Beam Analysis

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Figure 1: Example of a Shear and moment diagrams for a beam

Structural beams are an integral part of most structural projects. They are used for floors, columns, structural frames, and pretty much any part of a structure that needs added strength and stability to resist failure. Beams can be made of concrete, steel, wood, or other materials depending on the project.[1] To assure that the beams meet the specifics needed to resist failure, a structural analysis of the beam is completed.

The structural analysis of a beam examines what stresses occur in a beam when it is exposed to loading. In a real case these load effects can come from environmental effects such as snow, wind, and earthquakes, as well as the structures self weight and stresses induced from its use. There are three types of stress resultants that can occur in a beam; axial forces, shear forces, and bending moments which can occur on the internal of the beam or the external at the supports.[2] It is very important to understand how all these forces effect the beam so analysis must be completed and a shear and bending moment diagram can be drawn for that beam.

This article will describe how to calculation these three stress forces and draw the shear and bending moment diagram for a beam from these forces.


Stress Forces

There are three fundamental stress forces that need to be understood in order to explain how shear and bending moment diagrams are constructed. They are as follows:

Axial or Normal
An axial (normal) force can be denoted by (N)[3], (Q)[2] or another letter depending on convention. It is a force that acts parallel and along the center line of a member. If a member is “cut” the axial force will act in equal magnitude but opposite direction on each side of the cut, causing the beam to act in compression (push) or in tension (pull)[3]. For shear and moment diagram, the axial force is often zero unless an external horizontal force is applied to the beam.
A shear force, denoted by (V), acts perpendicular to a member and often results from a horizontal load being applied to a beam. Shear forces, like axial forces, act in equal magnitude but opposite direction to one another but cause the two segments of a “cut” beam to “slide against each other[3]. There is almost always a shear force acting on a beam in analysis and a diagram helps to visualize how these forces act on the beam.
A moment, denoted by (M), are forces that try to make the beam turn. Therefore any external force that finds a form of restriction can result in a moment within a beam, such as a fixed end with no allowed rotation. Moments are common in beams that have restricted rotation based on their design.

Common Types of Reaction Supports

The supports for a beam vary and are not always located at the beams ends. Each type of support exhibits different combinations of stress forces when exposed to a loaded beam:

A fixed support is susceptible to all three stress forces, axial, shear, and moment. This is because a fixed support is not permitted any free movement. It cannot rotate, or move vertically or horizontally. Thus, when a load is applied to a beam with a fixed support, three reaction forces will result at the support (Refer to Figure 2).

Figure 2: Fixed Support

A pin support can freely rotate thus no moment is induced when a load is applied to a beam with this type of support. However an axial force and shear force are present because a pin restricts horizontal and vertical movement (Refer to Figure 3).

Figure 3: Pinned Support

Roller are restrictive to only one type of stress force, shear. They can move horizontally without restriction and no moment is caused when a load is applied to a beam with these supports on it (Refer to Figure 4). However, one does not want a beam that only has rollers since there would be nothing stopping it from moving horizontally.

Figure 4: Roller Support

Hinges are located within a beam and allow that point in the beam to rotate without restriction meaning there is no moment at a hinge. A hinged location on beam still experiences axial and shear forces however, as if the beam were cut at that point. This means each side of the hinge will experience an axial and shear force that has the same magnitude but opposite direction (Refer to Figure 5).

Figure 5: Hinge Support

Common Types of Beams

Beams are important structural components that are usually very long in length but short in width[3]. Beams are essential to support floors and decks in structures and a lot more. Beams can support applied loads that are perpendicular (shear) to the length of the beam. The axial loads which are forces applied parallel to the beam are almost always negligible which is a result of the dimensions of a standard beam [4]. The general types of beams that will be introduced are the simply supported, cantilevered and the overhanging beams:

Simply Supported Beam
A simply supported beam has a roller support on one end and a pin connection on the other end (Refer toFigure 6). Figure 6: Simply Supported Beam
Cantilevered Beam
A cantilevered beam has one fixed end and one free end (Refer to Figure 7). Figure 7: Cantilevered Beam
Overhanging Beam
An overhanging beam can have one or both sides of the beam extending over the supports freely[3] (Refer to Figure 8). Figure 8: Overhanging Beam

Simplified Loading Cases

Beams will be subjected to external loading. In order to properly analyze beams, it's important to know and understand these types of loading cases:

Point load
A point load is a load exerted on the beam at a specific location (Refer to Figure 9). On the shear diagram, the magnitude of the force will be drawn either positively or negatively depending on the direction of the point load.

