by introducing an
appropriate restraint to consider this condition, only half frame can be analyzed. It
is solved by the displacement method.
⎧
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
Kinematics
• ϕ1 = 1
4E I
7E I
3E I
+
=
l
l
l
6E I
= 2
l
m 11 =
r21
88
2 Solution of Simple Reference Cases
• δ2 = 1
6E I
l2
12E I
=
l3
m 12 =
r22
• External load
3 pl2
4 12
3 pl
=
4 2
m 10 =
r20
Calculation of Unknowns
⎧
m 11 ϕ1 + m 12 δ2 + m 10 = 0
m 21 ϕ1 + m 22 δ2 + m 20 = 0
pl3
EI
pl3
δ2 = −0.04688
EI
ϕ1 = 0.03125
2.11 Problem E(III): Symmetry and Anti-symmetry
Summary of Bending Moments
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
89
90
2 Solution of Simple Reference Cases
Qualitative Deformed Shape
Moment, Shear and Axial Force Diagrams
The diagrams of the internal forces and moments and the deformation of the initial
structure—characterized by a symmetrical structure in geometry but generic loads—
are obtained by simply summing the results obtained from the analysis of the symmetrically loaded structure with those obtained from the study of the anti-symmetrical
loaded structure.
2.11 Problem E(III): Symmetry and Anti-symmetry
Qualitative Deformed Shape
91
92
2 Solution of Simple Reference Cases
2.12 Problem F(I): Imposed Displacement/Rotation
• E I = constant
• EA → ∞
1 pl4
• δ = 100
EI
Solution
The structure results to be a sway frame.
The displacement method is used to solve the structure by adopting the rotations
ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3 of node B as kinematic
unknowns.
2.12 Problem F(I): Imposed Displacement/Rotation
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
4E I
6E I
4E I
+
=
2l
l
l
EI
m 21 =
l
6E I
r31 = − 2
l
m 11 =
• ϕ2 = 1
93
94
2 Solution of Simple Reference Cases
EI
l
4E I
6E I
4E I
+
=
m 22 =
2l
l
l
6E I
r32 = − 2
l
m 12 =
• δ3 = 1
6E I
l2
6E I
m 23 = − 2
l
24E I
r33 =
l3
m 13 = −
• External load
2.12 Problem F(I): Imposed Displacement/Rotation
95
4 pl2
12
2
4
pl
m '20 =
12
'
r30
=0
m '10 = −
• Imposed displacement
6E I
4 pl2
209 2
−
pl
δ̄ = −
12
4l2
600
6E I
4 pl2
191 2
''
−
pl
m 20 =
δ̄ =
2
12
4l
600
''
r30 = 0
''
m 10 = −
Due to the superposition of effects, it is possible to calculate m 10 , m 20 and r30 .
6E I
4 pl2
209 2
−
pl
δ̄ = −
2
12
4l
600
6E I
4 pl2
191 2
'
''
−
pl
m 20 = m 20 + m 20 =
δ̄ =
12
4l2
600
'
''
r30 = r30 + r30 = 0
'
''
m 10 = m 10 + m 10 = −
96
2 Solution of Simple Reference Cases
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
⎧
⎡ 6E I
⎡
⎡ 209 ⎤
⎤
⎤
pl3
EI
I
⎪
− 6E
ϕ1
⎨ ϕ1 = 0.0704 E I
l
l
l2
600
3
I ⎦ ⎣
6E I
191 ⎦
⎣ EI
· ϕ2 ⎦ = ⎣ − 600
pl2 ⇒ ϕ2 = −0.0629 pl
− 6E
l
l
l2
E4 I
⎪
I
I 24E I
⎩
δ3
0
− 6E
− 6E
δ3 = 0.0019 pl
l2
l2
l3
EI
Summary of Bending Moments
Summary of Shear Forces
2.12 Problem F(I): Imposed Displacement/Rotation
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
97
98
2 Solution of Simple Reference Cases
2.13 Problem F(II): Imposed Displacement/Rotation
• E I = constant
• EA → ∞
1 pl3
• θ = 12
EI
Solution
The structure results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3
of node B. It is worth noting that the load p is introduced as a dimensionless variable
and for formal consistency with the other examples.
2.13 Problem F(II): Imposed Displacement/Rotation
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
4E I
6E I
4E I
+
=
l
2l
l
EI
m 21 =
l
6E I
r31 = − 2
l
m 11 =
• ϕ2 = 1
99
100
2 Solution of Simple Reference Cases
EI
l
4E I
6E I
4E I
+
=
m 22 =
2l
l
l
6E I
r32 = − 2
l
m 12 =
• δ3 = 1
6E I
l2
6E I
m 23 = − 2
l
24E I
r33 =
l3
m 13 = −
• Imposed rotation
2.13 Problem F(II): Imposed Displacement/Rotation
m 10 = −
101
pl2
2E I
θ =−
l
6
m 20 = 0
6E I
pl
r30 = 2 θ =
l
2
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎤ ⎡ ⎤
⎡
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
⎡
r31 r32 r33
6E I
l
⎣ EI
l
I
− 6E
l2
EI
l
6E I
l
I
− 6E
l2
I
− 6E
l2
I
− 6E
l2
24E I
l3
δ3
⎤ ⎡
Summary of Bending Moments
Summary of Shear Forces
r30
⎡
⎧
⎤
pl3
pl2
⎪
ϕ1
⎨ ϕ1 = 0.0063 E I
6
3
⎥
⎦ · ⎣ ϕ2 ⎦ = ⎢
⎣ 0 ⎦ ⇒ ϕ2 = −0.0271 pl
E4I
⎪
⎩
δ3
− pl
δ3 = −0.026 pl
2
⎤
EI
102
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
2 Solution of Simple Reference Cases
2.14 Problem G(I): Infinitely Rigid Bending Rod
103
2.14 Problem G(I): Infinitely Rigid Bending Rod
E I BC → ∞
E I AB = E IC D = E I
EA → ∞
Solution
The structure results to be a sway frame.
The presence of the infinitely rigid top member, which can only do rigid rototranslations, together with the beam-column continuity and the kinematics of the
system, ensures that the two columns at the top do not experience any rotation.
Consequently, no rotational unknowns are introduced at the intersection of beam
and columns. Furthermore, it would not have been possible to introduce a rotation
unknown at each of the two nodes B and C, since these rotations could never have
been activated in the cases ϕ(B) = 1 and ϕ(C) = 1. The displacement method is used
to solve the structure by adopting the horizontal translation δ1 of node B as the only
kinematic unknown.
104
2 Solution of Simple Reference Cases
r11 δ1 + r10 = 0
Kinematics
• δ1 = 1
r11 =
12E I
12E I
24E I
+
=
l3
l3
l3
• External load
r10 = −
pl
2
2.14 Problem G(I): Infinitely Rigid Bending Rod
Calculation of Unknowns
r11 δ1 + r10 = 0
24E I
pl
pl4
δ1 −
= 0 ⇒ δ1 =
3
l
2
48E I
Summary of Bending Moments
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
105
106
2 Solution of Simple Reference Cases
Qualitative Deformed Shape
2.15 Problem G(II): Infinitely Rigid Bending Rod
2.15 Problem G(II): Infinitely Rigid Bending Rod
107
• E I AB → ∞
• E I BC = E IC D = E I
• EA → ∞
Solution
The rod AB has infinite flexural stiffness and, as such, can only and exclusively
undergo rigid roto-translations. The presence of a clamping at node A imposes that the
translations and rotations of that point are zero and consequently that the translations
and rotations of the whole rod AB are zero.
Therefore, node B cannot neither rotate nor translate and, in accordance with the
kinematic analysis, if node B does not move, the whole rod BC also cannot make
the horizontal translation granted by the kinematic analysis.
Therefore, the structure results to be fixed-joint frame.
Because of what was noted above, node B cannot undergo any rotation, so no
rotation unknown is introduced at node B. The only unknown used is therefore the
rotation ϕ1 of node C.
m 11 ϕ1 + m 10 = 0
108
2 Solution of Simple Reference Cases
Kinematics
• ϕ1 = 1
m 11 =
4E I
6E I
4E I
+
=
2l
l
l
• External load
m 10 =
4 pl2
12
2.15 Problem G(II): Infinitely Rigid Bending Rod
109
Calculation of Unknowns
m 11 ϕ1 + m 10 = 0
6E I
pl2
pl3
ϕ1 +
= 0 ⇒ ϕ1 = −
l
3
18E I
Summary of Bending Moments
Summary of Shear Forces
The values of bending moment and shear force on rod AB not explicitly shown
in the previous figures are calculated by writing the equilibrium equations.
110
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
2 Solution of Simple Reference Cases
2.16 Problem H: Concentrated Load
111
2.16 Problem H: Concentrated Load
• EA → ∞
• P = pl
Solution
The structure results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3
of node B.
112
2 Solution of Simple Reference Cases
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
4(2E I )
10E I
4E I
+
=
2l
l
l
EI
m 21 =
l
12E I
r31 = − 2
l
m 11 =
• ϕ2 = 1
2.16 Problem H: Concentrated Load
113
EI
l
4E I
6E I
4E I
+
=
m 22 =
2l
l
l
6E I
r32 = − 2
l
m 12 =
• δ3 = 1
12E I
l2
6E I
m 23 = − 2
l
36E I
r33 =
l3
m 13 = −
• External load
m 10 = 0
m 20 = 0
r30 = −P
114
2 Solution of Simple Reference Cases
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎤ ⎡ ⎤
⎡
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
⎡
r31 r32 r33
10E I
l
⎣ EI
l
I
− 12E
l2
EI
l
6E I
l
I
− 6E
l2
δ3
12E I ⎤ ⎡
r30
⎧
3
⎡ ⎤
⎪ ϕ1 = 0.063218 pl
− l2
ϕ1
0
⎨
E I3
I ⎦ ⎣
· ϕ2 ⎦ = ⎣ 0 ⎦ ⇒ ϕ2 = 0.045977 pl
− 6E
l2
E I4
⎪
36E I
⎩
δ3
P
δ3 = 0.056513 pl
l3
Summary of Bending Moments
Summary of Shear Forces
⎤
EI
2.16 Problem H: Concentrated Load
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
115
116
2 Solution of Simple Reference Cases
2.17 Problem I: Statically Determined Portion
• E I = constant
• EA → ∞
Solution
To the left of node A and to the right of node B can be identified two statically
determined parts (appendages), which can be eliminated in the solution of the system
as long as they are replaced by the reactions they exchange with the remaining part
of the structure.
Specifically, the cantilever on the left will be replaced by a shear force and moment
concentrated at node A, while the triangle BDC will be replaced by two forces (one
horizontal and one vertical) at both nodes B and C. For the determination of these
reactions, the structures related to statically determined appendages should be solved;
therefore, the cantilever and the triangle BDC are analyzed in the present case.
2.17 Problem I: Statically Determined Portion
117
The structure results to be a sway frame. The displacement method is used to
solve the structure by adopting as kinematic unknowns the rotations ϕ1 of node A,
ϕ2 of node B and the horizontal translation δ3 of node A.
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
118
2 Solution of Simple Reference Cases
Kinematics
• ϕ1 = 1
4E I
2E I
6E I
+
=
l
l
l
EI
m 21 =
l
6E I
r31 = − 2
l
m 11 =
• ϕ2 = 1
EI
l
4E I
6E I
4E I
+
=
m 22 =
2l
l
l
6E I
r32 = − 2
l
m 12 =
2.17 Problem I: Statically Determined Portion
119
• δ3 = 1
6E I
l2
6E I
m 23 = − 2
l
24E I
r33 =
l3
m 13 = −
• External load
pl2
2
9 pl2
m 20 = −
32
9 pl
9 pl 9 pl 9 pl
−
−
=−
r30 =
4
4
8
8
m 10 =
It is emphasized that the moment pl2 /2 applied to node A, with counterclockwise
sign, is precisely a nodal moment. The convention used in this text always considers
the moments applied as a reaction moment on the rigid block, which in this case
results to be of opposite sign and therefore clockwise. For this reason, the sign of the
moment m 10 is positive.
