Formula for the moment of resistance W(x)

[Westerwolde]

Hi,

 

 

I need to derive a formula for the moment of resistance W (x), at a distance x from the left bearing.
This W (x) satisfies the formula: W(x)= b(x)^3 / 6

Given:
applies for a rectangular cross-section: W= b*h^2
The function b (x) : b(x) = b(l) + ((l-x)/l)* (b(0)-b(l))

How do I do this?

[Westerwolde]

Can someone help me?

[studiot]

You need to provide sufficient information such as the relationship between l and b, and the loading conditions.

Short stubby beams behave differently from long thin ones.

Since the beam is tapering the loadings also play their part.

[Westerwolde]

There has been given that the dimensions are determined at the end of the beam by the relationship : b(l) = α b(0) with α of less than or equal to 1

Edited March 24, 2017 by Westerwolde

[zztop]

Hi,

 

 

I need to derive a formula for the moment of resistance W (x), at a distance x from the left bearing.

This W (x) satisfies the formula: W(x)= b(x)^3 / 6

 

Given:

applies for a rectangular cross-section: W= b*h^2

The function b (x) : b(x) = b(l) + ((l-x)/l)* (b(0)-b(l))

 

How do I do this?

Hint:

 

You know that [math] b(x)=b(l) + ((l-x)/l)* (b(0)-b(l))[/math]

 

Then:

 

[latex]b(x+dx)=b(l) + ((l-x-dx)/l)* (b(0)-b(l))=b(x)-dx/l (b(0)-b(l)) [/latex]

 

Then, the volume element is:

 

[latex]dV=\frac{1}{3} (b^2(x)+b^2(x+dx)+b(x)b(x+dx))dx[/latex]

 

You are left with integrating [latex]x^2 dV[/latex] from 0 to l. (you will need to ignore the powers of dx higher than 2)

Edited March 24, 2017 by zztop

[studiot]

Hi,

 

 

I need to derive a formula for the moment of resistance W (x), at a distance x from the left bearing.

This W (x) satisfies the formula: W(x)= b(x)^3 / 6

 

Given:

applies for a rectangular cross-section: W= b*h^2

The function b (x) : b(x) = b(l) + ((l-x)/l)* (b(0)-b(l))

 

How do I do this?

 

I am going to start this again to correct some errors you have made right at the beginning.

 

Is this homework or are you trying to understand something you copied (wrongly) in class?

 

Or are you struggling with English?

 

These questions are meant to help since your first equation is impossible.

 

If W is the moment of resistance then it does not have the dimensions of L^3, which the equation you have written does.

 

I will derive the correct equation, using the attachment which shows a section taken through the body at .

 

The moment of resistance is defined as the moment resisting the couple generated by the compression © and tension (T) forces acting longitudinally within the body, due to applied stresses (S_c and S_t).

 

By horizontal equilibrium C = T and the couple = M_R = Ce = Te where E is the distance between the forces C and T.

 

Since the stresses increase linearly with distance from the neutral axis C and T act through the centroid of the stress triangle as shown.

This is at a distance 2/3 of the way from the neutral axis to the outside edge in each case.

 

That is 2/3 of b_(x)/2 since the neutral axis is halfway between the edges.

 

So e is twice this distance.

 

Now S_c and S_t are the maximum stresses, found at the edges. The average stress over the whole area is half of this ie (0 + S_c)/2 and (0 + S_t)/2

 

The Forces, C and T are found by multiplying the average stress by the area, which is shown in the right hand diagram as b_(x)/2 *b_(x) in each case.

 

Thus we correctly find that the moment of resistance equals the max stress times (b_(x))³/6.

 

Please confirm you have understood this so that we can continue.

 

[Westerwolde]

Hello,

 

Sorry for my late response, I was busy my other homework, this assignment I have to hand in a few weeks.

This is a task that I have to make for mathematics (homework)

 

I use google translate, I'm from Holland.

 

 

I understand what you write above.

 

 

First let me mention some:

 

The first question was: lead a formula for the dimension b (x) at a distance x from the left bearing.

 

 

Given:

 

The beam has a varying cross-section.

It is given that the dimensions at the end of the beam is determined by the relationship:
b (l) = αb (0), with α less than or equal to 1.

 

 

To help us showed our teacher this task 1, but he said clearly at; this should yourself be able to prove.

 

The following formula is derived by our teacher;
b(x) = b(l) + ((l-x)/l)* (b(0)-b(l))

 

I think the formula is valid for a distance from the right bearing. Therefore he said clearly defined ; this should yourself be able to prove.

 

 

I would say that the function of the left bearing: b(x) = (b(l)-b(0)/l)*x + b(0)

 

What do you think of this ?

Okay now returning to the moment of resistance W (x)

The function of the moment of resistance must also apply to the cross-section b (x) at a distance x from the left bearing.

Can we then the formula for the resistance moment substitute in the function b (x)?