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Okonkwo V. O.1, Aginam C. H.
2, and Chidolue C. A.
3
123Department of Civil Engineering Nnamdi Azikiwe University, Awka Anambra State-Nigeria
ABSTRACT: In this work the stiffness equations for
evaluating the internal stress of rigidly fixed portal frames
by the displacement method were generated. But obtaining
the equations for the internal stresses required non-scalar or parametric inversion of the structure stiffness matrix. To
circumvent this problem, the flexibility method was used
taking advantage of the symmetrical nature of the portal
frame and the method of virtual work. These were used to
obtain the internal stresses on rigidly fixed portal frames
for different cases of external loads when axial deformation
is considered. A dimensionless constant s was used to
capture the effect of axial deformation in the equations.
When it is set to zero, the effect of axial deformation is
ignored and the equations become the same as what can be
obtained in any structural engineering textbook. These
equations were used to investigate the contribution of axial deformation to the calculated internal stresses and how
they vary with the ratio of the flexural rigidity of the beam
and columns and the height to length ratio of the loaded
portal frames.
Keywords: Axial deformation, flexural rigidity, flexibility
method, Portal frames, stiffness matrix,
I. INTRODUCTION Structural frames are primarily responsible for
strength and rigidity of buildings. For simpler single storey
structures like warehouses, garages etc portal frames are
usually adequate. It is estimated that about 50% of the hot-
rolled constructional steel used in the UK is fabricated into
single-storey buildings (Graham and Alan, 2007). This shows the increasing importance of this fundamental
structural assemblage. The analysis of portal frames are
usually done with predetermined equations obtained from
structural engineering textbooks or design manuals. The
equations in these texts were derived with an underlying
assumption that deformation of structures due to axial
forces is negligible giving rise to the need to undertake this
study. The twenty first century has seen an astronomical use
of computers in the analysis of structures (Samuelsson and
Zienkiewi, 2006) but this has not completely eliminated the
use of manual calculations form simple structures and for
easy cross-checking of computer output (Hibbeler, 2006). Hence the need for the development of equations that
capture the contribution of axial deformation in portal
frames for different loading conditions.
II. DEVELOPMENT OF THE MODEL
Figure 1: A simple portal frame showing its dimensions and
the 6 degrees of freedom
The analysis of portal frames by the stiffness method
requires the determination of the structureβs degrees of
freedom and the development of the structureβs stiffness
matrix. For the structure shown in Figure 1a, the degrees of
freedom are as shown in Figure 1b. I1 and A1 are
respectively the second moment of inertia and cross-
sectional area of the columns while I2 and A2 are the second
moment of inertia and cross-sectional area of the beam
respectively. The stiffness coefficients for the various
degrees of freedom considering shear deformation can be obtained from equations developed in Ghali et al (1985)
and Okonkwo (2012) and are presented below
The structureβs stiffness matrix can be written as:
πΎ =
π11 π12 π13
π21 π22 π23
π31 π32 π33
π14 π15 π16
π24 π25 π26
π34 π35 π36
π41 π42 π43
π51 π52 π53
π61 π62 π63
π44 π45 π46
π54 π55 π56
π64 π65 π66
. (1)
Where kij is the force in coordinate (degree of freedom) i
when there is a unit displacement in coordinate (degree of
freedom) j. They are as follows:
π11 =πΈπ΄1
β
π12 = 0
Contribution of Axial Deformation in the Analysis of Rigidly
Fixed Portal Frames
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π13 =6πΈπΌ2
π2
π14 = β12πΈπΌ1
β3
π15 = 0
π16 =6πΈπΌ2
π2
π22 =12πΈπΌ1
β3 +πΈπ΄2
π
π23 =6πΈπΌ1
β2
π24 = 0
π25 = βπΈπ΄2
π . . (2)
π26 = 0
π33 =4πΈπΌ2
π+
4πΈπΌ1
β
π34 = β6πΈπΌ2
π2
π35 = 0
π36 =2πΈπΌ2
π
π44 =12πΈπΌ2
π3 +πΈπ΄1
β
π45 = 0
π46 = β6πΈπΌ2
π2
π55 =12πΈπΌ1
β3 +πΈπ΄2
π
From Maxwellβs Reciprocal theorem and Bettiβs Law kij =
kji ( Leet and Uang, 2002).
When there are external loads on the structure on the
structure there is need to calculate the forces in the
restrained structure Fo as a result of the external load.
The structureβs equilibrium equations are then written as
{πΉ} = {πΉπ} + πΎ {π·} . . . . (3)
{πΉπ} =
π10
π20
π30
π40
π50
π60
. . . . . (4)
Where kio is the force due to external load in coordinate i
when the other degrees of freedom are restrained.
{π·} =
π1
π2
π3
π4
π5
π6
. . . . . (5)
Where di is the displacement in coordinate i.