120
2 Solution of Simple Reference Cases
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
⎧
⎤
⎡
⎡ 6E I
⎡
⎤
⎤
2
pl3
EI
I
⎪
− 6E
ϕ1
− pl2
⎨ ϕ1 = −0.0352 E I
l
l
l2
3
2
⎥
⎢
I ⎦ ⎣
6E I
⎣ EI
· ϕ2 ⎦ = ⎣ 9 pl ⎦ ⇒
− 6E
ϕ2 = 0.121 pl
l
l
l2
E I4
32
⎪
I
I 24E I
⎩
9 pl
δ3
− 6E
− 6E
δ3 = 0.0684 pl
l2
l2
l3
8
EI
Summary of Bending Moments
Summary of Shear Forces
2.17 Problem I: Statically Determined Portion
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
121
122
2 Solution of Simple Reference Cases
2.18 Problem J: Unusual Restraints
• E I = constant
• EA → ∞
Solution
The structure results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3
of node B.
2.18 Problem J: Unusual Restraints
123
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
6E I
l
EI
m 21 =
l
6E I
r31 = − 2
l
m 11 =
• ϕ2 = 1
EI
l
3E I
m 22 =
l
r32 = 0
m 12 =
124
2 Solution of Simple Reference Cases
• δ3 = 1
m 13 = −
6E I
l2
m 23 = 0
12E I
r33 =
l3
• External Load
pl2
4 pl2
=−
12
3
pl2
m 20 =
3
r30 = 0
m 10 = −
2.18 Problem J: Unusual Restraints
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
⎧
⎡ 2 ⎤
⎡ 6E I E I
⎡
⎤
⎤
pl3
pl
I
⎪
− 6E
ϕ1
⎨ ϕ1 = 0.1666 E I
l
l
l2
3 2
⎥
⎢
⎣ E I 3E I 0 ⎦ · ⎣ ϕ2 ⎦ = ⎣ − pl ⎦ ⇒ ϕ2 = −0.1666 pl3
l
l
E4 I
3
⎪
I
I
⎩
δ3
0 12E
− 6E
0
δ3 = 0.0833 pl
l2
l3
EI
Summary of Bending Moments
Summary of Shear Forces
125
126
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
2 Solution of Simple Reference Cases
2.19 Problem K(I): Distributed Loads/Moments
127
2.19 Problem K(I): Distributed Loads/Moments
• EA → ∞
Solution
The structure results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotations ϕ1 of node B, ϕ2 of node C and the horizontal translation δ3
of node B.
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
128
2 Solution of Simple Reference Cases
Kinematics
• ϕ1 = 1
8E I
10E I
4E I
+
=
2l
l
l
EI
2E I
=
m 21 =
2l
l
12E I
r31 = − 2
l
m 11 =
• ϕ2 = 1
EI
l
4E I
6E I
4E I
+
=
m 22 =
2l
l
l
6E I
r32 = − 2
l
m 12 =
2.19 Problem K(I): Distributed Loads/Moments
• δ3 = 1
12E I
l2
6E I
m 23 = − 2
l
24E I
12E I
36E I
r33 =
+
=
l3
l3
l3
m 13 = −
• External load
m 10 = 0
m 20 = 0
r30 = − pl
129
130
2 Solution of Simple Reference Cases
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎤
⎡
ϕ1
m 10
m 11 m 12 m 13
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
⎧
⎡ 10E I
⎡
⎡ ⎤
⎤
⎤
pl3
EI
I
⎪
− 12E
ϕ1
0
⎨ ϕ1 = 0.06322 E I
l
l
l2
3
I ⎦ ⎣
6E I
⎣ EI
· ϕ2 ⎦ = ⎣ 0 ⎦ ⇒ ϕ2 = 0.04598 pl
− 6E
l
l
l2
E I4
⎪
I
I 36E I
⎩
δ3
pl
− 12E
− 6E
δ3 = 0.05651 pl
l2
l2
l3
EI
Summary of Bending Moments
Summary of Shear Forces
2.19 Problem K(I): Distributed Loads/Moments
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
131
132
2 Solution of Simple Reference Cases
2.20 Problem K(II): Distributed Loads/Moments
• E I = constant
• EA → ∞
Solution
The structure results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotation ϕ1 of node C and the horizontal translation δ2 of node B.
⎧
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
2.20 Problem K(II): Distributed Loads/Moments
Kinematics
• ϕ1 = 1
3E I
11E I
4E I
+
=
l
2l
2l
6E I
=− 2
l
m 11 =
r21
• δ2 = 1
6E I
l2
3E I
12E I
15E I
= 3 +
=
3
l
l
l3
m 12 = −
r22
133
134
2 Solution of Simple Reference Cases
• External load
m 10 = 0
The following is a demonstration of using the force method, assuming the moment
X at node C as static unknown, to solve the case at hand.
2.20 Problem K(II): Distributed Loads/Moments
Writing the compatibility equation at node C we obtain:
ϕ11 X + ϕ10 = 0 → X = 0
r20 = 0
Calculation of Unknowns
⎧
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
] [ ]
[
]
[
ϕ1
m 10
m 11 m 12
·
=−
r21 r22
δ2
r20
[ 11E I
]
[
]
[ ]
⎧
I
− 6E
ϕ1
0
ϕ1 = 0
2l
l2
·
=
−
⇒
I 15E I
− 6E
δ
δ2 = 0
0
2
2
3
l
l
Summary of Bending Moments
Summary of Shear Forces
135
136
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
2 Solution of Simple Reference Cases
2.21 Problem L: Rod with Finite Axial Stiffness
137
2.21 Problem L: Rod with Finite Axial Stiffness
•
•
•
•
E I = constant
E A ABC D → ∞
EI
E A B D = 10l
∗2
√
∗
l =l 2
Solution
The structure results to be a fixed-joint frame.
The rod with finite axial stiffness can be transformed into a translational spring
parallel to the axis of the rod. The stiffness of the spring results to be equal to k = El∗A
where l∗ is the length of the rod with finite axial stiffness.
138
2 Solution of Simple Reference Cases
⎧
m 11 ϕ1 + m 12 η2 + m 10 = 0
r21 ϕ1 + r22 η2 + r20 = 0
Kinematics
• ϕ1 = 1
3E I
7E I
4E I
+
=
l
l
l
6E I √
= 2 2
l
m 11 =
r21
• η2 = 1
m 12 =
6E I
δ
l2
2.21 Problem L: Rod with Finite Axial Stiffness
139
√
The force in the spring (Fspring ) along the direction of η2 (l∗ = l 2) is equal to:
Fspring
∑
√
2 EI
EA
EI
EI
= kη2 = k · 1 = ∗ =
=
=
√
∗3
l
10l
40 l3
10( 2l)3
FH = 0
√
√
2 3E I √
2
12E I √
r22
− 3 2−
=0
2 − Fspring
3
2
l
l
2
√
1200 + 2 E I
r22 =
40
l3
• External load
m 10 =
pl2
8
r20 = 0
Calculation of Unknowns
⎧
m 11 ϕ1 + m 12 η2 + m 10 = 0
r21 ϕ1 + r22 η2 + r20 = 0
] [ ]
[
]
[
ϕ1
m 10
m 11 m 12
·
=−
r21 r22
η2
r20
[
]
⎧
√
] [ pl2 ]
[
3
7E I
6 2E I
ϕ1 = −0.02716 pl
ϕ
−
2
1
l√
E
I
8
√l
=
·
⇒
4
6 2E I 1200+ 2 E I
η2
0
η2 = 0.007672 pl
EI
l2
40
l3
140
Summary of Bending Moments
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
2 Solution of Simple Reference Cases
2.21 Problem L: Rod with Finite Axial Stiffness
Qualitative Deformed Shape
141
Chapter 3
Solution of Complex Frames
3.1 Worked Example 1
•
•
•
•
•
E I BC = E I D E → ∞
E I AB = E IC D = E I E F = E I
EA → ∞
k = E2lI
αΔ T
t
=
pl2
100E I
Solution
The structure is symmetrically loaded, so it can be simplified as shown in the figures
below. The structure results to be a sway frame. Both the displacement method and
the mixed method can be used to solve the frame.
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
I. G. Colombo et al., Structural Analysis of Plane Frames, Springer Tracts in Civil
Engineering, https://doi.org/10.1007/978-3-031-35267-6_3
143
144
3 Solution of Complex Frames
Displacement Method
The displacements method is used by adopting as kinematic unknowns the rotation
ϕ1 and the horizontal translation δ2 . No rigid blocks are inserted at nodes B and
C because, since the flexural stiffness of rod BC is infinite, it will only be able to
perform rigid roto-translations dictated by the kinematics of the system. The rod BC
will not be able to assume any flexural deformation.
In the case where the horizontal displacement of node B is prevented (δ2 = 0), rod
BC will not undergo any roto-translation; consequently, due to the flexural continuity
between rods AB and BC, rod AB will also not be able to perform any rotation at
endpoint B. As a consequence, in case δ2 is zero, the static scheme of the rod AB
corresponds to that of a rod clamped at both ends.
3.1 Worked Example 1
145
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
Kinematics
• ϕ1 = 1
m 11 =
4E I
EI
9E I
4E I
+k =
+
=
l
l
2l
2l
r21 = −
8E I
l2
It should be noted that at node B, since there is no rotational unknown, the actions
shown in the figure all represent internal forces and moments and therefore the
equilibrium at the rotation of the node must be respected; consequently, on the rod
BC at node B a moment equal in modulus and opposite in direction to that acting at
the top of the rod AB will act. A purely static representation of this equilibrium is
given below.
146
3 Solution of Complex Frames
M B2+3 = 0 → NCC ' =
2E I
l2
FH1+2+3 = 0 → r21 = −
8E I
l2
• δ2 = 1
The figure shows a superposition of effects between a unit horizontal translation
and a counterclockwise rotation of magnitude α of node B of rod AB.
As a result of the translation of node B, the infinitely rigid rod BC makes a rigid
rotation of amplitude α with respect to the CIR. All points of rod BC are rotating by
α and, due to the flexural continuity at point B between rods AB and BC, the top of
the column (i.e. rod AB) will also undergo a counterclockwise rotation α.
3.1 Worked Example 1
147
m 12 = −
r22 =
8E I
l2
30E I
l3
Similarly to what has been described for node B, also at node C the continuity
between rods BC and CC’ generates, within rod CC’, a flexural deformation related
to the rotation α of end C. In particular, it is possible to write an equilibrium equation
at rotation.
• External load
M B2+3 = 0 → NCC ' =
FH = 0 → r22 =
12E I
l3
18E I
12E I
30E I
+
=
3
3
l
l
l3
148
3 Solution of Complex Frames
m '10 = 0
'
r20
=
7
pl
12
• Constant thermal variation
m ''10 = 0
''
r20
=0
• Linear thermal variation
3.1 Worked Example 1
149
m '''
10 = −
pl2
E I αΔ t
=−
t
100
'''
r20
=−
pl
100
The reaction r'''
20 is calculated by writing static equilibrium equations, on the structure made statically determined by downgrading the fixed end restraints and the
flexural continuity points.
M B2+3 = 0 → NCC ' = −
'''
=−
FH = 0 → r20
pl
100
pl
100
By applying the superposition of effects it is possible to calculate m 10 and r20 .
m 10 = −
r20 =
pl2
100
1
43
688
7
−
pl =
pl
pl =
12 100
1200
75
Calculation of Unknowns
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
m 11 m 12
ϕ1
m 10
·
=−
r21 r22
δ2
r20
9 EI
1
−8 El2I
pl2
ϕ1
2 l
100
·
=
−8 El2I 30 El3I
δ2
− 43
pl
75
⎧
pl3
⎪
⎪
⎨ ϕ1 = −0.06038
EI
⇒
4
⎪
pl
⎪
⎩ δ2 = −0.03521
EI
Summary of Bending Moments
A summary of the bending moments acting on the structure is given in the figure,
written with concordant or discordant sign, with respect to the moments defined in
the convention.
150
3 Solution of Complex Frames
Summary of Shear Forces
A summary of the shear forces acting on the structure is given in the figure,
written with concordant or discordant sign, with respect to the forces defined in
the convention.
3.1 Worked Example 1
Moment, Shear and Axial Force Diagrams
151
152
3 Solution of Complex Frames
Qualitative Deformed Shape
Mixed Method
The mixed method is used by adopting as static unknowns the moments X1 , X2 and
X3 and as kinematic unknown the horizontal translation δ4 .
3.1 Worked Example 1
153
⎧
ϕ11 X 1 + ϕ12 X 2 + ϕ13 X 3 + ϕ14 δ4 + X 10 = 0
⎪
⎪
⎨
ϕ21 X 1 + ϕ22 X 2 + ϕ23 X 3 + ϕ24 δ4 + X 20 = 0
⎪
ϕ X + ϕ32 X 2 + ϕ33 X 3 + ϕ34 δ4 + X 30 = 0
⎪
⎩ 31 1
r41 X 1 + r42 X 2 + r43 X 3 + r44 δ4 + δ40 = 0
Kinematics
The rod BC, due to its infinite flexural stiffness, following the application of the
moment at node C has no bending deformation.
