# Bending moments and deflections, safety factors and inertia which to calculate?

1. Aug 27, 2011

### RichMortimer

1. The problem statement, all variables and given/known data

Q: A beam has a solid square cross section of 100 mm and is simply supported by two supports 3 m apart. Calculate the dead load that can be safely supported when applied to the middle of the beam.

* The question doesn't state material but I'm informed that it is mild steel. *

Modules of Elasticity - E = 204 000 N/mm^2 (Stated in appendix of course book)
Safety Factor - k = 4 (Stated in appendix of course book)
Square beam - d = 100 mm (Stated in question)
Distance between supports - L = 3 M (Stated in question)
Average ultimate strength in bending = 480 N/mm^2 (Stated in appendix of course book)

2. Relevant equations

Moments of Inertia - I = 1/12d^4
Moments of Inertia - y = 1/2d
Maximum Deflection - Y = wL^3/48EI
Maximum Bending Moment - M = wL/4

3. The attempt at a solution

Ok, so my understanding of this question is that by calculating at what force the beam will fail then using the safety factor, divide this by 4.
This problem has been covered in thread here www.physicsforums.com/showthread.php?t=463768 however there seemed to be a lot of confusion over the problem, without a resolve.

Moment of Inertia to start then...

I = 1/12d^4 1/12 = 0.0833 x 100^4 I = 8 330 000 mm^4

Maximum Deflection as far as can be solved...

Y = wL^3/48EI wL^3 = w27 48EI = 48 x 204 000 x 8 330 000 = 8.156736 x 10^13

Y = w27/8.156736 x 10^13

It is around now that I loose myself and hit a wall. I don't understand how to extract w from the above without having Y, but can't get Y without w!!!

Is there another way to calculate Deflection without a load?? I've searched the net and come in empty handed.

I would like to solve this myself and therefore don't want an answer to the problem, just a kick in the right direction would be fantastic.

Richard.

Last edited: Aug 27, 2011
2. Aug 27, 2011

### PhanthomJay

Re: Bending moments and deflections, safety factors and inertia... which to calculate

I have not checked your numbers, but you seem to looking at max deflection (which is not required for this problem) when instead you should be looking at max moment and max stress produced by that moment, using the bending stress formula. What is the formula for max bending stress? Also note that the y = d/2 term is not the moment of inertia, that is just the distance from the neutral axis of the beam to the far fibers of the beam.

3. Aug 27, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

Good evening PhantomJay,

Unless I'm missing something, I cannot solve this as the formula requires a load and it is this value that I'm looking for.

The formula though, for a simple beam loaded in the middle is: M = wL/4

As for y = 1/2d, this formula is in a table in my course book headed "Moments of Inertia" with a sub heading of I = 1/12d^4 and another heading of y = 1/2d

I'm afraid there are a lot of inconsistencies in the book. Is that another one??

4. Aug 27, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

Ok, So I've been looking harder and found another formula for M.

M = stressb x I / k x y

stressb = 480 N/mm^2 (Stated in book for mild steel)

So..

M = 480 x 8 330 000 / 0.25 x 50

M = 319872000 N/mm^2 = 319872000 / 9.81 x 1000

M = 32606.727 tonnes/mm^2

Now that is obviously very wrong!!! It's late and I'll try again in the morning.

Thanks again.

5. Aug 27, 2011

### PhanthomJay

Re: Bending moments and deflections, safety factors and inertia... which to calculate

I'm again not checking your numbers, just your formulas. You now seem to have found the correct ones, M = wL/4, and also M = (stressb)I/(ky). But you used k = .25 whereas k is given as 4. And watch units! Moments are in units of force times length.

It is good that you were able to find the correct formulas, but you should first understand why these are the correct formulas to use, and the principles behind the development of the equations. Without much understanding of this, you'll be at a disadvantage, and you'll just be doing basic algebra, not engineering. Perhaps you may need a better book.

Last edited: Aug 27, 2011
6. Aug 28, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

Hi Phantom Jay.

It's a distance learning course I'm using in order to get onto a HNC/D at a "Real" college. The company are a little lacking shall we say. The frustrating thing is that I've now been accepted so it would be easy to give up, but I don't want to! I agree to the need to understand the formula rather than just using them. Hopefully I'll improve massively with an actual tutor at hand! In the mean time, are there any books you feel are worth picking up regards understanding equations?