{πΉ} =
πΉ1
πΉ2
πΉ3
πΉ4
πΉ5
πΉ6
. . . . . . (6)
Where Fi is the external load with a direction coinciding
with the coordinate i (McGuire et al, 2000).
By making {D} the subject of the formula in equation (3)
π· = πΎ β1{πΉ β πΉπ} . . . . (7)
Once {D} is obtained the internal stresses in the frame can
be easily obtained by writing the structureβs compatibility
equations given as
π = ππ + π1π1 + π2π2 + π3π3 . . (8)
Where M is the internal stress (bending moment) at any
point on the frame, Mr is the internal stress at the point
under consideration in the restrained structure while Mi is
the internal stress at the point when there is a unit
displacement in coordinate i.
To solve equation (8) there is need to obtain {D}. {D} can
be obtained from the inversion of [K] in equation (7).
Finding the inverse of K parametrically (i.e. without
substituting the numerical values of E, h, l etc) is a difficult
task. This problem is circumvented by using the flexibility
method to solve the same problem, taking advantage of the symmetrical nature of the structure and the principle of
virtual work.
III. APPLICATION OF THE FLEXIBILITY
MODEL The basic system or primary structure for the structure in
Figure 1a is given in Figure 2. The removed redundant
force is depicted with X1.
Figure 2: The Basic System showing the removed
redundant forces
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The flexibility matrix of the structure can be determined
using the principle of virtual work.
By applying the unit load theorem the deflection in beams
or frames can be determined for the combined action of the internal stresses, bending moment and axial forces with
π· = π π
πΈπΌππ +
π π
πΈπ΄ππ . . . (9)
Where π and π are the virtual internal stresses while M and N are the real/actual internal stresses.
E is the modulus of elasticity of the structural material
A is the cross-sectional area of the element (McGuire et al,
2000; Nash, 1998)
If dij is the deformation in the direction of i due to a unit
load at j then by evaluating equation (9) the following are
obtained:
π11 =πΌ1πΊπ΄1 π3+6πΌ2π΄1π2β+12πΌ1πΌ2β
12πΈπΌ1πΌ2π΄1 . . (10a)
π12 = 0 . . . . . (10b)
π13 = 0 . . . . . . (10c)
π22 =2π΄2β3+3πΌ1 π
3πΈπΌ1π΄2 . . . (10d)
π23 =ββ2
πΈπΌ1 . . . (10e)
π33 =ππΌ1 +2βπΌ2
πΈπΌ1πΌ2 . . . . (10f)
From Maxwellβs Reciprocal theorem and Bettiβs Law dij =
dji.
The structureβs compatibility equations can be written thus
π11 π11 π11
π21 π21 π21
π31 π31 π31
π1
π2
π3
+ π10
π20
π30
= 0 . (11a)
i.e. ππΉ + ππ = 0 . . . (11b)
Where F is the vector of redundant forces X1, X2, X3 and do
is the vector deformation d10, d20, d30 due to external load
on the basic system (Jenkins, 1990).
πΉ = πβ1(βππ) . . . . (12)
πβ1 =π΄ππ π
πππ‘ π . . (13)
(Stroud, 1995)
πβ1 =
1
π110 0
0π33
π22π33βπ232
βπ32
π22π33βπ232
0βπ23
π22π33βπ232
π22
π22π33βπ232
. (14)
From equations 10a β 10f
1
π11=
12πΈπΌ1πΌ2π΄1
πΌ1π΄1π3 +6πΌ2π΄1π2β+24πΌ1πΌ2β . . . (15a)
π33
π22π33βπ232 =
3πΈπΌ1π΄2(ππΌ1 +2βπΌ2 )
2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2+6βππΌ1πΌ2
. (15b)
π32
π22π33βπ232 =
β3πΈπΌ1πΌ2πΊπ΄2β2
2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2+6βππΌ1πΌ2
(15c)
π22
π22π33βπ232 =
πΈπΌ1πΌ2 2π΄2β3+3πΌ1 π
2π΄2β3ππΌ1 +3πΌ12π2 +π΄2β4πΌ2 +6βππΌ1πΌ2
. (15d)
Equation (12) is evaluated to get the redundant forces and
these are substituted into the structures force equilibrium
(superposition) equations to obtain the internal stresses at
any point.
π = ππ + π1π1 + π2π2 + π3π3 . . (16)
Where M is the required stress at a point, Mo is the stress at
that point for the reduced structure, Mi is stress at that point
when the only the redundant force Xi =1 acts on the reduced
structure.
For the loaded portal frame of Figure 3, the deformations of
the reduced structure due to external loads are
π10 = 0 . . . . .
. (17a)
π20 =π€β2π2
8πΈπΌ1 . . . . .