• X1 = 1
ϕ11 =
l
2E I
(concordant with the direction of X1 )
ϕ21 = ϕ31 = 0
154
3 Solution of Complex Frames
r41 = − 1l (discordant with the direction of δ4 )
• X2 = 1
ϕ12 = 0
ϕ22 = 3El I (concordant with the direction of X2 )
ϕ32 = 6El I (concordant with the direction of X3 )
r42 = 2l (concordant with the direction of δ4 )
• X3 = 1
ϕ13 = 0
ϕ23 = 6El I (concordant with the direction of X2 )
ϕ33 = 3El I + k1 = 3El I + E2lI = 73 ElI (concordant with the direction of X3 )
3.1 Worked Example 1
155
r43 = − 1l (discordant with the direction of δ4 )
• δ4 = 1
ϕ14 =
1
l
2
1 1
ϕ24 = − − = −
l l
l
ϕ34 =
1
l
r44 = 0 (all the restraining reactions are zero because the rods rotate and
translate without flexural deformation)
• External load
156
3 Solution of Complex Frames
'
ϕ10
=
pl3
24E I
'
ϕ20
=0
'
ϕ30
=0
'
r40
=
pl
2
• Constant thermal variation
''
ϕ10
=0
''
ϕ20
=0
''
ϕ30
=0
''
r40
=0
3.1 Worked Example 1
157
• Linear thermal variation
'''
ϕ10
=0
'''
ϕ20
=
pl3
200E I
'''
ϕ30
=
pl3
200E I
'''
r40
=0
Due to the superposition of effects, it is possible to calculate ϕ10 , ϕ20 , ϕ30 , r40 .
ϕ10 =
pl3
24E I
ϕ20 =
pl3
200E I
ϕ30 =
pl3
200E I
r40 =
pl
2
158
3 Solution of Complex Frames
Calculation of Unknowns
⎧
ϕ11 X 1 + ϕ12 X 2 + ϕ13 X 3 + ϕ14 δ4 + X 10 = 0
⎪
⎪
⎨
ϕ21 X 1 + ϕ22 X 2 + ϕ23 X 3 + ϕ24 δ4 + X 20 = 0
⎪
ϕ X + ϕ32 X 2 + ϕ33 X 3 + ϕ34 δ4 + X 30 = 0
⎪
⎩ 31 1
r41 X 1 + r42 X 2 + r43 X 3 + r44 δ4 + δ40 = 0
⎡
⎡
⎤ ⎡ ⎤
⎤
X1
ϕ10
ϕ11 ϕ12 ϕ13 ϕ14
⎢ ϕ20 ⎥
⎢ ϕ21 ϕ22 ϕ23 ϕ24 ⎥ ⎢ X 2 ⎥
⎢
⎢
⎥ ⎢ ⎥
⎥
⎣ ϕ31 ϕ32 ϕ33 ϕ34 ⎦ · ⎣ X 3 ⎦ = −⎣ ϕ30 ⎦
r41 r42 r43 r44
δ4
r40
⎡ pl3 ⎤
⎡ l
⎤ ⎡ ⎤
0 0 1l
X1
I
2E I
⎢ 24E
pl3 ⎥
⎢ 0 l l − 2 ⎥ ⎢ X2 ⎥
⎥
3E I 6E I
l ⎥·⎢
⎢
⎥ = −⎢
200E
⎢ pl3 I ⎥
⎣ 0 l 7l 1 ⎦ ⎣ X 3 ⎦
⎣
⎦
6E I 3E I l
200E I
pl
δ4
− 1l 2l − 1l 0
EI
By adopting the mixed method, the stiffness/flexibility matrix results to be
symmetric in absolute value.
⇒
⎧
⎪
X 1 = −0.013 pl2
⎪
⎪
⎪
⎪
2
⎪
⎨ X 2 = −0.241 pl
X 3 = 0.030 pl2
⎪
⎪
⎪
⎪
⎪
pl4
⎪
⎩ δ4 = −0.035
EI
For diagrams of internal forces and moments, refer to the graphs obtained with
the displacement method.
3.2 Worked Example 2
3.2 Worked Example 2
159
•
•
•
•
E I = constant
EA → ∞
αΔ T l∗ = δ
∗
t = l5
/
2 /
• l∗ = l2 + 2l = 45 l =
• cosα =
l
l∗
=
√2 l
5l
=
√
√
5
l
2
2 5
5
Solution
The structure results to be a sway frame. A horizontal roller should be added at node
A; however, the frame is not subject to external loads (but only to thermal variation
and imposed displacement), so the vertical rod AB is merely subjected to axial force.
To solve the problem, it is therefore sufficient to consider rotation ϕ1 as the unique
displacement unknown.
Displacement Method
m 11 ϕ1 + m 10 = 0
160
3 Solution of Complex Frames
Kinematics
• ϕ1 = 1
m 11
/
4 EI
4E I
8√ E I
= ∗ =4
5
=
l
5 l
5
l
• Imposed displacement
m '10 =
√
6E I
24 E I
12 5 E I
6E I
δ
δ
δ
=
=
=
·
·
·
δ
√ 2
cos α
5 l2 cos α
5 l2
(l∗ )2 cos α
5
l
2
3.2 Worked Example 2
161
• Thermal variation
m ''10 = −
m '''
10 =
12 E I
6E I αΔ T l∗
=−
·
δ
2
∗
2
5 l2
(l )
10E I
2E I αΔ T
8E I
= ∗ 2 ·δ = 2 ·δ
t
l
(l )
Due to the superposition of effects, it is possible to calculate m 10 .
162
m 10 =
3 Solution of Complex Frames
m '10
+
m ''10
+
m '''
10
√
√
12 5 E I
12 E I
8E I
12 5 + 28 E I
=
δ−
δ+ 2 δ=
δ
5 l2
5 l2
l
5
l2
Calculation of Unknowns
m 11 ϕ1 + m 10 = 0
√
12 5 + 28 E I
8E I
δ
· [ϕ1 ] = −
√
5
l2
5l
⇒ ϕ1 = −3.065
Summary of Bending Moments
Summary of Shear Forces
δ
l
3.2 Worked Example 2
Moment, Shear and Axial Force Diagrams
163
164
3 Solution of Complex Frames
Qualitative Deformed Shape
3.3 Worked Example 3
•
•
•
•
•
•
E I = constant
EA → ∞
k = 2El I
P = 2 pl
pl2
αΔ T = 10E
t=
I
t = 3l
pl3
30E I
Solution
The structure results to be a fixed-joint frame.
3.3 Worked Example 3
165
Force Method
The force method is used to solve the structure by adopting as static unknowns the
moment X 1 and the moment transmitted by the internal rotational spring X2 .
ϕ11 X 1 + ϕ12 X 2 + ϕ10 = 0
ϕ21 X 1 + ϕ22 X 2 + ϕ20 = 0
Kinematics
• X1 = 1
ϕ11 =
3l
25l
4l
+
=
3E I
4E I
12E I
166
3 Solution of Complex Frames
ϕ21 =
4l
2l
=
6E I
3E I
ϕ12 =
2l
4l
=
6E I
3E I
• X2 = 1
ϕ22 =
5l
1
7l
4l
+
+ =
3E I
3E I
k
2E I
• External load
'
ϕ10
=
p(3l)3
9 pl3
2 pl3
41 pl3
P(4l)2
+
=
+
=
16E I
48E I
16E I
EI
16E I
'
ϕ20
=
2 pl3
p(4l)2
=
16E I
EI
3.3 Worked Example 3
167
• Linear thermal variation
''
ϕ10
=0
''
ϕ20
=0
• Constant thermal variation
αΔ T =
δV =
pl3
30E I
3
αΔ T · 4l = 6αΔ T l
2
δV =
pl4
6 pl4
=
30E I
5E I
δH =
3
3 pl4
δV =
4
20E I
δ=
5
pl4
δV =
4
4E I
168
3 Solution of Complex Frames
γ =
α=
1 pl4
pl3
δ
=
·
=
5l
5l 4E I
20E I
β=
δV
1 pl4
pl3
=
·
=
4l
4l 5E I
20E I
1 3 pl4
3 pl3
δH
δH
=
·
=
=
2l
2l 20E I
40E I
(2/3)3l
'''
ϕ10
= −β + γ = −
3 pl3
pl3
pl3
+
=
20E I
40 E I
40E I
'''
ϕ20
=β −α =
pl3
pl3
−
=0
20E I
20E I
Due to the superposition of effects, it is possible to calculate ϕ10 and ϕ20 .
'
''
'''
ϕ10 = ϕ10
+ ϕ10
+ ϕ10
=
1
41
+0+
16
40
'
''
'''
ϕ20 = ϕ20
+ ϕ20
+ ϕ20
= (2 + 0 + 0)
pl3
207 pl3
=
EI
80 E I
pl3
pl3
=2
EI
EI
Calculation of Unknowns
ϕ11 X 1 + ϕ12 X 2 + ϕ10 = 0
ϕ21 X 1 + ϕ22 X 2 + ϕ20 = 0
ϕ11 ϕ12
X1
ϕ10
·
=−
ϕ21 ϕ22
X2
ϕ20
25l 2l 207 pl3
X
1
12E I 3E I
·
= − 80 plE3 I
2l
7l
X2
2 EI
3E I 2E I
X 1 = −1.517 pl2
⇒
X 2 = 0.860 pl2
3.3 Worked Example 3
169
Summary of Bending Moments
Calculation of Shear Forces by Writing the Equilibrium Equations
V (x) = R A − p · x = 1.1 p · l − p · x = 0
The shear force is zero at a distance x from point A equal to:
x = 1.1l
M(x) = R A · 1.1l − 0.316 pl2 − p
(1.1l)2
= 0.289 pl2
2
M BAB = 0 → R A · 3l + 1.517 pl2 − 0.316 pl2 − p · 3l · 1.5l = 0 → R A = 1.1 pl
F→ = 0 → R A + R B A − p · 3l = 0 → R B A = 1.9 pl
170
3 Solution of Complex Frames
MCBC = 0 → R BC · 4l − 1.517 pl2 − 0.86 pl2 − 2 pl · 2l = 0 → R BC = 1.59 pl
F↑ = 0 → RC B + R BC − 2 pl = 0 → RC B = 0.41 pl
MCC D = 0 → R D =
M AAC = 0 → RC A =
1
pl
30
0.86
pl = 0.172 pl
5
F→ = 0 → 1.9 pl − 0.033 pl − N AC cosα + 0.172 plsinα = 0 → N AC = 2.46 pl
F↑ = 0 → 0.41 pl + NC D + 2.46 plsinα + 0.172cosα = 0 → NC D = −2.024 pl
3.3 Worked Example 3
Moment, Shear and Axial Force Diagrams
171
172
3 Solution of Complex Frames
Qualitative Deformed Shape
3.4 Worked Example 4
• EA → ∞
pl3
• θ = 12E
I
• k = El3I
pl2
• αΔ T
= 24E
t
I
Solution
The structure is symmetric and symmetrically loaded and can be simplified as shown
in the figure. In simplifying the structure, the stiffness of the spring is doubled,
since the spring in the whole structure would be subjected to twice the displacement
evaluated in the simplified structure.
3.4 Worked Example 4
173
The structure results to be a sway frame. The displacement method is used to
solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and
the horizontal translation δ2 of node B.
