# Portal Method

## Contents

## 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]}

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:

- 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 mid-height 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 *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 $ \normalsize F_1 $.

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 40 kN + 60 kN - (F_1 + 2F_1 + F_1) \\ 0 &= 100 kN - 4F_1\\ F_1 &= 25 kN\\ \end{align} $

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 $ \normalsize F_2 $.

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 40 kN - (F_2 + 2F_2 + F_2) \\ 0 &= 40 kN - 4F_2 \\ F_2 &= 10 kN \\ \end{align} $

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 $ \normalsize F_{By} $, which was determined from the global equilibrium analysis of the top floor to be $ \normalsize F_2 = 10 kN $. 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 $ \normalsize F_{Ax} $.

- 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 $ \normalsize F_{Ay} $.

- Use the sum of the forces in the
*y*direction to find the remaining vertical force, $ \normalsize F_{By} $.

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 40 kN + F_{Ax} + F_{Bx} \\ F_{Ax} &= - 40 kN - F_{Bx} \\ F_{Ax} &= - 40 kN - (-F_2) \\ F_{Ax} &= - 40 kN - (-10kN) \\ F_{Ax} &= - 30 kN \\ \end{align} $

$ \normalsize \sum{M_B}= 0 \\ 0 = -(40 kN)(2.5m) -(-F_{Ax})(2.5m) + (2.5m)(F_{Ay}) \\ 0 = -(40 kN)(2.5m) -(-30 kN)(2.5m) + (2.5m)(F_{Ay}) \\ F_{Ay} = \frac{(25 kNm)}{(2.5m)}\\ F_{Ay} = 10 kN \\ $

$ \normalsize \begin{align} \sum{F_y}&= 0 \\ 0 &= F_{Ay} - F_{By} \\ F_{By} &= F_{Ay} \\ F_{By}&= 10 kN \\ \end{align} $

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, $ \normalsize F_{Bx} $, and $ \normalsize F_{By} $. Like the case for the first section, we also know the horizontal force in the hinge at D $ \normalsize F_{Dx} $, from our global equilibrium of the entire structure, to be $ \normalsize F_2 = 25 kN $. 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.

$ \sum{F_x}= 0 \\ 0 = 60 kN - F_{Bx} + F_{Cx} + F_{Dx} \\ F_{Cx} = - 60 kN + F_{Bx} - F_{Dx} \\ F_{Cx} = - 60 kN + (-10 kN) - (-F_1) \\ F_{Cx} = - 60 kN + (-10 kN) - (-25 kN) \\ F_{Cx} = - 45 kN \\ $

$ \sum{M_D}= 0 \\ 0 = (5m)(F_{Bx}) - (60 kN)(2.5m) - (F_{Cx})(2.5m) + (2.5m)(F_{Cy}) \\ 0 = (5m)(-10 kN) - (60 kN)(2.5m) - (-45 kN)(2.5m) + (2.5m)(F_{Cy}) \\ $

$ F_{Cy} = \frac{(87.5 kNm)}{(2.5m)} \\ F_{Cy} = 35 kN \\ $

$ \normalsize \begin{align} \sum{F_y}&= 0 \\ 0 &= F_{By} + F_{Cy} - F_{Dy} \\ F_{Dy} &= F_{By} + F_{Cy} \\ F_{Dy}&= 10 kN + 35 kN \\ F_{By}&= 45 kN \\ \end{align} $

Now we will continue to move downwards to our bottom section. In the last part we had already used the fact that $ \normalsize F_{Dx} = F_1 = 25 kN $, 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.

$ \normalsize \begin{align} \sum{M_D}&= 0 \\ 0 &= (2.5m)(F_{Dx}) - M_E \\ M_E &= (2.5m)(F_{Dx}) \\ M_E &= (2.5m)(-25 kN) \\ M_E &= -62.5 kNm \\ \end{align} $

$ \normalsize \begin{align} \sum{F_y}&= 0 \\ 0 &= F_{Dy} - F_{Ey}\\ F_{Ey} &= F_{Dy} \\ F_{Dy}&= 45 kN\\ \end{align} $

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 variable$ \normalsize F_1 $. In this case we have two interior columns with bear twice the force of the exteriors.

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 20 kN + 10 kN - (F_1 +2F_1 + 2F_1 + F_2) \\ 0 &= 30 kN - 6F_1\\ F_1 &= 5 kN\\ \end{align} $

Now since we have found the horizontal forces at the base $ \normalsize F_1 $, 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 $ \normalsize F_2 $. 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 $ \normalsize F_2 $ will be formed.

