Portal Method
Contents
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1 Introduction:
2 General Overview
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2.1 Assumptions
 2.1.1 First Assumption
 2.1.2 Second Assumption
2.2 General Solution
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Step 1
Step 2
Step 3
Step 4
Step 5
3 Example Problem 1
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3.1 Problem 1
3.2 Solution
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2.2.1
2.2.2
2.2.3
2.2.4
2.2.5
3.2.1 Solving the determinate structure
3.2.2 Axial Force, Shear Force and
Bending Moment Diagrams
4 Example Problem 2
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4.1 Problem 2
4.2 Solution
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4.3
4.4
4.5
4.6
4.7
4.8
4.2.1 Solving the determinate structure
Step 1
Step 2
Step 3
Section E-D
Section C-B-K-F
Section F-L
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5 Example Problem 3
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5.1 Problem 3
5.2 Solution
5.3 Solving the Determinate Structure
5.4 Axial Force, Shear Force, and Bending
Movement Diagrams
6 References
Introduction:
A portal frame is often used in a structure to transfer the laterally
directed loads applied along the sides, to the supports at the base
of the frame.[1] Portal frames are often designed such that they
are able to confidently withstand lateral loads. This results in
many portal frames being statically indeterminate externally
(fixed supports for several columns at the bases) ;because of the
frames ability to support horizontal loading, this type of frame is
commonly used in structures like buildings, factories, and
bridges. [2]
A Steel Frame.
Photo Credit: Dwight Burdette at Wikimedia Commons
Link:http://commons.wikimedia.org/wiki/File:Steel_Frame_C
ommercial_Building_Under_Construction,_Ann_Arbor_Town
ship,_Michigan.JPG
The behavior of these frames under lateral loads can be observed
in the General Overview.
The approximate analysis of portal frames can be investigated
through the portal method. Before the analysis, there are
necessary assumptions to be made:
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