Figure 9: Point loads

Moment point load
A moment point load is similar to a point load in which a moment load is being applied on a specific part of the beam (Refer to Figure 10). Moment loads do not affect the shear diagram, however it will be taken into consideration for the moment diagram. If the direction of the moment load is counterclockwise (based on the standard engineering convention is a positive moment[2]), then there will be a jump downwards (subtraction) in the moment diagram by the magnitude of the moment at the specific location. If the moment load direction is clockwise (negative moment), then this would result in jump upwards (addition) on the moment diagram by the magnitude of the moment load.

Figure 10: Moment point loads

Uniformly distributed load
A Uniformly distributed load is a load that is evenly distributed over the length of the beam[3] or a section of a beam which means each unit length of the member has the same amount of stress exerted on it (Refer to Figure 11). When solving for reactions, the distributive loads can be converted to a point load by simply calculating the area of the distributive load and placing it at location of the centroid of the distributive load. After the conversion, the concentrated load can be used in the equations of equilibrium to solve for unknown reactions. However, when drawing the shear diagrams, the point load is no longer acceptable and we must refer back to the distributive loading.

Figure 11: Distributed load

Linearly varying distributed load
Linearly varying distributed loading is distribution of the loading that varies per unit length of the beam[3] (Refer to Figure 12). The same procedure as the uniformly distributed load which was previously mentioned will be applied. When solving for reactions the area of the distributive load can be converted to a point load, and it will be placed at the location of the centroid of the loading shape. There could be combinations of both uniform and varying loading in one example ( A triangular loading shape on top of a rectangular distributive load). In this case it must be broken up into two sections and instead of one point force from the combined load there would be two concentrated loads applied at the location of the cetroid of each loading shape.

Figure 12: Triangle distributed load

Conventions and Equations

Sign Convention for Stress Forces

Axial, shear, and moment forces have a standard sign convention to describe how they react on a beam. This convention can also be referred to as the beam convention[2] and is the action of each stress force acting with equal magnitude and opposite direction upon a beam. These coupled forces can either occur between a cut in the beam or on the ends and have a district effect on the beam.

Axial Convention – positive axial forces “pull” on a beam placing the beam in tension. Negative axial “pushes” and puts the beam in compression (Refer to Figure 13). Figure 13: Axial sign convention for a beam

Shear Convention – shear causes two parts on a beam to “slide” against one another. Positive convention is upward on the right, downward on the left (Refer to Figure 14). Figure 14: Shear sign convention on a beam

Moment Convention – moment cause bend within a beam from the ends. An upwards bend indicates a positive moment while a downwards bend is a negative moment (Refer to Figure 15). Figure 15: Moment sign convention on a beam

An illustration of the beam convention for all three forces acting upon a beam is shown in Figure 16 below:

Figure 16: Positive sign convention of Shear, Axial, and Moment on a beam

Side Note

If you do not want follow the standard sign convention previously mentioned you could draw the internal forces in which ever direction you'd like; however please note that if any negative values are obtained in the calculations it means that forces was drawn in the incorrect directions.

Equations of Equilibrium

$ \sum Fy= 0 $;

$ \sum Fx= 0 $;

$ \sum Mo= 0 $;

The equations of equilibrium are very important when it comes to beam analysis. Using these three equations you can find out unknown values for the support reactions by knowing the forces and length segments of the beam. Generally for simply supported beams, if the loads are all perpendicular there will be no forces in the x direction. The moment equation is really helpful because it allows you to solve for unknows without factoring certain forces, any force that acts in the same line that you are taking the moment of is zero. So if you have three unknown forces but 2 of them pass through the same point then taking the moment at this point would neglect those forces, and isolating the third one into a solvable equation.

Step by Step Beam Analysis

The following steps[2] provide a guideline on how to calculate internal forces at specified locations on a beam:

Step 1
Draw the free body diagram of the beam and specify the sign convention that will be used throughout the solution.
Step 2
Compute the unknown support reactions by using the 3 equations of equilibrium. You can refer to the "Equations of Equilibrium" section of this wiki.