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 20 kN - (F_2 + F_2) \\ 0 &= 20kN - 2F_2\\ F_2 &= 10 kN\\ \end{align} $

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 $ \normalsize F_{y} $ and $ \normalsize F_{x}, $. The external load of $ \normalsize F_{Bx} $ is known as $ \normalsize F_2 = 10 kN $. 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 $ \normalsize F_{Ax} $.

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 20 kN - F_{Ax} - F_{Bx} \\ F_{Ax} &= 20 kN - F_{Bx} \\ F_{Ax} &= 20 kN - (F_2) \\ F_{Ax} &= 20 kN - (10kN) \\ F_{Ax} &= 10 kN \\ \end{align} $

### 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 $ \normalsize F_{By} $.

$ \normalsize \sum{M_A}= 0 \\ 0 = -(10 kN)(6m) + (10m)(F_{Ay}) \\ F_{By} = \frac{(60 kNm)}{(10m)}\\ F_{By} = 6 kN \\ $

### Step 3

- Use the sum of the forces in the
*y*direction to find the remaining vertical force,$ \normalsize F_{Ay} $.

$ \normalsize \begin{align} \sum{F_y}&= 0 \\ 0 &= F_{Ay} - F_{By} \\ F_{Ay} &= F_{By} \\ F_{Ay}&= 6 kN \\ \end{align} $

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 $ \normalsize F_1 = 5 kN $.

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 10 kN - F_{Cx} - F_{Dx} \\ F_{Cx} &= 10 kN - F_{Dx} \\ F_{Cx} &= 10 kN - (F_1) \\ F_{Cx} &= 10 kN - (5kN) \\ F_{Cx} &= 5 kN \\ \end{align} $

$ \normalsize \sum{M_C}= 0 \\ 0 = (5kN)(6m) - (10m)(F_{Dy}) \\ F_{Dy} = \frac{(30 kNm)}{(10m)}\\ F_{Dy} = 3 kN \\ $

$ \normalsize \begin{align} \sum{F_y}&= 0 \\ 0 &= F_{Dy} - F_{Cy} \\ F_{Cy} &= F_{Dy} \\ F_{Cy}&= 3 kN \\ \end{align} $

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.

### Section E-D

$ \normalsize \begin{align} \sum{F_y}&= 0 \\ 0 &= F_{Dy} - F_{Ey} \\ F_{Ey} &= F_{Dy} \\ F_{Ey}&= 3 kN \\ \end{align} $

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= F_{Dx} - F_{Ex} \\ F_{Ex} &= F_{Dx} \\ F_{Ex}&= 5 kN \\ \end{align} $

$ \normalsize \sum{M_E}= 0 \\ 0 = (5kN)(6m) - ({M_E}) \\ \sum{M_E}= 30 kNm\\ $

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

$ \normalsize \sum{M_K}= 0 \\ 0 = -(3 kN)(20m) + (10 kN)(6m) + (6kN)(10m)+(10kN)(6m)- (10m)(F_{Fy}) \\ F_{Fy} = \frac{(120 kNm)}{(10m)}\\ F_{Fy} = 12 kN \\ $

$ \normalsize \begin{align} \sum{K_y}&= 0 \\ 0 &= F_{Ky}-(3kN)-(12kN)+(6kN) \\ F_{Ky}&= 9 kN \\ \end{align} $

### Section F-L

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= F_{Fx} - F_{Lx} \\ F_{Lx} &= F_{Fx} \\ F_{Lx}&= 10 kN \\ \end{align} $

$ \normalsize \begin{align} \sum{F_y}&= 0 \\ 0 &= F_{Fy} - F_{Ly} \\ F_{Ly} &= F_{Fy} \\ F_{Ly}&= 12 kN \\ \end{align} $

$ \normalsize \sum{M_L}= 0 \\ 0 = (10kN)(6m) - ({M_E}) \\ \sum{M_E}= 60 kNm\\ $

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.

### 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.

$ \sum{M_5}=0 \\ y_4(-2)+12.5(1)=0 \\ y_4=6.25 \\ \sum{F_y}=0 \\ 6.25-y_5=0 \\ y_5=6.25 \\ \sum{F_x}=0 \\ 12.5-x_5=0 \\ x_5=12.5 $ | $ \sum{F_y}=0 \\ y_3=50 $ | $ \sum{M_1}=0 \\ 12.5(1)-y_2(2) \\ y_2=6.25M \\ \sum{F_y}=0 \\ -6.25+y_1=0 \\ y_1=6.25 \\ \sum{F_x}=0 \\ x_1+12.5-50=0 \\ x_1=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.

