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Simply Supported Beam Safe load

  • Thread starter rad10k
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  • #1
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Homework Statement



A beam of mild structual steel 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 safely be supported when applied to the middle of the beam?

Homework Equations





The Attempt at a Solution



I am very confused about this could someone please offer me some guidance on what I should be looking to calculate please. Should I be calculating the maximum bending moment / by the material saftey fraction in my book?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
rock.freak667
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In order to do this you will need to get the maximum stress that you want to have in the beam. Without that, you can't really solve it. Were you given the yield point and a factor of safety or anything like that?

EDIT:

A relevant equation should be the simple bending equation of

[tex]\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}[/tex]

You should know how to get I for a square section and M for a simply supported beam with a load acting at the center.
 
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  • #3
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Ok I have an attempt at a solution:

Moment of interia
I = i/y

i = 1/12d^4 = 10*10*10*10*0.0833 = 8333.3 y = 1/2d = 0.5*10 = 5

I= i / y = 166.66 cm^4

k = 4 (factor of safety) Ob = 480 N/ mm^2 ( Table in book for mild steel forgings)

M = ObI/ky

480 * 100 * 166.66 / 4*5 = 399984 / 100 = 3999.84 N M

W = safe load in kg/m e=length of beam

W = M4/e

3999.84*4/3 = 5333.12 NM / 9.81 = W

W = 544 Kg/m

Answer is 544 Kg /m

how does this look ? thanks
 
  • #4
rock.freak667
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Ok I have an attempt at a solution:

Moment of interia
I = i/y

i = 1/12d^4 = 10*10*10*10*0.0833 = 8333.3 y = 1/2d = 0.5*10 = 5

I= i / y = 166.66 cm^4

I am not sure what you mean by I=i/y.

I=8333.3 mm4 and y= 5 mm, but I am not sure what 'i' is.

Though your safe load would in in Newtons not kg/m
 
  • #5
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imnot sure think maybe just confusing myself but for that equationimean I/y =166.66cm^4 and the final answer 5333 NM but just divided it by 9.81 to convert to kg. Is the final answer correct ? thanks
 
  • #6
rock.freak667
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imnot sure think maybe just confusing myself but for that equationimean I/y =166.66cm^4 and the final answer 5333 NM but just divided it by 9.81 to convert to kg. Is the final answer correct ? thanks
The final answer would be 5333 N, not with the units Nm (newton-meter, units for a bending moment).

Also, you don't necessarily have to convert it to kg if you don't want to.
 
  • #7
740
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The elastic section modulus is often called Z (which you have mistakenly called i/y) and has units of length^3. In your notation, i would be the second moment of area with units of length^4. What you have called moment of interia, you mean moment of inertia, which is often regarded as the same as second moment of area. As you are in a muddle, I suggest you make a glossary of relevant terms, their usual symbols, and the units they carry. You need I, y, and Z. What you have called a 'safety fraction', I assume you mean 'safety factor'. What do you suggest for that? Although you could could divide the maximum moment by that, it is usual to reduce the yield stress by dividing it by the safety factor to give you a working stress. Then use that to determine the maximum safe moment.
 
  • #8
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thanks for both your advice.

pongo : thanks for the advice I will make a glossary of terms hopefully stop me getting confused. I used a safety factor of 0.25 as a table from a book in my coursework suggested for mild steel forgings.
 
  • #9
740
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Your 0.25 is probably the reciprocal of the factor of safety I was talking about, which is always greater than 1. Your 0.25 would be then be a multiplicative factor, rather than a dividing one.
 
  • #10
nvn
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rad10k: Your current answer is incorrect. Your current moment of inertia appears incorrect. You must use consistent units. Convert all of your values to N, mm, and MPa, and try again.
 
  • #11
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Ok
d = 100 mm

I = 1/12 * d^4 = 0.0833 * 100 *100 * 100 * 100 = 833000 y = 1/2 * d = 0.5 * 100 = 50

I / y = 8330000 / 50 = 166600 mm^4 is that correct? thanks
 
  • #12
nvn
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rad10k: Your first line in post 11 is correct. Your second line is incorrect. Your third line is almost correct, but it has the wrong units.

You seem to sometimes be forgetting what I mentioned in item 2 of post https://www.physicsforums.com/showthread.php?t=462319#post3078566". Please reread item 2.

Secondly, you are rounding numbers too much. Remember what I told you in the last paragraph of post https://www.physicsforums.com/showthread.php?t=462319#post3078958", which is explained in the first chapter of Beer and Johnston, Vector Mechanics for Engineers. Please reread the following.

Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits, unless the first significant digit of the final answer is 1, in which case round the final answer to four significant digits.

