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Solid Mechanics II
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Solid Mechanics II
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C7: Deflection of Beams and Shafts
7.1 Superposition Method for Statically Indeterminate Beams and Shafts
- Theory - Example - Question 1 - Question 2 - Question 3 - Question 4

C7.1 Superposition Method for Statically Indeterminate Beams and Shafts

In the real world, beams and shafts are often given more support than necessary which causes it to be indeterminate. A common example is the beam in steel frames of buildings, where it’s fixed on both ends:


Steel frame being fixed on both ends (statically indeterminate) to provide more support for use in structural frames

C7.1 Superposition Method for Statically Indeterminate Beams and Shafts

In the real world, beams and shafts are often given more support than necessary which causes it to be indeterminate. A common example is the beam in steel frames of buildings, where it’s fixed on both ends:


Steel frame being fixed on both ends (statically indeterminate) to provide more support for use in structural frames

Adding supports in this manner is actually good (even though its indeterminate), because it makes structures stronger. Therefore knowing how to analyse statically indeterminate beams/shafts will be important for you as future engineers.

In indeterminate scenarios we have more forces than necessary to ensure static equilibrium. To overcome this, we use the superposition method, which states that the net displacement on a point is just the sum of the displacements due to the individual forces acting at that point.

Sounds complicated? Not to worry, let’s look at a simple example to understand this:


statically indeterminate beam with redundant pin-support

The beam above has an extra support at B which is redundant. To fully analyse the beam, we need to reactions Ay, By and Cy. Using our standard equations of equilibrium, we can get Ay and Cy in terms of By but we still need to solve for By.

Here’s where superposition comes in. We isolate the support B as such:


Isolation of support B to apply superposition method

That is, for (i) we remove the redundant support at B and consider the resulting displacement (ΔBi↓). Then in (ii) we consider the displacement caused by the reaction By (ΔBii↑). Since in the real scenario there is a pin at B (i.e. ΔB = 0), we have our compatibility equation:

Compatibility equation

Using this, we can solve for By and then get our Ay and Cy.

The displacements ΔBi↓ and ΔBii↓ can be obtained either using double integration or virtual work. For simplicity, we provide here a list of displacement expressions for common loading conditions:


Displacement formulas for common loading scenarios

Alright! Looks like we’re ready. Let’s look at an example now.

Adding supports in this manner is actually good (even though its indeterminate), because it makes structures stronger. Therefore knowing how to analyse statically indeterminate beams/shafts will be important for you as future engineers.

In indeterminate scenarios we have more forces than necessary to ensure static equilibrium. To overcome this, we use the superposition method, which states that the net displacement on a point is just the sum of the displacements due to the individual forces acting at that point.

Sounds complicated? Not to worry, let’s look at a simple example to understand this:


statically indeterminate beam with redundant pin-support

The beam above has an extra support at B which is redundant. To fully analyse the beam, we need to reactions Ay, By and Cy. Using our standard equations of equilibrium, we can get Ay and Cy in terms of By but we still need to solve for By.

Here’s where superposition comes in. We isolate the support B as such:


Isolation of support B to apply superposition method

That is, for (i) we remove the redundant support at B and consider the resulting displacement (ΔBi↓). Then in (ii) we consider the displacement caused by the reaction By (ΔBii↑). Since in the real scenario there is a pin at B (i.e. ΔB = 0), we have our compatibility equation:

Compatibility equation

Using this, we can solve for By and then get our Ay and Cy.

The displacements ΔBi↓ and ΔBii↓ can be obtained either using double integration or virtual work. For simplicity, we provide here a list of displacement expressions for common loading conditions:


Displacement formulas for common loading scenarios

Alright! Looks like we’re ready. Let’s look at an example now.

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