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Solid Mechanics II
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Solid Mechanics II
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C3: Transverse Shear
3.1 Shear Flow
- Theory - Example - Question 1 - Question 2 - Question 3
3.2 Shear Centre
- Theory - Example - Question 1 - Question 2

C3.2 Shear Centre

In all our previous examples, we’ve looked only at cross-sections that are symmetrical. But what happens when we have a shear force acting through a centroid of a non-symmetrical cross-section? Here’s what happens:


Twisting of non-symmetrical cross-section due to shear force applied at centroid

C3.2 Shear Centre

In all our previous examples, we’ve looked only at cross-sections that are symmetrical. But what happens when we have a shear force acting through a centroid of a non-symmetrical cross-section? Here’s what happens:


Twisting of non-symmetrical cross-section due to shear force applied at centroid

It twists! The reason for this is because of the internal shear flow which causes a net torque on the cross-section:


Reaction shear flow causing twisting in cross-section

Note that we usually look at shear flow in terms of the action shear flow (same direction as V), but it is the reaction shear flow (opposite direction) that causes the twisting. For consistency, we will present the following shear flow in the action shear flow direction.

Fortunately there is a way to overcome this. We can apply the shear force at an offset, such that the torque caused by the shear force balanced the torque caused by the reaction shear flow. The position of this offset is what we call our shear centre.


Applying shear force at shear centre to prevent twisting

To calculate the position of our shear centre e, all we do is take the sum of moments about O for the forces acting due to q, and equate that to the moment caused by P acting at e:

Shear centre formula

For other types of cross-section, the same principle applies: taking sum of moments due to other shear flow forces and use that to find the shear centre e.

Let’s look at an example now.

It twists! The reason for this is because of the internal shear flow which causes a net torque on the cross-section:


Reaction shear flow causing twisting in cross-section

Note that we usually look at shear flow in terms of the action shear flow (same direction as V), but it is the reaction shear flow (opposite direction) that causes the twisting. For consistency, we will present the following shear flow in the action shear flow direction.

Fortunately there is a way to overcome this. We can apply the shear force at an offset, such that the torque caused by the shear force balanced the torque caused by the reaction shear flow. The position of this offset is what we call our shear centre.


Applying shear force at shear centre to prevent twisting

To calculate the position of our shear centre e, all we do is take the sum of moments about O for the forces acting due to q, and equate that to the moment caused by P acting at e:

Shear centre formula

For other types of cross-section, the same principle applies: taking sum of moments due to other shear flow forces and use that to find the shear centre e.

Let’s look at an example now.

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