# Homework Help: Area of moment formula

1. Jul 21, 2016

### foo9008

1. The problem statement, all variables and given/known data
i'm having problem of understanding the formula of area of moment of uniformly distributed load and uniformly varying load... the shape of graph for moment of uniformly distributed load and uniformly varying load.are similar,right?
http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/moment-diagrams-by-parts

2. Relevant equations

3. The attempt at a solution
So, the formula of area of moment should be the same? why for uniformly distributed load , it is (1/6)(w)(L^3),while for uniformly varying load.is (1/24)(w)(L^3) ?

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2. Jul 22, 2016

### collinsmark

They might be "similar" in the lay-person's vernacular of the word. But they are not "similar" in the geometric sense.

Before getting into A, let's first explore what Mx is.

For a given value of x, Mx is simply the contribution of the moment to the left of the point defined by x, (I chose "left" because in the figure, the horizontal beam is affixed at the far right).

Let's start with the uniformly distributed load. Pick some arbitrary point on the beam. The value for x is the distance from the leftmost* side of the load to that point. The "x" as it is drawn in the diagram should work nicely; the selected point is a distance x away from the load's leftmost* point. Now, what is the moment around the point defined by x if you only consider the load to the left* of that point (and ignore the load to the right of that point)? That's Mx.

Can you derive that?

*(Again, I say "left" here because the horizontal beam's support is to the right, as depicted in the figure.)

Last edited: Jul 22, 2016
3. Jul 22, 2016

### foo9008

that is -(wx / L)(0.5x)(2x / 3 ) ? but , i got -w(x^3) / 3L instead of (1/6)(w)(L^3) ....

4. Jul 22, 2016

### collinsmark

What is the total force of the load to the left of the point defined by x?

What is the distance from the centroid of the load to the point defined by x (i.e., where is the center of force if you only consider the load to the left of the specified point)?

5. Jul 22, 2016

### foo9008

What is the total force of the load to the left of the point defined by x?
it's (wx)(0.5x) / L

What is the distance from the centroid of the load to the point defined by x (i.e., where is the center of force if you only consider the load to the left of the specified point)?
(2x/ 3)
is it correct ?

6. Jul 22, 2016

### collinsmark

(I'll come back your most recent post in my next post.)

Ah, I think I see your source of confusion about the uniformly varying load. And I think it's an inconsistency in the coursework. It's a big enough inconsistency to call it a mistake.

The problem is the way the coursework defines w0. It defines it differently for its A calculation than it does for its Mx calculation. That difference is inconsistent. That's what's wrong.

Let's concentrate on the total force to the left of the point defined by x (and let's only discuss the uniformly varying load case):

In the case of the Mx calculation, the total force (to the left) is:

$F = \frac{1}{2} w_0 \frac{x}{L}$

The $\frac{1}{2}$ comes from the fact that the force is triangle shaped (i.e., 1/2 the force if it was rectangle [uniform] shaped). Then, $w_0$ is scaled by $\frac{x}{L}$ because the triangle is smaller the smaller x is. So in this case, $w_0$ is defined such that if x = L, the total force is simply $\frac{1}{2} w_0$.

But the case of the A calculation, the total force (to the left) is:

$F = \frac{1}{2} w_0 \frac{x}{L} x$

Like before, the $\frac{1}{2}$ comes from the fact that the force is triangle shaped. Also like before, $w_0$ is scaled by $\frac{x}{L}$ because the triangle is smaller the smaller x is. However, under this definition this resulting force density is then multiplied by x to get the total force. So in this case, $w_0$ is defined such that if x = L, the total force is $\frac{1}{2} w_0 L$.

So it's really the definition of $w_0$. It's a matter if whether $w_0$ is defined as a force, or a force density. Arguably, both are fine choices in how one defines $w_0$. But the fact that it's used inconsistently in the same example is a problem. I'd call that a mistake in the coursework.

Last edited: Jul 22, 2016
7. Jul 22, 2016

### collinsmark

In the case of the uniformly distributed load, I'd define $w_0$ such that if x = L the total force is $F = w_0 L$. That's the way I've seen it defined before and I'm pretty sure that's the way your coursework defines $w_0$, at least for the case of the uniformly distributed load case.

So the force to the left of an arbitrary point defined by x is simply $F = w_0 x$.

At least your coursework is consistent with this definition in the case of the uniformly distributed load example.

It's almost correct. It's a distance of $\frac{2}{3}x$ from the leftmost point of the load, but how far is that from the point defined by x?

So what is the moment about the point defined by x? (Multiplying the two answers together, with a possible negative sign indicating into or out of the page)

8. Jul 22, 2016

### foo9008

ok , i got it . I assume it 's 2x/3 because i am taking moment about the leftmost end.... x/3 is taking moment about point x . Thanks!

9. Jul 22, 2016

### foo9008

sorry , why the force not equal to w(x^2) / L ?
I think the the force = w(x^2) / L because wx / L is the force per unit length of beam , so Force = force per unit length x length = (wx / L) x (x) = w(x^2) / L ...