As for the units, again, I've followed an example in the course book, but I'll research this online.

Thanks again for your help. I'll post my findings and maybe you could have a look?

Cheers,

Richard.

7. Aug 28, 2011

### PhanthomJay

Re: Bending moments and deflections, safety factors and inertia... which to calculate

I don't have any particular book in mind, but you need a good Intro Physics background, some basic calculus, and books on 'Engineering Statics' and "Strength of Materials". You also need a tutor since it is not easy to self teach these topics. (I don't know what is an HNC/D).

8. Aug 28, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

Right then,

I'm a little confused again as I've now got 3 formulas for M, one which cannot be solved without w, and the other 2 give different results. I need to make sense of these formulas!!!

1) M = wL/4 (Need w)

2) M = (stessb) I / k x y 480 x 8 330 000 / 4 x 50 M = 19 992 000 N/mm^2

3) M = (stressb) x (I/y) 480 x ( 8 330 000 / 50 ) M = 79 968 000 N/mm^2

Right, by showing as so, I've realised that formula 3 is 4 times formula 2, which uses the safety factor of 4!!! How does this help me??? Doe's this mean that formula 2 is the safe bending moment where as 3 is the maximum bending moment?

I know I'm typing out load, but in doing so someone may spot my mistaken thinking!!!

Ok then. If I take the 2nd option as the safe moment, do I just need to transpose the formula of:

M = wL/4 into... w = 4M/L resolved is .... w = 4 x 19 992 000 / 3 w = 26 656 000 N

So if this is the case, my answer to the question should be.....

The safe dead load maintainable on a simply supported mild steel beam of 100 mm cross section at a length of 3 meters is: 26 656 kN's

This seam completely off the mark!!!! So where am I going wrong... My assumption that I can extract w so easily!? Or that my value of M is correct/incorrect?

Back to researching Bending Moments I think...

Any input anyone, feel free!

EDIT

Still typing out loud.. Why have I not needed / used E yet?...

Last edited: Aug 28, 2011
9. Aug 28, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

I'll take a look at some books, thanks.

As for the course, It's one year short of a Degree with the course I'm taking now being minimum entry!

Thanks again.

10. Aug 28, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

Ok, with some more reading, I have come across section modulus and this statement came up.

"for a given cross section, the moment of inertia tells us what kind of stiffness to expect, and the section modulus tells us what kind of strength to expect.

So, by this, if I calculate the section modulus to give me the beams strength... can I then use this to calculate the load???

* * * * * * * * Head dizzy!!! Seeing stars! * * * * ** *

11. Aug 28, 2011

### pongo38

Re: Bending moments and deflections, safety factors and inertia... which to calculate

There is plenty of scope for confusion. The modulus of Elasticity, E, is also known as Young's Modulus, and could be used as part of the deflection formula. The elastic section modulus (sometimes confusingly shortened to elastic modulus) is usually given the symbol Z. However, if you are on an hnc course, you may in the future get involved with eurocodes which use a different symbol for Z. Z is simply I/y. So the stress is given by f=My/I, and this is what you need to calculate strength, in this example.

12. Aug 28, 2011

### PhanthomJay

Re: Bending moments and deflections, safety factors and inertia... which to calculate

There is no way to understand section modulus without studying how it is derived and why it is used. In your example, the section modulus is I/ymax = 2I/d. Forget the section modulus for a moment.

The problem is asking for w. You know that M = wL/4. But you don't yet know M or w. You look up the the properties of mild steel and find what its allowable stress is using a given safety factor. Then since stress is My/I , set the stress equal to the allowable stress and solve for the allowable moment. Then plug that back into the first equation to solve for w.

13. Aug 30, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

Good evening,

Once again, thanks again for the replies, they're very much appreciated!

Ok, so, right then, well, hmmm....

PhantomJay, You've said "You look up the the properties of mild steel and find what its allowable stress is using a given safety factor."
Is this the same as the Ultimate strength in bending? Its the only relevant table I can find within the course book! Also titled (stressb) and measured in N/mm^2.
If this is the same thing, I have (stressb) at 480 N/mm^2 and my safety factor of 4. Would I be right to assume an allowable stress of 120 N/mm^2?

