Understanding differential equations with integrals

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Tygra

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Dear all,

I have been doing maths for some years now and have graduated university with a Civil Engineering Bachelors degree. I have hit some topics within my field that have some complicated maths. One thing that is confusing me in when I have equations mixed with differential equations and integrals. I understand what to do when I have an integral, and I understand what I have to do when I have a differential equation, but when they are mixed is throws me off!

For example, in a book of mine when considering the interaction between rigid frames and shear walls, the following equations are given:

[math]-EI \frac{d^3y}{dz^3} = \int_{z}^{H} (w(z) - q(z))dz - Q_H[/math]
[math]GA \frac{dy}{dz} = \int_{z}^{H} q(z)dz + Q_H[/math]

I'm not asking you to solve this. The book already does. But I get confused by this type of equation. How can I understand this?

Many thanks!
 
It often helps me to use the FTC and think of [imath] I=\int_z^H q(z)\,dz [/imath] as [imath] I=Q(H)-Q(z) [/imath] with [imath] Q'=q. [/imath] This results in a purely differential equation.
 
Tygra said:
But I get confused by this type of equation. How can I understand this?
Click to expand...
If someone has given you this differential equation:

\(\displaystyle \frac{d^2y}{dz^2} - y = \sin z\)

you would understand how to solve it perfectly because it has a known structure.

Now if we analyze and combine the two equations you have given, they are nothing more than this differential equation:

\(\displaystyle EI\frac{d^4y}{dz^4} - GA\frac{d^2y}{dz^2} = w(z)\)

where \(\displaystyle w(z)\) is the lateral load per unit height. And \(\displaystyle EI\) & \(\displaystyle GA\) are just constants.

\(\displaystyle w(z)\) can be anything. For example, if \(\displaystyle w(z) = \sin z\), we have a fourth order linear nonhomogeneous ordinary differential equation.

\(\displaystyle EI\frac{d^4y}{dz^4} - GA\frac{d^2y}{dz^2} = \sin z\)

And it can be solved easily by ordinary methods.

I myself solved a lot of beam deflection from a load problems using this differential equation:

\(\displaystyle EI\frac{d^4y}{dz^4} = w(z)\)

And when we have combined deflection from load shared between bending and shear, we use your equation:

\(\displaystyle EI\frac{d^4y}{dz^4} - GA\frac{d^2y}{dz^2} = w(z)\)

In other words, it is like

\(\displaystyle \textcolor{blue}{\bold{Simple \ beam}}\) (my problem) vs \(\displaystyle \textcolor{red}{\bold{Frame \ Wall \ System}}\) (your problem)

Or

\(\displaystyle \textcolor{blue}{\bold{Single \ element}}\) vs \(\displaystyle \textcolor{red}{\bold{Coupled \ System}}\)

So basically, I am calculating the deflection of a beam while taking into account the load, the shear force, and the bending moment.

And you are calculating the deflection of a frame wall system while taking into account the load, the shear force, and the bending moment as well as the slope and curvature.
 
Thank you for your responses, guys.

@logistic_guy

Yes, that is what they do in the book - differentiating and summing the equations together. So that you get:

[math]EI\frac{d^4y}{dz^4} - GA\frac{d^2y}{dz^2} = w(z)[/math]
So when you differentiate the equations does this eliminate the integral? Because evidently the above equation does not contain any integral?

I have also used often the equation: [math]EI\frac{d^4y}{dz^4} = w(z)[/math]

For me this is easy to understand because to get the deflection you integrate 4 times, for the slope 3 times, for the curvature 2 times and the shear 1 time.

Saying this though I do not understand very well the shear with the equation:

[math]GA\frac{dy}{dz}[/math]
Other than GA is the racking shear rigidity:

Logistic_guy, as you appear to have experience with structural engineering could you explain what the above equation means? I have looked at Timishenko's theory, but I haven't quite got my head around it.

Thank you in advance.
 
Tygra said:
Saying this though I do not understand very well the shear with the equation:

[math]GA\frac{dy}{dz}[/math]
Click to expand...
\(\displaystyle GA\frac{dy}{dz}\) is just the total internal shear force.

Tygra said:
I have looked at Timishenko's theory, but I haven't quite got my head around it.
Click to expand...
In Timoshenko beam theory, the total deflection of a point consists of two parts:

\(\displaystyle \textcolor{blue}{\bold{A}} \longrightarrow\) bending deflection due to curvature.
\(\displaystyle \textcolor{red}{\bold{B}} \longrightarrow\) shear deflection due to shear strain.

The bending deflection is related to the first equation you have given in the OP post while the shear deflection
is related to the second equation.

Say we have a shear strain \(\displaystyle \gamma\).

In Timoshenko theory, the shear strain is given by:

\(\displaystyle \gamma = \frac{dy}{dz} - \theta(z)\)

where \(\displaystyle \theta(z)\) is the rotation of the beam’s cross-section.

Say we have a shear stress \(\displaystyle \tau\), then it is given by:

\(\displaystyle \tau = G\gamma\)

Say we have a shear force \(\displaystyle V\). This shear force is just the integration of the shear stress \(\displaystyle \tau\) over the cross section area.

Or

\(\displaystyle V = \int_A \tau \ dA = \tau A = G\gamma A = GA \left(\frac{dy}{dz} - \theta(z)\right)\)

In simplified frame wall systems, they sometimes ignore \(\displaystyle \theta(z)\), so the shear force becomes

\(\displaystyle V = GA \frac{dy}{dz}\)

Therefore, \(\displaystyle GA \frac{dy}{dz}\) is just the internal shear force, sometimes called the racking shear force. it resists the lateral deformation.