. (17b)
π30 =βπ€π2(ππΌ1 +6βπΌ2 )
24πΈπΌ1πΌ2 . . (17c)
By substituting the values of equations (17a) β (17c) into
equations (12)
π1 = 0 . . . . . (18a)
π2 =βπ΄2πΌ1π€β2π3
4 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. (18b)
π3 =π€π2 2π΄2β3ππΌ1 +3πΌ1
2π2+3π΄2β4πΌ2 +18βππΌ1πΌ2
24 2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2+6βππΌ1πΌ2
. (18c)
Evaluating equation (16) for different points on the
structure using the force factors obtained in equations (18a)
β (18c)
ππ΄ =2π€π3πΌ1 β3π΄2β3ππΌ1
12 2π΄2β3ππΌ1 +3πΌ12π2 +π΄2β4πΌ2 +6βππΌ1πΌ2
. . (19)
ππ΅ =β2π€π3πΌ1 2β3π΄2+3ππΌ1
24 2π΄2β3ππΌ1 +3πΌ12π2 +π΄2β4πΌ2 +6βππΌ1πΌ2
. . (20)
ππΆ =β2π€π3πΌ1 2β3π΄2+3ππΌ1
24 2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. (21)
ππΆ =βπ€π3πΌ1 8πΌ2β3+π 2π3πΌ1
12 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
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ππ· =2π€π3πΌ1 β3π΄2β3ππΌ1
12 2π΄2β3ππΌ1 +3πΌ12π2 +π΄2β4πΌ2 +6βππΌ1πΌ2
. . (22)
For the loaded portal frame of Figure 4, the deformations of
the reduced structure due to external loads are
π10 =βπ€π π3πΌ1π΄1+8βπ2πΌ2π΄1 +64βπΌ1πΌ2
128πΈπΌ1πΌ2π΄1 . (23a)
π20 =π€β2π2
16πΈπΌ1 . . . (23b)
π30 =βπ€π2(ππΌ1 +6βπΌ2 )
48πΈπΌ1πΌ2 . . (23c)
By substituting the values of equations (23a) β (23c) into
equations (12)
π1 =3π€π π3πΌ1π΄1 +8βπ2πΌ2π΄1+64βπΌ1πΌ2
32 πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 . . (24a)
π2 =β2π€β2π3π΄2πΌ1
16 2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2+6βππΌ1πΌ2
. (24b)
π3 =π€β3π2π΄2 3βπΌ2 +2ππΌ1 +3π€π3πΌ1 ππΌ1 +6βπΌ2
48 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. (24c)
Let π½ =π 2
π 1 . . . . . . (25)
Evaluating equation (16) for different points on the
structure using the force factors obtained in equations (24a)
β (24c)
ππ΄ =π€π2 β3π΄2 3βπΌ2+8ππΌ1 +3ππΌ1 ππΌ1 +6βπΌ2
48 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
βπ€π4π΄1 5ππΌ1 +24βπΌ2
64 πΌ1π΄1 π3+6πΌ2π΄1π2β+24βπΌ1πΌ2
. . . . (26)
ππ΅ =
βπ€π4π΄1 5ππΌ1 +24βπΌ2
64 πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 +
π€π2 β3π΄2 3βπΌ2+2ππΌ1 +3ππΌ1 ππΌ1 +6βπΌ2
48 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. . . (27)
ππΆ =
β3π€π2 π3πΌ1π΄1 +8βπ2πΌ2π΄1+64βπΌ1πΌ2
64 πΌ1π΄1 π3+6πΌ2π΄1 π2β+24βπΌ1πΌ2 +
π€π2 β3π΄2 3βπΌ2+2ππΌ1 +3ππΌ1 ππΌ1 +6βπΌ2
48 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. . . (28)
ππ· =
β3π€π2 π3πΌ1π΄1 +8βπ2πΌ2π΄1+64βπΌ1πΌ2
64 πΌ1π΄1 π3+6πΌ2π΄1 π2β+24βπΌ1πΌ2 +
π€π2 β3π΄2 3βπΌ2+8ππΌ1 +3ππΌ1 ππΌ1 +6βπΌ2
48 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. . . (29)
For the loaded portal frame of Figure 5, the deformations of
the reduced structure due to external loads are
π10 = βπ€π β3
12πΈπΌ1 . . . . (30a)
π20 =π€β4
8πΈπΌ1 . . . . (30b)
π30 =βπ€β3
6πΈπΌ1 . . . . . (30c)
By substituting the values of equations (30a) β (30c) into
equations (12)
π1 =π€β3ππΌ2π΄1
πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 . . . (31a)
π2 =βπ€β4π΄2 3ππΌ1 +2βπΌ2
8 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. (31b)
π3 =π€β3πΌ2 ββ3π΄2+12ππΌ1
24 2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2+6βππΌ1πΌ2
. . (31c)
Evaluating equation (16) for different points on the
structure using the force factors obtained in equations (31a)
β (31c)
ππ΄ =
βπ€β2 πΌ1π΄1π3+5π΄1π2βπΌ2 +24βπΈπΌ1πΌ2
2 πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 +
π€β3 9β2ππΌ1π΄2+5π΄2β3πΌ2 +12ππΌ1πΌ2
24 2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. . . . (32)
ππ΅ =π€β3π2πΌ2π΄1
2 πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 +
π€β3πΌ2 ββ3π΄2 +12ππΌ1
24 2π΄2β3ππΌ1 +3πΌ12π2 +π΄2β4πΌ2 +6βππΌ1πΌ2
.
