Simply Supported Beam Safe load

  • Thread starter rad10k
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In summary: MM in some places.- using space between digits, instead of no space between digits in some numbers.Otherwise, all your unit symbols are correct.Your calculations are still incorrect. You need to multiply by the gravity acceleration, g, or divide by the acceleration due to gravity, g, depending on which equation you use.You are missing a multiplication sign between 4 and y, in the denominator of the second equation. It should be 4*y, not 4y. Also, you must multiply by g, or divide by g, depending on which equation you use.You are missing a multiplication sign between 399 840 and 4, in the numerator of the third equation
  • #1
rad10k
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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?
 
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  • #2
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
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
rad10k said:
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
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
rad10k said:
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
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
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
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
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
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
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
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 into the rest of the formula for finding the strength of the beam ? thanks for everyone help I am finding this quite hard !
 
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  • #14
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
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
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
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
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 that's the units in the table of my course work book .
 
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  • #19
rad10k: Your answer for M is correct. Now compute W.
 
  • #20
W = 4M/e

1 999 9.92 * 4 / 3 = 26 667 N
 
  • #21
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
thanks nvn for your help ... and patience!
 
  • #23
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
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 textbook. 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
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 I am back to square one .

The max deflection formula in the textbook 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.
 
  • #26
rad10k said:
A big problem I have is not understanding how or the correct way of changing formulae around to give correct answers.

Could you ellaborate on this? Then, you might get appropriate help. Can you give an example of what you mean?

Looking at the whole thread suggests that you just want to 'plug numbers into formulas' rather than understand what the formulas are telling you. Sometimes, it helps to state formulas in such a form as: stress f = My/I (MPa), or deflection D = WL^3/48EI (mm) or whatever...
 
  • #27
rad10k: In post 25, your values and/or units for E, I, M*e^2, and 12*E*I are wrong. E.g., E should be 204e9 Pa.

When you group digits in long numbers, always start from the decimal point and go outward. E.g., write M = 19 999.92 N*m, not 1 999 9.92 N*m.

You should not be using 19 999.92 N*m for your formula in post 25. You must set your formula in post 25 equal to the deflection limit, such as L/250, or whatever deflection limit your professor wants you to use. Then solve for W.
 
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  • #28
Pongo38 : Well for instance the formula in post 20 : w = 4M/e. That was originally in my textbook as a formula for maximum bending moment : M = we/4 for a simply supported beam with point load in the middle but had been swapped around to solve to be used to solve W. I don't understand why the 4 was moved to be multiplied by M and then divided by e ?? Hope this ellaborates more on what I mean :eek:s
 
  • #29
nvn: Ok thanks , I shall go and study deflection some more as I seem to have got completely lost in this question at the moment.
 
  • #30
I have a new attempt at a solution using formulae in post 23. The safe load comes out again at 26.7kN so I'm thinking it must be correct if two different formulae have given the same answer. Using these formulae gives the answer to the delfection as well.

ML^2/12EI = Allowable Deflection = D, then DEI48/L^3 = W

As advised took Mild Structural Steel as material

Ob = 480 N , FOS = 4 I=1/12*d4= 8 333 000 , y= 50 mm Z=I/y E = 204 000

M=ObZ/k= 480*166 660/4= 1 999 9200 N mm

ML^2/12EI = 1 999 9200* 90^6 / 20 399 841 000 000 = 8.8235 mm

Max Deflection = 8.8 mm

DEI48/L^3 = 8.8235 * 204 000 * 8 333 000 * 48/27^9 = 26 665.51111466 N

Safe Load = 26.7kN

How does this look ?
 
  • #31
rad10k: You obtained the same answer because you used your previous applied load to compute the deflection. Then you used that deflection to again compute the applied load. Perhaps talk to your tutor.
 
  • #32
rad10k said:
Pongo38 : Well for instance the formula in post 20 : w = 4M/e. That was originally in my textbook as a formula for maximum bending moment : M = we/4 for a simply supported beam with point load in the middle but had been swapped around to solve to be used to solve W. I don't understand why the 4 was moved to be multiplied by M and then divided by e ?? Hope this ellaborates more on what I mean :eek:s

w = 4M/e is an equation where the left hand side is equal to the right hand side. If you divide both sides of the equation by 4, then the equality should hold. So, w/4 = M/e because 4/4 is the same as 1.
Then if you multiply both sides of the equation by e, then the equality still holds. So, we/4 = M.
Does that cover the point you are making?
Incidentally it is conventional to distinguish between upper case W and lower case w. So you should be careful to use the same symbol throughout.
 
  • #33
thanks pongo38 that helps.

nvn : I spoke to my tutor and explained that I was getting the same answer using the formula in post 23. Apparently I am solving ObZ/k = M incorrectly and should be using EL = M
then using the formula in post 23.

E = 204 000
 
  • #34
hi i am have the same problems with this quetion you mentioned that your tutor said use EL= M i asume that E is modules of elastcity in N/mm2 and i assume L is length of beam in mms and you also say that the formula is post 23 but i cannot work out which formula you used any help and i would be very greatfull
 

1. What is a simply supported beam?

A simply supported beam is a type of structural element that is supported at two points, usually at its ends. It is commonly used in construction and engineering projects to support loads and distribute weight.

2. How do you determine the safe load for a simply supported beam?

The safe load for a simply supported beam is determined by calculating the maximum bending moment and shear stress that the beam can withstand without experiencing failure. This calculation takes into account the material properties of the beam, its dimensions, and the support conditions.

3. What factors affect the safe load of a simply supported beam?

The safe load of a simply supported beam is affected by several factors, including the material properties of the beam, its dimensions, the type of load applied, and the support conditions. Other factors such as temperature, humidity, and environmental conditions can also impact the safe load.

4. Can the safe load of a simply supported beam be increased?

Yes, the safe load of a simply supported beam can be increased by using stronger materials, increasing the beam's dimensions, or changing the support conditions. However, it is important to consult with a structural engineer to ensure the changes do not compromise the overall structural integrity of the beam.

5. Are there any safety guidelines for using simply supported beams?

Yes, there are safety guidelines that should be followed when using simply supported beams. These include ensuring proper support conditions, not exceeding the safe load, and regularly inspecting the beam for any signs of damage or wear. It is also important to follow any specific guidelines or regulations set by local building codes or industry standards.

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