In this simplified version, the slope \(\displaystyle \frac{dy}{dz}\) is the shear strain and \(\displaystyle GA\) acts like a spring constant converting the strain into a shear force.

To put all this together, suppose that we have several columns connected by beams, then

\(\displaystyle GA \frac{dy}{dz}\) is the total racking shear force for the entire system. In other words, it is the sum of all the internal shear forces in the connected vertical elements.
 
Thanks @logistic_guy, that has very much helped me.

I was wondering if I could ask you a further question?

Currently I am working on the deflection of frames with the central bay reinforced with cross bracing, like below:

Braced Frame.png


Given your previous explanation, I have approximately calculated the deflection of such frames by modelling them as a cantilever beam, with both shear and bending.

First I have tried a purely rigid frame (without cross-bracing) and I have achieved good approximations - like within 2% accuracy. This is how I did it:

For the shear component as a cantilever beam under a udl (w):

[math]V = GA\frac{dy}{dz}[/math]
[math]V - wz = GA\frac{dy}{dz}[/math]
Integrate once:

[math]Vz - \frac{wz^2}{2} + C_1 = GAy[/math]
[math]y = 0 , z = 0, C_1 = 0[/math]
Substituting the total height of the structure H into z we get:

[math]y = \frac{V(H) - \frac{w(H)^2}{2}}{GA}[/math]
From Stafford-Smith and Coull Tall Building Structures GA is given for rigid frames as:

[math]GA = \frac{12E}{h(\frac{1}{C} + \frac{1}{G})}[/math]
Where:

[math]C = \frac{\Sigma I_c}{h}[/math]
[math]G = \frac{\Sigma I_b}{L}[/math]
Next, consider the deflection due to bending. This is a very common formula and is given as:

[math]\frac{wH^4}{8EI_g}[/math]
Where taking the moment of inertia of the frame about its common centroid as:

[math]I_g = \Sigma A c^2[/math]
And by simply summing the shear and flexural components you get the maximum deflection.

Now getting back to the frame with cross-bracing. I do the same procedure, but use a different equation for GA. GA for cross-bracing is given as:

[math]GA = \frac{2Eh}{\frac{d^3}{L^2A_d} + \frac{L}{A_b}}[/math]
However, this is given a large under deflection of 27%.

So, I am wondering if you know why this is? I thought if anything I should be getting an over deflection because I am just considering the braced bay at the frame centre for GA, and not taking into account the unbraced columns.

Thanks in advance.
 
Tygra said:
Currently I am working on the deflection of frames with the central bay reinforced with cross bracing, like below:
Click to expand...
I have never solved problems in frame wall systems or braced bay systems. My work was only done on single beams where we don't calculate the shear deflection, only the bending deflection.

But I understand the math of Timoshenko beam theory and frame shear wall analogies pretty good because these topics are very related to the classical beam theory (single beams).

Therefore, now when I look at your math calculations, I can see and tell that everything you have done was correct.

Tygra said:
However, this is given a large under deflection of 27%.
Click to expand...
I don't pretend to be an expert in this topic but here is my opinion.

I think that the \(\displaystyle 27\%\) difference in the braced case comes from the assumed shear rigidity \(\displaystyle GA\) being too high. Usually, in a braced frame, factors like brace axial flexibility, joint deformation, and interaction with the surrounding bays reduce the real stiffness. The simplified formulas (from Stafford Smith & Coull) assume ideal conditions, so they make the model stiffer and give a smaller deflection than reality.
 
Hi @logistic_guy

Sorry for not responding sooner.

Thank you for your help and valid opinion. I was wondering though, why is it that the surrounding bays reduce the real stiffness?
 
Tygra said:
Hi @logistic_guy

Sorry for not responding sooner.

Thank you for your help and valid opinion. I was wondering though, why is it that the surrounding bays reduce the real stiffness?
Click to expand...
The reason the other sections of the frame make it less stiff is because the middle part doesn’t carry all the load by itself. All the sections work together, and the columns and braces can bend or stretch a little. So the frame is not as perfectly stiff as we might think.
 
Hi @logistic_guy ,

I have been looking at the equation:

[math]V = GA(\frac{dy}{dz} - \theta(z))[/math]
I want to see how including theta changes the results. You said that theta is the rotation of the beam’s cross-section in post #5.

Thing is when I include it I am left with theta as an unknown with no boundary conditions left to solve it.

[math]y = C_1 + \theta z + \frac{(Vz - \frac{wz^2}{2})}{GA}[/math]
As you can see there are two unknowns and there is one boundary condition to be used:

[math]y=0, z = 0[/math]
So I can eliminate C1, but I'm left with theta.

Do you know how to get around this at all?

Thanks
 
When you use \(\displaystyle \theta\) in the shear force equation, you need to bring up two or more equations to have a complete system that can cover all the unknowns.

The bending moment, for example, can give us one equation related to \(\displaystyle \theta\):

\(\displaystyle M = EI\frac{d\theta}{dz}\)

Another one is the equilibrium equation:

\(\displaystyle \frac{dM}{dz} + V = q\)

Since you are missing a lot in this field, my advice for you is not to create random problems (And get stuck!). Instead, try to solve already available problems. Read a book or rich website that explains this topic and try to solve the problems there. In this way, you will learn many new equations and formulas that helps you build up a strong foundation in this field.

And since mechanics of materials is a very difficult topic, solving some of its complete problems correctly will level up your math skills to the top in no time.
 
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