# Moment Area Theorem

## Contents

- 1
**Introduction** - 2
**Definitions** - 3
*Symbols* - 4
**First Moment Area Theorem** - 5
**Second Moment Area Theorem** - 6
*Referring to figure 2* - 7
**Example Including Both Theorems**- 7.1 Calculate the slope at point A and deflection at B of the given beam in Figure 3 using the Moment Area Theorems:
- 7.1.1 1. Solve Reactions using method taught in "Calculations of Support Reactions"
- 7.1.2 2. Draw Shear, Bending Moment, and Displaced Shape diagrams
- 7.1.3 3. Use similar triangles to solve for intersection along x-axis on moment diagram (Figure 4) between point B and C
- 7.1.4 4. First solve $ \Delta_{CA} $ use the Second Moment Area Theorem:=
- 7.1.5 5. Use $ \Delta_{CA} $ to solve for $ \theta_{C} $ using the First Moment Area Theorem:
- 7.1.6 6. Solve for $ \Delta_{BC} $ the deflection between the point B and the tangent of C:
- 7.1.7 7. Input $ \Delta_{BC} $ and $ \theta_{C} $ to solve for $ \Delta_{B} $ deflection at point B:
- 7.1.8 8. Input $ \Delta_{BC} $ and $ \Delta_{B} $ to solve for $ \theta_{A} $ slope at A:

- 7.1 Calculate the slope at point A and deflection at B of the given beam in Figure 3 using the Moment Area Theorems:
- 8
**Example with Distributed Load**- 8.1 Calculate the deflection at B using the Moment Area Theorems:
- 8.1.1 1. Solve Reactions using method taught in "Calculations of Support Reactions"
- 8.1.2 2. Draw Shear, Bending Moment, and Displaced Shape diagrams
- 8.1.3 3. Use similar triangles to solve for intersection along x-axis on moment diagram (Figure 4)
- 8.1.4 4. First solve $ \Delta_{D/A} $ use the Second Moment Area Theorem:
- 8.1.5 5. Use $ \Delta_{D/A} $ to solve for $ \theta_{A} $ using the First Moment Area Theorem:
- 8.1.6 6. Solve for $ \Delta_{B/A} $ the deflection between the point B and the tangent of A:
- 8.1.7 7. Input $ \Delta_{B/A} $ and $ \theta_{A} $ to solve for $ \Delta_{B} $ deflection at point B:

- 8.1 Calculate the deflection at B using the Moment Area Theorems:
- 9
**Summary** - 10
**References**

**Introduction**

All structures experience loads, either by nature itself or loads caused by other objects. When a load is applied, even if one cannot see it, the structure will always bend or deflect in the direction of the applied load if there is no support reaction at that location. When a structure begins to bend because of a load, moment is introduced. The moment area method is then also introduced. This method is a tool that is used to solve problems dealing with structures experiencing a bending affect or deflection. ^{[1]} One may use this tool to find the actual deflection height, the slope or angle of the bending structure at a point, the rotation of the bend or even the moment of inertia when not given. There are many theories in Structural Analysis that pertain to determinate beam deflections (such as the Bernoulli - Euler Beam Theory). However, the moment area method is composed from the moment area theorem, which will be focused on. This theorem is unique because unlike others, it is composed of two "sub-theorems" if you will: The First Moment Area Theorem which solves for the slope of a beam and the Second Moment Area Theorem which solves for the deflection height at a point when required.

**Definitions**

- Moment
- The tendency to cause rotation about a point or an axis.
^{[2]}

- Moment of Inertia
- the tendency of a body to resist angular acceleration, expressed as the sum of the products of the mass of each particle in the body and the square of its perpendicular distance from the axis of rotation. Symbol "I".
^{[2]}

- Young's Modulus
- A modulus of elasticity, applicable to the stretching of a wire etc., equal to the ratio of the applied load per unit area of cross section to the increase in length per unit length. Symbol "E".
^{[2]}

*Symbols*

- Slope - $ \theta $
- Deflection - $ \Delta $
- Moment - $ M $
- Flexural Rigidity - $ EI $
- Length - $ x $

**First Moment Area Theorem**

The first Moment-Area Theorem is used to determine the slope which represents the angle between the tangents at two points on the elastic curve. This is equal to the area under the moment divided by the flexural rigidity diagram (M/EI) between the two points.
^{[3]}

**Procedure for Analysis**

The following are some few steps to adequately use this method:^{[1]}.^{[4]}
Take the figure below as a sample assuming that EI is constant:

- First, analyze the beam and find all the reactions at the supports. Refer to Equilibrium and Compatibility for a better understanding of how to use equilibrium to find the reactions.
- For our beam, we end up with $ A_y = B_y= {\frac{P}{2}} $

- Next, we can draw the bending moment over EI diagram which represents the curvature diagram of the beam ( see figure below). Refer to Beam Analysis if help needed to solve for the moment diagram.

- As the beam represented above is subjected to bending ( the whole point of the moment area method),a deflection occurs.Qualitative Deflected Shapes section actually explains how to precisely represent the deflection.