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


A point of inflection is located at the center of each
member of the portal frame,
For each story of the frame, the interior columns bear
twice as much shear as the exterior columns,
Lateral forces resisted by frame action,
Inflection points at mid-height of columns,
Inflection points at mid-span of beams,
Column shear is based on tributary area,
Overturn is resisted by exterior columns only.[1]
General Overview
To solve for a multi-storied building under lateral loads we are
able to use what is called an Assumption Method. The
Assumption Method consists of two sub categories; the Portal
and Cantilever Method [1]. Both these Methods have there own
assumptions to help solve for all unknowns and reactions. On this
wiki we will be focusing on understanding the assumptions of a
Portal Frame and how to solve a Portal Frame example. But,
before we can solve any examples we must have a general
understanding for the assumptions made in a Portal Frame and
obtain a general knowledge for solving a Portal Frame example.
Assumptions
In the Portal Method there are a two main assumptions that must
be made before we are able to solve for any unknowns or
reactions. The first assumption is, there must be hinges at midheight of every column and mid-length of every beam [1][2][3].
The second assumption is that the interior column must have
twice the base shear force than the two exterior columns [1][2]
[3]. Both assumptions will be explained with greater detail in the
sub sections to follow.
First Assumption
The first assumption that will be made applies to the Cantilever
Method as well. It states that under lateral loadings, a portal
frame will have deflect in such a manner that its moment
diagrams that will resemble the following [1]:
That
is to
say
that
it is
being assumed that there is zero moment at the mid-span and
mid-height of each member. Combining this with our previous
knowledge of analyzing structures, we are able to make the
assumption that since the frame has multiple zero moments we
are able to replace all zero moments with an internal hinge [1][2].
This assumption is true for all Portal Frames regardless of the
amount of storeys [1]. For example, a two-storey building, as can
be seen below, there will be 10 hinges present:
Therefore, the general assumption for all Portal Frames is that
there is a hinge at mid-height of every column and mid-length of
every beam [1][2][3].
Second Assumption
Even with the first assumption the Portal Frame is still
indeterminate to 2 degrees. Therefore, we must have another
assumption that makes our frame determinate.
This second assumption is unique to the Portal method, and is
essentially the main difference between the two approximate
methods for indeterminate structures subjected to lateral loads.
This assumption will be that the interior columns base shear will
have double the reaction force to the two exterior
columns [1] [2]. Since the column in the middle generally takes
more shear force and is generally more stiff, this results in double
the base shear force. Refer to the diagram below for a visual.
With this final assumption in place our frame now becomes
determinate and it is able to be solved for in a couple of general,
very easy steps.
General Solution
Now that we have an understanding and some general knowledge
about the assumptions and what a Portal Frame is. We will now
obtain the skills to solve an example. For every Portal Frame
there are a basic couple of steps that you must follow to solve for
all unknown and reaction forces. If you follow these 5 general
steps than you will be able to solve for most Portal Frame
questions. To begin a Portal Method question, you must know
what has to be solved for, in most cases you will be asked to
solve for all unknowns and reaction forces in the entire frame. In
that case you will follow all 5 steps, if you are not to solve for all
unknowns and reaction forces it is up to you to decide whether to
deviate from the 5 steps or to shorten your method. In this
general case we will be solving all unknowns and reaction forces.
It shall be known that for these general 5 steps the method was
derived from the steps in the A. Kassimali Structure Analysis
textbook [1].
Step 1
For your first step you must solve the base shear reactions in all
