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This question is regarding the following problem from Bending stresses in beams

Problem:

enter image description here

enter image description here

My solution and hence the answer doesn't match with the answer in the textbook.

My Solution: The planks A will experience water pressure which will vary along the length of the wooden posts. The pressure at any depth d from the surface will be,

$$P=P_{atm} + \rho g d$$ or at any distance x from the bottom (in the figure below)

$$P=P_{atm} + \rho g (h-x)$$

If we talk about any one the wooden posts, the pressure at any distance x from the bottom will be,

$$P=\rho g (h-x)$$

since the pressure also acts on the post from the other side equal to atmospheric

At any distance x, I consider an area on the wooden post $dA= b\,dx$

enter image description here

The pressure P at this x can be written as

$$P = \frac{\delta F}{dA}$$ $$Pb=\frac{\delta F}{dx}$$

where $\delta F$ is the distributed force over area dA

The quantity $\frac{\delta F}{dx}= q_x$ represents the intensity of the load at any x. Thus, $$q_x = Pb= \rho g (h-x) b$$ maximum intensity occurs at x=0,

$$q_{max}= q = \rho g hb$$

The post is like a cantilever beam subjected to linearly varying load with a maximum intensity at the fixed end, for such a beam the maximum bending moment is given by $qL^2 /6$

Thus maximum bending moment in the post will be

$$M_{max}=\frac{\rho g hb \, h^2}{6}=\frac{\rho g b h^3}{6}$$

Finally, maximum bending stress

$$\sigma_{max} = \frac{\frac{\rho g b h^3}{6}. b/2}{\frac{b^4}{12}}$$

$$\sigma_{max}= \frac{\rho g h^3}{b^2} $$

This maximum stress has to be less than or equal to the allowable bending stress

$$\sigma_{max}= \frac{\rho g h^3}{b^2} \leqslant \sigma_{allow}$$

$$b \geqslant (\frac{\rho g h^3}{\sigma_{allow}})^{1/2}$$

substituting the values gives $b \geqslant 99.04mm $


However the answer in the textbook is 199mm

Where am I going wrong?

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2 Answers 2

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Try solving this. I think you forgot to plug in the spacing to come up the force.

enter image description here enter image description here

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  • $\begingroup$ I don't get why we need to account for that s= 0.8. The water pressure at any distance x from the bottom will be same in the horizontal plane (parallel to the ground). Which means that the pressure on each wooden post (on an area $bdx$) will also be the same. So I can simply multiply the pressure at a distance x on the post by the differential area considered on that post to get the total force acting on the area. Then I can divide this force with dx to get the intensity at that x. Can't I? $\endgroup$ Feb 3 at 5:35
  • $\begingroup$ S is the width of the load that is to be resisted by the post. At any depth from the free water surface, the force per unit height = rhdA = rhS*dx (r is the unit weight of water). $\endgroup$
    – r13
    Feb 3 at 16:46
  • $\begingroup$ "S is the width of the load that is to be resisted by the post.". isnt the Water is extending beyond length s on either side, so that posts have to resist the load of not just length s but entire length of the plank? $\endgroup$ Feb 3 at 18:14
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    $\begingroup$ Each post has a duty (tributary) area that, for an interior post, is consisted of half of the space/length between two posts, and for the end post, half of the post spacing. For the purpose of design, usually, only the interior post is considered. If a post only supports the pressure directly exerted on its width (b), then who is responsible for the remaining pressure exerted on the wood board (wall)? I can't see you have missed the concept of summing the pressure on the wall to an equivalent concentrated load on the post.,please review the sketch and formula carefully. $\endgroup$
    – r13
    Feb 3 at 18:31
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Everything you've done is correct. except you need to consider pressure of 0.8 meter span of water.

So the hydrostatic pressure on each post is $$2* 1000kg/2=1000kg*0.8 \ and\ its \ moment, M= 1000*0.67m*0.8 $$

$\sigma_{max} = M*6/b^3 $

I leave the units conversion to you.

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  • $\begingroup$ I don't get why we need to account for that s= 0.8. The water pressure at any distance x from the bottom will be same in the horizontal plane (parallel to the ground). Which means that the pressure on each wooden post (on an area bdx) will also be the same. So I can simply multiply the pressure at a distance x on the post by the differential area considered on that post to get the total force acting on the area. Then I can divide this force with dx to get the intensity at that x. Can't I? $\endgroup$ Feb 3 at 5:37
  • $\begingroup$ @ Harshit Rajput, The planks are actually resisting the water pressure,. if you remove the planks water will spill and posts wouldn't support anything. $\endgroup$
    – kamran
    Feb 3 at 6:07

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