$ \sum{M_7}=0 \\ -y_6(2)+18.75(1)+12.5(1)+6.25(2)=0 \\ y_6=21.875 \\ \sum{F_y}=0 \\ 21.875--6.25-y_7=0 \\ y_7=15.625 \\ \sum{F_x}=0 \\ 18.75-12.5-x_2=0 \\ x_2=6.25 $ | $ \sum{F_y}=0 \\ 15.625-50+y_{10}-15.625=0 \\ y_{10}=50 $ | $ \sum{M_9}=0 \\ y_8(2)+12.75(1)+12.5(1)+6.25(2)=0 \\ y_8=21.875 \\ \sum{F_y}=0 \\ y_9=15.625 \\ \sum{F_x}=0 \\ x_9=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 |
---|---|---|

## Problem 4

### Problem

The portal method is used as an approximate in order to solve for the axial, shear, and moment diagrams for the frame shown below. The frame consist of two horizontal forces 20 kN and 10 kN acting on the left side of the frame.

### Solution

### Step 1

The first step in order to solve this problem using the portal method is to add a hinge at the midpoint of each member, this is allowed because of the first assumption stated above. Another assumption that must be considered is the stiffness of the middle column is twice the exterior column. This means that the horizontal force on the interior column is twice the exterior columns.

### Step 2

The next step is to solve for the horizontal force acting on each column on the lower level and upper level. This is done by using the equations of equilibrium using the following free body diagram for the lower level.

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 20 kN + 10 kN - F_1 - 2F_1 - F_1 \\ 0 &= 30 kN - 4F_1\\ F_1 &= 7.5 kN\\ \end{align} $

We must do this again for the upper level this time using the free body diagram shown below.

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 20 kN - F_2 - 2F_2 - F_2 \\ 0 &= 20 kN - 4F_1\\ F_2 &=5 kN\\ \end{align} $

### Step 3

The next step is to disassemble the entire structure at the hinges. This seperates the frame into nine different sections. We know that the moment at the hinges is zero which simplifies each section. This means at there will always be one to three unknown forces meaning the three equations of equilibrium are enough to find all unknown forces. In this example, I will solve for the three sections on the left side of the disassembled frame. The same procedure is done for the interior and right sections.

### Step 4

looking at the top left section, there are three unknowns (FAB(x), FAB(y), and FAD(y)). All three can be solved using the three equations of equilibrium (as shown below).

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 20 kN - 5 kN - Fab_x \\ 0 &= 15 kN - Fab_x\\ Fab_x &=15 kN\\ \end{align} $

$ \normalsize \begin{align} \sum{M_L}&= 0 \\ 0 &= (20 kN)\times2 - (15 kN)\times2 - (Fab_y)\times2 \\ 0 &= 5 kN - Fab_y\\ Fab_y &=5 kN\\ \end{align} $

$ \normalsize \begin{align} \sum{F_y}&= 0 \\ 0 &= 5 kN - Fad_y\\ Fad_y &=5 kN\\ \end{align} $

Then looking at the middle section, There is three unknowns (FDE(x), FDE(y), and FDG(y)). This again means that the three equations of equilibrium are enough to solve for all three unknown forces (shown below).

$ \normalsize \begin{align} \sum{F_x}&= 0 \\ 0 &= 5 kN + 10 kN - 7.5 kN - Fde_x \\ 0 &= 7.5 kN - Fde_x\\ Fde_x &=7.5 kN\\ \end{align} $

$ \normalsize \begin{align} \sum{M_d}&= 0 \\ 0 &= (5 kN)\times2 + (7.5 kN)\times2 - (Fde_y)\times2 \\ 0 &= 12.5 kN - Fde_y\\ Fde_y &=12.5 kN\\ \end{align} $

$ \normalsize \begin{align} \sum{F_y}&= 0 \\ 0 &= 12.5 kN + 5 kN - Fdg_y\\ Fdg_y &= 17.5 kN\\ \end{align} $

Now looking at the last section of the left side, We have three unknowns (FG(x), FG(y), and MG). And just like before we use the three equations of motion and solve for each unknown force/moment.

$ \normalsize \begin{align} \sum{F_y}&= 0 \\ 0 &= 17.5 kN - Fg_y\\ Fg_y &= 17.5 kN\\ \end{align} $

$ \normalsize \begin{align} \sum{M_d}&= 0 \\ 0 &= (7.5 kN)\times2 - M_g \\ 0 &= 15 kN - M_g\\ M_g &=12.5 kN\\ \end{align} $

Axial Force Diagram | Shear Force Diagram | Bending Moment Diagram |
---|---|---|

## References

- ↑
^{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. - ↑
^{2.0}^{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.uap-bd.edu/ce/anam/Anam_files/Structural%20Engineering%20II.pdf> - ↑
^{3.0}^{3.1}^{3.2}Professor Schierle (2012). "Portal Method". Nov 22nd, 2013. <http://www-classes.usc.edu/architecture/structures/Arch613/lectures/05-portal.pdf>