columns, keeping in mind assumption number two (interior
column has double base shear force than the exterior).
Step 2
Once your first step is complete, you will take the frame given to
you and place hinges at all mid-height and mid-lengths of the
frame.
Step 3
Once all base shear forces have been calculated and hinges
placed, you must then split the frame up at all hinge locations.
Step 4
Once the Frame is split into pieces you must then take the
moment about the hinges, preferably you should start with the
piece where the axial force was applied. This same process must
be repeated on the other side of the Frame. Once this is solved for
you are able to balance out every piece by equilibrium and solve
for the moments at the fixed column ends.
Step 5
Once all values have been found you are able to fill in and back
solve any unknowns remaining within your frame. Once this is
completed you have successfully complete a Portal Frame
example.
Example Problem 1
Problem 1
The portal method will be used as an approximate method to
generate the axial, shear and bending moment diagrams for the
building frame shown below. The building is 2 storeys tall, and is
divided into 4 equal sized bays, each with dimensions of 5m x
5m. The building is exposed to two lateral loadings of 40 kN and
60 kN, acting at the top of the second storey and first storey
respectively.
Solution
To begin analyzing this 12 degree indeterminate structure, we
must first make use of our simplifying assumptions. We will
begin by placing hinges at the mid-span and mid-height of each
member, as this has been determined to be the approximate
location of zero moment. This first assumption has reduced the
degree of indeterminacy to 2. The second assumption that must
now be made is taking the stiffness of the interior columns to be
twice that of the exterior columns. This assumption allows us to
take the horizontal reaction force of the middle column as being
double the force at either of the leftmost or rightmost column.
Now we have a relationship which binds 3 of our unknowns to a
single unknown, which has removed our once indeterminate
structure, leaving a statically determinate one in its place.
Solving the determinate structure
Now that the issue of resolving the building's indeterminacy has
been overcome, all that remains is solving a complex, but
determinate system. To do so, the first step is to sum the forces in
the xdirection, for global equilibrium to solve for the horizontal
reactions at the base of the structure, which are all given in terms
of the variable F1.
∑Fx00F1=0=40kN+60kN−(F1+2F1+F1)=100kN−4F1=25kN
After this is done, a similar procedure will be used to analyse the
second storey of the building. The two storey frame will be
separated at an arbitrary location through the cross sections of the
columns to yield something that resembles the figure to the
bottom left. In this case, the assumption stands that the interior
columns will bear twice the force of the exterior ones, so we can
make a new equation in terms of F2.
∑Fx00F2=0=40kN−(F2+2F2+F2)=40kN−4F2=10kN
At this point, we will begin to disassemble the entire structure at
the hinges. The implied condition that there is no moment at the
location of the hinges, still stands, and allows us to solve the
forces in each member of the structure by separating it into 9
individual sections. This is shown in the figure to the right.
Depending on which piece we are looking at, there may be
anywhere from 1 to 3 unknown forces acting on it, so our three
equations of equilibrium will be sufficient to find each of them.
This example will go through the process explicitly for the three
sections which contain the left column of the figure to the right.
The procedure will be the exact same for the remaining 6
sections. The figure to the left shows the pieces that we will be
looking at now.
Starting with the top section, we have an external load, and 4
internal forces, being a horizontal and vertical component force
acting at both hinges. The external load is known, and as you
may recall, so is the force labelled FBy, which was determined
from the global equilibrium analysis of the top floor to
be F2=10kN. Our procedure for solving for the three remaining
unknowns is as follows:

Use the sum of the forces in the x direction to find the
remaining unknown horizontal force FAx.

