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 x direction, 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=−60kN+(−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=45kN

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 xforces 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 F2will 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 andFx,. The external load

of FBx is known asF2=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=(120kN

m)(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 mid-height 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=6.2

5∑Fy=06.25−y5=0y5=6.25∑Fx=

012.5−x5=0x5=12.5

∑Fy=0y3=50

∑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=018.75−12.5−x2=0x2=6

.25

∑Fy=015.625

−50+y10−15.6

25=0y10=50

∑M9=0y8(2)+12.75(1)+12.

5(1)+6.25(2)=0y8=21.875∑

Fy=0y9=15.625∑Fx=0x9=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