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Statics- Bending Moment Diagram

  • Thread starter Kalus
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  • #1
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I'm trying to construct a bending moment diagram for the situation as below:

[PLAIN]http://img42.imageshack.us/img42/633/bmdiag.jpg [Broken]

Where Ra and Rb are reaction forces, L1 and L2 are lengths between Ra and N and N and Rb respectively, N is a force acting perpendicular to the beam and P is a force acting at length L3 (sorry forgot to label this) parallel to the beam. The beam has fixed ends so cannot hinge.

The shear force diagram is fairly straight forward, but i'm having issues trying to decide whether to include P in the first "cut" between Ra and N.

What i've done so far is:

If you take x from the left hand side, then between Ra and N:

M= Ra*x

between N and Rb:

M= Ra*x - L3*P - N*(x-0.1)

However, when this is plot it looks very strange as suddenly the magnitude changes drastically when L3*P is included.

What have I done wrong?

Thanks,
Kalus
 
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Answers and Replies

  • #2
PhanthomJay
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You will have to clarify the problem. Are the ends fixed or are they supported at one end by a hinged (pinned) support and at the other end by a hinged roller support? I assume that there is a vertical member between point P and the horizontal beam?
 
  • #3
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As I said in the OP both ends of the beam are fixed. Also there is a vertical member of length L3 that force P is applied at.

Thanks,
Kalus
 
  • #4
PhanthomJay
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If the beam is fixed at both ends, it is statically indeterminate to the 3rd degree, that is, there are 6 unknowns (1 moment and 2 forces at each end), but only 3 static equilibrium equations, so the solution requires the use of other methods based on strain compatability and boundary conditions, unless you are allowed to use beam tables based on fixed fixed end conditions with a load and couple applied at a point. Your solution neglects the fixed end moments at each end. In general, the couple P*L3 will cause a sharp change in moment at the point of its application. This is a time consuming problem without the use of tables, so that is why I questioned the end conditions and your use of the term 'fixed' ends rather than 'hinged' ends. The solution ia a lot easier if the ends are hinged, because the beam is allowed to rotate at its ends and there will be no fixed end moments at the ends.
 
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  • #5
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Thanks for your reply Jay. Sorry, to clarify P, N, L1, L2 and L3 are all known, I was working algebraically on the problem, as its the general solution I'm interested in. This 'beam' is actually a shaft supported by two bearings.
 
  • #6
PhanthomJay
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Thanks for your reply Jay. Sorry, to clarify P, N, L1, L2 and L3 are all known, I was working algebraically on the problem, as its the general solution I'm interested in. This 'beam' is actually a shaft supported by two bearings.
OK, then the ends may be assumed as hinged (free to rotate but not translate). So, I assume you have correctly calculated the vertical end reactions Ra and Rb, and drawn the shear diagram correctly (which does not include the horizontal force P). Then when you draw the moment diagram, the couple P*L3 abruptly (discontinuity) changes the moment at its point of application on the beam. Watch plus/minus signs.
 
  • #7
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Plugging in the numbers below gives the following bending moment diagram:

L1=0.1 m
L2=0.05 m
L3=0.1 m
N= 1770 N
P= 1690 N

[PLAIN]http://img413.imageshack.us/img413/8440/bmdiag2.jpg [Broken]

Notice on the graph (excuse the lack of labels) I plot the same point, 0.1 m, using both equations that it fits into. I guess because of the discontinuity that occurs due to P3*N I first thought it was wrong, because the single point has two different bending moments. However, if it is a discontinuity then I can understand it.

Does the graph look right to you and what would the bending moment at the point 0.1m actually be? Surely it can't exist in those two states at once so is it simply zero?

Thanks,

Kalus
 
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  • #8
PhanthomJay
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A few points here. First, your bending moment diagram is correct.

Second, it is useful to draw a shear diagram before drawing the moment diagram; you should find it helpful in not only determining the bending moment diagram, but also as a check of your values for Ra and Rb .

Thirdly, the moment changes abruptly at the 0.1 m mark. Both moments are valid...one exists infinitesimally to the left of the point, the other infinitesimally to the right of the point, but beam design would be governed by the higher value. In reality, there is no such such thing as a concentrated point load or couple, it acts over some small length greater than 0, but mathematically, treating it as acting at a point is no problem, and the discontinuity accounts for the reson of the infinite change over zero distance..

Edit: I missed a minus sign, and have edited my response. Your diagram is correct. Sorry for any confusion.
 
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  • #9
PhanthomJay
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See my edited response above. I misread the direction of the P load..your diagram is correct, sorry for error!
 

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