• Welcome to engineeringclicks.com
• # approximate tensile strength on simple angled container

Discussion in 'The main mechanical design forum' started by oemodm, Apr 18, 2015.

1. ### oemodmActive Member

Joined:
May 2014
Posts:
25
0
Hello everyone,

I have an angled container, carrying 20kg, its bottom support includes a section of steel wire (A). I would really appreciate an easy way to approximate the tensile stress on this part. Any tips on ways to approximate stress in approximate terms give or take 10 or 20% would be great. Thanks so much / simon

2.
3. ### LochnagarWell-Known Member

Joined:
Feb 2011
Posts:
157
0
Hi Oemodm,

The problem you have is what is called an indeterminate problem. However, if you change the problem to a determinate problem, by "assuming" that the "triangle" you have on the right hand side of the picture are three links with pins joining them - then the problem can be solved - without resorting to the use of FEA which would typically be used these days to solve an indeterminate problem. (Also whether this assumption is reasonable - or approximate - would need to be reviewed upon more details than you have provided above).

Since the 20Kg force acts vertically downwards - due to gravity - then there are no horizontal components of force in the system - and so the force in the "link" A - will theoretically be zero. However, in practice it would have a "nominal" non zero value. Additionally, if the container is subjected to any kind of wind load - or the two vertical forces you have drawn are not entirely vertical - then this would result in a horizontal component of force - which will need to be reacted at the two support points - and will therefore make the force in "link" A - non zero too.
I would not however remove "link" A since it is also possible that in addition to the above points I have made - that when you are filling the container - that there will also be horizontal components of force too - which would need to be reacted by "link" A.

Hope this helps.

4. ### oemodmActive Member

Joined:
May 2014
Posts:
25
0
Hi Lochnagar, you are right. I should have defined this better. B = Swing out leg, hinged at C to the main body
A = Steel wire

Disregarding horizontal forces for now, I'm assuming the tension on A will be somewhere in the region of 8kg? Complete guess. I wonder if there are any simple methods for guesstimating as I'll simply double the strength rating of A linkage in real anyway.

(Horizontal forces will be a definite fact, and will be calculated from actual testing in real).

thanks / simon

5. ### LochnagarWell-Known Member

Joined:
Feb 2011
Posts:
157
0
Hi Simon,

Firstly, your sketch is not to scale - by a long way - and if it was to scale then link B would be inclined in the opposite direction to which you have drawn it.
(Your 10cm link A is almost as long as the 50cm side of the box, and the 90cm side of the box is about half the length it should be if scaled from link A). So I wonder, if the image is meant to depict the real layout - or if the dimensions are correct - and if so if they depict the real layout?
I guess it would be helpful if the sketch was updated - and drawn to scale - then we could see the real problem. (It is good engineering practice to draw sketches approximately to scale - because otherwise it can mislead other engineers).

Hope this helps.

6. ### oemodmActive Member

Joined:
May 2014
Posts:
25
0

Totally understand. But I'm just after the gist rather than precise detail.

For arguments sake, will the tensile force on A (steel wire) be approximately 13kg? I guess as most of the weight is take by B, pulled by A. I'm guessing it will be around the 13kg, but no more than 20kg? If so I could just rate the steel wire to 40kg. Pretty please ;-)

7. ### LochnagarWell-Known Member

Joined:
Feb 2011
Posts:
157
0
Hi Simon,

The magnitude of the force in link "A" and whether this force is tensile or compressive is going to depend on two things:-

1. The angle of inclination of the link "B" to the vertical. In the sketch you have posted (which as I have said previously is completely out of scale) - but if by chance the angle is say 10 degrees to the vertical - then the force in link "A" will be very small. If the angle to the vertical was say 80 degrees - then the force in link "A" will be large.

2. Whether the angle of inclination of the link "B" is negative as shown (using standard mathematical sign convention) - will put a tensile force in link "A" - however if the angle of inclination of the link "B" is positive - will put a compressive force in link "A".

You need to be careful with compressive forces - since failure can occur well below yield strength - due to buckling if the link is slender.

So you can perhaps see why it is important to draw the "free body diagram" to scale - so that we can see at a glance what is going on.

Hope this helps.

8. ### oemodmActive Member

Joined:
May 2014
Posts:
25
0
Hi Lochnagar,

Ok, so there are Lochnagar named files on my PC now Drawn to scale. A is in tensile stress for sure. Noted about risk of compressive strength if B is too slender. Could you help guestimate the tensile stress on A? thanks / simon

9. ### LochnagarWell-Known Member

Joined:
Feb 2011
Posts:
157
0
Hi Simon,

See the numbers in the pictures below:-  Hope this helps.

10. ### oemodmActive Member

Joined:
May 2014
Posts:
25
0
Hi Lochnagar

Really excellent, thank you SO much. Working backwards I can begin to understand practically how you are working this out. This is a great way for learning.

On top of that, the force on A is higher than I thought. And I now totally understand the relevance for scale drawings. On top of that, I can see how to prepare in future for such equations which I don't really understand how to do fully (yet).

thank you so much again Lochnagar 11. ### LochnagarWell-Known Member

Joined:
Feb 2011
Posts:
157 