Principle of Superposition

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Introduction

The Principle of Superposition is a method used to solve complex problems with multiple loads and/or reactions acting on the member. Superposition helps us solve these problems by breaking the member down as many times as necessary for each force acting on it. Once all the stresses or deflections for the point of interest are found, they can then be added all together to get a final answer.

Important Terms

Deflection:

The deviation of a point of interest, from it's original position, due to an external force acting on the member.

Distributed Load:

Uniform external forces that acts on the surface of a member over a specific length.

External Load:

Are the forces acting on the surface of a member. These can include support reactions,applied forces, normal force etc.

Hooke's Law of Elasticity:

For relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load. [1]

Internal Load:

Are the forces that act on a member from the inside. It is the forces that hold the member together when external forces are being applied.

Point Load:

Is a force acting on a single stationary point on a given structure.

Summary

"The principle of superposition simply states that on a linear elastic structure, the combined effect of several loads acting simultaneously is equal to the algebraic sum of the effects of each load acting individually." [2]

Hibbeler states this as "The total displacement or internal loadings (stress) at a point in a structure subjected to several external loads, which can be determined by adding together the displacements or internal loadings (stress) caused by each external load acting separately." [3]

In other words, for a linearly elastic structure, the effect of several loads acting on a member is equal to the summation of the loads acting separately.

Figure 1: Principle of Superposition


Simplified usage of Superposition shown in FIGURE 1:

A given beam and its loadings can be split into simpler beams and loading. As shown in figure 1, the beam with the distributed load and the point load can be split into two beams. One beam with the distributed load and the other with the point load.

Conditions

There are two conditions in which a structure must satisfy in order for the principle to be valid. Structures which do satisfy these two conditions are referred to as Linearly Elastic Structures.

The two conditions are as follows:

i. The equations of equilibrium must be based off of the undeformed shape of the structure:

If the size of the deformations occurring to the structure are small enough to be considered negligible, it can be determined that the undeformed shape of the structure can be used as a basis for the equilibrium equations[2].


ii. The material used in the structure must be Linearly Elastic:

In order for a material to be considered this way, It's stress-strain relationship must relate to Hooke's Law of elasticity, referring to the stiffness of the materials[2].

Equation

The right side of the equation is the algebraic sum of the left side of the equation.

  • $ \ f(z_1 + z_2 + z_3 + ... + z_n) = f(z_1) + f(z_2) + f(z_3)+ ... + f(z_n) $

Sign Convention

In order to prevent confusion and maintain clarity throughout a complex analysis, an arbitrary axis system is established and positive directions indicate tensional or compressive forces, whichever is established. Negative values in a positive tension orientation would indicate a compressive force and a negative force in a positive compression orientation would indicate a tensional force. Typically, an X-Y coordinate system would be labelled upwards and rightwards with tension being represented as a positive structural force. Thus any negative force intuitively is compressive.

Trigonometry

Since Superposition states that complex plane forces can be analyzed by respective axis’s, how the forces are decomposed is important and must follow convention. To break these directionally-complex forces into simple axis forces, trigonometry must be extensively used.


Any force acting on a support structure acts on some direction to the beam which is supporting it. The angle between the acting force and the plane of the beam is typically used as the basis angle to which the angle acting on the beam will be broken down into parallel and perpendicular components of the force acting on the beam.

Application

There are many different ways that superposition can be used to analyze structures. Listed below are a couple of applications that it can be used for.

Method of Sections

The application of the principle of superposition in the method of sections is applied to a section of a frame and rather than analyzing a single joint at a time, a selection of members are cut and equilibrium is established using the internal forces of the cut members.

Superposition is used to determine internal forces of the cut member by the equilibrium equations where; $ \Sigma F_x = 0, $ $ \Sigma F_y = 0, $ $ \Sigma M = 0 $.

Where $ \Sigma F_x = 0 $ means the total forces (external and internal) that lie in the x-axis is equal to zero. $ \Sigma F_y = 0 $ is the total forces (external and internal) that lie in the y-axis is equal to zero. $ \Sigma M_p = 0 $ is the the total moment about point p is also equal to zero.

Equilibrium and Superposition

Since the principle of superposition states that "the resultant stress or displacement (or force) can be determined by algebraically summing the stress or displacement (or force) caused by each load component applied separately to the member" [4]. With equilibrium stated as a balance of forces, the two can be equated mathematically as follows:

Superposition:

$ \Sigma F = \Sigma F_n $

$ \Sigma M = \Sigma M_n $

Equilibrium:

$ \Sigma F = 0 $

$ \Sigma M = 0 $

Therefore we can see that:

$ \Sigma F = \Sigma F_n = 0 $

$ \Sigma M = \Sigma M_n = 0 $

Which means that for a given static system with applied external loads, the total sum of the forces acting on the member(s) is the applied external forces plus the reaction forces or moments.

Example

Given Figure 2, use superposition to find the resulting deformation at point E due to the loads applied. $ E=250 $ $ GPa $ and $ I = 125 \times 10^6 mm $.


Figure 2: Example
Figure 3: Free Body Diagram

First we will use our equilibrium equations to find our support reactions.

$ \Sigma M = 0 $

$ 0 = -50 (10) + C_y (20) - 12 \cos 45 (25) - 75 $

$ C_y = 39.4 kN $


$ \Sigma F_y = 0 $

$ 0 = -50 +39.4 - 12 \cos 45 +A_y $

$ A_y = 19.1 kN $


$ \Sigma F_x = 0 $

$ 0 = A_x - 12 \cos 45 $

$ A_x = 8.5 kN $







We can now see our converted beam which is ready for our superposition method.