(Note: When dealing with cantilever beams the fixed end calculations can be avoided by selecting the cut section of the free end when computing internal shear and moment[2])

Step 4
Apply an imaginary cut (perpendicular to the beam) at a location of interest where the calculations of the internal shear and moment are needed. The beam is now split into two cut sections. Since either side of the two cut sections have equal and opposite reactions,only one cut section needs to be analyzed.

(Note 1: Pick the cut section with the least amount of external forces and reactions for a simpler analysis)

(Note 2: Make sure to be consistent with the sign convention as you go through Steps 5-7) You can refer to "Conventions and Equations" section of this wiki.

Step 5
Solve for the internal axial force of the cut section by summing up all the forces parallel to the beam.
Step 6
Solve for the internal shear force of the cut section by summing all the forces perpendicular to the beam.
Step 7
Solve for the internal moment force of the cut section by summing up the external force moments as well as any point moments on the cut section.
Step 8
If you wanted to check your solutions of the internal forces, use the abandoned cut section from Step 4 and compute the internal forces from Steps 5-7. You should get the same answer.[2]

Constructing Shear and Moment Diagrams

Method 1: Cut Sections

The following steps[2] go through the general method for constructing shear and moment diagrams for a beam. The method used in this section is similar to the "Step by Step Beam Analysis" of determining internal reactions at specified points. The difference here is that when the cut sections are made, it's in terms of the variable'x'[5].

Step 1
Draw the free body diagram of the beam and specify the sign convention that will be used throughout the solution.
Step 2
Solve for all the unknown external reactions using the three equations of equilibrium.
Step 3
Determine how many cut sections your beam needs. For example if a simply supported beam consisted of only one point force then there would be two cut sections needed, one before the point load and one somewhere after the point load[5] (this is the same for support reactions). For a distributed load, a cut section must be made right before, somewhere in the middle and after the load distribution[5]. You want to analyze the parts before and after supports and loads (and sometimes in the middle) by cut sections; so depending on how complicated the structure is will depict the number of cut sections needed for a proper analysis of the beam.
Step 4
Once you've determined the number of cut sections, proceed to the first cut in the beam in terms of 'x' (make sure to specify the boundaries) and calculate the internal reactions in terms of x, using the same method as in the "Step By Step Beam Analysis" section Steps 4-7. Do this for all the cut sections required for your beam.
Step 5
Now you will use the equations that you calculated in Step 4 to plot for the shear and moment diagram by plugging in values for x and finding the shear or moment at those points.

Method 2: Graphical Analysis

The following steps[2] introduce the graphical method of constructing shear and moment diagrams:

Step 1
Draw the free body diagram of the beam and specify the sign convention that will be used throughout the solution.
Step 2
Solve for all the unknown external reactions using the three equations of equilibrium.
Step 3
Once all the external reactions are calculated, you can proceed to your shear diagram. Look at the free body diagram starting from the left side of the beam. If there is no load at the end then you could proceed to the next step since the shear at the point is zero[2]. However, if there is a force located, then a line can be drawn from zero all the way to the magnitude of the concentrated load. The direction of the point depends on whether the force is positive (upward force) or negative (downward force).
Step 4
Now keep looking to the right of the beam while drawing a horizontal line (meaning there is no change in loading) until another value for a load is shown, if it's another point load then you would continue from the point you left off from the previous section and can add or subtract the value of the force depending on it's direction. You would ultimately do this until you get to the other side of the beam. If there was a uniformly distributive load on the beam depending on the length of the distribution, a line would be drawn diagonally connecting the positive and negative value of the area of the distributed load. By the time you get to the right end of the beam, the diagram should come back to zero ( with the exception of rounding errors).
Step 5
It's good to note that any line that crosses the axis to get from positive to negative (inflection points) should be kept in mind when drawing the moment diagram since it becomes a max or min value with the slope at the point being zero.
Step 6
In order to draw the moment diagram you must first check the left end of the beam to see if there is any point moment. If so, a jump from zero to the magnitude of the point load would occur. If there were no point moment on the free body diagram that you will be starting your diagram from zero.
Step 7
The area under the shear diagram is the slope of the moment in the moment diagram. Therefore to make this simple, split up the shear diagram into common shapes and calculate the area of each one and the value you get can be plotted in the moment diagram. You would do this moving along to to the right side of the beam. If any point moments are found along the beam, you add or subtract the magnitude of the point load (straight line) from the point you left off.
Step 8
By the time you get the right end of the beam it should end at zero (perhaps close to zero if there are rounding errors[2]).


