Understanding Bending Moment in a Suspended Pile

This question asks about the maximum bending moment experienced by a pile that is supported or "suspended" at two points along its length. The pile has a length $L$ and a uniform load per unit length, let's call it $W$. The phrasing "length L per meter length" is a bit unusual, but it's standard practice to assume $W$ represents the load per unit length over the total length $L$. The key is to find the maximum bending moment, which occurs either at the support points or at the center of the pile.

Scenario: Pile Suspended at Two Points

Imagine a long pile resting on two supports. If the supports are placed at the very ends, it acts as a simply supported beam. However, the question implies the supports are *within* the length, suspending the central portion and leaving overhangs at both ends. To minimize the maximum bending moment, the supports are typically placed symmetrically inwards from the ends.

Let's define:

• Total length of the pile = $L$
• Uniform load per unit length = $W$
• Distance from each end to the suspension point = $a$

The two suspension points are located at a distance $a$ from each end, meaning the distance between the supports is $L - 2a$.

Calculating Bending Moments

The pile experiences bending due to the downward uniform load $W$.

1. Bending Moment in the Overhangs:

Consider an overhang of length $a$. The bending moment at any distance $x$ from the free end (where $0 \le x \le a$) is caused by the uniform load over that length. The moment is negative (hogging).

$$M(x) = - W \frac{x^2}{2}$$

The maximum bending moment in the overhang occurs right at the support (at $x=a$):

$$M_{support} = - W \frac{a^2}{2}$$

The magnitude of this moment is $W \frac{a^2}{2}$.

2. Bending Moment Between Supports:

The section between the supports (length $L-2a$) acts like a beam subjected to the uniform load $W$ and supported by upward reactions at the suspension points. By symmetry, each support reaction ($R$) is half of the total load on the pile:

$$R = \frac{W \times L}{2}$$

The bending moment in the section between supports is generally positive (sagging). The maximum positive bending moment occurs at the center of the pile ($x = L/2$).

Let's calculate the bending moment at the center ($x=L/2$) considering the left support is at $x=a$. We can take moments about the center. The load on the left half ($L/2$) is $W \times (L/2)$. The left reaction $R$ is at $x=a$.

$$M_{center} = R \times (\frac{L}{2} - a) - W \times (\frac{L}{2}) \times (\frac{L}{4})$$ (Load on half span $W(L/2)$ acts at $(L/4)$ from center)

$$M_{center} = \frac{WL}{2} (\frac{L}{2} - a) - \frac{WL^2}{8}$$

$$M_{center} = \frac{WL^2}{4} - \frac{WLa}{2} - \frac{WL^2}{8}$$

$$M_{center} = \frac{2WL^2 - 4WLa - WL^2}{8}$$

$$M_{center} = \frac{WL^2 - 4WLa}{8} = \frac{WL}{8}(L-4a)$$

The maximum bending moment on the pile will be the larger of the magnitude of the support moment $|M_{support}|$ and the center moment $M_{center}$ (assuming $M_{center}$ is positive for appropriate 'a').

Minimizing Maximum Bending Moment

The question implies that the maximum bending moment occurs at the center *or* at the points of suspension. This is the specific case where the magnitudes of the maximum positive and negative moments are equal:

$$|M_{support}| = M_{center}$$

$$W \frac{a^2}{2} = \frac{WL}{8}(L-4a)$$

Assuming $a < L/4$ (which is necessary for the center moment to be positive and equal to the support moment magnitude in this setup) and $W \ne 0$ and $L \ne 0$, we can simplify:

$$\frac{a^2}{2} = \frac{L(L-4a)}{8}$$

$$4a^2 = L(L-4a)$$

$$4a^2 = L^2 - 4aL$$

$$4a^2 + 4aL - L^2 = 0$$

This is a quadratic equation for $a$. Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2c}$ with $a=4$, $b=4L$, $c=-L^2$:

$$a = \frac{-4L \pm \sqrt{(4L)^2 - 4(4)(-L^2)}}{2(4)}$$

$$a = \frac{-4L \pm \sqrt{16L^2 + 16L^2}}{8}$$

$$a = \frac{-4L \pm \sqrt{32L^2}}{8}$$

$$a = \frac{-4L \pm 4L\sqrt{2}}{8}$$

$$a = \frac{-L \pm L\sqrt{2}}{2}$$

Since the distance $a$ must be positive, we take the positive root:

$$a = \frac{(\sqrt{2}-1)L}{2}$$

This distance $a$ from the ends to the supports minimizes the maximum bending moment along the pile. At this configuration, the magnitude of the bending moment at the supports and at the center is equal.

Calculating the Maximum Bending Moment Value

Now, substitute this value of $a$ back into the expression for the magnitude of the support moment $|M_{support}|$:

$$|M_{support}| = W \frac{a^2}{2} = W \frac{1}{2} \left( \frac{(\sqrt{2}-1)L}{2} \right)^2$$

$$|M_{support}| = W \frac{1}{2} \frac{(\sqrt{2}-1)^2 L^2}{4}$$

$$|M_{support}| = \frac{WL^2}{8} (\sqrt{2}-1)^2$$

Expand $(\sqrt{2}-1)^2$:

$$(\sqrt{2}-1)^2 = (\sqrt{2})^2 - 2(\sqrt{2})(1) + 1^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}$$

So, the maximum bending moment (magnitude) is:

$$M_{max} = \frac{WL^2}{8}(3 - 2\sqrt{2})$$

Let's evaluate the numerical value:

$$M_{max} = \frac{WL^2}{8}(3 - 2 \times 1.4142)$$

$$M_{max} = \frac{WL^2}{8}(3 - 2.8284)$$

$$M_{max} = \frac{WL^2}{8}(0.1716)$$

$$M_{max} \approx 0.02145 WL^2$$

Now let's compare this value to the given options:

The calculated maximum bending moment, approximately $0.02145 WL^2$, is closest to $WL^2/47$, which is approximately $0.0213 WL^2$. The slight difference is due to using the approximated value of $\sqrt{2}$ or possibly rounding in the options. The formula $\frac{WL^2}{8}(3 - 2\sqrt{2})$ is the exact value for the minimum possible maximum bending moment in this configuration.

Therefore, the maximum bending moment at the center of the pile or at the points of suspension, for optimal support placement, is approximately $WL^2/47$.

Conclusion

For a pile of length $L$ with uniform load $W$ per unit length, suspended at two symmetric points, the minimum value for the maximum bending moment occurs when the magnitude of the bending moment at the supports equals the bending moment at the center. This maximum bending moment value is $\frac{WL^2}{8}(3 - 2\sqrt{2})$, which is approximately equal to $WL^2/47$.

Revision Table: Key Concepts for Suspended Pile Bending Moment

Additional Information: Beam Bending and Support Types

Understanding different beam support types is crucial for analyzing bending moments. A beam's behavior under load depends heavily on how it is supported.

• Simply Supported Beam: Supported by a pin at one end and a roller at the other. Bending moment is zero at supports. Maximum bending occurs somewhere between supports (e.g., at center for uniform load).
• Cantilever Beam: Fixed at one end and free at the other. Maximum bending moment occurs at the fixed support.
• Overhanging Beam: A beam with supports not at the ends, creating overhangs. This type has both positive and negative bending moments. Negative moments occur over the supports (due to overhangs), and positive moments occur between supports.

The pile in this question is an example of an overhanging beam supported at two points. The load distribution and support locations determine the bending moment diagram. The goal is often to design structures such that the maximum stress due to bending is minimized, which relates directly to minimizing the maximum bending moment.

The formula $\frac{WL^2}{8}(3 - 2\sqrt{2})$ is a standard result for the minimum possible maximum bending moment in a uniformly loaded beam of length $L$ supported symmetrically at two points. It is achieved when the supports are located at a distance $a = \frac{(\sqrt{2}-1)L}{2}$ from each end.