- Remember that while sketching the deflected shape, the beam curves: concave up when the moment is
*positive*and concave down when the moment is*negative*. From the theory above, we now know that the beam above will deform, concaving up because its moments is positive( see figure 1). - Let consider two points on the elastic curve, C and D. Their tangents create two angles when they intercept. One going from C to D, and the other one from D to C (see figure 1).
- The first moment area theorem is basically stating that:,

$ \theta_{D/C} = \int_C^D ({\frac{M}{EI}})dx $ $ = \theta_{C}-\theta_{D} $ (change in slope from point C to D) = area shaded on the $ {\frac{M}{EI}} $ diagram

where

$ \theta_{D/C} $ is the angle change from C to D $ {\frac{M}{EI} $ is the moment over the flexural rigidity $ \theta_{C} $ is the slope at C $ \theta_{D} $ is the slope at D

This method is also easily to use when we are dealing with a cantilever beam ( a beam having a fixed end support).At a fixed support, the slope is zero. Therefore, when asked to find the slope at the other end of the beam, we will just have to find the area of the curvature diagram between the two points.

**Second Moment Area Theorem**

The second moment area theorem is used to determine the vertical distance to point(B) from the tangent line at another point(A), which is equal to the first moment of area under the $ M/EI $ diagram. (refer to figure 2).^{[3]}

The equation used in the second moment theorem is^{[3]}:

- $ \Delta_{BA} = \int_A^B ({\frac{M}{EI})}\bar{x}dx $

where

- $ \Delta_{BA} $ = deviation of tangent at point B with respect to the tangent at point A
- $ \bar{x} $ = centroid of M/EI diagram measured horizontally from point A
- B,A =Points on the diagram
- $ M $ = Moment
- $ EI $ = flexural rigidity

*Procedure for Analysis*

*In other for you to be able to ease through this method, you have to be comfortable with solving reaction forces of a structure and drawing the M/EI diagram of the structure.*

*Referring to figure 2*

#### The steps used to find deflections using Second moment area (referring to figure 2)^{[3]} ^{[6]}

*First solve for all the unknown reactions(Ax,Ay and By)**Draw the Bending Moment $ M/EI $ diagram also Considering the external force. see Beam Analysis for better understanding on how to solve for the moments.**By looking at the $ M/EI $ diagram in figure 2, you can now draw the deflected shape (remember the sign conversion "concave up shows a positive moment while concave down shows a negative moment")**Since we have a positive moment, our deflected shape has a concave up shape**Before you find the deflection, we have to find the $ \bar{x} $ (be careful because you calculate the $ \bar{x} $ at the $ M/EI $ diagram but starting from the point(B) to the centroid)**In figure 2 since our moment is in triangular shape, you can divide the diagram to make it easier to find the $ \bar{x} $**Using the above equation, you can now find the deflection*

##### Tips on how to draw the $ M/EI $ diagram^{[3]}

*Once you have concentrated loads on the structure, the $ M/EI $ diagram would result in a series of triangular shapes, or if you have both distributed and concentrated loads, your $ M/EI $ diagram would result in parabolic curves, cubic and etc*

**Example Including Both Theorems**

### Calculate the slope at point A and deflection at B of the given beam in Figure 3 using the Moment Area Theorems:

- EI = constant
- E = 200GPa
- I = 900 x 10^6 mm^4

#### 1. Solve Reactions using method taught in "Calculations of Support Reactions"

#### 2. Draw Shear, Bending Moment, and Displaced Shape diagrams

#### 3. Use similar triangles to solve for intersection along x-axis on moment diagram (Figure 4) between point B and C

- $ ({\frac{(237.51+120)kNm}{7m}}) = ({\frac{237.51kNm}{x}}) $

- $ x = 4.65m $

#### 4. First solve $ \Delta_{CA} $ use the Second Moment Area Theorem:=

- Now use Figure 5 to solve for the missing values.

- $ \Delta_{CA} = ({\frac{1}{EI}})(({\frac{2.35m*-120kNm}{2}})*({\frac{2.35m}{3}})+({\frac{4.65m*237.51kNm}{2}})*(2.35m+{\frac{2*4.65m}{3}})+({\frac{7m*237.5kNm}{2}})*(7m+{\frac{7m}{3}})) $

- $ \Delta_{CA} = ({\frac{10657.76kNm^3}{EI}}) $

#### 5. Use $ \Delta_{CA} $ to solve for $ \theta_{C} $ using the First Moment Area Theorem:

- $ \theta_{C} = ({\frac{\Delta_{CA}}{x_{CA}}}) $

- $ \theta_{C} = ({\frac{10657.76kNm^3}{14m*EI}}) = ({\frac{761.28kNm^2}{EI}}) $

#### 6. Solve for $ \Delta_{BC} $ the deflection between the point B and the tangent of C:

- $ \Delta_{BC} = ({\frac{1}{EI}})(({\frac{4.65m*237.5kNm}{2}})*({\frac{4.65m}{3}})-({\frac{2.35m*120kNm}{2}})*(4.65m+{\frac{2*2.35m}{3}})) $