. . (33)
ππΆ =
βπ€β3π2πΌ2π΄1
2 πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 +
π€β3πΌ2 ββ3π΄2 +12ππΌ1
24 2π΄2β3ππΌ1 +3πΌ12π2 +π΄2β4πΌ2 +6βππΌ1πΌ2
. . . . (34)
ππ· =
βπ€β3π2πΌ2π΄1
2 πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 +
π€β3 9β2ππΌ1π΄2+5π΄2β3πΌ2+12ππΌ1πΌ2
24 2π΄2β3ππΌ1 +3πΌ12π2 +π΄2β4πΌ2 +6βππΌ1πΌ2
. . . . (35)
For the loaded portal frame of Figure 6, the deformations of
the reduced structure due to external loads are
π10 = βππ2π
2πΈπΌ1 . . . (36a)
π20 = 0 . . . . (36b)
π30 = 0 . . . (36c)
By substituting the values of equations (36a) β (36c) into
equations (12)
π1 =6ππ2ππΌ2π΄1
π΄1πΌ1 π3 +6π΄1βπ2πΌ2 +24βπΌ1πΌ2 . . . (37a)
π2 = 0 . . . . (37b)
π3 = 0 . . . . . . (37c)
Evaluating equation (16) for different points on the
structure using the force factors obtained in equations (37a) β (37c)
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ππ΄ = βππ 1 β3ππ2πΌ2π΄1
π΄1πΌ1 π3 +6π΄1βπ2πΌ2 +24βπΌ1πΌ2 . (38)
ππ΅ =3ππ2π2πΌ2π΄1
π΄1πΌ1 π3+6π΄1βπ2πΌ2 +24βπΌ1πΌ2 . . (39)
ππΆ = β3ππ2π2πΌ2π΄1
π΄1πΌ1 π3+6π΄1βπ2πΌ2 +24βπΌ1πΌ2 . . (40)
ππ· = ππ 1 β3ππ2πΌ2π΄1
π΄1πΌ1 π3+6π΄1βπ2πΌ2 +24βπΌ1πΌ2 . . (41)
For the loaded portal frame of Figure 7, the deformations of the reduced structure due to external loads are
π10 = βππ2π
4πΈπΌ1 . . . (42a)
π20 =ππ2 3ββπ
6πΈπΌ1 . . (42b)
π30 = βππ2
2πΈπΌ1 . . . . (42c)
By substituting the values of equations (42a) β (42c) into
equations (12)
π1 =3ππ2ππΌ2π΄1
πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 . . . (43a)
π2 =βππ2π΄2 ππΌ1 3ββπ +βπΌ2 3ββ2π
2 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. (43b)
π3 =βππ2πΌ2 ββ3π΄2 +πβ2π΄2+3ππΌ1
2 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. (43c)
Evaluating equation (16) for different points on the
structure using the force factors obtained in equations (43a)
β (43c)
ππ΄ =
βππ +3ππ2π2πΌ2π΄1
2 πΌ1π΄1π3+6πΌ2π΄1 π2β+24βπΌ1πΌ2 +
ππ2 π΄2πΌ1 π 3β2βπβ +πΌ2 2π΄2β3βππ΄2β2+3πΌ1 π
2 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. . (44)
ππ΅ =3ππ2π2πΌ2π΄1
2 πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 +
ππ2πΌ2 ββ3π΄2 +πβ2π΄2+3ππΌ1
2 2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2 +6βππΌ1πΌ2
.
