# Cantilever Method

## Contents

## History

The cantilever method was described by A. C. Wilson ^{[1]}. It is used to approximately determine the effect of lateral forces experienced by tall buildings.^{[1]} The cantilever method uses the assumption that a tall building with a horizontal load such as wind will behave like a cantilevered beam. Cantilever Method is grouped under approximate analysis of structures, which also includes Portal Method ^{[2]}. Portal Method is very similar to cantilever method in that is makes the same assumptions, except that it begins by determining horizontal forces rather than vertical forces ^{[1]}.

The significant implications from this assumption are: inflection points are located at the middle of each beam and column in the structure, and the axial forces experienced by the columns on each level of the building will be proportional to the distance from the centroid of that level ^{[3]}.

## Overview

#### Important Terms and Background

- Determinacy, Indeterminacy and Stability
- Determining whether a system is stable, and is solvable with the equations of condition.

- Loads and Load Combinations
- how loads can be combined and how they interact with a member or structure.

- Moment Equilibrium
- States that the sum of the moments at any given point will be zero. Determines the proportionality of forces with respect to distance. $ \sum {M} = 0 $
^{[1]}

- Equations of Equilibrium
- With any given section, joint, or structure, the equations fo equilibrium must be statisfied. They include, Forces in x; $ \sum {F_x} = 0 $, Forces in y; $ \sum {F_y} = 0 $, and Moments;$ \sum {M} = 0 $.
^{[1]}

- Calculation of Support Reactions
- How to calculate Support Reactions

- Method of Sections
- How to calculate reactions using the Method of Sections

- Determinate Frame Analysis
- How to solve for a frame if it is determinant.

- Centroid
- the centre of mass of an object.
^{[1]}

#### Assumptions and Theory

The figures above show the behaviour of cantilever beams and columns with respect to shear, moment, and deflected shape. In Approximate Analysis using the Cantilevered Beam Method, we assume that a large frame behaves much the same way a cantilevered beam would, experiencing no moment at its centroid ^{[4]}. We also then assume that there is no moment in the center of each member of that frame. These assumptions change an indeterminate frame into one that is determinate.^{[3]}^{[1]}

#### Equations

Summary of the formulas we will be using.

Equations of Equilibrium $ \sum Moment about a point =0 $ $ \sum Forces in x = 0 $

$ \sum Forces in Y = 0 $

$ \sum M =0 $ $ \sum F_{x} = 0 $

$ \sum F_{y} = 0 $

Equation of Centroid $ \bar{x}=\frac{\sum Area of each column * distance from reference}{\sum{Area of each column}} $ $ =\frac{\sum {A*x}}{\sum{A}} $

Relative Column Forces $ Force_{y}= \frac{distance from Horizontal Force to centroid}{distance from Reference to centroid} (Ref_{y}) $ $ F_{y}= \frac{X_{f}}{X_{Ref}}*Ref_{y} $

### Method

[Adapted from Kassimali and Erochko]^{[1]} ^{[5]}

The cantilever method features the following steps:

**1.** Sketch a free-body diagram with a hinge at the center of each member. This is our first assumption;all moments at the center of the members are zero.^{[1]}

**2.** Find the centroid of the structure.

**3.** For each level starting with the bottom, cut through the hinges, find all axial forces in the columns with respect to a reference column of your choice. Solve for the forces in Y using the Moment equilibrium equation. This is our second assumption that all forces in Y are proportional to their distance from the centroid. ^{[1]}

**4.** Starting at the top cut each level into sections at the hinge, and solve using Method of Sections.

**5.** Re-draw the diagram with all forces and moments.

## Examples

### Example 1

We will determine the approximate axial forces of the frame shown below using the cantilever method:

1. Firstly, we will make the assumption that **the moment at the midpoint of each member is zero.** ^{[1]} To represent this in our diagram we place hinges at the center of each member in the frame, which release the moment. The frame with hinges will look like the figure below:

2. Then, we will find the centroid of the frame starting from the left to right. We determine the centroid of the frame using:

$ \bar{x}=\frac{\sum Area of each column * distance from reference}{\sum{Area of each column}} \\=\frac{\sum {A*x}}{\sum{A}} $

In this example, we are going to assume that the area of each column and beam is the same, but there are different cases where the area varies for each column and beam. The equation is then:

$ \bar{x}=\frac{ \left (A (0m) \right) + \left (A (5m) \right) + \left (A (10m) \right) }{{3(A)}} =\frac{A(0m+5m+10m) }{3(A)} =\frac{15m}{3}=5m $

The centroid is 5 m from the left and right of the frame.