Now we can solve our individual deflections due to each force acting on the member.

We know with our second moment area theorem that:

Figure 4: Superposition of Beam using 50 kN force
Figure 5: Moment/EI diagram for beam using 50 kN force



$ \delta = area \times \bar{x} $

$ \delta_E_1 = \frac{-500 \times 10 \times 0.5}{EI} \times 26.7 $

$ \delta_E_1 = -213 mm $






Figure 6: Superposition of Beam using 39.4 kN force
Figure 7: Moment/EI diagram for beam using 39.4 kN force

'


$ \delta_E_2 = \frac{788 \times 20 \times 0.5}{EI} \times 23.3 $

$ \delta_E_2 = 587 mm $





Figure 8: Superposition of Beam using 12 kN force
Figure 9: Moment/EI diagram for beam using 12 kN force



$ \delta_E_3 = \frac{-212.5 \times 25 \times 0.5}{EI} \times 21.7 $

$ \delta_E_3 = -184 mm $





Figure 10: Superposition of Beam using 75 kNm moment
Figure 11: Moment/EI diagram for beam using 75 kNm moment


$ \delta_E_4 = \frac{-75 \times 30}{EI} \times 15 $

$ \delta_E_4 = -108 mm $


$ \delta_E = \delta_E_1 + \delta_E_2 + \delta_E_3 + \delta_E_4 $

$ \delta_E = 82 mm $

Example

Example 2: Determine the maximum deflection caused by the applied loads on the cantilever beam shown below in figure 12 using the principle of superposition. $ EI=100MN.m^2 $

Figure 12: Cantilever Beam with the applied loads

Using the universal equilibrium equations, the reactions at support A is calculated as shown below and is depicted in Figure 13:

Figure 13: Cantilever Beam with the reaction forces solved applied loads
↑ + ∑Fy = 0 ;         -70kN - 40(8)kN + Ay = 0     →    Ay = 390 kN 

→ + ∑Fx = 0 ;          0 + Ax = 0          →   Ax = 0 kN 

+ ∑Ma = 0 ;            -70(4) - 40(8)(4) + Ma = 0     →   Ma = 1560 kNm    
 

Deflection caused by the distributed load of 40 kN/m on the beam and deflection caused by the point load of 70 kN is analyzed separately as explained in the steps below.

Step #1) Distributed load of 40 kN/m on the beam is analyzed by first finding the support reactions at point A.

Figure 14: Cantilever Beam with the reaction forces solved for the distributed load of 40kN/m

Support Reactions are found using the equilibrium equations as shown below;

↑ + ∑Fy = 0 ;        - 40(8)kN + Ay = 0     →    Ay = 320 kN 

→ + ∑Fx = 0 ;          0 + Ax = 0          →   Ax = 0 kN 

+ ∑Ma = 0 ;        - 40(8)(4) + Ma = 0     →   Ma = 1280 kNm   

The moment diagram is drawn for the above beam as seen in Figure 15.


Figure 15: Moment Diagram for the distributed load of 40kN/m on the Cantilever beam


The deflection caused by the distributed load is calculated using the following formula that can be obtained from any slope-deflection charts [5] and is as follows:


$ Maximum Defelction,Va = \frac{Distributed load*length^4}{8EI} = \frac{40*8^4}{8EI} = \frac{20480kNm^3}{EI} $


Figure 16: Deflection caused by the distributed load of 40kN/m on the Cantilever beam


Step #2) Point load of 70 kN on the beam is analyzed to find the support reactions at point A.


Figure 17: Cantilever Beam with the reaction forces solved for the Point load of 70 kNm acting on the beam


Support Reactions are found using the equilibrium equations as shown below;

↑ + ∑Fy = 0 ;        - 70 kN + Ay = 0     →    Ay = 70 kN 

→ + ∑Fx = 0 ;          0 + Ax = 0          →   Ax = 0 kN 

+ ∑Ma = 0 ;          -70(4) + Ma = 0     →   Ma = 280 kNm   


The moment diagram is drawn for the above beam as seen in Figure 18.


Figure 18: Moment Diagram for the point load of 70kN on the Cantilever beam


The deflection caused by the point load is calculated using the following formula that can be obtained from any slope-deflection charts [5] and is as follows:


$ Maximum Deflection,Vb = \frac{-5*Load*length^3}{48EI} = \frac{5*70*8^3}{48EI} = \frac{3733kNm^3}{EI} $


Figure 19:Maximum Deflection, Vb caused by the point load of 70kN on the Cantilever beam


The algebraic sum of deflections Va and Vb is equivalent to the deflection Vc, caused by the loads acting simultaneously on the beam as seen in the figure below;


Figure 19:Maximum Deflection, Vc caused by the point load of 70kN and the distributed load 40kN/m on the Cantilever beam


$ Maximum Deflection,Vc = Vb + Vc = \frac{3733kNm^3}{EI} + \frac{20480kNm^3}{EI} = 0.24213m $

References

  1. http://www.britannica.com/EBchecked/topic/271336/Hookes-law
  2. 2.0 2.1 2.2 Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.), Pg 78. Stamford, CT: Cengage Learning.
  3. Hibbeler, R.C. (2012). Structural Analysis (8th ed.), Pg 46. Prentice Hall. Pearson Education, Inc.
  4. Hibbeler, R.C. (2011). Mechanics of Materials (8th ed.), Pg 136. Prentice Hall. Pearson Education, Inc.
  5. 5.0 5.1 Hibbeler,R.C.(2015).Mechanics of Materials(10th ed.),Appendix C. Hoboken,NJ:Pearson Education, Inc.