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This is the question

I tried to solve this question I arrived at this diagram but taking the moments about the center of the beam I arrived at x=L/2. Can someone tell me what I need to do to get the correct answer?

Diagram

This is my answer

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    $\begingroup$ If you show your working so far - we may be able to help... $\endgroup$
    – Solar Mike
    Commented Feb 3, 2018 at 16:37
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    $\begingroup$ Welcome to Engineering! This looks like a homework question. As mentioned by Solar Mike, in order for such questions to be answered in this site, we need you to show how you have tried to solve this yourself. Please edit your question to include this information. $\endgroup$
    – Wasabi
    Commented Feb 3, 2018 at 16:58
  • $\begingroup$ Following your edit, I have recommended this question to be reopened. That being said, see my answer to a similar question for the solution. $\endgroup$
    – Wasabi
    Commented Feb 7, 2018 at 18:02

1 Answer 1

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Second Attempt:

After reading your comments, I approached the problem using a different process.

Step 1: Find the y-forces using $\Sigma$$F_y = 0$.

$(2\cdot Tsin(\theta))$ - $(W\cdot L)$ $=0$

$2Tsin(\theta)$ = $WL$

Step 2: Treat the beam as a collection of beams.

I decided to break the beam into three parts. The outside edges can be thought of as cantilever beams that are attached by the supports.

Step 3: Making the first cut at the midpoint to the left support.

Using the formula for bending moment of a simple beam under a uniformly distributed load, we can determine that $M_{mid} = \frac{WS^2}{8}$

Step 4: Making the second cut at the left sling.

Using the formula for bending moment of a cantilever beam under a uniformly distributed load, we can determine that $M_{cantilever} = \frac{Wx^2}{2}$

Step 5: Setting equations equal and solving.

From left end, to get the sum of moments to equal zero, $M_{cantilever} = \frac{M_{mid}}{2}$

Therefore:

$\frac{WS^2}{16} =\frac{Wx^2}{2}$

$WS^2=8Wx^2$

$S^2=8x^2$

$S = 2\sqrt{2}x$

Final:

As you can see, the center span $S = 2\sqrt{2}x$. Because $S = L-2x$, you can substitute to find the equation $L-2x = 2\sqrt{2}x$

$L = 2\sqrt{2}x + 2x$

$L \approx 4.8284x$

$x \approx \frac{1}{4.8284L} \approx 0.2071L$

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Where $x$ is the distance from a support on the beam to the left end of the beam.

Where $S$ is the span of the beam.

Where $L$ is the length of the beam.

To clarify, the beam of length $L$ is divided into three sections; the left cantilever of length $x$, the center span $S$ of length $L-2x$, and the right cantilever of length $x$.

Where $Tsin(\theta)$ is y-component of the tension force.

Where $W$ is the weight of the beam.

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  • $\begingroup$ The final answer given in the book is 0.207L $\endgroup$
    – kamalesh
    Commented Feb 7, 2018 at 1:51
  • $\begingroup$ What other information is provided in the problem? $\endgroup$
    – Cardinal
    Commented Feb 7, 2018 at 3:39
  • $\begingroup$ Nothing else is there $\endgroup$
    – kamalesh
    Commented Feb 7, 2018 at 6:31
  • $\begingroup$ A different approach led to the solution above. Less intuition is needed and more practice, I guess. $\endgroup$
    – Cardinal
    Commented Feb 8, 2018 at 1:01

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