Displacement Method
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
174
3 Solution of Complex Frames
Kinematics
• ϕ1 = 1
m 11 =
8E I
12E I
4E I
+
=
l
l
l
r21 =
12E I
l2
m 12 =
12E I
l2
• δ2 = 1
r22 =
24E I
26E I
+ 2k =
l3
l3
3.4 Worked Example 4
175
• External load
m '10 =
pl2
12
'
r20
=0
• Imposed rotation
m ''10 = −
''
r20
=−
4E I
pl2
·θ =−
l
3
12E I
· θ = − pl
l2
176
3 Solution of Complex Frames
• Linear thermal variation
m '''
10 =
pl2
4E I αΔ T
=
t
6
'''
r20
=0
Due to the superposition of effects, it is possible to calculate m 10 and r20 .
m 10 = m '10 + m ''10 + m '''
10 =
pl2
pl2
pl2
pl2
−
+
=−
12
3
6
12
'
''
'''
r20 = r20
+ r20
+ r20
= − pl
Calculation of Unknowns
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
ϕ
m
m 11 m 12
· 1 = − 10
r21 r22
δ2
r20
12E I 12E I pl2
ϕ
−
2
1
l
l
12
·
=−
12E I 26E I
δ2
− pl
l2
l3
⎧
pl3
⎪
⎪
⎨ ϕ1 = −5.853 · 10−2
EI
⇒
4
⎪
pl
⎪
⎩ δ2 = 6.548 · 10−2
EI
3.4 Worked Example 4
Summary of Bending Moments
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
177
178
Qualitative Deformed Shape
3 Solution of Complex Frames
3.5 Worked Example 5
179
3.5 Worked Example 5
•
•
•
•
E I = constant
E A → 3∞
pl
θ = 24E
I
I
k = 6E
l3
• l∗ =
√
5l
2
Solution
The structure is symmetric and symmetrically loaded; it can be simplified as shown
in the figure below.
The structure results to be a sway frame. The displacement method is used to
solve the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and
the horizontal translation δ2 of node B.
180
3 Solution of Complex Frames
Displacement Method
Kinematics
• ϕ1 = 1
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
3.5 Worked Example 5
181
m 11 =
3E I
4E I
EI
+ ∗ = 6.683
l
l
l
Writing the rotational equilibrium at point O, we obtain:
r21 ·
3E I
6E I 3
l 4E I
2E I
+
+ ∗ +
− 2 l=0
2
l
l
l
l 2
r21 =
EI
6E I
3E I 2
− ∗ · = 0.633 2
2
l
l
l
l
• δ2 = 1
2
1
=
l/2
l
√
√
5
2
l= 5
αl∗ = ·
l 2
α=
m 12 =
r22 =
6E I
3E I √
EI
− ∗ 2 · 5 = 0.633 2
2
l
(l )
l
EI
12E I
3E I √ 2
+ 2k + ∗ 2 · 5 · = 34.733 3
3
l
(l )
l
l
182
3 Solution of Complex Frames
• External load
pl = p ∗ l∗ → p ∗ =
2
pl
2l
= p √ = √ p = 0.894 p
l∗
5l
5
pl2
8
2
pl
pl2
3 pl
2
−
=
· =−
8
2
l
4
m '10 = −
'
r20
• Imposed rotation
m ''10 = −
2E I
pl2
·θ =−
l
12
3.5 Worked Example 5
183
''
r20
=−
6E I
pl
·θ =−
2
l
4
Due to the superposition of effects, it is possible to calculate m 10 and r20 .
m 10 = −
pl2
5 pl2
pl2
−
=−
8
12
24
r20 = −
Calculation of Unknowns
pl
3 pl
−
= − pl
4
4
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
m 11 m 12
ϕ
m
· 1 = − 10
r21 r22
δ2
r20
6.683 ElI 0.633 El2I
ϕ1
0.208 pl2
·
=
0.633 El2I 34.733 El3I
δ2
pl
⎧
pl3
⎪
⎪
⎨ ϕ1 = 2.849 · 10−2
EI
⇒
4
⎪
pl
⎪
⎩ δ2 = 2.827 · 10−2
EI
Summary of Bending Moments
184
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.6 Worked Example 6
Qualitative Deformed Shape
3.6 Worked Example 6
185
186
3 Solution of Complex Frames
• E I = constant
• EA → ∞
1 pl3
• αΔ T = 104
EI
/
• l1 = 45 l
/
• l2 = 13
l
4
Solution
The structure is symmetric and symmetrically loaded; it can be simplified as shown
in the figure below.
The structure results to be a fixed-joint frame and it is solved with the force method
by adopting as static unknown the moment X 1 applied at node B.
3.6 Worked Example 6
187
Force Method
ϕ11 X 1 + ϕ10 = 0
Kinematics
• X1 = 1
ϕ11 =
l1
1
l
l
+
=
(l + l1 ) = 0.706
3E I
3E I
3E I
EI
188
3 Solution of Complex Frames
• External load
'
ϕ10
=
• Constant thermal variation
p ∗ (l1 )3
24E I
3.6 Worked Example 6
189
ω=
αΔ tl2 1
δc
=
l
cosβ l
3
l
3
l → cosβ = /2
= 0.832
2
13
l
4
/
/
pl3
13
13 2 1
pl3
l
=
ω=
104E I 4
4 3 l
48E I
l2 cosβ =
''
ϕ10
= −ω −
3
pl3
ω
=− ω=−
2
2
32E I
'
''
ϕ10 = ϕ10
+ ϕ10
= 0.0153
pl3
48E I
Calculation of Unknowns
ϕ11 X 1 + ϕ10 = 0
0.706
l
pl3
X 1 + 0.0153
=0
EI
EI
X 1 = −0.022 pl2
Summary of Bending Moments
190
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.7 Worked Example 7
Qualitative Deformed Shape
3.7 Worked Example 7
• EA → ∞
pl2
)
• α(Δ T
= 6E
t
I
191
192
3 Solution of Complex Frames
• l = 9t
• k = 2El I
Solution
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotations ϕ1 of node B and ϕ2 of node C.
Displacement Method
m 11 ϕ1 + m 12 ϕ2 + m 10 = 0
m 21 ϕ1 + m 22 ϕ2 + m 20 = 0
3.7 Worked Example 7
193
Kinematics
• ϕ1 = 1
m 11 =
12E I
(2E I ) 3(2E I ) 4E I
+
+
=
l
l
l
l
m 21 =
2E I
l
m 12 =
2E I
l
• ϕ2 = 1
m 22 = k +
6E I
4E I
=
l
l
194
3 Solution of Complex Frames
• External load, linear and constant thermal variation
m 10 = −
=−
αΔ T
3
pl2
αΔ T
6E I
−
− 2
(2E I ) + (2E I )
12
t
2
t
l
3
3(2E I )
αΔ T l +
(3αΔ T l) =
2
l2
pl2
pl2
pl2
pl2
+
+
=
12
6
6
4
m 20 =
pl2
pl2
pl2
−
=−
12
6
12
3.7 Worked Example 7
195
Calculation of Unknowns
m 11 ϕ1 + m 12 ϕ2 + m 10 = 0
m 21 ϕ1 + m 22 ϕ2 + m 20 = 0
m 11 m 12
ϕ1
m 10
·
=−
m 21 m 22
ϕ2
m 20
2 12E I 2E I pl
ϕ1
l
l
4 2
·
=
−
2E I 6E I
pl
ϕ
−
2
l
l
12
⎧
pl3
⎪
⎪
⎨ ϕ1 = −2.451 · 10−2
EI
⇒
3
⎪
pl
⎪
−2
⎩ ϕ2 = 2.206 · 10
EI
Summary of Bending Moments
Summary of Shear Forces
196
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.8 Worked Example 8
Qualitative Deformed Shape
3.8 Worked Example 8
•
•
•
•
E I = constant
EA → ∞ 3
pl
αΔ T = 100E
I
2E I
k= l
Solution
The structure results to be a sway frame.
197
198
3 Solution of Complex Frames
The mixed method is used to solve the structure by adopting as kinematic
unknowns the rotation ϕ1 of node B and the horizontal translation δ3 of node A
and as static unknown the moment X2 at node D.
Mixed Method
⎧
⎨ m 11 ϕ1 + m 12 X 2 + m 13 δ3 + m 10 = 0
ϕ ϕ + ϕ22 X 2 + ϕ23 δ3 + ϕ20 = 0
⎩ 21 1
r31 ϕ1 + r32 X 2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
3.8 Worked Example 8
199
m 11 =
3E I
6E I
13E I
4E I
+
+
=
l
l
l
l
ϕ21 =
r31 = −
1
2
6E I
l2
• X2 = 1
m 12 = −
ϕ22 =
l
1
7 l
l
+
+ =
3E I
6E I
k
8 EI
r32 = −
• δ3 = 1
1
2
3
l
200
3 Solution of Complex Frames
6E I
l2
/ 2
l
2
3
24E I
=
=
l3
2E I
l
m 13 = −
ϕ23
r33 =
12E I
24E I
36E I
+
=
l3
l3
l3
• External load and constant thermal variation
m 10 = −
ϕ20 =
3 pl2
3 2 9 pl2
pl +
=−
32
100
800
3 pl3
7 pl3
pl3
−
=
48E I
200E I
1200E I
r30 = 0
3.8 Worked Example 8
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 X 2 + m 13 δ3 + m 10 = 0
ϕ ϕ + ϕ22 X 2 + ϕ23 δ3 + ϕ20 = 0
⎩ 21 1
r31 ϕ1 + r32 X 2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎡
⎤
m 11 m 12 m 13
ϕ1
m 10
⎣ ϕ21 ϕ22 ϕ23 ⎦ · ⎣ X 2 ⎦ = −⎣ ϕ20 ⎦
r31 r32 r33
δ3
r30
⎤
⎡
⎡ 13E I
⎤
3
I ⎤ ⎡
− 800
pl2
− 21 − 6E
ϕ1
l
l2
⎥
7 pl3
7l
3
⎣ 1
⎦ · ⎣ X 2 ⎦ = −⎢
⎦
⎣ 1200E
2
8E I
l
I
I
3 36E I
δ
− 6E
−
0
3
2
3
l
l
l
⎧
pl3
⎪
⎪
⎪
ϕ1 = −0.00011
⎪
⎪
EI
⎨
⇒
X 2 = −0.0051 pl2
⎪
⎪
⎪
⎪
pl4
⎪
⎩ δ3 = −0.00044
EI
Summary of Bending Moments
201
202
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.9 Worked Example 9
Qualitative Deformed Shape
3.9 Worked Example 9
• E I = constant
• EA → ∞
EI
• k = 2l
3
203
204
3 Solution of Complex Frames
Solution
Initially, a roller is introduced at node C to control the displacement at the translational
spring. We then proceed with the kinematic analysis of the structure.
The structure results to be a sway frame and it is solved by the displacement
method. The introduction of a roller in B with a vertical constraint reaction allows
to prevent any movement of the structure. The rotation ϕ1 of joint C, the translations
η2 of joint C and η3 of joint B are adopted as kinematic unknowns.
Displacement Method
⎧
⎨ m 11 ϕ1 + m 12 η2 + m 13 η3 + m 10 = 0
r ϕ + r22 η2 + r23 η3 + r20 = 0
⎩ 21 1
r31 ϕ1 + r32 η2 + r33 η3 + r30 = 0
3.9 Worked Example 9
205
Kinematics
• ϕ1 = 1
m 11 =
12E I
27 E I
3E I
+
=
l
5l
5 l
3
3E I
3E I
3E I
3
+ 5l − 2 l + l = 0
M BBC D = 0 → r21 · l +
4
l
l
4
4
r21 = −
F↑ = 0 → r31 + r21 −
r31 =
• η2 = 1
1 EI
5 l2
16 E I
5 l2
3E I
=0
l2
206
3 Solution of Complex Frames
m 12 =
3E I
80 E I
1 EI
−
=−
2
2
l
25 l
5 l2
3E I 7
3E I
3
3E I 5
∗
· l − 3 l + 2 − 2 = 0
M BBC D = 0 → r22
5l
4
l 4
l
3
4
'
r22
109 E I
=
15 l3
'
''
'
r22 = r22
+ r22
= r22
+k =
r32 = −
• η3 = 1
64 E I
15 l3
233 E I
30 l3
3.9 Worked Example 9
207
m 13 =
16 E I
5 l2
r23 = −
r33 =
64 E I
15 l3
109 E I
15 l3
Remark: the relative displacement between the two ends of rod BC, in the direction
orthogonal to the rod is 5/3.
