Continuing from this example, let's learn how to find the point of zero shear.

*The solution presented is in SI. The author will update the post soon to reflect English units.*

The point of zero shear is the position along the beam in which the shear is zero. Let's explore different ways to calculate this position using functions or graphs.

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Learn moreContinuing from this example, let's learn how to find the point of zero shear.

*The solution presented is in SI. The author will update the post soon to reflect English units.*

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There are many ways to find the point of zero shear:

**Analytical Approach.**Create the shear equation of a segment. Then, equate the shear to zero and solve for the position \(x\).**Graphical Approach.**Using geometry, plot the points and find the intersection point of the graph and the x-axis.

We will focus on the analytical approach to solve for the positions of zero shear.

The first thing is to find which beam segment will have the point of zero shear occur.

We do this by analyzing the shear at both endpoints of each segment using equations or relating the diagrams. Afterward, we perform a simple test:

- If the shear changed from positive to negative or vice versa between its endpoints, then a point of zero shear occurred.
- If the shear remains the same (positive to positive or negative to negative) for both endpoints, there is no point of zero shear.

In our example, there is a change from positive to negative when analyzing the shear at segment \(BC\):

\(V_B=90kN\)

\(V_C=-11.25kN\)

These results mean that there is a point of zero shear between these two points \(B\) and \(C\).

The point of zero shear is where the maximum moment will occur. In our example, it occurs between points \(B\) and \(C\). We want to answer the question: "where in between these two points is the location of zero shear?"

To answer that, we apply algebra. We formulate the shear function for segment \(BC\), substitute \(V_{BC}=0\), and solve for \(x\):

\(V_{BC}=-5x^2+20x+70\)

\(0=-5x^2+20x+70\)

Since this is a quadratic function, we can apply the quadratic formula:

\(x=\frac{-(-20) \pm \sqrt{(-20)^2-4 (5) (-70)}}{2 (5)}\)

\(x=6.25\)

The point of zero shear is at a distance \(6.25m\) from the origin at \(A\).

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Created On

June 5, 2023

Updated On

February 23, 2024

Contributors

Edgar Christian Dirige

Founder

References

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Revision

1.00

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