Since the centroid coincides with member BE: $ B_{y}= 0 kN $

3. We will use the relative distances from the centroid as well as the rule of similar triangles. For the centroid, we need to make our next assumption:

**The axial load in the columns is linearly proportional to their individual distances from the centroid.** ^{[1]}

Using the centroid (see the above figure on the right):

Select your reference column - here, AD is used - and find the forces in the other column with respect to AD.

$ C_{y}= \frac{distance from CF to centroid}{distance from AD to centroid} (A_{y}) = \frac{5 m}{5 m}(A_y) \\C_{y}=A_{y} $

Using the rule of similar triangles (see the above figure on the left):

$ \frac{C_{y}}{5}=\frac{A_{y}}{5}\\ C_{y}=A_{y} $

4. If we applied the moment at the centroid in the above given frame, we get:

$ \sum {M_B}=0 \\ A_{y}(5 m)+C_{y}(5 m)-300 kN\times 2.5 m = 0\\ A_{y}=c_{y}\\ A_{y}(5 m)+A_{y}(5 m)-750 kN.m=0\\ A_{y}=75 kN\\ C_{y}=75 kN\\ $

In conclusion, we found a relative equations between the $ A_{y} $ and $ C_{y} $ reactions that will help us in finding one force. Finding this force will change the frame from an indeterminate structure to a determine structure.

*Note*: there are cases where we have more than one level; we would have to repeat steps 3 and 4 for each level of the frame by cutting each section at the hinges. ^{[1]}

### Example 2

We will determine the approximate axial forces of the frame shown below using the cantilever method.

**1.** First we make an assumption: **the moment at the midpoint of each member is zero.** ^{[1]} To apply this to our frame we place hinges at the center of each member in the frame.

**2.** Then, we will find the centroid of the frame starting from the left to right. We determine the centroid of the frame using:

$ \bar{x}=\frac{\sum Area of each column (distance from reference)}{\sum{Area of each column}}=\frac{\sum {A(x)}}{\sum{A}} $

In this example, we are going to assume that the area of each column and beam is the same; our equations are then:

$ \bar{x}=\frac{ \left (A (0m) \right) + \left (A (5m) \right) + \left (A (7m) \right) }{{3(A)}} =\frac{A(0m+5m+7m) }{3(A)} =\frac{12m}{3}=4m $

The centroid is 4 m from the left to right of the frame.

This means that each of our distances to centroid are 4 m, 1 m, and 3 m, respectively.

**3.** Our next assumption: **the axial load in the columns is linearly proportional to their individual distances from the centroid.**^{[1]} We pick a reference column - here, A is used - and find the forces in the other column with respect to A.

$ B_{y}= \frac{distance from B to centroid}{distance from A to centroid} (A_{y})= \frac{1m}{4m}(A_y) \\ \\C_{y}=\frac{3m}{4m}(A_{y}) $

**4.** Use the Moment-Equilibrium equations to solve for the reference force.

*Note: We are not using a standard moment here - instead, we are using the principle of moment equilibrium, simply translating our external applied loads proportionally into y, ignoring all forces in x. This is based on the assumption in the previous step. *^{[1]}

$ \sum{M_{C}}=0=40kN(6m) + 20kN(2m) + B_{y}(2m) - A_{y}(7m) \\ \\0= 40kN(6m) + 20kN(2m) + \frac{1m}{4m}(A_y)(2m) - A_{y}(7m) \\ \\A_{y}=43.1 kN downwards $

**5.** Solve for the other forces in Y.

$ B_{y}= \frac{1m}{4m}(A_y) =\frac{1m}{4m}(43.1kN) = 10.8 kN upwards $

$ C_{y}=\frac{3m}{4m}(A_{y}) = \frac{3m}{4m}(43.1kN) = 32.3 kN upwards $

**6.** Repeat Steps 4 and 5 for each level of the frame by cutting at the hinges on that level.

$ \sum{M_{C}}=0=40kN(2m) + B_{y}(2m) - A_{y}(7m) \\ \\0= 40kN(2m) + \frac{1m}{4m}(A_y)(2m) - A_{y}(7m) \\ \\A_{y}=12.3 kN downwards $

$ B_{y}= \frac{1m}{4m}(A_y) =\frac{1m}{4m}(43.1kN) = 3.1 kN upwards $

$ C_{y}=\frac{3m}{4m}(A_{y}) = \frac{3m}{4m}(43.1kN) = 9.2 kN upwards $

**7.** Using Method of Sections start at the top and solve for all unknown forces and moments, starting with the top corner.