. . (45)
ππΆ =
β3ππ2π2πΌ2π΄1
2 πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 +
ππ2πΌ2 ββ3π΄2+πβ2π΄2+3ππΌ1
2 2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2+6βππΌ1πΌ2
. . . (46)
ππ· =
β3ππ2π2πΌ2π΄1
2 πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 +
ππ2 π΄2πΌ1 π 3β2βπβ +πΌ2 2π΄2β3βππ΄2β2+3πΌ1 π
2 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. . (47)
For the loaded portal frame of Figure 8, the deformations of
the reduced structure due to external loads are
π10 = βπ π2πΌ1π΄1 3πβ2π +6βπΌ2 πππ΄1 +2πΌ1
12πΈπΌ1πΌ2π΄1 . (48a)
π20 =ππβ2
2πΈπΌ1 . . . (48b)
π30 = βππ ππΌ1 +2βπΌ2
2πΈπΌ1πΌ2 . . . (48c)
By substituting the values of equations (48a) β (48c) into
equations (12)
π1 =π π2πΌ1π΄1 3πβ2π +6βπΌ2 ππ π΄1+2πΌ1
πΌ1π΄1π3 +6πΌ2π΄1π2β+24βπΌ1πΌ2 . (49a)
π2 =β3ππβ2πΌ1π΄2 πβπ
2 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
(49b)
π3 =ππ β3π΄2 2ππΌ1 +βπΌ2 +3ππΌ1 ππΌ1 +2βπΌ2
2 2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2 +6βππΌ1πΌ2
(49c)
Evaluating equation (16) for different points on the
structure using the force factors obtained in equations (49a)
β (49c)
ππ΄ = βππ +ππ π2πΌ1π΄1 3πβ2π +6βπΌ2 πππ΄1 +2πΌ1
2 πΌ1π΄1 π3+6πΌ2π΄1π2β+24βπΌ1πΌ2 +
ππ π΄2β3 βππΌ1 +βπΌ2 +3ππΌ1 +3ππΌ1 ππΌ1 +2βπΌ2
2 2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2+6βππΌ1πΌ2
. . (50)
ππ΅ = βππ +π π2πΌ1π΄1 3πβ2π +6βπΌ2 πππ΄1 +2πΌ1
2(πΌ1π΄1π3+6πΌ2π΄1 π2β+24βπΌ1πΌ2 )+
ππ β3π΄2 2ππΌ1 +βπΌ2 +3ππΌ1 ππΌ1 +2βπΌ2
2 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. . . (51)
ππΆ =
βππ π2πΌ1π΄1 3πβ2π +6βπΌ2 ππ π΄1+2πΌ1
2 πΌ1π΄1π3+6πΌ2π΄1 π2β+24βπΌ1πΌ2 +
ππ β3π΄2 2ππΌ1 +βπΌ2 +3ππΌ1 ππΌ1 +2βπΌ2
2 2π΄2β3ππΌ1 +3πΌ12π2+π΄2β4πΌ2 +6βππΌ1πΌ2
. . . . (52)
ππ· =
βππ π2πΌ1π΄1 3πβ2π +6βπΌ2 ππ π΄1+2πΌ1
2 πΌ1π΄1π3+6πΌ2π΄1 π2β+24βπΌ1πΌ2 +
ππ π΄2β3 βππΌ1 +βπΌ2 +3ππΌ1 +3ππΌ1 ππΌ1 +2βπΌ2
2 2π΄2β3ππΌ1 +3πΌ12 π2+π΄2β4πΌ2+6βππΌ1πΌ2
. . (53)
IV. DISCUSSION OF RESULTS The internal stress on the loaded frames is summarized in
table 1. The effect of axial deformation is captured by the
dimensionless constant s taken as the ratio of the end
translational stiffness to the shear stiffness of a member.
π 1 =12πΈπΌ1
β3 ββ
πΈπ΄1=
12πΌ1
β2π΄1 . . . (54)
π 2 =12πΈπΌ2
π3 βπ
πΈπ΄2=
12πΌ2
π2π΄2 . . (55)
When the axial deformation in the columns is ignored
π 1 = 0 and likewise when axial deformation in the beam is
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ignored π 2 = 0 . If axial deformation is ignored in the
whole structure, π 1 = π 2 = 0.
The internal stress equations enable an easy investigation into the contribution of axial deformation to the internal
stresses of statically loaded frames for different kinds of
external loads.
For frame 1 (figure 3), the moment at A, MA is given by
equation (19). The contribution of axial deformation in the
column βππ΄ is given by
βππ΄ = ππ΄(ππππ πππ’ππ‘πππ 19) β ππ΄(ππππ πππ’ππ‘πππ 19 π€βππ π 2=0)
=π€π3πΌ1
12
4β3πΌ2 +π 2π3πΌ1
4β3πΌ2 2ππΌ1 +βπΌ1 +π 2π3πΌ1 ππΌ1 +2βπΌ2 β
1
2ππΌ1 +βπΌ2 .
. . (56)
Equation (56) gives the contribution of axial deformation to MA as a function of h, l, I1, I2 and s
By considering the case of a portal frame of length l = 5m,
h = 4m. Equation 56 was evaluated to show how the
contribution of axial deformation varied with πΌ1 πΌ2 . The
result is shown in Table 2. When plotted on a uniform scale
(Figure 9) the relationship between βππ΅ and πΌ1 πΌ2 is seen
to be linear. This was further justified by a linear regression
analysis of the results in Table 2 which was fitted into the
model βππ΄ = π1πΌ1
πΌ2+ π2 to obtain π1 = β0.004815 and
π2 = 0.0001109 and the fitness parameters sum square of
errors (SSE), coefficient of multiple determination (R2 ) and
the root mean squared error (RMSE) gave 1.099 x 10-7,
0.999952 and 0.0001105 respectively. From Table 2 when
πΌ1 πΌ2 = 0 , βππ΄ β 0 and when πΌ1 πΌ2 = 10; βππ΄ =β0.0479 which represent only a 5% reduction in the
calculated bending moment.
