# Simply supported beams?

1. Aug 11, 2009

### Emzielou83

simply supported beams???

1. Can anybody help me? please. This seems like I have gone completely off track.

A simply supported beam of length 6m supports a vertical point load of 45kN a distance of 4m from one end. If the maximum allowable bending stress is 120MPa:

(there were 4 previous questions on point loads, which I have completed)

recalculate the reaction forces at either end, taking into account the actual weight of the beam as a UDL.

Determine the actual maximum bending stress.

I'm not sure if the UDL would be a point load equal to 180kN acting 2m from one end, I have worked Rb=60kN and Ra=165kN. I have the final answers and they should be Ra=16.15kN and Rb=31.15kN. I really can't see how to get to this?

Thanks to anyone that can help!

2. Aug 11, 2009

### PhanthomJay

Re: simply supported beams???

When calculating end reactions (but not when calculating stress), the UDL (units: kN/m) may be assumed as a resultant point load (units: kN) acting at the center of gravity of the beam (at it's midpoint). But besides that, you appear to have miscalculated the magnitude of the resultant force of that UDL, which you did not state. According to the answers given, it appears that the UDL amounts to much less. Where'd the 180 kN come from??

3. Aug 13, 2009

### Emzielou83

Re: simply supported beams???

Hey,

thanks for the quick response

The 180kN i got by multiplying the 45kN by 4m as the question states it is 4m from one end. I am not sure if the UDL is taken across the whole of the beam or just the 4m. I then took this 180kN to act as a point load 2m from one end. I assume the formula to find the reaction forces is the same as the formula for a point load?

4. Aug 13, 2009

### PhanthomJay

Re: simply supported beams???

It appears you are not reading the problem correctly, or there is missing information. The 45 kN load is a point load applied at 4 m from one end. Based on that load alone, you should have come up with a 30 kN reaction at one end and a 15 kN reaction at ther other. I assume you did, because you said you answered the other parts correctly, so its the same method. The applied 45 kN point load has nothing to do with the beam weight UDL load. Then the problem goes on to ask what those end reactions would be if you add in the weight of the beam as a UDL across the entire length of the beam. Does the problem state the weight of the beam, either in kN/m, or kN?? Whatever that value is, in kN, it splits evenly to each end reaction point.

5. Aug 13, 2009

### Emzielou83

Re: simply supported beams???

Jay,

I'm sorry if it seems like im being stupid, but i'm really struggling to get my head around this.

The earlier part of the question asked me to calculate the reaction forces for the 45kN as a point load. I did this and came out with the correct answers. I thought this part of the question is asking me to now recalculate the reaction forces but using the 45kN as a UDL. I know that to work this out I need to calculate what the equivelant point load would be, the thing is i'm not sure at what point it would act. i.e would it be 2m in as the original load was 4m from one end, or would it be 3m because the whole beam is 6m long.

Again i'm sorry if it seems like i'm being stupid but I really am struggling to grasp this.

Thanks

Emma

6. Aug 13, 2009

### Emzielou83

Re: simply supported beams???

Jay,

sorry the problem does not give me the weight of the actual beam. Just the UDL

Thanks

Emma

7. Aug 13, 2009

### PhanthomJay

Re: simply supported beams???

No need to apologize, especially since i'm struggling, too. But something is not right here. Are you sure you are presenting the problem exactly as written? The UDL (Uniformly Distributed Load) is the beam weight per unit length. It is uniformly distributed along the entire length of the uniform beam.
In the first part, with the application of the 45 kN point load 4 m from one end, and without any UDL, you get end reactions of 15 kN amd 30 kN, right? They sum to 45 kN, as they must, for equilibrium (sum of forces in vertical direction = 0).
Now for the 2nd part, you have given the answers for the end reactions as 16.15 kN and 31.15 kN. Those reactions sum to 47.3 kN. That indicates to me that the beam weighs 47.3 -45 = 2.3 kN (or 2.3/6 = 0.383 kN/m as the UDL). Are you sure that the UDL is not given?? Otherwise, i don't understand the problem. And when determining bending stresses, you must know the properties of the beam.

8. Aug 14, 2009

### Emzielou83

Re: simply supported beams???

Jay,

I think I see what you mean. In an earlier part of the question I had to select a universal beam section from a worksheet that will carry the 45kN load. This beams mass unit per length is 39kg/m. As you worked the UDL to be 0.383 (or 2.3kN) is this what you are missing?

Thanks

Emma

9. Aug 14, 2009

### PhanthomJay

Re: simply supported beams???

Yes, that's the missing piece. If 39kg/m is the beams mass per unit length, then its weight per unit length is m*g (where g, the acceleration of gravity, is 9.81m/sec^2). That works out to 39(9.81) = 383 N/m , or 0.383 kN/m. Since the beam is 6 m long, it weighs (.383)(6) = 2.3 kN. Now apply that as a concentrated load at the mid point of the beam, and together with the additional 45 kN concentrated load applied at 4m from one end, you should be able to get the correct values for the end reaction forces.

10. Aug 14, 2009

### Emzielou83

Re: simply supported beams???

Jay,

Thank you very much for your help. I see now what needs doing. Hopefully I will not be needing any more help.

Thanks very much

Emma

11. Aug 14, 2009

### PhanthomJay

Re: simply supported beams???

Emma, my pleasure.

12. Aug 15, 2009

### Emzielou83

Re: simply supported beams???

Jay,

sorry to bother you again but im stuck..........again!

The next part of the question asks me to ......

'Determine the actual maximum bending stress of the selected beam, taking into account the actual weight of the beam as a udl and check whether your selection is still appropriate by comparing this stress with the maximum allowable bending stress'.