• External load
m 10 = −
7
pl2
128
r20 = −
35
pl
32
r30 = −
21
pl
32
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 η2 + m 13 η3 + m 10 = 0
r ϕ + r22 η2 + r23 η3 + r20 = 0
⎩ 21 1
r31 ϕ1 + r32 η2 + r33 η3 + r30 = 0
208
3 Solution of Complex Frames
⎡
⎤ ⎡
⎤
⎡
⎤
m 11 m 12 m 13
ϕ1
m 10
⎣ r21 r22 r23 ⎦ · ⎣ η2 ⎦ = −⎣ r20 ⎦
r31 r32 r33
η3
r30
⎤
⎡
⎡
⎡ 27 E I
⎤
⎤
EI
7
− 15 El2I 16
pl2
ϕ1
− 128
5 l
5 l2
⎣ − 1 E2I 233 E3I − 64 E3I ⎦ · ⎣ η2 ⎦ = −⎣ − 35 pl ⎦
5 l
30 l
15 l
32
21
16 E I
E I 109 E I
η3
pl
− 32
− 64
5 l2
15 l3
15 l3
⎧
pl3
⎪
⎪
ϕ
=
−0.2044
⎪
1
⎪
EI
⎪
⎪
⎨
pl4
⇒ η2 = 0.3463
⎪
EI
⎪
⎪
⎪
4
⎪
pl
⎪
⎩ η3 = 0.3837
EI
Summary of Bending Moments
Summary of Shear Forces
3.9 Worked Example 9
Moment, Shear and Axial Force Diagrams
209
210
Qualitative Deformed Shape
3.10 Worked Example 10
• E I = constant
• EA → ∞
pl3
• αΔ T = 24E
I
3 Solution of Complex Frames
3.10 Worked Example 10
211
Solution
The structure results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotation ϕ1 of node C and the horizontal translation δ2 of node B. The
addition of a horizontal roller in B prevents any movements of the structure.
Displacement Method
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
212
3 Solution of Complex Frames
Kinematics
• ϕ1 = 1
m 11 =
10E I
l
r21 = −
• δ2 = 1
3E I
l2
3.10 Worked Example 10
213
3E I
l2
3 EI
= 18 + √
2 l3
m 12 = −
r22
• External load and constant thermal variation
m 10 = −
r20 =
pl2
8
pl
8
214
3 Solution of Complex Frames
Calculation of Unknowns
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
m 11 m 12
ϕ1
m 10
·
=−
r21 r22
δ2
r20
10E I
pl2 3E I
ϕ
l
− l2 8
· 1 =
I
EI
√3
18
+
− 3E
δ2
− pl
2
3
l
8
2 l
3
ϕ1 = 0.0111 pl
EI 4
⇒
δ2 = −0.0046 pl
EI
Summary of Bending Moments
Summary of Shear Forces
3.10 Worked Example 10
Moment, Shear and Axial Force Diagrams
215
216
3 Solution of Complex Frames
Qualitative Deformed Shape
3.11 Worked Example 11
•
•
•
•
E I = constant
EA → ∞ 3
pl
αΔ T = 8E
I
k = ElI
Solution
It is observed that the right-hand portion of the structure (C-D-E) constitutes an
isostatic appendage; the effect of thermal variation on the rod DE simply results in
a rigid kinematic motion of the appendage itself, caused by the axial elongation of
the oblique rod, and will therefore only be taken into account in the deformation
tracking, since it does not generate any internal forces and moments in the structure
under consideration. We then proceed to solve the isostatic appendage.
3.11 Worked Example 11
217
The structure can be simplified as:
We then proceed to analyze the simplified structure, which results to be a sway
frame.
The mixed method is used to solve the structure by adopting as kinematic unknown
the horizontal translation δ2 of node B and as static unknown the moment X1
transmitted by the spring.
218
3 Solution of Complex Frames
Mixed Method
ϕ11 X 1 + ϕ12 δ2 + ϕ10 = 0
r21 X 1 + r22 δ2 + r20 = 0
Kinematics
• X1 = 1
ϕ11 =
l
1
7 l
l
19 l
l
+
+ =
+
=
3E I
4E I
k
12 E I
EI
12 E I
r21 =
3
2l
3.11 Worked Example 11
219
• δ2 = 1
3
1 3E I l
+ 2
=−
ϕ12 = −
l
l 6E I
2l
r22 =
3E I
6E I
3E I
+
=
l
l
l
• External load
ϕ10 =
pl3
pl2 l
pl3
−
=
24E I
8 6E I
48E I
220
3 Solution of Complex Frames
r20 = −
5 pl
7 pl
pl
− pl +
=−
2
8
8
Calculation of Unknowns
ϕ11 X 1 + ϕ12 δ2 + ϕ10 = 0
r21 X 1 + r22 δ2 + r20 = 0
ϕ11 ϕ12
X1
ϕ10
·
=−
r21 r22
δ2
r20
19 l
3
pl3
− 2l
X1
− 48E
12 E I
I
·
=
7 pl
3
6E I
δ2
2l
l3
8
X 1 = 0.101 pl2
⇒
4
δ2 = 0.1205 pl
EI
Summary of Bending Moments
3.11 Worked Example 11
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
221
222
Qualitative Deformed Shape
3.12 Worked Example 12
• EA → ∞
ql4
• δ = 24E
I
ql3
• αΔ T = 12E
I
l
• h = 10
3 Solution of Complex Frames
3.12 Worked Example 12
223
Solution
The structure results to be a fixed-joint frame and it is solved with the displacement
method, by adopting as kinematic unknown the rotation ϕ at node B.
Displacement Method
m 11 ϕ1 + m 10 = 0
Kinematics
• ϕ1 = 1
224
3 Solution of Complex Frames
m 11 =
13E I
l
• External load
m '10 = −
ql2
8
• Imposed displacement
m ''10 =
3E I
ql2
δ
=
l2
8
3.12 Worked Example 12
225
• Constant thermal variation
m '''
10 = −
3
9E I
αΔ T = − ql2
l
4
• Linear thermal variation
m ''''
10 = −
5
2E I αΔ T
= − ql2
h
2
6
Due to the superposition of effects, it is possible to calculate m 10 .
1 1 3 5
19
''''
+
−
−
ql2 = − ql2
m 10 = m '10 + m ''10 + m '''
+
m
=
−
10
10
8 8 4 6
12
226
3 Solution of Complex Frames
Calculation of Unknowns
m 11 ϕ1 + m 10 = 0
l
ql3
m 10
19 ql2
= 0.1217
⇒ ϕ1 = −
=− −
m 11
12 E I 13E I
EI
Summary of Bending Moments
Summary of Shear Forces
3.12 Worked Example 12
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
227
228
3 Solution of Complex Frames
3.13 Worked Example 13
• E I = constant
• EA → ∞
pl2
)
• 2α(Δ T
= 24E
h
I
Solution
The structure results to be a sway frame. The displacement method is used to solve
the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the
horizontal translation δ2 of node B.
3.13 Worked Example 13
229
The geometric quantities l∗ and l1 are calculated:
/
l2 +
l2
= 1.118l
4
9l2 +
9l2
= 3.354l
4
∗
l =
/
l1 =
Displacement Method
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
Kinematics
• ϕ1 = 1
m 11 =
4E I
4E I
EI
+
= 7.578
l∗
l
l
230
3 Solution of Complex Frames
6E I
6E I
6E I
− ∗ + ∗ 2 · l1 = 0
l
l
(l )
1
6E I
6E I
6E I
EI
=
+
−
l1
= −2.367
l∗
l
1.25l
2l
l
M O = 0 → r21 · 2l −
r21
• δ2 = 1
√
EI
5
3E I
6E I
= −2.367
−
·
l2
l
(l∗ )2 2
√
5
6E I
M O = 0 → r22 · 2l + 2
·
+
(l∗ )2 2
√
12E I
5
6E I 1
− ∗ 3 ·
· l1 = 0
·
−2
l2 2
2
(l )
m 12 = +
r22 =
• External load
12E I
12E I
EI
+ 2 ∗ = 13.73 3
3
4l
l l
l
3.13 Worked Example 13
231
• Linear thermal variation
m 10
2α(Δ T )
pl2
pl2
− EI
=−
=−
12
h
8
r20 = −
Calculation of Unknowns
pl
4
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
m 11 m 12
ϕ1
m 10
·
=−
r21 r22
δ2
r20
7.578E I
2
I
− 2.367E
ϕ1
− pl8
l
l2
·
=−
I 13.73E I
− 2.367E
δ2
− pl
l2
l3
4
⎧
3
pl
⎪
⎪
⎨ ϕ1 = 0.0229
EI
⇒
4
⎪
⎪
⎩ δ2 = 0.0222 pl
EI
Summary of Bending Moments
232
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.14 Worked Example 14
233
Qualitative Deformed Shape
3.14 Worked Example 14
•
•
•
•
E I AB = E I
E I BC → ∞
EA → ∞
I
k = 4E
l3
Solution
The structure results to be a sway frame. The mixed method is used to solve the
structure by adopting as static unknown the moment X1 of node B and as kinematic
unknowns the vertical translations δ2 of node A and δ3 of node B.
234
3 Solution of Complex Frames
Mixed Method
⎧
⎨ ϕ11 X 1 + ϕ12 δ2 + ϕ13 δ3 + ϕ10 = 0
r X + r22 δ2 + r23 δ3 + r20 = 0
⎩ 21 1
r31 X 1 + r32 δ2 + r33 δ3 + r30 = 0
Kinematics
• X1 = 1
ϕ11 =
l
3E I
r21 = −
r31 =
• δ2 = 1
1
l
1
l
3.14 Worked Example 14
235
ϕ12 =
r22 = k =
1
l
4E I
l3
r32 = 0
• δ3 = 1
ϕ13 = −
1
l
r23 = 0
r33 = 0
• External load
ϕ10 =
pl3
24E I
r20 = −
pl
2
236
3 Solution of Complex Frames
r30 = −
pl
2
Calculation of Unknowns
⎧
⎨ ϕ11 X 1 + ϕ12 δ2 + ϕ13 δ3 + ϕ10 = 0
r X + r22 δ2 + r23 δ3 + r20 = 0
⎩ 21 1
r31 X 1 + r32 δ2 + r33 δ3 + r30 = 0
⎡
⎡
⎤ ⎡ ⎤
⎤
X1
ϕ10
ϕ11 ϕ12 ϕ13
⎣ r21 r22 r23 ⎦ · ⎣ δ2 ⎦ = −⎣ r20 ⎦
r31 r32 r33
δ3
r30
⎤
⎡
⎡ l 1
⎤
1⎤ ⎡
pl3
−
X
1
3E I
l
l
24E I ⎥
⎣ − 1 4E3 I 0 ⎦ · ⎣ δ2 ⎦ = −⎢
⎦
⎣ − pl
l l
2
pl
1
δ
0
0
−
3
l
2
⎧
2
⎪
⎨ X 1 = 0.5 pl4
pl
⇒ δ2 = 0.25 E I
⎪
⎩ δ = 0.4583 pl4
3
EI
Summary of Bending Moments
Summary of Shear Forces
3.14 Worked Example 14
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
237
238
3 Solution of Complex Frames
3.15 Worked Example 15
• E I = constant
• EA → ∞
• k = EI
l
Solution
The structure is a sway frame. The statically determined part CD is eliminated and
replaced by a resultant force and moment.