In like manner by evaluating the axial contribution in the
beam, βππ΅ for varying πΌ1 πΌ2 of the portal frame, Table 3 was produced. This was plotted on a uniform scale in
Figure 10. From Figure 10 it would be observed that there
is also a linear relationship between βππ΅ and πΌ1 πΌ2 . When
fitted into the model βππ΅ = π3πΌ1
πΌ2+ π4 it gave π3 =
β0.006502 and π4 = β0.0003969 for the fitness
parameters sum square of errors (SSE), coefficient of
multiple determination (R2 ) and the root mean squared
error (RMSE) of 6.51 x 10-7, 0.9999 and 0.000269
respectively.
By pegging πΌ1 πΌ2 to a constant value of 0.296 and the
variation of βππ΄ with respect to β πΏ investigated, Table 4
was generated. A detailed plot of Table 4 was presented in
Figure 11. From Table 4 when β πΏ = 0; βππ΄ = β3.125
(about 30% drop in calculated bending moment value)
while at β πΏ > 0, βππ΄ dropped in magnitude exponentially
to values below 1. In like manner, when the variation of
βππ΅ with respect to β πΏ was investigated, Table 5 was
generated. A detailed plot of Table 5 was presented in
Figure 12. From Table 5 when β πΏ = 0; βππ΄ = β4.1667
(about 40% drop in calculated bending moment value)
while at β πΏ > 0, βππ΄ dropped in magnitude exponentially
to values below 1.
V. CONCLUSION The flexibility method was used to simplify the analysis
and a summary of the results are presented in table 1. The
equations in table 1 would enable an easy evaluation of the
internal stresses in loaded rigidly fixed portal frames
considering the effect of axial deformation. From a detailed analysis of frame 1 (Figure 3), it was
observed that the contribution of axial deformation is
generally very small and can be neglected for reasonable
values of πΌ1 πΌ2 . However, its contribution skyrockets at
very low values of β π i.e. as β π => 0. This depicts the
case of an encased single span beam and a complete
departure from portal frames under study. This analysis can
be extended to the other loaded frames (Figures 4 β 8) using
the equations in Table 1. These would enable the
determination of safe conditions for ignoring axial deformation under different kinds of loading.
REFERENCES
[1] Ghali A, Neville A. M. (1996) Structural Analysis: A Unified Classical and Matrix Approach (3rd Edition)
Chapman & Hall London
[2] Graham R, Alan P.(2007). Single Storey Buildings:
Steel Designerβs Manual Sixth Edition, Blackwell
Science Ltd, United Kingdom
[3] Hibbeler, R. C.(2006). Structural Analysis. Sixth Edition, Pearson Prentice Hall, New Jersey
[4] Leet, K. M., Uang, C.,(2002). Fundamental of
Structural Analysis. McGraw-Hill ,New York
[5] McGuire, W., Gallagher R. H., Ziemian, R.
D.(2000). Matrix Structural Analysis,Second
Edition, John Wiley & Sons, Inc. New York
[6] Nash, W.,(1998). Schaumβs Outline of Theory and
Problems of Strength of Materials. Fourth Edition,
McGraw-Hill Companies, New York
[7] Okonkwo V. O (2012). Computer-aided Analysis of
Multi-storey Steel Frames. M. Eng. Thesis,
Nnamdi Azikiwe University, Awka, Nigeria.
[8] Reynolds, C. E.,and Steedman J. C. (2001). Reinforced Concrete Designerβs Handbook, 10th
Edition) E&FN Spon, Taylor & Francis Group,
London
[9] Samuelsson A., and Zienkiewicz O. C.(2006),
Review: History of the Stiffness Method.
International Journal for Numerical Methods in Engineering. Vol. 67: 149 β 157
[10] Stroud K. A.,(1995). Engineering Mathematics.
Fourth Edition, Macmillan Press Ltd, London.