Equations for Top Corner Section

$ \sum M_{f} = 0 = 12.3kN(2.5m) - E_{x}(2m) \\ E_{x} = 15.4kN \\ \sum F_{x}=0 =40kN-15.4kN-F_{x} \\F_{x}= 24.6 $

**8.** Repeat this process moving across the frame, working section by section.

**9.** Solve the frame by each level, using similar techniques to Determinate Frame Analysis. The complete frame will look like the following figure:

### Example 3

Determine the approximate axial and shear forces for all the members, and the reaction forces of the frame shown below by using the cantilver method. Note area of first column is twice the other two.

**1.**Place hinges at midspan of each member in frame. Based on the assumption discussed before.

Exception: no hinges for pinned members.

**2.** Find the neutral axis of frame. Since the frame is not symmetrical it will be different for top storey and bottom storey.

For top story:

$ \bar{x}=\frac{ \left (A (0m) \right) + \left (A (6m) \right) }{{2(A)}} = 3m $

**3.** Determine the axial stress of all columns. In terms of the stress in the left column.

FBD: of top storey

$ x_{1}=3m \\ x_{2}=3m $

$ \ \sigma_{j}=\frac{3}{3}(\sigma_{i})= \sigma $

$ F=A*{\sigma_{i}} $

$ \sum{M_{i}}=0=-35kN(2m) + A*{\sigma_{i}}(6m) $

$ \sigma_{i}=\frac{11.67}{A} $

$ F_{i}= A_{C}*\frac{11.67}{A_{C}} = 11.67 kN $

$ F_{j}= A_{C}*\frac{11.67}{A_{C}} = 11.67 kN $

**4.** Using method of sections solve for unknown forces and moments.

$ \sum F_{y} = 0 $

$ F_{y}=11.67 kN $

$ \sum M_i =0 = -35(2)+F_{kx}(2)+11.67(3) $

$ \sum F_x=0 \\F_{kx} = 17.5 kN $

$ F_{ix} = 35-17.5=17.5kN $

$ \sum F_x=0 \\F_{jx}=17.5 kN $

**5.** Repeat steps 2 to 4 for second storey.

$ \bar{x}=\frac{ \left (2A_{C} (0m) \right) + \left (A _{C}(9m) \right) + \left (A_{C}(15m) \right) }{{4(A_{C})}} = \frac{24A_{C}}{4A_{C}} = 6m $

$ x_{1}=6m $

$ x_{2}=3m $

$ x_{3}=9m $

$ \sigma_{e}=\sigma_{d}0.5 \sigma_{d} $

$ \sigma_{f}=\sigma_{d}=1.5 \sigma_{d} $

$ \sum M_d =0 = -35(6)-70(2)+(A_{c}*0.5 \sigma_{d})+(A_{c}*1.5 \sigma_{d})(15) $

$ 27*A_{c}*\sigma_{d}=350 $

$ \sigma_{d}=\frac{12.96}{A_{c}} kPa $

$ F_{d}=2*A_{c}*(\frac{12.96}{A_{c}})=26 kN $

$ F_{e}=12.96*0.5=6.48 kN \\F_{f}=1.5*12.96=19.44kN $

$ \sum F_y=0\\ F_{hy}=19.44-11.67= 7.8 kN $

$ \sum M_f =0 = -17(4)+7.8(3)-F_{hx}(2) \\F_{hx} =23.3 kN $

$ \sum F_x=0 \\F_{fx}=-17.5-23.3=5.8 kN $

$ \sum F_y=0 \\F_{gy}=7.8+6.48+11.67=26 kN \\ \sum M_e=0 = -23.3(2)+7.8(3)+26(4.5)-F_{gx}(2)-17.5 \\F_{gx}=11.9 kN $

