What is Tangential Deviation?

Deriving the Tangential Deviation

To solve deflection components using this method, we need to find an expression to determine the tangential deviation \(t\) from position \(x\).

Consider the simple beam \(AB\) and its deflected shape \(A'B'\) as shown. 

Say we are interested in finding deviation \(t_{B/A}\).

We can think of a line or curve as points that obey a specific mathematical condition. For example, a circle is a set of points equidistant to a fixed point called the center.

Using the same analogy, we can say that the tangential deviation is a line segment composed of multiple points; however, instead of using one-dimensional points, we can also say that deviation is composed of tiny pieces \(dt\)

We have an idea of what composes our deviation, but what makes \(dt\)? To answer that, consider point \(C^{\prime}\) and an adjacent point \(C^{\prime \prime}\). Draw tangent lines and extend them until it intersects the deviation. 

Since we are obeying the Bernoulli-Euler model, the distance between \(C^{\prime}\) and \(C^{\prime \prime}\) relative to each other is approaching zero. In that way, we can say that points \(C^{\prime}\) and \(C^{\prime \prime}\) come from the same position \(\bar{x}\) - the distance of \(C\) to the deviation. 

Let \(d\theta\) be the central angle between these two tangent lines. If you look closely, we can think of it as an intercept (arc length) of \(\bar{x}\) and \(d\theta\):

\(dt=\bar{x}d\theta\)

Since deviation is the sum of minuscule intercepts, we integrate the previous expression to get \(t_{B/A}\):

Equation 1: \(t_{B/A}=\int\bar{x}d\theta\)

Recall in the derivation of the Bernoulli-Euler Beam Equation that the beam's curvature is:

\(d\theta=\frac{M}{EI}dx\)

Substituting this to Equation 1 will lead to an expression for \(t\):

\(t_{x_2/x_1}=\int_{x_1}^{x_2}\frac{M}{EI}\bar{x}dx\)

This expression is the second theorem of the area moment method: "the tangential deviation of point \(x_2\) from \(x_1\) is equal to the first moment of the area of the \(\frac{M}{EI}\) diagram."

We will use this expression to find the tangential deviation at position \(x_2\) from a reference tangent at position \(x_1\). It is essential to note that \(\bar{x}\) refers to a relative distance to position \(x_2\).

Sign Convention

Like the change in slope, we also have sign conventions for the tangential deviation:

• The tangential deviation is positive if the point along the deflected shape is above the reference tangent.
• Otherwise, it is negative if the point along the deflected shape is below the reference tangent.

‍Summary

Let's summarize:

Any perpendicular distance starting from any point of the deflected shape to the "reference" tangent line is called the tangential deviation.

The second theorem of the area moment method states that "the tangential deviation of point \(x_2\) from \(x_1\) is equal to the first moment of the area of the \(\frac{M}{EI}\) diagram."

\(t_{x_2/x_1}=\int_{x_1}^{x_2}\frac{M}{EI}\bar{x}dx\)

When we have a positive tangential deviation, the point along the deflected shape is above the reference tangent. Otherwise, it is negative if the point along the deflected shape is below the reference tangent.