As you know in an earlier part I had to select a beam. All of the info I have for the beam is:

UDL 0.383kNm
39kg/m
Second moment of area=10 087
Elastic modulus = 571.8
Plastic modulus= 653.6
Max allowable bending stress=120MPa

The final answer is 512.8 cm^3.

I think the formula that I need to be using is

Max bending stress/max working stress = I(2nd moment of area)/Y

If im honest, i don't really have a clue where to start. Do I need to find out what 'Y' is before I can work this out? If so can you tell me what 'Y' is and what I need to do to find it?

Thanks once again

Emma

13. Aug 15, 2009

### nvn

Emzielou83: Please include the units on your numeric values in post 12. You can hit the Edit button. By the way, the international standard for writing units (ISO 31-0) states that there should always be a space between a numeric value and its following unit symbol. E.g., 39 kg/m, not 39kg/m. You can see this by looking in any text book.

The question you quoted in post 12 asks you to compute maximum bending stress. Therefore, the formula you listed is currently incorrect. Look in your text book for the formula to compute bending stress. In that formula, use the maximum moment on your beam. Parameter y is the cross section extreme fiber distance.

14. Aug 17, 2009

### Emzielou83

Re: simply supported beams???

nvn:

Sorry that I did not put the units in....this was just me being sloppy!!

Radius of gyratio = 14.29 cm
Second moment of area = 10 087 cm^4
Elastic modulus = 571.8 cm^3
Plastic modulus = 653.6 cm^3

I also have the depth of the beam = 352.8 mm
and the width = 126.0 mm.

I have got my maximum bending moment as 64.6 kNm.

The formula I keep coming back to is

Bending stress=MY/Ix

This just looks like a transposed version of the original equasion I thought it was.

Im sorry to keep bothering you but im so confused its unreal.

Thanks

Emma

15. Aug 17, 2009

### Emzielou83

Re: simply supported beams???

nvm:

Also the correct answer should be

Elastic modulus Z=512.8 cm^3

I dont really understand what i am supposed to be working out....Elastic modulus (which im given!)
or bending stress.

Thanks

Emma

16. Aug 17, 2009

### PhanthomJay

Re: simply supported beams???

Emma, a few things to consider here. Your max bending moment is close but slightly in error. The max moment in the beam occurs at the point of zero shear. Did you draw a shear diagram? Max moment occurs at the point where the 45 kN load is applied (2 m from one end). You then have to draw a free body diagram at that cut section to calculate it.
Now apparently you initially chose a beam size based on the 45 kN acting alone, without considering the beams weight as a UDL, and the problem is asking if that beam is still sized properly to carry both the 45 kn load and the UDL beam weight. The beam you initially chose had an "I" of 10,087 cm^4, and an elastic section modulus "Z" of 571.8 cm^3. Z is just I/c, where 'c' is the distance from the neutral axis of the beam to the outer fibers of the beam, in other words, c is 1/2 the depth of the beam (assumed symetrical). That's the value you use for "Y" in your max stress = MY/I formula (or max stress = M/Z). Now set the max allowable stress equal to 120 MPa, and calcualte the requiredZ. If the required Z is less than the actual Z of the beam you chose, the beam is OK. The actual max bending stress, which you must determine, will be less than 120 MPa.

Last edited: Aug 17, 2009
17. Aug 17, 2009

### Emzielou83

Re: simply supported beams???

Jay,

I have drawn a bending moment diagram and a free body diagram (how correct they are I have no idea!). I have my max bending moment happening 1m from one end being 75.96. Th emax bending moment 2m from the one end where the 45kN is works out to be 61.53kNm. I have worked this from both ands and they work out the same. The formula I am using for this is Ra*x-(UDL*x^2)/2.

From what you are telling me to do, my c should be 0.1764m (half the depth of my beam)

My formula with the values entered should be:

61.53*0.1764/10087

The answer for this is 0.001076 (I think in m).

Where does this fit with the answer of 512.8 cm^3 that I should be getting?

Thanks Jay.

Emma

18. Aug 17, 2009

### nvn

Emzielou83: If you go past x = 4 m, then the 45 kN point load comes into your free-body diagram. Therefore, for x > 4 m, the 45 kN point load would need to be included in your moment equation, thus decreasing the moment. For x < 4 m, your equation is correct. And you computed the maximum moment correctly, which occurs at x = 4 m. Nice work. The question in post 12 asks you to compute the maximum bending stress, and compare it to the allowable bending stress, 120 MPa.

You must use consistent units. Because you decided to use 61.53 kN*m and 0.1764 m, then don't you also need to convert I to m^4? Afterwards, your computed bending stress will be in kN/m^2, which is called kPa. After this, convert kPa to MPa, and compare your answer to 120 MPa.

I don't know why they say the correct or final answer is Z = 512.8 cm^3. We would need to know the question for which that is the answer. 512.8 cm^3 is just the minimum allowable elastic section modulus.

19. Aug 18, 2009

### Emzielou83

Re: simply supported beams???

nvm:

my bending moment diagram goes up to maximum point at 5 m. I know it should start decreasing at 4 m, because max moment occurs at the 4 m point where the 45kN is acting upon the beam. Whatever way, (from right end or left end), I work out the moments at the 5 m point it works out to be 75.96 kNm meaning that this is my max moment, although I know its not.

How do I show this correctly in my bending moment diagram?

Emzielou83

20. Aug 18, 2009

### nvn

It sounds like you might be forgetting to subtract (45 kN)(x - 4 m) from your moment when x > 4 m, mentioned in post 18. Your bending moment diagram would be almost a straight line (slightly parabolic) from your moment value at x = 4 m down to zero moment at x = 6 m.