Find the sum of the moments about one of the hinges to
solve for one of the unknown vertical forces (we will take
the sum of the moments about B to solve for FAy.

Use the sum of the forces in the y direction to find the
remaining vertical force, FBy.
∑Fx0FAxFAxFAxFAx=0=40kN+FAx+FBx=−40kN−FBx=−40kN
−(−F2)=−40kN−(−10kN)=−30kN
∑MB=00=−(40kN)(2.5m)−(−FAx)(2.5m)+(2.5m)(FAy)0=−(40kN)
(2.5m)−(−30kN)(2.5m)+(2.5m)(FAy)FAy=(25kNm)
(2.5m)FAy=10kN
∑Fy0FByFBy=0=FAy−FBy=FAy=10kN
Now that we have solved all of the forces at this section we will
move on to the next. At this point we're going to have to decide
which section we will analyse next, and we have some options
here. Ideally we would progress in some orderly manner, and
solve for one of the adjacent sections (either immediately to the
right or directly below) but we could go to any section which
contains three or less unknown forces. We will proceed
downwards. This section has three hinges corresponding to 6
internal forces, as well as another external lateral load.
From Newton's Third Law of Motion, we know that the forces
which we found at hinge B in the above section will have equal
and opposite reaction forces on this system at B, thus we already
know two of our internal forces, FBx, and FBy. Like the case for
the first section, we also know the horizontal force in the hinge at
D FDx, from our global equilibrium of the entire structure, to
be F2=25kN. We now have a system with three unknowns as
before, and we will follow the same procedure as we are faced
with the same issue of one unknown horizontal force and two
vertical forces.
∑Fx=00=60kN−FBx+FCx+FDxFCx=−60kN+FBx−FDxFCx=−60
kN+(−10kN)−(−F1)FCx=−60kN+(−10kN)−(−25kN)FCx=−45kN
∑MD=00=(5m)(FBx)−(60kN)(2.5m)−(FCx)(2.5m)+(2.5m)
(FCy)0=(5m)(−10kN)−(60kN)(2.5m)−(−45kN)(2.5m)+(2.5m)
(FCy)
FCy=(87.5kNm)(2.5m)FCy=35kN
∑Fy0FDyFDyFBy=0=FBy+FCy−FDy=FBy+FCy=10kN+35kN=4
5kN
Now we will continue to move downwards to our bottom section.
In the last part we had already used the fact that FDx=F1=25kN,
and of course that relationship still stands. Because we know the
forces at the hinge, D, we are left with one unknown vertical
force and for the first time, a moment. In each of the other
sections there were no moments to be calculated, which is the
result of us choosing to break the sections at the hinge locations.
We will use our equations of equilibrium to solve for the two
remaining unknowns as always.
∑MD0MEMEME=0=(2.5m)(FDx)−ME=(2.5m)(FDx)=(2.5m)
(−25kN)=−62.5kNm
∑Fy0FEyFDy=0=FDy−FEy=FDy=45kN
We would now return to the middle section of the top storey and
follow work our way down again, then go up to the rightmost
section of the top storey and go downwards until all of the
unknown forces are resolved. After going through all 9 individual
sections, all of the pin reactions will have been found. These pin
reactions, as you may have realized, correspond to the internal
shear and axial force that exists in the according member. These
forces are summarized in this image.
Axial Force, Shear Force and Bending Moment
Diagrams
The next step is to find the axial, shear and bending moment
diagrams. Once again this will be done explicitly for the two
members which make up the left column of the structure, and the
remainder will be summarized below. This part is quite simple.
To find the shear and axial force in a member, one would
normally be required to make a cut along the member and then
solve for these internal forces, however since this procedure
required us to place hinges at the mid-spans and mid-heights of
the members, we can take the reaction forces at these hinges as
the internal forces. All loads are assumed to be applied to the
joints and thus the shear force is constant along the length of the
member, and accordingly the slope of the moment will also be
constant. This gives simply that for the top leftmost column, the
shear force is simply given by 10 kN, and the axial force is 10 kN
(in tension). Recall that these reaction forces were found at a pin
so that there would be no internal moment at that point, and thus
simplifying our analysis. Since we know that the shear is constant
over the member, the moment at the member's end can be
calculated by multiplying the shear by the half length of the
member. This would result in a moment of 25kNm for at the top
of the member in question and -125 kNm at the bottom.
The same procedure is used to find the axial force, shear force
and bending moment in the bottom left column. Once again we
find that the axial force in the member is 45 kN (in tension) shear
in the member is 25 kN and accordingly the internal moment at
the member's ends are of magnitude 62.5 kNm. This bending
moment can be confirmed to be correct by comparing it with the
support moment reaction at the base of the column, which was
obtained in our analysis of the determinate structure.
This corresponds to the following axial force diagram, shear
diagram and bending moment diagram.
Axial Force
Diagram
Shear Force
Diagram
Bending Moment
Diagram
Example Problem 2
Problem 2
The portal method will be used to construct the shear force and
moment diagram for girder EFGH. The building structure is two
stories high, with 3 bays located on first floor and one subsequent
floor on second level, each with dimensions 20m x 12m. The
building is exposed to two lateral loadings of 20 kN and 10 kN,
acting at the top of the second storey and first storey respectively.
Solution
To analyse this indeterminate structure, we will calculate the
internal loads at the influence points. We will place hinges at the
mid way of each beam where it has zero moment. Similar to
problem 1 above the same assumptions of taking the interior
column stiffness to be twice of the exterior. This assumption
allows us to have one unknown in the structure and therefore the
other internal forces can easily be calculated.
Solving the determinate structure
We can now solve the determinate structure, we do this by
summing all the x forces for equilibrium to solve for horizontal
reaction at the base of the structure. We do this for the entire
structure to find our variableF1. In this case we have two interior
columns with bear twice the force of the exteriors.
∑Fx00F1=0=20kN+10kN−(F1+2F1+2F1+F2)=30kN−6F1=5kN
Now since we have found the horizontal forces at the base F1, we
can focus on the second level storey. The same method is used to
calculate the horizontal force at the base cut of the second storey
to find variable F2. In this case there is only one bay located at
the second level and therefore there is only exterior columns.
Therefore a new equation in terms of F2 will be formed.
∑Fx00F2=0=20kN−(F2+F2)=20kN−2F2=10kN
With any structure you always want to start at the top to begin
solving your unknowns. On the top floor we have an external
load of 20 kN, and 4 internal forces of Fy and Fx,. The external
load of FBx is known as F2=10kN. Now we can solve for the
three unknows as follows:
Step 1