Keep trying.
 
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  • #13
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d= 100 MM

I/Y = Z

I = 1/12d^4 = 83 330 00 MM^4 y = 1/2d = 0.5*100 MM = 50 MM

Z = I/Y 16660 MM^3

This is giving me the section modulus which plugs in to the rest of the formula for finding the strength of the beam ? thanks for everyone help im finding this quite hard !
 
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  • #14
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to give this..

k = 4 (factor of safety) Ob = 480 N/ mm^2 ( Table in book for mild steel forgings)

M = ObI/ky

480 * 16 660 * 100 / 4*5 = 399 984 / 100 = 3 998 400 N M

W = safe load in Newtons e=length of beam

W = M4/e

399 840 * 4 / 3 000 = 5 331.2 N
 
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  • #15
nvn
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rad10k: Now you misspelled some of the unit symbols, by using the wrong capitalization. Please reread post https://www.physicsforums.com/showthread.php?t=462319#post3079135". E.g., use mm, not MM. Use N*m or N*mm, not N*M. Also, newton, not Newton.

The third line of post 13 is correct (except mm is misspelled). The fourth line of post 13 is incorrect. Z = I/y is correct, but you made a mistake computing its value. However, delete the line Z = I/y, because you do not even use Z in your post 14 solution. You instead used I/y.

Hint 1: I = (1/12)*d^4 = 0.083 333*(100 mm)^4 = 8 333 300 mm^4, y = 0.5*d = 0.5*(100 mm) = 50 mm

In post 14, your I and y values are wrong. Try again.
 
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  • #16
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Here I go again ..
d=100 mm k = 4 y Ob = 480 mm^2

I/Y

I==1/12*100^4= 8 333 300 mm^4 y = 1/2*d = 50 mm


I/Y = 166 666 mm^4

M = ObI/ky = 480 * 10 * 166 666 / 200 = 3 999 98.4 /1000 = 3 999.984 N m

W = 4M/e

3 999 98.4 * 4 / 3000 = 5 333 N


:eek:s driving me mad!
 
  • #17
nvn
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rad10k: Your units on Ob in post 16 are wrong, and should be MPa, not mm^2. Your I and y values are correct. Your units on I/y are incorrect, and should be mm^3, not mm^4.

Your answer for moment M is currently wrong. Hint 2: M = Ob*Z/k. Ob = 480 MPa = 480 N/mm^2, Z = I/y, and k = 4. Now compute M.
 
  • #18
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M = ObZ/k

480 * 166 666 / 4 = 1 999 992 0 / 1000 = 1 999 9.92 N m

I used N/mm^2 for Ob as thats the units in the table of my course work book .
 
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  • #19
nvn
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rad10k: Your answer for M is correct. Now compute W.
 
  • #20
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W = 4M/e

1 999 9.92 * 4 / 3 = 26 667 N
 
  • #21
nvn
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rad10k: Your answer is correct. Just one minor comment. It can be written as follows.

W = 4*M/e = 19 999.92*4/3 = 26 667 N

And in post 12, I mentioned you can round the final answer, which, in this case, would be W = 26 700 N = 26.7 kN.
 
  • #22
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thanks nvn for your help .... and patience!
 
  • #23
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The answer 26.7 kN has come back as incorrect. I have been advised to use formulae
WL3/48EI = ML2/12EI i am totally lost now ??
 
  • #24
nvn
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rad10k: It appears they might want you to also check deflection. Or it appears deflection might govern. Therefore, you need to look up the maximum allowable deflection limit. Check your notes or your text book. After you look up the deflection limit, then use the deflection limit and your formula in post 23 to compute W.

However, the question in post 1 asks for the safe load, which you computed. Generally, deflection is not a safety issue. Is the person who is telling you the information in post 23 a credible source? It sounds somewhat questionable.

Perhaps 26.7 kN failed because they want 26.67 kN (?). Also, this assumes your Ob and k values are correct. Are you sure you are using the correct Ob and k values?
 
  • #25
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I was thinking about if I substitute the Max bending moment formula in post 20 w = 4M/e and instead use Me^2/12EI would this give w ?

M = 1 999 9.92 N m, E = 204 N m, I = 8 333.3 m^4 e = 3

I have divided E and I by 1 000 to keep in same units as M

Me^2 = 17 999.28 , 12EI = 203 999 18.4 , 17 999.28 / 203 999 18.4 = 0.008823

W = 0.008823 Which is obviously not correct so now Im back to square one .

The max deflection formula in the text book is 1/48*we^3/EI could this be substituted to give w ? A big problem I have is not understanding how or the correct way of changing formulae around to give correct answers.
 

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