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module:Beam calculations

page:Expressions for shearing force and bending moment

So here we have it: expressions for both the shearing force and the bending moment acting within the beam's left-hand half.

To reiterate:

f = x – Fa = x - 15

M = 0.5x2 – 15x

Let’s see what these expressions suggest about maximum and minimum values of f and M:


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Questions
    Write down the magnitude of the shearing force at A (when x = 0).
    Now determine the magnitude of the shearing force in the middle of the beam (when x = 5).
    Where is the shearing force greatest?
    Write these answers down, as you will need to refer to them later on.
    Repeat for the bending moment.
    Reflect on the value of M at the end of the beam, where it is 'free' to flex. Can you see why this makes sense?
    (The value for f in the middle of the beam might surprise you a little, if you were expecting an answer of zero.)

 

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Answers
    When x = 0, f = - 15
    When x = 5, f = -10
   
f is greatest at the end of the beam.
   
When x = 0, M = 0. There is no constraint on the beam at its end,and hence no tendency to deform it.
   
When x = 5, M = - 62.5

 

Let's repeat these calculations as x becomes greater than 5 (and we are examining the nature of the internal forces in the right-hand half of the beam). We need a new diagram, as the forces involved are now slightly different to those in the left half alone.

Bear in mind that the moveable load V will now enter the equations, acting at the mid-point of the beam. Don't forget: we are taking its value as 20 units for the time being. The weight of the segment AP will still be of magnitude x – it's just that now x > 5.


Dealing with the whole beam

Resolving vertically:

f + Fa – 20 – x = 0, where Fa = 15 in this case
So, f = x + 5

Taking clockwise moments about P:

M + Fa.x – [x.(x/2)] – 20.(x - 5) = 0
M = 0.5x2 – 15x + 20x – 100
So, M = 0.5x2 + 5x – 100


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Questions
    Write down what the shearing force will be at B (when x = 10).
    Now calculate the shearing force in the middle of the beam (when x = 5).
    Look back to your previous answers for the shearing force in the middle and at the left-hand end of the beam.
    What has happened at the mid-point of the beam, where the 'point' load of magnitude 20 is acting?
    Calculate the magnitude of the bending moment where x = 5 and where x = 10.
    How do these values compare with those you found for x = 0 and x = 5 earlier on?

 

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Answers
    When x = 10, f = 15
    When x = 5, f = 10
   
The value of f appears to have jumped from –10 to 10 in the centre of the beam, where the point load of size 20 is acting.
   
When x = 5, M = - 62.5 again, and when x = 10, M = 0, again.
   
These values exactly mirror those on the left hand side of the beam.

 

The relationships between the left and right ends of the beam probably need a bit of unpacking, so the next section will examine these results and explain some of their significance in terms of the stresses on the beam.

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