3.15 Worked Example 15
239
Displacement Method
⎧
⎪
m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0
⎪
⎪
⎨ m ϕ + m ϕ + m ϕ + m δ + m = − pl2
21 1
22 2
23 3
24 4
20
pl2
⎪
m
ϕ
+
m
ϕ
+
m
ϕ
+
m
δ
+
m
31 1
32 2
33 3
34 4
30 = 2
⎪
⎪
⎩r ϕ + r ϕ + r ϕ + r δ + r = 0
41 1
42 2
43 3
44 4
40
Kinematics
• ϕ1 = 1
m 11 =
4E I
EI
7E I
EI
4E I
+k =
+
=
= 2.33
3l
3l
l
3l
l
m 21 =
EI
2E I
= 0.67
3l
l
m 31 = 0
r41 = −
6 EI
2 EI
EI
=−
= −0.67 2
9 l2
3 l2
l
240
3 Solution of Complex Frames
• ϕ2 = 1
m 12 =
m 22 =
EI
2 EI
= 0.67
3 l
l
4 EI
32 E I
EI
4 EI
+
=
= 2.13
5 l
3 l
15 l
l
m 32 =
r42 = −
EI
2 EI
= 0.4
5 l
l
6 EI
2 EI
EI
=−
= −0.67 2
9 l2
3 l2
l
• ϕ3 = 1
m 13 = 0
m 23 =
EI
2 EI
= 0.4
5 l
l
3.15 Worked Example 15
241
m 33 =
1 EI
EI
4 EI
+
= 1.3
5 l
2 l
l
r43 = 0
• δ4 = 1
m 14 = −
6 EI
EI
= −0.67 2
9 l2
l
m 24 = −
6 EI
EI
= −0.67 2
9 l2
l
m 34 = 0
r44 =
12 E I
4 EI
EI
=
= 0.44 3
3
3
27 l
9 l
l
• External load
m 10 = 0
242
3 Solution of Complex Frames
m 20 = −
m 30 =
25 2
pl
12
25 2
pl
12
r40 = 0
Calculation of Unknowns
⎧
⎪
m 11 ϕ1 + m 12 ϕ2 + m 13 ϕ3 + m 14 δ4 + m 10 = 0
⎪
⎪
⎨ m ϕ + m ϕ + m ϕ + m δ + m = − pl2
21 1
22 2
23 3
24 4
20
2
⎪ m 31 ϕ1 + m 32 ϕ2 + m 33 ϕ3 + m 34 δ4 + m 30 = pl2
⎪
⎪
⎩r ϕ + r ϕ + r ϕ + r δ + r = 0
41 1
42 2
43 3
44 4
40
⎡
⎤ ⎡ ⎤
⎤
⎡
m 11 m 12 m 13 m 14
ϕ1
m 10
⎢ m 21 m 22 m 23 m 24 ⎥ ⎢ ϕ2 ⎥
⎢ m 20 ⎥
⎢
⎥ ⎢ ⎥
⎥
⎢
⎣ m 31 m 32 m 33 m 34 ⎦ · ⎣ ϕ3 ⎦ = −⎣ m 30 ⎦
r41 r42 r43 r44
δ4
r40
⎡
⎤ ⎡ ⎤ ⎡
0
0 −0.67 El2I
2.33 ElI 0.67 ElI
ϕ1
25
2
⎢ 0.67 E I 2.13 E I 0.4 E I −0.67 E2I ⎥ ⎢ ϕ2 ⎥ ⎢
pl
− pl2
⎢
l
l
l
l ⎥·⎢
⎢
⎥ = ⎢ 12
2
EI
EI
25
2
⎣
⎦
⎦
⎣
0
0.4 l 1.3 l
0
ϕ3
⎣ − 12 pl + pl2
EI
EI
EI
δ4
0
0.44 l3
−0.67 l2 −0.67 l2
0
⎧
pl3
⎪
⎪
⎪
ϕ1 = 0.4236
⎪
⎪
EI
⎪
⎪
⎪
3
⎪
pl
⎪
⎪
⎨ ϕ2 = 1.6943
EI
⇒
⎪
pl3
⎪
⎪
ϕ
=
−1.7393
⎪
3
⎪
⎪
EI
⎪
⎪
⎪
4
⎪
pl
⎪
⎩ δ4 = 3.1769
EI
⎤
⎥
⎥
⎥
⎦
3.15 Worked Example 15
Summary of Bending Moments
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
243
244
Qualitative Deformed Shape
3 Solution of Complex Frames
3.16 Worked Example 16
245
3.16 Worked Example 16
•
•
•
•
•
E I = constant
EA → ∞
1 pl4
δ = 10
EI
k = ElI
P = pl
Solution
The ABCD structure is the statically indeterminate frame, while the remaining
portion of the structure constitutes an isostatic appendage.
The three-hinged arch FEC results loaded by a thermal variation that deforms the
rod, but since FEC is statically determined this does not induce any thermal co-action.
246
3 Solution of Complex Frames
The statically determined portion BEFGC can then be easily analyzed by the node
method; globally it transmits to the statically indeterminate portion ABCD the nodal
actions highlighted in the following figure. We then proceed to solve this structure
using the displacement method.
The structure results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotations ϕ1 of node B, ϕ2 of node D and the horizontal translation δ3
of node C.
3.16 Worked Example 16
247
Displacement Method
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
7E I
3 EI
=
m 11 = 2 +
2 l
2l
m 21 = 0
248
3 Solution of Complex Frames
r31 = −
6E I
3E I
=− 2
2l
(2l)2
• ϕ2 = 1
m 12 = 0
m 22 =
4E I
3E I
+k =
l
l
r32 = −
• δ3 = 1
3E I
l2
3.16 Worked Example 16
249
m 13 = −
6E I
3E I
=− 2
2
2l
(2l)
m 23 = −
3E I
l2
r33 =
12E I
3E I
9E I
+ 3 =
l
2l3
(2l)3
m 10 =
3 2
3E I 1 pl4
=
pl
2
4l 10 E I
40
• Imposed displacement
m 20 = 0
r30 = 0
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎡
⎤ ⎡ ⎤
⎤
m 11 m 12 m 13
ϕ1
m 10
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
250
3 Solution of Complex Frames
⎤
⎡
⎤
⎡ 7
3 ⎤ ⎡
0 − 2l
ϕ1
0.075 pl2
EI ⎣ 2
⎦
0 4 − 3l ⎦ · ⎣ ϕ2 ⎦ = −⎣
0
l
9
3
3
δ3
0
− 2l − l 2l2
⎧
3
pl
⎪
⎨ ϕ1 = −0.030 E I
pl3
⇒ ϕ2 = −0.015 E I
⎪
4
⎩
δ3 = −0.020 pl
EI
Summary of Bending Moments
Summary of Shear Forces
3.16 Worked Example 16
Moment, Shear and Axial Force Diagrams
251
252
Qualitative Deformed Shape
3 Solution of Complex Frames
3.17 Worked Example 17
253
3.17 Worked Example 17
• EA → ∞
pl2 h
• αΔ T = 16E
I
l
• h = 10
Solution
The structure results to be a sway frame. The displacement method is used to solve
the structure by adopting as kinematic unknowns the rotations ϕ1 of node B, ϕ2 of
node C and the horizontal translation δ3 of node B.
254
3 Solution of Complex Frames
Displacement Method
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
Kinematics
• ϕ1 = 1
m 11
4
EI
EI
8+
= 10.53
=
l
1.581
l
m 21 =
EI
2 EI
= 1.265
1.581 l
l
3.17 Worked Example 17
r31 = −
255
12E I
6 sin α E I
EI
+
= −11.241 2
2
2
2
l
1.581 l
l
• ϕ2 = 1
m 12 = 1.265
r32
• δ3 = 1
EI
l
4
4
EI
EI
+
= 5.060
m 22 =
l 1.581 1.581
l
6 sin α
6
EI
EI
· cos α +
= −1.581 2
= 2 −
2
l
1.581
1.581
l
256
3 Solution of Complex Frames
m 13 = −11.241
m 23 = −1.581
r33 =
EI
l2
EI
l2
EI
EI
(24 + 2.733 + 0.304) = 27.037 3
3
l
l
• External load and linear thermal variation
m 10 = −
m 20 =
3 2
1 2
1
pl +
pl = − pl2
16
16
8
3 2 1 2
1 2
pl − pl −
pl = 0
16
8
16
3.17 Worked Example 17
257
r30 = 0.375 pl
Calculation of Unknowns
⎧
⎨ m 11 ϕ1 + m 12 ϕ2 + m 13 δ3 + m 10 = 0
m ϕ + m 22 ϕ2 + m 23 δ3 + m 20 = 0
⎩ 21 1
r31 ϕ1 + r32 ϕ2 + r33 δ3 + r30 = 0
⎡
⎤ ⎡ ⎤
⎡
⎤
m 11 m 12 m 13
ϕ1
m 10
⎣ m 21 m 22 m 23 ⎦ · ⎣ ϕ2 ⎦ = −⎣ m 20 ⎦
r31 r32 r33
δ3
r30
⎤
⎡
⎤
⎡ 1 2 ⎤
⎡
10.53 1.265 − 11.241
ϕ1
− 8 pl
l
EI ⎣
⎦ · ⎣ ϕ2 ⎦ = −⎣
⎦
1.265 5.060 − 1.518
0
l
l
11.241
1.518 27.037
δ3
0.375 pl
− l − l
l2
⎧
3
⎪
−3 pl
⎪
⎪ ϕ1 = −4.888 · 10
⎪
EI
⎪
⎪
⎨
3
pl
⇒ ϕ2 = −3.610 · 10−3
⎪
EI
⎪
⎪
⎪
4
⎪
pl
⎪
−2
⎩ δ3 = −1.610 · 10
EI
Summary of Bending Moments
258
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.18 Worked Example 18
Qualitative Deformed Shape
3.18 Worked Example 18
• E I = constant
• EA → ∞
259
260
3 Solution of Complex Frames
Solution
The structure results to be a sway frame. The displacement method is used to solve
the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the
horizontal translation δ2 of node B. Member CD represents a statically determined
portion.
Displacement Method
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
3.18 Worked Example 18
261
Kinematics
• ϕ1 = 1
m 11 = (3 + 4)
r21 = (−6 + 3)
7E I
EI
=
l
l
3E I
EI
=− 2
l2
l
• δ2 = 1
m 12 = (−6 + 3)
r22 = (12 + 3)
EI
3E I
=− 2
2
l
l
15E I
EI
=
3
l
l3
262
3 Solution of Complex Frames
• External load
1
m '10 = − pl2
8
'
r20
=
3
pl
8
m ''10 =
1 2
pl
4
''
r20
=
3
pl
4
3.18 Worked Example 18
263
m '''
10 = 0
'''
r20
= pl
Due to the superposition of effects, it is possible to calculate m 10 and r20 .
1 2 1 2
1
pl = pl2
m 10 = m '10 + m ''10 + m '''
10 = − pl +
8
4
8
'
''
'''
r20 = r20
+ r20
+ r20
=
Calculation of Unknowns
3
3
17
pl + pl =
pl
8
4
8
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
m 11 m 12
ϕ
m
· 1 = − 10
r21 r22
δ2
r20
pl2
E I 7 − 3l
ϕ1
·
= − 178pl
δ2
l − 3l 15
l2
8
⎧
pl3
⎪
⎪
⎨ ϕ1 = −8.5937 · 10−2
EI
⇒
4
⎪
pl
⎪
⎩ δ2 = −1.5885 · 10−1
EI
Summary of Bending Moments
264
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.19 Worked Example 19
Qualitative Deformed Shape
3.19 Worked Example 19
• E I = constant
• E A BC = E AC D → ∞
I
• E A AC = 768E
l2
pl2
αΔ T
• h = 6E I
• m = pl
2
265
266
3 Solution of Complex Frames
Solution
Node C can move vertically due to the axial deformability of rod AC. The structure
results to be a sway frame. The displacement method is used to solve the structure
by adopting as kinematic unknowns the rotation ϕ1 of node C and the horizontal
translation δ2 of node A.
The BC rod, because of the particular loading condition, can be considered an
isostatic appendage. It transfers to the frame ACD a vertical load equal to pl/2.
F↑ = 0 → VB = VC
M BBC = 0 → VC · l − m · l = 0 → VC = 0.5 pl
VB = VC = 0.5 pl
3.19 Worked Example 19
267
Displacement Method
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
Kinematics
• ϕ1 = 1
m 11 =
4E I
8E I
4E I
+
=
l
l
l
r21 =
6E I
l2
268
3 Solution of Complex Frames
• δ2 = 1
m 12 =
r22 = k +
6E I
l2
12E I
780E I
=
3
l
l3
• External load
m '10 = −
pl2
12
'
r20
=−
pl
2
3.19 Worked Example 19
269
m ''10 = 0
''
r20
=−
pl
2
• Linear thermal variation
m '''
10 =
2E I αΔ T
h
'''
r20
=0
Due to the superposition of effects, it is possible to calculate m10 and r20 :
m 10 = m '10 + m ''10 + m '''
10 = −
2E I αΔ T
pl2
pl2
+0+
=
12
h
12
'
''
'''
r20 = r20
+ r20
+ r20
=−
pl
pl
−
+ 0 = − pl
2
2
270
3 Solution of Complex Frames
Calculation of Unknowns
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
m 11 m 12
ϕ1
m 10
·
=−
r21 r22
δ2
r20
2 pl
E I 8 6l
ϕ1
12
·
=
−
δ
l 6l 780
−
pl
2
2
l
3
ϕ1 = −1.1444 · 10−2 pl
E
I
⇒
4
δ2 = 1.37 · 10−3 pl
EI
Summary of Bending Moments
Summary of Shear Forces
3.19 Worked Example 19
Moment, Shear and Axial Force Diagrams
271
272
3 Solution of Complex Frames
Qualitative Deformed Shape
3.20 Worked Example 20
• E I = constant
• EA → ∞
pl2
• αΔ T
= 10E
h
I
• l = 5h
Solution
Taking advantage of symmetry, the frame is simplified as shown in the following
figure for the kinematic analysis.
3.20 Worked Example 20
273
The simplified structure results to be a fixed-joint frame and it is solved by the
displacement method adopting as kinematic unknown the rotation ϕ1 of node B.