Therefore...

stress = My/I M = stress / (y/I) M = 120 / (50 / 8 330 000) M = 8.3300 N/mm^2

M = wL/4 w = 4M/L w = 4 x 8.3300/3 w = 11.11 What? N?

This sounds far far too low a figure for a steel beam!!!!

Will try again...

14. Aug 30, 2011

### PhanthomJay

Re: Bending moments and deflections, safety factors and inertia... which to calculate

Being from the US, I'm not so good with SI units, but doing a conversion it seems that 480 N/mm^2 is an ultimate stress as opposed to a yield stress and that the Safety Factor of 4 is thus reasonable....so your 120 N/mm^2 seems good as an allowable stress.
yes
Better do the math(s) again...and please watch units... I get M = 20 000 000 N-mm, roughly
make your correction for M and please watch your units! that 3 meters has to be converted to 3000 millimeters to get the answer for w in Newtons.

15. Aug 30, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

So I'm typing out loud here!

What am I doing wrong?.... Is the allowable stress of mild steel an available number? Based on online searches, "Mild steel" is very general!

What have I missed in my text book??? Appendix Tables...

1) Tensile strength. Gives range. Tensile not useful now. (Although beam is under tension and compression during bending.)

2) Compressive strength. (as above.)

3) Shearing and torsional. Not req.

4) Working Stress. "Fraction of the ultimate strength to regard as the working strength" All fractions, same values as later "Safety" table! Why?

5)Compositions of alloy steels. Not req.

6) Linear expansion. (Not req. I think!)

7) Average ultimate strength in bending - stressb. THIS IS IMPORTANT RICH! how does it relate to "allowable" stress? Research more.

8) Values of constants of columns. Not req.

9) Factors of safety. Same values as working stress table.

10) Bending moments and deflection calculations. - Gives me M = wL/4 and deflection Y = wL^3/48EI --- HOW CAN DEFLECION HELP? Can it help?

11) Last table.. Moments of inertia. I = 1/12d^4 and 1/2d Needed!

Right, next step, look into deflection and continue searching "Allowable" stress.

What else have I missed?...

16. Aug 30, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

PhantomJay...

Thank you!

17. Aug 30, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

Is the end in site...

Ok Phantom Jay, what do you think?

"Better do the math(s) again...and please watch units... I get M = 20 000 000 N-mm, roughly"

I can get 19 992 000 N/mm if I transpose as M = Stress x I / y but that seems to be wrong as the original formula was stress = M(I/y)

As for the the 19 992 000 N/mm, I have actually already calculated this previously in post #8 with the formula:

M = (stessb) I / k x y 480 x 8 330 000 / 4 x 50 M = 19 992 000 N/mm^2

So by that we're heading the right way!

So to clarify, M is the maximum bending moment and is in N/mm not N/mm^2

Carrying on then...

M = wL/4

w = 4M/L

w = 4 x 19 992 000 / 3000

w = 26 656 N

The final answer then is: 26 656 N ???

It just shows how close you can be in error... post #8

So what do you think? Thanks again for all your help.

18. Aug 30, 2011

### PhanthomJay

Re: Bending moments and deflections, safety factors and inertia... which to calculate

correct
No-oo, you transposed some terms
Indeed, sir, you did!
yes...
Moment has the units of force times length...in the States, it is often stated as ft-lb, where the dash in between ft and lb is really a multiplictaion sign. In your case , the moment is in
Newton millimeters, and I'm not sure how that is abbreviated.
Yes , you have it right, maybe for the second time, nice work. By the way, you don't need to calculate the deflection using Young's modulus. If it was asked for, then you would need to use it. Or if the problem was more complex, it might also come into play.

My ears are ringing from all that typing out loud!

19. Aug 31, 2011

### RichMortimer

Re: Bending moments and deflections, safety factors and inertia... which to calculate

Woooooooooooooooooooooooo Hoooooooooooooooooooooooo!!!!!

Mr PhantomJay, you have been great!!!!

That felt like a long trip, however I feel like I might have actually enjoyed it!!!

Thank you again for the help,

have a great day,

Rich.

P.s. sorry about the ears!!

20. Aug 31, 2011

### PhanthomJay

Re: Bending moments and deflections, safety factors and inertia... which to calculate

My pleasure. You are a gentleman.