One more thing , why the area of moment is -(w)(L^3) / 24 ?

10. Jul 22, 2016

### collinsmark

You'll still need the factor of 1/2 in there somewhere, because force is distributed in the shape of a triangle. The factor of 1/2 wouldn't be necessary if the force distribution was rectangular shaped, but since it's a triangle, it gets a factor of 1/2.

That's the way the coursework describes it in the case case of its A calculation. In that particular case, the force to left of the point defined by x is $F = \frac{1}{2} w_0 \frac{x}{L}x = \frac{w_0}{2L} x^2$.

First let's calculate the moment. The moment Mx is the cross product of the force F and the displacement from the point defined by x (since they are perpendicular, it's just multiplying them together with a possible negative sign). Calculate that first.

To find the area, integrate the moment Mx over x from 0 to L.

[Edit: By the way, keep in mind, the coursework treats $w_0$ differently when it calculates its Mx. So when you calculate your own Mx above, it's not going match your coursework's Mx. That's because of the inconsistency I was talking about earlier.]

Last edited: Jul 22, 2016
11. Jul 22, 2016

### foo9008

Why the moment is -(w)(x^2)/ 6L ? the moment is force x distance , right ? since the force is $F = \frac{1}{2} w_0 \frac{x}{L}x = \frac{w_0}{2L} x^2$ , then moment should have x^3 , right ? why moment has x^2 , not x^3 ?

12. Jul 22, 2016

### collinsmark

In our calculations here (consistent with your coursework's A calculation) The force has an $x^2$ in it, not the moment. In order to get the moment $M_x$, you need to multiply the force times the distance $\frac{x}{3}$

So, multiply force $\frac{w_0}{2L}x^2$ by the distance $\frac{x}{3}$ to get your $M_x$, with a negative sign to indicate direction (into or out of the page). That will have an $x^3$ in it like you suggest.

(Note that this $M_x$ is different than your coursework has listed for $M_x$. That's because your coursework used inconsistent definitions for $w_0$. In it's own $M_x$ calculation it treated $w_0$ as a force magnitude itself, not as a force density [force per unit length] as we are doing here. So you can expect the $M_x$ calculation you are doing here to be different than the one in the coursework by a multiple of x).

Once you find your $M_x$, integrate that to find the "area under the curve."

13. Jul 22, 2016

### collinsmark

Just to be clear, I'd like to reiterate that there is a "mistake" in your coursework. The mistake is an inconsistent definition of what $w_0$ means.

In the example, Uniform Varying Load, it has

$A = \frac{1}{24} w_0 L^3$
which is consistent with $w_0$ being a force density, in other words a force per unit length. This is consistent with what we've been discussing in the last couple of posts.

But it also has

$M_x = \frac{w_0}{6L}x^2$
which is consistent with $w_0$ being a force magnitude, independent of unit length.

Arguably, either interpretation of $w_0$ is okay on its own, but the fact that both interpretations are used, each in a different section of the same example, is a mistake in my opinion. So either way your look at it, one way or the other, one of them is wrong; the inconsistency being the mistake in the coursework.

[Edit: corrected another typo.]

[Another edit: if the example in the coursework had just said, $M_x = -\frac{w_0}{6L} x^3$, that would have cleared up the inconsistency.]

Last edited: Jul 22, 2016
14. Jul 22, 2016

### foo9008

ok , let us make some correction here ... Mx = (1/6)(w)(x^3)/ L , so , A should be moment multiply by the location of centroid , am i right ? so Area of moment = (1/6)(w)(L^3)(L/5) = (1/30)(w)(L^4) ??

15. Jul 22, 2016

### collinsmark

Um, I not sure understand your approach. I think you just need to integrate.

Using our new moment, $M_x = -\frac{w_0}{6L}x^3$, find the area under the curve.

$$A = \int_0^L M_x dx$$

16. Jul 22, 2016

### collinsmark

I see now that your coursework gives you an alternative way to find the area, saying that

$A = \frac{1}{n+1}bh$

Noting that the Uniformly Varying Load example is third degree.

That will give you the same answer for simple load distribution functions like this one. I like the integral approach better since you can use it for arbitrarily shaped load distributions. The method in your coursework will work for load distributions that meet certain criteria, but you can't use it for any old load distribution.

But for here, use the one in the coursework if you like; it gives you the same answer. If you'd rather integrate over Mx, use that instead; it is more powerful and more flexible.

[Edit: either way, you should end up with the coursework's answer of $A = \frac{1}{24}w_0 L^3$.]

Last edited: Jul 22, 2016
17. Jul 22, 2016

### foo9008

by intergrating Mx with dx i get $A = \frac{1}{24}w_0 L^4$, not L^3

18. Jul 22, 2016

### collinsmark

Don't forget there is already an L in the denominator.

Recall, $M_x = -\frac{w_0}{6 \color{red}{L}}x^3$