The loads that the rod AC transfers to node C are initially calculated:
Displacement Method
m 11 ϕ1 + m 10 = 0
274
3 Solution of Complex Frames
Kinematics
• ϕ1 = 1
m 11 =
8E I
l
• External load, constant and linear thermal variation
3.20 Worked Example 20
275
6E I
3E I
E I αΔ T
δT −
−
l2
l2
h
m 10 =
15 2
pl +
64
m 10 =
pl2
359 2
15 2 3E I 3 pl2
pl + 2 ·
−
=
pl
64
l
100
10
1600
Calculation of Unknowns
m 11 ϕ1 + m 10 = 0
ϕ1 = −
l
pl3
m 10
359 pl2
·
= −0.02805
=−
m 11
1600 8E I
EI
276
Summary of Bending Moments
Summary of Shear Forces
3 Solution of Complex Frames
3.20 Worked Example 20
Moment, Shear and Axial Force Diagrams
277
278
Qualitative Deformed Shape
3 Solution of Complex Frames
3.21 Worked Example 21
279
3.21 Worked Example 21
•
•
•
•
E I = constant
EA → ∞
I
k = 6E
l3
Pl2
αΔ T = 12E
I
Solution
Taking advantage of symmetry, the frame is simplified as shown in the following
figure. The simplified structure results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotation ϕ1 of node B and the vertical translation δ2 of node C.
280
3 Solution of Complex Frames
Displacement Method
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
Kinematics
• ϕ1 = 1
m 11 =
8E I
l
r21 = −
6E I
l2
3.21 Worked Example 21
281
• δ2 = 1
m 12 = −
r22 =
6E I
l2
12E I
18E I
+k =
l3
l3
• External load
m '10 = −Pl
'
r20
= −P
282
3 Solution of Complex Frames
• Imposed displacement
m ''10 =
''
r20
=−
6E I
δT = Pl
l2
12E I
δT = −2P
l3
Due to the superposition of effects, it is possible to calculate m 10 and r20 .
m 10 = m '10 + m ''10 = 0
'
''
r20 = r20
+ r20
= −3P
Calculation of Unknowns
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
E I 8 − 6l
ϕ1
0
·
=−
δ2
−3P
l − 6l 18
l2
⎧
2
⎪
⎪ ϕ1 = 1 Pl
⎨
6 EI
⇒
3
⎪
2
⎪
⎩ δ2 = Pl
9 EI
3.21 Worked Example 21
Summary of Bending Moments
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
283
284
Qualitative Deformed Shape
3 Solution of Complex Frames
3.22 Worked Example 22
285
3.22 Worked Example 22
• E I = constant
• EA → ∞
I
• k = 10E
l
Solution
The structure is symmetrically loaded, and taking advantage of symmetry, the frame
is simplified as shown in the following figure. A fixed end constraint is inserted at
node C as the continuity constraint of the rod BD is recreated, while the presence of
the rod CF prevents any possible vertical displacement.
The kinematic analysis shows that the simplified structure is a fixed-joint frame—
as it represents a three-hinged arch—and it is solved with the force method by
adopting as static unknown the moment X1 of node B.
286
3 Solution of Complex Frames
Force Method
ϕ11 X 1 + ϕ10 = 0
Kinematics
• X1 = 1
ϕ11 =
4l
1
9l
l
47l
5l
+
+ =
+
=
4E I
4E I
k
4E I
10E I
20E I
3.22 Worked Example 22
287
• External load
ϕ10 =
5 pl3
13 pl3
2 p(4l)3
+
=
48E I
3 EI
3 EI
Calculation of Unknowns
ϕ11 X 1 + ϕ10 = 0
X1 = −
260 2
ϕ10
13 pl3 20 E I
·
=−
pl
=−
ϕ11
3 E I 47 l
141
⇒ X 1 = −1.8439 pl2
288
Summary of Bending Moments
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.22 Worked Example 22
Qualitative Deformed Shape
289
290
3 Solution of Complex Frames
3.23 Worked Example 23
•
•
•
•
•
E I AB = E IC D = E I
E I BC → ∞
EA → ∞
I
k1 = 2E
l3
3E I
k2 = l
Solution
The structure results to be a sway frame. The mixed method is used to solve the
structure by adopting as static unknown the moment X1 of node B and as kinematic
unknowns the vertical translations δ2 of node A and δ3 of node B.
Mixed Method
3.23 Worked Example 23
291
⎧
⎨ ϕ11 X 1 + ϕ12 δ2 + ϕ13 δ3 + ϕ10 = 0
r X + r22 δ2 + r23 δ3 + r20 = 0
⎩ 21 1
r31 X 1 + r32 δ2 + r33 δ3 + r30 = 0
Kinematics
• X1 = 1
ϕ11 =
1
l
2 l
+
=
3E I
k2
3 EI
r21 = −
1
l
r31 =
2
l
ϕ12 =
1
l
• δ2 = 1
r22 = 0 + k1 =
r32 = 0
2E I
l3
292
3 Solution of Complex Frames
• δ3 = 1
2
1 1
+
=−
ϕ13 = −
l l
l
r23 = 0
r33 = 0
• External load
ϕ10 =
pl3
24E I
r20 = −
pl
2
r30 = −
pl
2
3.23 Worked Example 23
Calculation of Unknowns
⎧
⎨ ϕ11 X 1 + ϕ12 δ2 + ϕ13 δ3 + ϕ10 = 0
r X + r22 δ2 + r23 δ3 + r20 = 0
⎩ 21 1
r31 X 1 + r32 δ2 + r33 δ3 + r30 = 0
⎡ 3 ⎤
⎡2 l
⎤
2 1 ⎤ ⎡
pl
−
X
1
3 EI
l l
24E I ⎥
⎣ 2 0 0 ⎦ · ⎣ δ2 ⎦ = −⎢
⎦
⎣ − pl
l
2
1
2E I
δ3
− l 0 l3
− pl
2
⎧
⎪
X 1 = 0.25 pl2
⎪
⎪
⎪
⎪
⎨
pl4
⇒ δ2 = 0.2917 E I
⎪
⎪
⎪
⎪
pl4
⎪
⎩ δ3 = 0.375
EI
Summary of Bending Moments
Summary of Shear Forces
293
294
Moment, Shear and Axial Force Diagrams
Qualitative Deformed Shape
3 Solution of Complex Frames
3.24 Worked Example 24
295
3.24 Worked Example 24
• E I = constant
• EA → ∞
• k = 3El I
Solution
The structure results to be a fixed-joint frame. The displacement method is used to
solve the structure by adopting as kinematic unknowns the rotations ϕ1 and ϕ2 of
node D.
Displacement Method
296
3 Solution of Complex Frames
m 11 ϕ1 + m 12 ϕ2 + m 10 = 0
m 21 ϕ1 + m 22 ϕ2 + m 20 = 0
Kinematics
• ϕ1 = 1
m 11 =
6E I
3E I
+k =
l
l
m 21 = −k = −
3E I
l
m 12 = −k = −
3E I
l
• ϕ2 = 1
m 22 =
3E I
9E I
3E I
+
+k =
l
l
l
3.24 Worked Example 24
297
• External load
m 10 =
pl2
8
m 20 = 0
Calculation of Unknowns
m 11 ϕ1 + m 12 ϕ2 + m 10 = 0
m 21 ϕ1 + m 22 ϕ2 + m 20 = 0
2
pl
E I 6 −3
ϕ1
=− 8
·
ϕ2
l −3 9
0
⎧
3
⎪
⎪ ϕ1 = −0.025 pl
⎨
EI
⇒
3
⎪
pl
⎪
⎩ ϕ2 = −0.008
EI
Summary of Bending Moments
298
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.25 Worked Example 25
Qualitative Deformed Shape
3.25 Worked Example 25
299
300
3 Solution of Complex Frames
• E I = constant
• EA → ∞
• k = 2El I
pl2
• αΔ T
= 10E
τ
I
• l = 5τ
Solution
The structure results to be a sway frame. In particular, the vertical displacement of
node D is due to the presence of the slider, while nodes B and C are fixed. Therefore,
the frame is solved by the displacement method by adopting as kinematic unknowns
the rotations ϕ1 and ϕ2 of node C.
Displacement Method
m 11 ϕ1 + m 12 ϕ2 + m 10 = 0
m 21 ϕ1 + m 22 ϕ2 + m 20 = 0
3.25 Worked Example 25
301
Kinematics
• ϕ1 = 1
m 11 =
3E I
9E I
4E I
+
+k =
l
l
l
m 21 = −k = −
• ϕ2 = 1
2E I
l
302
3 Solution of Complex Frames
m 12 = −k = −
m 22 =
2E I
l
3E I
EI
+k =
l
l
• External load, constant and linear thermal variation
3.25 Worked Example 25
m 10 = −
303
3E I 5 pl2
·
l2 2 10E I
m 20 = −
Calculation of Unknowns
l
3
l = − pl2
5
20
pl2
3
m 11 ϕ1 + m 12 ϕ2 + m 10 = 0
m 21 ϕ1 + m 22 ϕ2 + m 20 = 0
3
E I 9 −2
− 20
pl2
ϕ1
=−
·
2
ϕ2
l −2 3
− pl3
⎧
pl3
⎪
⎪
⎨ ϕ1 = 0.0486
EI
⇒
3
⎪
pl
⎪
⎩ ϕ2 = 0.1435
EI
304
Summary of Bending Moments
Summary of Shear Forces
3 Solution of Complex Frames
3.25 Worked Example 25
Moment, Shear and Axial Force Diagrams
305
306
Qualitative Deformed Shape
3 Solution of Complex Frames
3.26 Worked Example 26
307
3.26 Worked Example 26
• E I = constant
• E A ABC D E → ∞
• E AC F = 5 El2I
1 pl4
• δ = 100
EI
Solution
The frame is studied by exploiting symmetry. The structure results to be a sway frame.
The displacement method is used to solve the structure by adopting as kinematic
unknowns the rotation ϕ1 of node B and the vertical translation δ2 of node C. The
rod CF, with finite axial deformability, can be idealized as a translational spring of
stiffness k:
k=
5 EI
1 E AC F
=
2 l
2 l3
308
3 Solution of Complex Frames
Displacement Method
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
Kinematics
• ϕ1 = 1
m 11 =
4E I
8E I
4E I
+
=
l
l
l
r21 = −
6E I
l2
3.26 Worked Example 26
309
• δ2 = 1
m 12 = −
r22 =
6E I
l2
12E I
12E I
5 EI
EI
+k =
+
= 14.5 3
3
3
3
l
l
2 l
l
• External load
m '10 = −
pl2
pl2
+
12
16
'
r20
=−
pl
2
310
3 Solution of Complex Frames
• Imposed displacement
m ''10 = 0
''
r20
= −k · δ = −
5 pl
5 EI
1 pl4
=−
·
3
2 l
100 E I
200
In the adopted scheme, the forces on the spring are positive in the case of spring
shortening (see case δ2 = 1). For the case under study, the roller has no translation
(δ2 = 0) and the constraint at the base of the spring undergoes a lowering causing
''
the spring to elongate, with which a force r20 < 0 is then associated.
Due to the superposition of effects, it is possible to calculate m 10 and r20 .
m 10 = m '10 + m ''10 = −
'
''
r20 = r20
+ r20
=−
Calculation of Unknowns
pl2
pl2
pl2
+
=−
12
16
48
21 pl
pl 5 pl
−
=−
2
200
40
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
2
E I 8 − 6l
ϕ1
− pl
48
·
=−
δ2
l − 6l 14.5
− 2140pl
l2
⎧
pl3
⎪
⎪
⎨ ϕ1 = 0.043151
EI
⇒
4
⎪
pl
⎪
⎩ δ2 = 0.0541
EI
3.26 Worked Example 26
Summary of Bending Moments
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
311
312
Qualitative Deformed Shape
3 Solution of Complex Frames
3.27 Worked Example 27
313
3.27 Worked Example 27
• E I = constant
• EA → ∞
• k = E5lI
pl2
• αΔ T
= 10E
h
I
Solution
The frame is studied by exploiting symmetry. The structure results to be a fixedjoint frame. The displacement method is used to solve the structure by adopting as
kinematic unknowns the rotation ϕ1 of node B and the rotation ϕ2 of node D.
