Beam calculations
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 lefthand half.
To reiterate:
f = x – F_{a} = x  15
M = 0.5x^{2} – 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 righthand 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 midpoint 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.
Resolving vertically:
f + F_{a} – 20 – x = 0, where F_{a} = 15 in this case So, f = x + 5
Taking clockwise moments about P:
M + F_{a}.x – [x.(x/2)] – 20.(x  5) = 0 M = 0.5x^{2} – 15x + 20x – 100 So, M = 0.5x^{2} + 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 lefthand end of the beam. 


What has happened at the midpoint 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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