$ \sum F_x=0 \\F_{ex}=11.9+17.5+23.3=52.7kN $

$ \sum F_x=0 \\F_{dx}= 70-11.9=58.1 kN $

**6.** Find equations of shears as a function of axial for members A and C

$ \sum M_b=0 = -58.1(2)-26 (4.5) -A_{x}(4)+A_{y}(4.5) \\A_{x}=\frac{ \left (-233 \right) + \left (4.5(A_{y}) \right) }{{4}} \\A_{x}=1.125A_{y}-58.3 kN $

$ \sum M_c=0= 5.8(2)-19.44(3)+C_{y}(3)-C_{x}(4) \\C_{x}= \frac{ \left (-46.7 \right) + \left (C_{y}(3) \right) }{{4}} \\C_{X}= 0.75C_{y}-11.7 kN $

**7.** Repeat steps 2 to 4 for bottom storey

$ \bar{x}=6m \\ \Sigma_{c}= 1.5 \Sigma_{A} $

$ \sum M_a=0 = -35(10)-70(6)-C_{x}(2)+C_{y}(6)-A_{x}(2)+A_{y}(9) $

$ 770= -((2)(0.75*A_{C}\Sigma_{A}(1.5))-11.7) + (A_{C}\Sigma_{A}(1.5)(6)) - [ 2((1.25*2(A_{C}*\Sigma_{A})-58.3)] + (9(2A_{C}\Sigma_{A})) \\Sigma_{A}=\frac{31.1}{A_{C}} kPa \\A_{y}=2A_{C}(\frac{31.1}{A_{C}})=62.2 kN \\A_{x}=11.7 kN \\C_{y}=46.6 kN \\C_{x}=23.3 kN $

$ \sum F_y=0\\ F_{ay}=62.2-46.7=15.5 kN \\ \\ \sum F_x=0 \\F_{ax}=70 kN $

$ \sum F_y=0\\ F_{by}= 62.2-26 = 36.2 kN \\ \\ \sum F_x=0 \\F_{bx}=58.1-11.7=46.4 kN $

$ \sum F_y=0 \\F_{cy}=46.7-19.44=27.3 kN \\ \\ \sum F_{x}=0 \\F_{cx}=15.8+23.3=39.1 kN $

$ \sum F_{y}=0 \\B_{y}=15.5 kN\\ \\ \\ \sum F_{x}=0 \\B_{x}=70 kN \\ \\ \sum M_B=0=-m_{B}-70(2)=0 \\m_{B}=140 kNm $ (counter clockwise)

Summary:

Member | Axial (kN) | Shear (kN) |

AD | 62.2 | 11.7 |

BE | 15.5 | 70 |

CF | 46.7 | 23.3 |

DE | 46.4 | 36.2 |

EF | 39.1 | 27.3 |

DG | 26 | 58.1 |

EH | 6.48 | 52.7 |

FI | 19.4 | 5.8 |

GH | 11.9 | 26 |

HI | 23.3 | 7.8 |

HJ | 11.67 | 17.5 |

IK | 11.67 | 17.5 |

JK | 17.5 | 11.67 |

## Summary

A brief wrap-up, when using the cantilever method, a few important assumptions apply to a fixed supported frame: at the centre of each column and each girder of the frame a hinge is placed, since they are both assumed to be a zero-moment point.^{[1]}. Assume that the axial load in every column corresponds to the distance from the cross-sectional area of the structure. ^{[1]}.Building on those assumptions a statically determinate and stable frame will be attained.

The examples above show how to apply the Cantilever Method in lateral frames.

## References

- ↑
^{1.00}^{1.01}^{1.02}^{1.03}^{1.04}^{1.05}^{1.06}^{1.07}^{1.08}^{1.09}^{1.10}^{1.11}^{1.12}^{1.13}^{1.14}^{1.15}^{1.16}^{1.17}Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning - ↑ http://www.sut.ac.th/engineering/Civil/Courseonline/430332/pdf/09_Approximate.pdf
- ↑
^{3.0}^{3.1}Taranath, Bungale S. (2011). Structural Analysis and Design of Tall Buildings: Steel and Composite Construction. Boca Raton, FL: CRC Press. - ↑ http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT%20Kharagpur/Structural%20Analysis/pdf/m6l36.pdf
- ↑ Erochko, J. (2013). Intro to Structural Analysis: Class Notes. Ottawa, ON: Carleton University