The sum of all forces in the x direction to find the
remaining unknown horizontal force FAx.
∑Fx0FAxFAxFAxFAx=0=20kN−FAx−FBx=20kN−FBx=20kN−
(F2)=20kN−(10kN)=10kN
Step 2

Calculate the moment about one of the hinges to solve for
one of the unknown vertical forces (we will take the sum
of the moments about B to solve for FBy.
∑MA=00=−(10kN)(6m)+(10m)(FAy)FBy=(60kNm)
(10m)FBy=6kN
Step 3

Use the sum of the forces in the y direction to find the
remaining vertical force,FAy.
∑Fy0FAyFAy=0=FAy−FBy=FBy=6kN
Now we will continue to solve for another section. Ideally you
want to solve the section with external forces on them because
you can easily calculate your 3 internal forces. In this case you
can solve section with a external force of 10 kN to calculate your
3 unknowns. Using the same steps above with your external
load F1=5kN.
∑Fx0FCxFCxFCxFCx=0=10kN−FCx−FDx=10kN−FDx=10kN−
(F1)=10kN−(5kN)=5kN
∑MC=00=(5kN)(6m)−(10m)(FDy)FDy=(30kNm)(10m)FDy=3kN
∑Fy0FCyFCy=0=FDy−FCy=FDy=3kN
Now we will continue to proceed downwards at section E. From
Newton's Third Law of Motion, we know that the internal forces
at the hinges D are equal and opposite reactions forces on section
E. Since we already calculated the internal forces for hinge D we
can calculate horizontal, vertical and moment at point E.
Shear diagram
Moment diagram
Section E-D
∑Fy0FEyFEy=0=FDy−FEy=FDy=3kN
∑Fx0FExFEx=0=FDx−FEx=FDx=5kN
∑ME=00=(5kN)(6m)−(ME)∑ME=30kNm
Using the same steps we can continue to the right of the structure
to calculate the horizontal and vertical interior columns forces
and the moment at point M.
Section C-B-K-F
∑MK=00=−(3kN)(20m)+(10kN)(6m)+(6kN)(10m)+(10kN)(6m)
−(10m)(FFy)FFy=(120kNm)(10m)FFy=12kN
∑Ky0FKy=0=FKy−(3kN)−(12kN)+(6kN)=9kN
Section F-L
∑Fx0FLxFLx=0=FFx−FLx=FFx=10kN
∑Fy0FLyFLy=0=FFy−FLy=FFy=12kN
∑ML=00=(10kN)(6m)−(ME)∑ME=60kNm
Now with all the forces and moments calculated we can find the
shear and moment diagram for EFGH.
Example Problem 3
Problem 3
The Portal Method is an approximate analysis used for analyzing
building frames subjected to lateral of and vertical loading of 50
kN and 25 kN, acting at the top of the second storey and first
storey respectively. The two storey building divided into 4 equal
sized bays, each with dimensions of 4m x 2m. Determine the
approximate values of moment, shear and axial force in each
member of the frame.
Solution
In order to solve such problem using the portal method the
following assumptions are made:
1. Placing hinges (approximate location of zero moment) at midheight of each column and centre of each beam.
2. The horizontal shear is divided among all the columns on the
basis that each interior column takes twice as much as exterior
column
First, consider the upper part and place hinges at mid-height of
each column and centre of each beam. Obtain the shear in each
column from a free body diagram by assuming shear of the
interior column equal to twice the shear in exterior column.
Upper part of the structure
Solving the Determinate Structure
For simplicity, each node is given a number from 1 to 10.
Use the three equations of equilibrium to solve for the unknown
forces by:



Use the sum of the forces in x direction to find the
remaining horizontal forces.
Calculate the moment about one of the hinges to solve for
one of the unknown vertical forces.
Use the sum of the forces in the y direction to find the rest
of the vertical forces.
∑M5=0y4(−2)+12.5(1)
=0y4=6.25∑Fy=06.25−
y5=0y5=6.25∑Fx=012.
5−x5=0x5=12.5
∑Fy=0y3=5
0
∑M1=012.5(1)−y2(2)y2
=6.25M∑Fy=0−6.25+y1
=0y1=6.25∑Fx=0x1+12.
5−50=0x1=37.5
Now, consider the bottom part and place hinges at mid-height of
each column and centre of each beam. Obtain the shear in each
column from a free body diagram by assuming shear of the
interior column equal to twice the shear in exterior column.
Lower part of the structure
∑M7=0−y6(2)+18.75(1)+12.5
(1)+6.25(2)=0y6=21.875∑Fy
=021.875−
−6.25−y7=0y7=15.625∑Fx=0
18.75−12.5−x2=0x2=6.25
∑M9=0y8(2)+12.75(
∑Fy=015.
1)+12.5(1)+6.25(2)=
625−50+y
0y8=21.875∑Fy=0y
10−15.625
9=15.625∑Fx=0x9=
=0y10=50
6.25
The sum of the forces on the base of the structure shown in the
diagram below:
Axial Force, Shear Force, and Bending
Movement Diagrams
Axial Force
Diagram
Shear Force
Diagram
Bending Moment
Diagram
References
1. ↑ Jump up to:1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 Kassimali, A.
(2011). Structural Analysis: SI Edition (4th ed.).
Stamford, CT: Cengage Learning.
2. ↑ Jump up to:2.0 2.1 2.2 2.3 2.4 2.5 Dr. Iftekhar Anam (2011).
"Approximate Lateral Load Analysis by Portal Method".
Nov 25th, 2013. <http://www.uapbd.edu/ce/anam/Anam_files/Structural%20Engineering
%20II.pdf>
3. ↑ Jump up to:3.0 3.1 3.2 Professor Schierle (2012). "Portal
Method". Nov 22nd, 2013. <http://wwwclasses.usc.edu/architecture/structures/Arch613/lectures/0
5-portal.pdf>
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