314
3 Solution of Complex Frames
At point C, in order to consider symmetry, it would be necessary to introduce a
slider with a sliding plane parallel to the symmetry axis. In that case, a kinematic
analysis of the simplified structure shows that it results a fixed-joint frame, and
consequently from the kinematic point of view point C can neither rotate nor translate
and could be assimilated to a fixed end. Despite what has been said from the kinematic
point of view, the slider is retained since from the static point of view no forces parallel
to the axis of symmetry arise in that point.
Displacement Method
m 11 ϕ1 + m 12 ϕ2 + m 10 = 0
m 21 ϕ1 + m 22 ϕ2 + m 20 = 0
3.27 Worked Example 27
315
Kinematics
• ϕ1 = 1
m 11 =
4E I
4 · 2 EI
EI
4E I
+
+ √
= 11.58
l
l
l
5 l
m 21 =
2E I
l
Due to what was previously discussed, although there is a slider at point C, since
point C—based on the kinematic analysis—cannot translate, the stiffness coefficients
related to the fixed end—fixed end conditions are adopted for the rod BC.
• ϕ2 = 1
m 12 =
m 22 =
2E I
l
4E I
EI
21 E I
4E I
+k =
+
=
l
l
5l
5 l
316
3 Solution of Complex Frames
• External load
m 10 = −
2E I αΔ T
2 pl2
17
1 2
1 2
−
pl = −
−
pl = − pl2
h
12
10
12
60
m 20 = 0
Calculation of Unknowns
m 11 ϕ1 + m 12 ϕ2 + m 10 = 0
m 21 ϕ1 + m 22 ϕ2 + m 20 = 0
17 E I 8 + √85 2
− 60
ϕ1
=
−
pl2
·
21
ϕ
0
l
2
2
5
⎧
pl3
⎪
⎪
⎨ ϕ1 = 2.67 · 10−2
EI
⇒
3
⎪
⎪
⎩ ϕ2 = −1.27 · 10−2 pl
EI
3.27 Worked Example 27
Summary of Bending Moments
Summary of Shear Forces
317
318
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.28 Worked Example 28
319
Qualitative Deformed Shape
3.28 Worked Example 28
• E I = constant
• EA → ∞
Solution
Rods BEC represent a statically determined appendage (three-hinged arch), and
therefore only the actions that the rods BEC transmit to the rods ABCD at nodes B
and C can be carried over to the frame ABCD.
320
3 Solution of Complex Frames
The structure results to be a sway frame and it can be solved with the displacement
method.
The structure is symmetric and anti-symmetrically loaded.
Taking advantage of the anti-symmetry, the structure can be divided in half, and
a roller can be inserted at midspan of rod BC. At the roller location the moment
is zero and a shear force is present. The rotation ϕ1 of node B and the horizontal
displacement δ2 of joint B are adopted as kinematic unknowns.
3.28 Worked Example 28
321
Displacement Method
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
Kinematics
• ϕ1 = 1
m 11 =
4E I
7E I
3E I
+
=
l
l
l
322
3 Solution of Complex Frames
r21 = −
6E I
l2
• δ2 = 1
m 12 = −
r22 =
6E I
l2
12E I
l3
• External load
m 10 =
r20 = −
pl2
12
pl
pl
−
= − pl
2
2
3.28 Worked Example 28
323
Calculation of Unknowns
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
2 pl
E I 7 − 6l
ϕ1
·
= − 12
6 12
δ2
l − l l2
− pl
⎧
3
⎪
⎪ ϕ1 = 0.104 pl
⎨
EI
⇒
4
⎪
pl
⎪
⎩ δ2 = 0.135
EI
Summary of Bending Moments
Summary of Shear Forces
324
Moment, Shear and Axial Force Diagrams
3 Solution of Complex Frames
3.29 Worked Example 29
325
Qualitative Deformed Shape
3.29 Worked Example 29
•
•
•
•
E I AB → ∞
E I D BC = E I
E A ABC → ∞
1 EI
E A B D = 10
l2
• δ=
1 pl4
200 E I
Solution
The infinitely rigid rod AB is a statically determined appendage, so it can be simplified
with a roller with a horizontal axis applied at node B and a concentrated vertical load
also applied at node B.
326
3 Solution of Complex Frames
To consider the finite stiffness of the rod BD, a spring with stiffness k = E AlB D =
equal to the axial stiffness of the rod is introduced. The structure is then
simplified as follows:
1 EI
10 l3
The structure is solved with the displacement method by adopting the rotation ϕ1
of node B and the vertical displacement δ2 of node D as kinematic unknowns.
Displacement Method
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
3.29 Worked Example 29
327
Kinematics
• ϕ1 = 1
m 11 =
3E I
6E I
3E I
+
=
l
l
l
r21 =
3E I
l2
m 12 =
3E I
l2
• δ2 = 1
r22 =
3E I
31 E I
+k =
l3
10 l3
328
3 Solution of Complex Frames
• External load and imposed displacement
m 10 = −
r20 = −
3E I
3
pl2
pl2
7
− 2 δ=−
−
pl2 = − pl2
8
l
8
200
50
3E I
3
41
13
13
pl − 3 δ = − pl −
pl = − pl
8
l
8
200
25
Calculation of Unknowns
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
7 2
E I 6 3l
ϕ1
− 50 pl
·
=
−
31
δ2
− 41
pl
l 3l 10l
2
25
3.29 Worked Example 29
329
⇒
3
ϕ1 = −0.467 pl
EI
4
δ2 = 0.981 pl
EI
Summary of Bending Moments
Summary of Shear Forces
Moment, Shear and Axial Force Diagrams
330
Qualitative Deformed Shape
3 Solution of Complex Frames
3.30 Worked Example 30
331
3.30 Worked Example 30
• E I = constant
• EA → ∞
Solution
The structure results to be a sway frame. The displacement method is used to solve
the structure by adopting as kinematic unknowns the rotation ϕ1 of node B and the
horizontal translation δ2 of node C.
332
3 Solution of Complex Frames
Displacement Method
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
Kinematics
• ϕ1 = 1
m 11 =
3E I
7E I
4E I
+
=
l
l
l
r21 =
6E I
l2
3.30 Worked Example 30
333
• δ2 = 1
m 12 =
r22 =
6E I
l2
12E I
l3
• External load
m 10 =
pl2
8
m 20 = − pl
334
3 Solution of Complex Frames
Calculation of Unknowns
m 11 ϕ1 + m 12 δ2 + m 10 = 0
r21 ϕ1 + r22 δ2 + r20 = 0
1 2
E I 7 6l
pl
ϕ1
=− 8
6 12 ·
δ2
− pl
l l l2
⎧
3
⎪
⎪ ϕ1 = −0.1563 pl
⎨
EI
⇒
4
⎪
pl
⎪
⎩ δ2 = 0.1615
EI
Summary of Bending Moments
Summary of Shear Forces
3.30 Worked Example 30
Moment, Shear and Axial Force Diagrams
335
336
Qualitative Deformed Shape
3 Solution of Complex Frames
Appendix A
Table A.1 provides the influence coefficients related to the force method. Rotations
and displacements are referred to the extremities of the rods. Coefficients provided
in Table A.1 are associated to unitary forces and moments.
Table A.1 Influence coefficients for the force method
Case ID
Rotation/displacement
values
1
ϕ11 =
ϕ21 =
2
3
4
l
3E I
l
6E I
ϕ22 =
l
6E I
l
3E I
ϕ11 =
l
EI
δ11 =
l2
2E I
ϕ12 =
l
4E I
m 21 = 21
ϕ11 =
(continued)
© The Editor(s) (if applicable) and The Author(s), under exclusive license
to Springer Nature Switzerland AG 2023
I. G. Colombo et al., Structural Analysis of Plane Frames, Springer Tracts in Civil
Engineering, https://doi.org/10.1007/978-3-031-35267-6
337
338
Appendix A
Table A.1 (continued)
Case ID
Rotation/displacement
values
5
ϕ11 =
l2
2E I
δ11 =
l3
3E I
Pl2
16E I
6
ϕ10 = ϕ20 =
7
ϕ10 =
(
)
Pb l2 −b2
6lE I
ϕ20 =
Pab(2l−b)
6lE I
8
ϕ10 = ϕ20 =
9
ϕ10 =
pl
48E I
m 20 =
pl
8
10
11
pl3
24E I
2
2
ϕ10 =
7 pl3
360E I
ϕ20 =
pl3
45E I
ϕ10 = ϕ20 =
Ml
24E I
(continued)
Appendix A
339
Table A.1 (continued)
Case ID
Rotation/displacement
values
12
ϕ10 =
pl3
24E I
δ10 =
pl4
30E I
ϕ10 =
pl3
6E I
δ10 =
pl4
8E I
13
14
ϕ10 = ϕ20 =
15
ϕ10 =
16
3δ
2l
I
m 20 = 3E
l2
δ
l
δ
ϕ10 = ϕ20 = αlΔT
t
(α = thermal expansion
coefficient)
(t = beam height)
Table A.2 provides the influence coefficients related to the displacement method.
Forces and moments are referred to the extremities of the rods (reaction forces and
moments on constraints). Coefficients provided in Table A.2 are associated to unitary
displacements and rotations.
340
Appendix A
Table A.2 Influence coefficients for the displacement method
CASE ID
Force/moment values
1
m 11 =
m 21 =
4E I
l
2E I
l
r11 = r21 =
2
m 11 =
6E I
l2
3E I
l
m 21 = 0
r11 = r21 =
3
m 11 = m 21 =
r11 = r21 =
4
3E I
l2
m 11 =
6E I
l2
12E I
l3
3E I
l2
m 21 = 0
r11 = r21 =
5
3E I
l3
m 11 = m 21 =
r11 = r21 = 0
EI
l
(continued)
Appendix A
341
Table A.2 (continued)
CASE ID
Force/moment values
6
m 11 =
12μ(E I )2 +4E I l
l(l+4μE I )
m 21 =
6μ(E I )2 +2E I l
(l+4μE I )(l+3μE I )
I ) +2lE I
r11 = r21 = 3 6μ(E
l2 (l+4μE I )
2
l+2μE I
l+3μE I
(μ = rotational spring stiffness)
7
m 11 =
6lE I +12μ(E I )2
l2 (l+4μE I )
m 21 =
6E I
l(l+4μE I )
I l+μE I
r11 = r21 = 12E
l3 l+4μE I
(μ = rotational spring stiffness)
8
m 10 = m 20 =
r10 = r20 =
9
m 10 =
r10 =
r20 =
10
P
2
3Pl
16
11P
16
5P
16
m 10 = m 20 =
r10 = r20 =
11
m 10 =
Pl
8
pl2
12
pl
2
pl2
8
m 20 = 0
r10 =
r20 =
5 pl
8
3 pl
8
(continued)
342
Appendix A
Table A.2 (continued)
CASE ID
Force/moment values
12
m 10 = m 20 =
r10 = r20 =
13
m 10 =
M
8
r10 = r20 =
14
M
4
3M
2l
9M
8l
m 10 = m 20 =
2E I αΔT
t
n 10 = n 20 = αΔT E A
(A = beam cross-section)
(t = beam height)
15
m 11 = m 21 =
r10 = 0
16
m 10 =
pl2
3
m 20 =
pl2
6
EI
l
r10 = pl
17
3E I αΔT
t
r20 = 3E ItlαΔT
m 10 =
r10 =
Appendix B
Table B.1 summarizes, for each complex frame example (Chap. 3), the simple
reference cases covered in Chap. 2 present in it.
Table B.1 Simple reference cases covered in each complex frame example
Simple reference cases
Complex frame ID
A
1
•
2
•
3
•
4
B
C
D
E
•
•
•
•
•
•
5
•
6
•
9
•
10
•
•
•
8
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
16
•
•
L
•
15
•
K
•
•
•
18
J
•
•
17
I
•
12
14
H
•
•
11
13
G
•
•
7
F
•
•
•
•
•
•
•
•
19
•
20
•
•
•
•
© The Editor(s) (if applicable) and The Author(s), under exclusive license
to Springer Nature Switzerland AG 2023
I. G. Colombo et al., Structural Analysis of Plane Frames, Springer Tracts in Civil
Engineering, https://doi.org/10.1007/978-3-031-35267-6
(continued)
343
344
Appendix B
Table B.1 (continued)
Simple reference cases
Complex frame ID
A
21
22
B
•
•
24
E
F
•
•
G
H
I
J
K
L
•
•
•
•
•
•
•
25
•
26
•
•
•
28
•
•
•
•
•
•
•
•
•
29
30
D
•
•
23
27
C
•
•
•
•
•
•
•