Lecture 18 - Superposition for Statically Indeterminate Beams PDF

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Lecture 18

19/05/2021

Methods of Superposition for Analysing Statically Indeterminate Beams •

Readings → GG Chapter 10.4

What is a Statically Indeterminate Beam: • •

A beam with more than 2 displacement boundary conditions / constraints A beam with more than 2 reaction forces

Method of Superposition: A powerful method for solving statically indeterminate beams is to use the method of superposition. We follow a four-step process to solve these problems: 1. Remove enough supports to make the beam statically determinate (preferably we want to remove supports to leave either a simply supported beam or a cantilever beam, since we have beam deflection tables for these problems). 2. Replace each support with the reaction force applied as an external load. 3. To enforce removed constraints, add an appropriate displacement constraint condition (sometimes called a compatibility condition). 4. Break the resulting problem into simpler problems for which the solution is known, applying the principle of superposition. This is hard to understand in this abstracted way - we will illustrate it with a simple example.

Figure 1: Propped Cantilever Beam with Uniform Distributed Load q

Consider a propped cantilever beam with a uniform distributed load, as shown in figure 1. We can draw the FBD for this beam and write the boundary conditions.

This beam is statically indeterminate – we have only one force boundary condition. Step 1: Remove the support at B, resulting in a cantilever beam. Step 2: Replace the support with a reaction force, RB, applied as an external load.

Figure 2: Removing one Support and replacing it with a Reaction Force as an Equivalent Load

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Note that figure 2(a) and figure 2(b) are not equivalent yet, because in figure 2(b) we are not enforcing the requirement that there should be no deflection at B. We address this in the next step. Step 3: Add an appropriate displacement constrain, shown in figure 3.

Figure 3: Adding a Displacement Constraint makes (b) equivalent to (a)

Step 4: Break into simpler problems, solve by superposition.

Figure 4: Superposition of Known Solutions

Remember to enforce the displacement constraint!

𝛖(𝐋) = 𝛖𝟏 (𝐋) + 𝛖𝟐 (𝐋) = 𝟎 -----> (1)

Figure 5: Deflections from each case.

We can also write:

𝛅𝐁 = 𝛅𝐁𝟏 − 𝛅𝐁𝟐 = 𝟎 -----> (2) (Recall, 𝛿B is written without sign, that’s why we have a minus sign in this equation). From the deflection tables in Appendix H, we can find expressions for δB1 and δB2 . Table H-1: Case 1

𝛅𝐁𝟏 =

𝐪𝐋𝟒 𝟖𝐄𝐈

-----> (3)

Table H-1: Case 2

𝛅𝐁𝟐 =

𝐑𝐁 𝐋𝟑 -----> (4) 𝟑𝐄𝐈

(3) and (4) in (2): 𝐪𝐋𝟒

𝛅𝐁 = 𝟖𝐄𝐈 − 𝐪𝐋𝟒

= 𝟖𝐄𝐈

𝐑𝐁 =

𝐑 𝐁 𝐋𝟑 𝟑𝐄𝐈

=𝟎

𝐑 𝐁 𝐋𝟑 𝟑𝐄𝐈

𝟑𝐪𝐋 -----> (5) 𝟖

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We have now found the exact value of RB which will ensure that there is zero deflection at B. We can solve for the remaining reactions using N2.

∑ 𝐅𝐲: ∑ 𝐑 𝐀 + 𝐑 𝐁 = 𝐪𝐋 ⟹ 𝐑𝐀 = ∑ 𝐌𝐀 = 𝟎: 𝐌𝐀 + ⟹ 𝐌𝐀 =

𝟑𝐪𝐋𝟐 𝟖

𝟓𝐪𝐋 𝟖

𝐪𝐋𝟐 𝟐

−

− 𝐑𝐁 𝐋 = 𝟎

𝐪𝐋𝟐 𝟐

=−

𝐪𝐋𝟐 𝟖

We can also find the deflections everywhere in the beam by superposition.

𝛖(𝐱) = 𝛖𝟏(𝐱) + 𝛖𝟐(𝐱) 𝛖 ( 𝐱) = −

𝐪𝐱 𝟐 𝟐𝟒𝐄𝐈

(𝟔𝐋𝟐 − 𝟒𝐋𝐱 + 𝐱 𝟐) −

(−𝐑𝐁 )𝐱 𝟐 (𝟑𝐋 𝟔𝐄𝐈

− 𝐱)

𝐪𝐱 𝟐

𝛖(𝐱) = − 𝟒𝟖𝐄𝐈 (𝟑𝐋𝟐 − 𝟓𝐋𝐱 + 𝟐𝐱 𝟐) Let’s repeat this problem, but removing a different support - we will remove the moment reaction at A. Step 1: Remove the moment reaction at A, resulting in a simply supported beam. Step 2: Replace the support with a reaction moment, MA, applied as an external load.

Figure 6: Removing MA and replacing it as an External Load

In order for figure 6(a) and figure 6(b) to be equivalent, we need to enforce the requirement of zero slope at A. Step 3: Add an appropriate displacement constraint, shown in figure 7.

Figure 7: Adding a Displacement Constraint makes (b) equivalent to (a)

Step 4: Break into simpler problems and solve by superposition.