- $ \Delta_{BC} = ({\frac{-20.66kNm^3}{EI}}) $

#### 7. Input $ \Delta_{BC} $ and $ \theta_{C} $ to solve for $ \Delta_{B} $ deflection at point B:

- $ \Delta_{B} = (7*\theta_{C}) - (\Delta_{BC}) $

- $ \Delta_{B} = (7m*{\frac{761.28kNm^2}{EI}})+({\frac{20.66kNm^3}{EI}}) = ({\frac{5349.62kNm^3}{200*10^6MPa*900*10^6m^-4}})*({\frac{1000mm}{1m}}) $

- Therefore, $ \Delta_{B} = 29.72mm $

#### 8. Input $ \Delta_{BC} $ and $ \Delta_{B} $ to solve for $ \theta_{A} $ slope at A:

- $ \theta_{A} = ({\frac{\Delta_{CA}}{x_{AC}}) = ({\frac{10657.76kNm^3}{14m*EI}}) $

- Therefore, $ \theta_{A} = 0.004229 rad $

**Example with Distributed Load**

### Calculate the deflection at B using the Moment Area Theorems:

- EI = constant
- E = 200000 MPa
- I = 400 x 10^6 mm^4

#### 1. Solve Reactions using method taught in "Calculations of Support Reactions"

#### 2. Draw Shear, Bending Moment, and Displaced Shape diagrams

#### 3. Use similar triangles to solve for intersection along x-axis on moment diagram (Figure 4)

- $ ({\frac{(25+75)kNm}{10m}}) = ({\frac{25kNm}{x}}) $

- $ x = 2.5m $

#### 4. First solve $ \Delta_{D/A} $ use the Second Moment Area Theorem:

- $ \Delta_{D/A} = ({\frac{1}{EI}})[({\frac{5m*125kNm}{2}}*(15m+{\frac{5m}{3}}))+((5m*125kNm)*12.5m)+( {\frac{5m*125kNm}{2}}*(10m+{\frac{5m}{3}}))+((2.5m*250kNm)*(7.5m+1.25m))+({\frac{2*2.5m*31.25kNm}{3}}*(7.5m+({\frac{3*2.5m}{8}})))+({\frac{2*7.5m*281.25kNm}{3}}*(7.5m-({\frac{3*7.5m}{8}})))] $

- $ \Delta_{D/A} = ({\frac{29166.7kNm^3}{EI}}) $

#### 5. Use $ \Delta_{D/A} $ to solve for $ \theta_{A} $ using the First Moment Area Theorem:

- $ \theta_{A} = ({\frac{\Delta_{D/A}}{x_{D/A}}}) $

- $ \theta_{A} = ({\frac{291666.7kNm^3}{20m*EI}}) = ({\frac{1458.333kNm^2}{EI}}) $

#### 6. Solve for $ \Delta_{B/A} $ the deflection between the point B and the tangent of A:

- $ \Delta_{B/A} = ({\frac{1}{EI}})(312.5kNm^2*({\frac{5m}{3}})) $

- $ \Delta_{B/A} = ({\frac{520.833kNm^3}{EI}}) $

#### 7. Input $ \Delta_{B/A} $ and $ \theta_{A} $ to solve for $ \Delta_{B} $ deflection at point B:

- $ \Delta_{B} = (5*\theta_{A}) - (\Delta_{B/A}) $

- $ \Delta_{B} = (5m*{\frac{1458.33kNm^2}{EI}})+({\frac{520.833kNm^3}{EI}}) = ({\frac{6770.833kNm^3}{ EI }}) $

- Therefore, $ \Delta_{B} = 84.64mm $

**Summary**

In conclusion, the First Moment Area Theorem is used to solve for the slope or angle in the elastic curve using the change of slope equation $ \Delta\theta = \int_A^B ({\frac{M}{EI}})dx $ from one point to another. When finding the slope is more complex, the Second Moment Area Theorem is considered because it solves for the actual deflection height from a tangent line formed from the elastic curve and the original structure. Once the height is found using the equation $ \Delta = \int_A^B ({\frac{M}{EI})}xdx $ then it is just back to the basics finding the slope with the First Moment Area Theorem. One will have to use the Second Moment Area Theorem almost all the time to solve structures unless given a deflection already. However, as a tip, if you come across a cantilever, use the First Moment Area Theorem immediately because since a fixed end can never deflect, you already know that the slope at that point is zero. Solving the remainder of the structure should be easy after.

**References**

- ↑
^{1.0}^{1.1}Kassimali, A. (2011).*Structural Analysis: SI Edition (4th edition).*Stamford, CT: Cengage Learning. - ↑
^{2.0}^{2.1}^{2.2}http://www.thefreedictionary.com/moment - ↑
^{3.0}^{3.1}^{3.2}^{3.3}^{3.4}http://en.wikipedia.org/wiki/Moment-area_theorem - ↑ Hibbeler, R.C. (2010).
*Mechanics of Materials: 8th Edition.*Upper Saddle River, NJ: Pearson Education. - ↑
^{5.0}^{5.1}^{5.2}Credited by Amandine Amoussougbo - ↑ http://www.colincaprani.com/files/notes/SAIII/5%20-%20Moment-Area%20Method%20-%20R2.pdf