Example 1: A beam connected to a pin at point A and roller at point B gets 18kN and 6kN of force exerted on random segments along the beam, with a clockwise moment of 10 kN*m located where the 6 kN force is. Draw a free body diagram showing all the acting forces on the beam, then use the equilibrium equations to determine the values of Ax, Ay, By. Also draw the shear and moment diagram for the beam, as shown below in Figure 17.

Figure 17

Step #1) First, draw the free body diagram of the beam in the figure above and make sure to have the correct direction for the forces acting on the beam. (This FBD can be seen in the final Shear and Moment Diagram) Note that you can later on determine whether the direction of the forces is correct or not after determining the unknown forces acting on the beam by using the equations of equilibrium as shown in the next step.

From here on onward the problem could be solved by using two different methods known as the Graphical method and the Cut-Section method. We will begin by using the Cut-Section method, which is a type of approach to solving beams where we cut the beam into sections, hence the name Cut-Section method.

Cut-Section Method

Step #2) Apply the equations of equilibrium to the full beam to determine the unknown forces acting on it.

   + ∑Ma = 0 ;       -18kN(2m) + By(4m) – 6kN(6m) – 10 kN*m = 0    →    4By = 36 + 36 +10    →    By = 20.5 kN              
 ↑ + ∑Fy = 0 ;       -Ay – 18kN +20.5kN = 0    →    Ay = -18 + 20.5     →    Ay = 3.5 kN ;
 → + ∑Fx = 0 ;        Ax = 0kN ;

Step #3) After determining the unknown forces proceed by taking a cut section right before the 18 KN force, as shown below in Figure 18 and solve for the internal forces using the equations of equilibrium.

Figure 18

 ↑ + ∑Fy = 0 ;       3.5 kN – Vx = 0     →    Vx = 3.5 kN
   + ∑Ma = 0 ;      -3.5(x) + Mx = 0     →   Mx[1] = 3.5x kN*m     (1)
 → + ∑Fx = 0 ;       0 + Nx = 0          →   Nx = 0 kN

Step #4) Now we take a second cut, this time from the point right before the roller we have at Point B supporting the beam as shown below in Figure 19.

Figure 19

 → + ∑Fx = 0 ;       0 + Nx = 0               →    Nx = 0 kN
 ↑ + ∑Fy = 0 ;       3.5 kN – 18kN - Vx = 0   →    Vx = -14.5 kN
   + ∑Ma = 0 ;      -18kN(2m) – (-14.5(2+x)) + Mx = 0     →     -36 +29 +14.5x + Mx = 0     →     Mx[2] = (7-14.5x) kN*m       (2)

Step #5) For the final cut we take it from the end of the beam till right before the roller, as shown below in Figure 20.

Figure 20

 → + ∑Fx = 0 ;       0 + Nx = 0            →    Nx = 0 kN
 ↑ + ∑Fy = 0 ;       Vx – 6kN = 0          →    Vx = 6 kN
   + ∑Ma = 0 ;      -6kN(x) – 10KN*m = 0   →     x = -5/3 m         (3)

Step #6) After determining the value of x substitute it into the moment equations (1) and (2) as shown below.

  (1)  Mx[1] = 3.5x kN*m      = 3.5(-5/3 m)      = -5.83 kN*m 
  (2)  Mx[2] = (7-14.5x) kN*m = 7 – 14.5(-5/3 m) = 31.17 kN*m 

Another simpler method which allows us to skip the Cut-Section method to determine the shear and moment diagram is called the graphical method, where we determine the shear then the moment diagram as shown below.

Graphical Method

Step #2) Apply the equations of equilibrium to the full beam to determine the unknown forces acting on it.

   + ∑Ma = 0 ;       -18kN(2m) + By(4m) – 6kN(6m) – 10 kN*m = 0    →    4By = 36 + 36 +10    →    By = 20.5 kN              
 ↑ + ∑Fy = 0 ;       -Ay – 18kN +20.5kN = 0    →    Ay = -18 + 20.5     →    Ay = 3.5 kN ;
 → + ∑Fx = 0 ;        Ax = 0kN ;

Step #3) For shear diagram we place the values on an x-y plane, starting with a 3.5KN upward then continues in a staight line because there is no load on it until it reaches the 18KN force, which makes the shear drop down to -14.5KN and continue in a straight line until it reaches the hinge, which then pushes the shear back up with a value of 20.5KN leaving our shear at 6KN, continues in a straight line till it reaches the 6KN force pushing the shear back to zero.