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Table 1: Internal stresses for a loaded rigid frame
S/No LOADED FRAME REMARKS
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ππ΄ = ππ· =π€π3πΌ1 4β3πΌ2βπ 2 π3πΌ1
12 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2 π3πΌ1 ππΌ1 +2βπΌ2
ππ΅ = ππΆ =βπ€π3πΌ1 8πΌ2β3+π 2 π3πΌ1
12 4β3πΌ2 2ππΌ1+βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
ππ΄ = ππ· =π€π
2 π»π΄ = π»π· =
π€β2π3πΌ1πΌ2
4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
See equations (19) β (22)
ππ΄ =π€π2 4β3πΌ2 3βπΌ2+8ππΌ1 +π3πΌ1 π 1 ππΌ1 +6π 2βπΌ2
48 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3 ππΌ1 +2βπΌ2 β
π€π4 5βπΌ1 +24βπΌ2
64 π2πΌ1 +6βπ2πΌ2 +2π 1β3πΌ2
ππ΅ = βπ€π4 5βπΌ1 +24βπΌ2
64 π2πΌ1 +6βπ2πΌ2 +2π 1β3πΌ2 +
π€π2 4β3πΌ2 3βπΌ2 +2ππΌ1 +π 2 π3πΌ1 ππΌ1 +6βπΌ2
48 4β3πΌ2 2πβ πΌ1+βπΌ2 +π 2π3πΌ1 ππΌ1+2βπΌ2
ππΆ = βπ€π 3πΌ1 π3+24πΌ2π2β+16π 1β3πΌ2
64 πΌ1 π3+6πΌ2 π2β+2π 1β3πΌ2 +
π€π2 4β3πΌ2 3βπΌ2 +2ππΌ1 +π 2π3πΌ1 ππΌ1 +6βπΌ2
48 4β3πΌ2 2πβπΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
ππ· =π€π2 4β3πΌ2 3βπΌ2+8ππΌ1 +π3πΌ1 π 1 ππΌ1 +6π 2βπΌ2
48 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3 ππΌ1 +2βπΌ2 β
π€π 3πΌ1 π3+24πΌ2π2β+16π 1β3πΌ2
64 πΌ1 π3+6πΌ2 π2β+2π 1β3πΌ2
ππ· =π€π 3π3πΌ1 +24βπ2πΌ2 +16π 1β3πΌ2
32 π3πΌ1 +6βπ2πΌ2+2π 1β3πΌ2 ππ΄ =
π€π
2β ππ·
π»π΄ = π»π· =π€β2π3πΌ1πΌ2
2 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3 ππΌ1 +2βπΌ2
See equations (26) β (29)
ππ΄ = βπ€β2 πΌ1 π3 +5βπ2πΌ2 +2π 2β3πΌ2
2 πΌ1 π3+6βπ2πΌ2 +2π 1β3πΌ2 +
π€β3πΌ2 9β2ππΌ1 +5β3πΌ2 +π 2π3πΌ1
6 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
ππ΅ =π€β3π2πΌ2
2 πΌ1π3 +6βπ2πΌ2 +2π 1β3πΌ2 +
π€β3πΌ2 β4β3πΌ2 +π 2π3πΌ1
6 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
ππΆ = βπ€β3π2πΌ2
2 πΌ1 π3+6βπ2πΌ2 +2π 1β3πΌ2 +
π€β3πΌ2 β4β3πΌ2+π 2π3πΌ1
6 4β3πΌ2 2ππΌ1+βπΌ2 +π 2π3πΌ1 ππΌ1+2βπΌ2
ππ· = βπ€β3π2πΌ2
2 πΌ1 π3+6βπ2πΌ2 +2π 1β3πΌ2 +
π€β3πΌ2 9β2ππΌ1 +5β3πΌ2 +π 2 π3πΌ1
6 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2 π3πΌ1 ππΌ1 +2βπΌ2
ππ΄ = ππ΅ =π€β3ππΌ2
πΌ1 π3+6πΌ2 π2β+2π 1πΌ2β3
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π»π· =π€β4πΌ2 3ππΌ1+2βπΌ2
2 4β3πΌ2 2ππΌ1+βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2 π»π΄ = π€β β π»π·
See equations (32) β (35)
ππ΄ = βππ 1 β3ππ2πΌ2
πΌ1 π3+6βπ2πΌ2+2π 1β3πΌ2 ππ΅ =
3ππ2π2πΌ2
πΌ1π3 +6βπ2πΌ2 +2π 1β3πΌ2
ππΆ = β3ππ2π2πΌ2π΄1
π΄1πΌ1 π3 +6π΄1βπ2πΌ2 +24βπΌ1πΌ2 ππ· = ππ 1 β
3ππ2πΌ2
πΌ1 π3+6βπ2πΌ2 +2π 1β3πΌ2
ππ΄ = ππ΅ =6ππ2ππΌ2
π3πΌ1+6π2βπΌ2 +2π 1β3πΌ2 π»π΄ = π»π΅ = π
See equations (38) β (41)
ππ΄ = βππ +3ππ2π2πΌ2
2 πΌ1 π3 +6βπ2πΌ2 +2π 1β3πΌ2 +
ππ2 4πΌ1πΌ2βπ 3ββπ +πΌ2 8β3πΌ2β4πβ3πΌ2 +π 2π3πΌ1
2 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
ππ΅ =3ππ2π2πΌ2
2 πΌ1π3 +6βπ2πΌ2 +2π 1β3πΌ2 +
ππ2πΌ2 β4β3πΌ2 +4πβ2πΌ2+π 2π3πΌ1
2 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