Figure 8: Superposition of Known Solutions

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We enforce the displacement constraint:

𝛉(𝟎) = 𝛉𝟏 (𝟎) + 𝛉𝟐 (𝟎) = 𝟎 -----> (6) OR

𝛖′(𝟎) = 𝛖𝟏 ′(𝟎) + 𝛖𝟐 ′(𝟎) = 𝟎 -----> (7)

Figure 9: Deflections from each case

We could also write:

𝛉𝐀 = 𝛉𝐀𝟏 + 𝛉𝐀𝟐 = 𝟎 -----> (8) Deflection tables give us expressions for θA1 and θA2.

(𝛉𝐀 )𝟏 = (𝛉𝐀 )𝟐 = 𝐪𝐋𝟑 𝟐𝟒𝐄𝐈

=

⟹ 𝐌𝐀 = −

𝐪𝐋𝟑 𝟐𝟒𝐄𝐈 𝐌𝐀 𝐋 𝟑𝐄𝐈

𝐌𝐀 𝐋 𝟑𝐄𝐈

𝐪𝐋𝟐 𝟖

-----> (9)

This agrees with what we found from the first problem. Once we have found the unknown reactions, we can solve the problem as before. Notice that we could also have solved the problem by removing the support for RA as shown in figure 10.

Figure 10: An Alternative approach to solve the problem

Why didn’t we? → We want solutions that we can look up in the beam deflection tables – this isn’t a standard solution.

Example 10.3: Find all reactions for the beam shown in figure 11, using the method of superposition.

Figure 11: A two-space continuous beam ABC with a uniform distributed laod

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Conceptualise: FBD:

Additional displacement constraint: 𝛖𝐁 = 𝛖(𝐋) = 𝟎 Categorise: • The beam is statically indeterminate, and we will solve by superposition. • Unknowns: RA, RB, RC • Strategy: Remove supports at B and replace with applied force RB and displacement constraint 𝜐B = 0 Conceptualise: We are now solving this problem.

Break into two problems using superposition.

Expected deflections:

𝛅𝐁 = 𝛅𝐁1 − 𝛅𝐁2 = 𝟎 Be very careful using the table! Notice we have length 2L instead of L. The labelling of the figures in this problem and in Table H-2 are different!

(𝛅𝐁 )𝟏 = 𝛅𝐦𝐚𝐱 = (𝛅𝐁 )𝟐 = 𝛅𝐦𝐚𝐱 =

𝟓𝐪(𝟐𝐋)𝟒 𝟑𝟖𝟒𝐄𝐈 𝐑 𝐁 (𝟐𝐋)𝟑 𝟒𝟖𝐄𝐈

(𝛅𝐁 )𝟏 = (𝛅𝐁 )𝟐 𝟓𝐪(𝟐𝐋)𝟒 𝟑𝟖𝟒𝐄𝐈

𝐑𝐁 =

=

𝐑 𝐁 (𝟐𝐋)𝟑

𝟒𝟖𝟎𝐪𝐋 𝟑𝟖𝟒

𝟒𝟖𝐄𝐈

=

𝟓𝐪𝐋 𝟒

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We can find the other reactions from N2:

∑ 𝐅𝐲 = 𝟎: 𝐑 𝐀 + 𝐑𝐁 + 𝐑 𝐂 = 𝟐𝐪𝐋 ∑ 𝐌𝐀 = 𝟎: 𝐑𝐁 𝐋 + 𝐑𝐂 𝟐𝐋 − 𝐪(𝟐𝐋)𝐋 = 𝟎 𝐑 𝐂 = 𝐪𝐋 −

𝟓𝐪𝐋

𝐑 𝐀 = 𝟐𝐪𝐋 −

𝟖

=

𝟓𝐪𝐋 𝟒

𝟑𝐪𝐋 𝟖

=

𝟑𝐪𝐋 𝟖

Evaluate: • Check units • Check sign of terms • Check against similar solution Notice that RA = RC, as we would expect, since the problem is symmetric.

Many Supports: Suppose we have a problem like this:

Figure 12: A beam with many Supports

How would we approach it?

Figure 13: FBD for beam in figure 12

What are the unknowns? → MA, ME, RA, RB, RC, RD, RE Remove 5 supports and replace with forces.

Figure 14: Released problem, redundant supports replaced with external forces

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Add 5 compatibility equations, one for each support: • 𝛉𝐀 = 𝟎 • 𝛖𝐁 = 𝟎 • 𝛖𝐂 = 𝟎 • 𝛖𝐃 = 𝟎 • 𝛉𝐄 = 𝟎 Use superpositions to break the problem up into known solutions – nine of them! We will have 5 equations:

(𝛉𝐀 )𝟏 + (𝛉𝐀 )𝟐 + ⋯ + (𝛉𝐀 )𝟗 = 𝟎 ( 𝛖 𝐁 ) 𝟏 + ( 𝛖 𝐁 )𝟐 + ⋯ + ( 𝛖 𝐁 ) 𝟗 = 𝟎 …. … … This will give us a linear system of 5 equations in 5 unknowns, allowing us to solve for MA, ME, RB, RC, RD.

This is a process that can be automated!

Lecture 18

19/05/2021

Reading: GG Chapter 10.4...