Bending moment diagram equation method

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Homework Statement


Draw the bending moment diagram
upload_2017-6-28_16-20-41.png

the top part is the question, the bottom part is the answer.

Homework Equations




The Attempt at a Solution


There is a force at the wall in x and y direction and a moment ##F_x, F_y, M_w##
Sum of forces in x : ##F_x-1000=0, F_x=1000##
Sum of forces in y : ##F_y-2000=0, F_y=2000##
Sum of moments at wall, clockwise = positive : ##M_w+2000*10-5000=0,M_x=-15000##

So... I am not sure how to write the equations for moment here. I only managed to come up with this equation :
##M_x+2000(10-x)-1000*5=0## ---> ##M_x=-15000+2000x##
This equation... it works out for the entire beam... but it has to be wrong. I originally had 2 equations, one for the left and right side of the downward 2000 force, but when I realized the first equation covered the whole beam I dropped it.

I think the shear diagram is a 2000lb constant line from x=0 to x=16, so the moment diagram is the integral of that M_w+area under shear curve ##=-15000+2000*16=17000##, which is correct according to the given answer...

But how do I do this with an equation? I hate the graphical method.
 

Answers and Replies

  • #2
haruspex
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how do I do this with an equation? I hate the graphical method.
The main beam does not "know" about the two applied forces, except via their effects at its right hand end. So replace them by appropriate forces and moments applied there.
 
  • #3
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In my judgment, your original analysis was correct. Beyond x=10, the moment is constant at 5000 in-lb.
 
  • #4
haruspex
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In my judgment, your original analysis was correct. Beyond x=10, the moment is constant at 5000 in-lb.
Break the system into two components: the horizontal main beam and L-shaped bracket.
Where they meet, the forces on the bracket must be 2000lb up, 1000lb to the right, and a clockwise moment of 17000 in-lb.
Correspondingly, at that end, the main beam is subject to 2000lb down, 1000lb to the left, and an anticlockwise moment of 17000 in-lb.
 
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@Chestermiller constant after x=10? I thought I needed a second equation where instead of 2000(10-X) it would become 2000(X-10)
 
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@haruspex Say I do what you said and I pretend the beam isnt there and just write it as a force on the beam at the end... How should I write the equation? I'm still confused... Do I or do I not add the moments into the equation... (M_E=moment at right end of beam, F_Y is reactionary force on left side of beam in y direction=2000Lb))##-M_x+F_Y*x-M_E+2000(16-x)##-->##-15000-17000+2000(16)##
I am fundamentally confused. (did not add horizontal forces because assuming they are in plane of beam)
 
  • #7
haruspex
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when I write an equation for the moment, I should not add the components from the top beam unless I add them at a distance of x=16
Maybe... Not quite sure what you mean there, so let me be quite clear.
For the sake of illustration, suppose the beam continues past 16 in, i.e. out past the junction with the angled bracket.
If the vertical 2000lb were applied directly to the main beam then at a point x from the left you would count its moment as:
- 0 for x < 10 in
- 2000 (x-10) for x > 10 in.
With the 2000 lb as in the diagram:
- 0 for x < 16 in
- 2000 (x-10) for x > 16 in
Only the threshold value of x at which that moment is included changes. The numerical value of the moment when it is included does not change.

The method of breaking it into two components gets the same result by a slightly different method:
- 0 for x < 16 in
- 2000 (x-16) for x > 16 in
- plus a counterclockwise moment of 2000 (6) for x > 16

I prefer the component method because it also makes it clear what to do with the horizontal force on the bracket, namely:
- 0 for x < 16
- a counterclockwise moment of 1000 (5) for x > 16.
 
  • #8
haruspex
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@haruspex Say I do what you said and I pretend the beam isnt there and just write it as a force on the beam at the end... How should I write the equation? I'm still confused... Do I or do I not add the moments into the equation... (M_E=moment at right end of beam, F_Y is reactionary force on left side of beam in y direction=2000Lb))##-M_x+F_Y*x-M_E+2000(16-x)##-->##-15000-17000+2000(16)##
I am fundamentally confused. (did not add horizontal forces because assuming they are in plane of beam)
My last post crossed with yours. I'll give you a chance to read mine to see if it clears the matter up.
 
  • #9
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My last post crossed with yours. I'll give you a chance to read mine to see if it clears the matter up.
sorry late reply holiday weekend. Can you tell m
##M_X## is moment at left end of beam, M=moment at point in question
So... if I cut the beam in the middle ##M_X+2000x=M##
This 2000x is the force at the wall up. So I add the moment AND the force in y direction from the left side of the beam when I cut the beam down the middle? This works because it matches the moment diagram provided in the book.

This goes all the way to x=16. Now at x=16.... This equation actually still gives me the answer. ##-15000+(2000*16)=17000##
Is this because the 1000Lb force is considered to act at the beam, not at a distance 5" from the beam?
 
  • #10
haruspex
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Is this because the 1000Lb force is considered to act at the beam, not at a distance 5" from the beam?
Not exactly.
The 1000lb force generates a torque of 5000 in-lb which acts at the extreme right hand end of the beam. The analysis method chooses a "source" end of the beam and at each point only considers forces which act directly on the beam somewhere between that point and the source. If you choose the left hand end as the source then this 5000 in-lb torque never gets included. Neither does the downward 2000lb, for the same reason. If the beam were to project just a little further to the right, past the junction, then these would, suddenly, be included. This has the effect of neutralising all the torques from the left hand end, leaving no torque on this stub, as is correct.

This might seem odd, but it works. At a short distance x from the right hand end (i.e. from the junction) the torques from the left hand end are 2000(16-x)-Mx = 32000-2000x-15000=17000-2000x. Suppose we choose the right hand as the source instead. At distance x from the right the torque is 2000(6-x)+5000=17000-2000x.
Note how, with the right hand end as source, the 2000 down has a moment even about points which are to the right of its line of action. This is because the angled bracket shifts its point of application on the beam to the right. Again, I stress that this shift does not affect the calculation of the moment; that is still calculated in regard to the distance between the point on the beam under consideration and the line of action of the force (6-x here). It only affects the decision of whether to include the torque in the total.
 

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