ππΆ = β3ππ2π2πΌ2
2 πΌ1 π3+6βπ2πΌ2 +2π 1β3πΌ2 +
ππ2πΌ2 β4β3πΌ2 +4πβ2πΌ2 +π 2π3πΌ1
2 4β3πΌ2 2ππΌ1+βπΌ2 +π 2π3πΌ1 ππΌ1+2βπΌ2
ππ· = β3ππ2π2πΌ2
2 πΌ1 π3+6βπ2πΌ2 +2π 1β3πΌ2 +
ππ2 4πΌ1πΌ2βπ 3ββπ +πΌ2 8β3πΌ2β4πβ3πΌ2 +π 2π3πΌ1
2 4β3πΌ2 2ππΌ1+βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
ππ΄ = ππ΅ =3ππ2ππΌ2
π3πΌ1+6π2βπΌ2 +2π 1β3πΌ2
π»π· =2ππ2πΌ2 ππΌ1 3ββπ +βπΌ2 3ββ2π
4β3πΌ2 2ππΌ1 +π 2π3πΌ1 ππΌ1+2βπΌ2 π»π΄ = π β π»π·
See equations (44) β (47)
HD
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ππ΄ = βππ +ππ π2πΌ1 3πβ2π +βπΌ2 6ππ +π 1β2
2 πΌ1 π3+6βπ2πΌ2 +2π 1β3πΌ2 +
ππ 4β3πΌ2 βππΌ1 +βπΌ2 +3ππΌ1 +π 2π3πΌ1 ππΌ1 +2βπΌ2
2 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
ππ΅ = βππ +ππ π2πΌ1 3πβ2π +βπΌ2 6ππ +π 1β2
2 πΌ1 π3+6βπ2πΌ2 +2π 1β3πΌ2 +
ππ 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
2 4β3πΌ2 2ππΌ1+βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
ππΆ = βππ π2πΌ1 3πβ2π +βπΌ2 6ππ+π 1β2
2 πΌ1 π3 +6βπ2πΌ2 +2π 1β3πΌ2 +
ππ 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
2 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2 π3πΌ1 ππΌ1 +2βπΌ2
ππ· = βππ π2πΌ1 3πβ2π +βπΌ2 6ππ+π 1β2
2 πΌ1 π3+6βπ2πΌ2 +2π 1β3πΌ2 +
ππ 4β3πΌ2 βππΌ1 +βπΌ2 +3ππΌ1 +π 2π3πΌ1 ππΌ1 +2βπΌ2
2 4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
ππ· =π π2πΌ1 3πβ2π +6πβππΌ2 +π 1β3πΌ2
π3πΌ1 +6βπ2πΌ2 +2π 1β3πΌ2 ππ΄ = π β ππ·
π»π΄ = π»π΅ =6ππβ2πΌ1πΌ2 πβπ
4β3πΌ2 2ππΌ1 +βπΌ2 +π 2π3πΌ1 ππΌ1 +2βπΌ2
See equations (50) β (53)
Table 2: Axial deformation contribution βπ΄π¨ for different values of π°π
π°π
w =1kN/m L = 5m H = 0.4m h = 4m
π°ππ°π
0 1 2 3 4 5
βπ΄π¨ 0 -0.0045 -0.0095 -0.0144 -0.0192 -0.0241
π°ππ°π
6 7 8 9 10
βπ΄π¨ -0.0289 -0.0337 -0.0384 -0.0432 -0.0479
Table 3: Axial deformation contribution βπ΄π© for different values of π°π
π°π
w =1kN/m L = 5m H = 0.4m h = 4m
π°ππ°π
0 1 2 3 4 5
βπ΄π© 0 -0.0066 -0.0135 -0.0201 -0.0267 -0.0332
π°ππ°π
6 7 8 9 10
βπ΄π© -0.0396 -0.0460 -0.0524 -0.0587 -0.0651
Table 4: Axial deformation contribution βπ΄π¨ for different values of π π
w =1kN/m L = 5m πΌ1
πΌ2 = 0.296
ππ 0 0.1 0.2 0.3 0.4 0.5
βπ΄π¨ -3.1250 -0.5409 -0.0823 -0.0239 -0.0096 -0.0047
ππ 0.6 0.7 0.8 0.9 1.0
βπ΄π¨ -0.0026 -0.0015 -0.0010 -0.0006 -0.0004
Table 5: Axial deformation contribution βπ΄π© for different values of π π
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w =1kN/m L = 5m πΌ1
πΌ2 = 0.296
ππ 0 0.1 0.2 0.3 0.4 0.5
βπ΄π¨ -4.1667 -0.7668 -0.1208 -0.0359 -0.0146 -0.0072
ππ 0.6 0.7 0.8 0.9 1.0
βπ΄π¨ -0.0040 -0.0024 -0.0015 -0.0010 -0.0007
Figure 9: A graph of βπ΄π¨ against π°π
π°π
Figure 10: A graph of βπ΄π© against π°π
π°π
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Figure 11: A graph of βπ΄π¨ against π π³
Figure 12: A graph of βπ΄π© against π π³
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