Step #4) For moment diagram we place the values on an x-y plane and begin by taking the area of the first rectangle in our shear diagram and then continue doing that for the rest of the rectangles, where the area is:

  0<x<2 :    A1 = 3.5*(2)  =  7.5kN*m
  2<x<4 :    A2 = -14.5*(2) = -29 kN*m
  4<x<6 :    A3 = 6*(2)    =  12 kN*m

Shear and Moment Diagram

The shear and moment diagram for Example 1 is shown below in Figure 21.

Figure 21

Note: You can check whether your answer is correct or not by comparing the shear values aquired in the cut section method to the shear diagram values, if both values are equal then the solution is correct and error free.


Example 2: Determine the equations of the axial force, shear force and bending moment as a function of the position along the length of the beam. Draw the shear force diagram and the bending moment diagram for the beam below in figure 22.

Screen Shot 2016-11-07 at 2.25.26 PM.png

Step #1) Draw the free body diagram, and find the reaction forces using the 3 equations of equilibrium. Figure 23.

Screen Shot 2016-11-03 at 2.10.15 AM.png

 + ∑Ma = 0 ;       -20kN(5m)(2.5m) - 50kN(8m) – Cy(10m) = 0    →    10Cy = 250 + 400    →    By = 65 kN              
 ↑ + ∑Fy = 0 ;       Ay – 20kN(5m) - 50kN + 65kN = 0    →    Ay = 100 - 50 + 65     →    Ay = 85 kN ;
 → + ∑Fx = 0 ;        Ax = 0kN ;

Step #2) Determine how many cut section the beam needs, in this particular beam we need 3 cuts. Then determine the unknown forces for each cut using the 3 equations of equilibrium.

cut 1: 0<x<5 Screen Shot 2016-11-03 at 2.35.35 AM.png

 ↑ + ∑Fy = 0 ;       85 kN – 20kN(x) - Vx1 = 0     →    Vx1 = 85 - 20x kN
   + ∑Ma = 0 ;      -85(x) + 20(x)(x/2) + Mx = 0     →   Mx1 = 85x - 10x^2 kN*m     
 → + ∑Fx = 0 ;       0 + Ax = 0          →   Ax1 = 0 kN

cut 2: 0<x<8 Screen Shot 2016-11-03 at 2.44.51 AM.png

 ↑ + ∑Fy = 0 ;       85 kN - 100 kN – Vx2 = 0     →    Vx2 = -15 kN
   + ∑Ma = 0 ;      100(x-2.5) - 85(x) + Mx = 0     →   Mx2 = 250 - 15x kN*m    
 → + ∑Fx = 0 ;       0 + Ax2 = 0          →   Ax2 = 0 kN

cut 3: 0<x<10 Screen Shot 2016-11-03 at 2.52.04 AM.png

 ↑ + ∑Fy = 0 ;       85 kN - 100 kN – 50 - Vx3 = 0     →    Vx3 = -65 kN
   + ∑Ma = 0 ;      100(x-2.5) + 50(x-8) - 85(x) + Mx3 = 0     →   Mx3 = 650 - 65x kN*m     
 → + ∑Fx = 0 ;       0 + Ax2 = 0          →   Ax3 = 0 kN

Step #3) Plug in the values of x into the shear and moment diagram, then use the values calculated to draw the shear and moment diagrams.

for x=0               Vx1 = 85 kN                            Mx1 = 0 kN*m
for x=5               Vx1 = -15 kN                          Mx1 = 175 kN*m
for x=8               Vx2 = -15 kN                          Mx2 = 130 kN*m
for x=10              Vx3 = -65 kN                          Mx3 = 0 kN*m

Step #4) Draw the shear force diagram and the bending moment diagram using the values calculated. Screen Shot 2016-11-03 at 3.23.39 AM.png

Screen Shot 2016-11-03 at 3.23.46 AM.png


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  3. 3.0 3.1 3.2 3.3 3.4 3.5 3.6 Hibbeler, R.C. (2011). Mechanics of Materials: SI Edition (8th ed.). Pearson Education, Inc: Prentice Hall. pp.8, pp.255-265
  4. Analysis of Beams | Shear Force & Bending Moment Diagram Nov. 26, 2013
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