Last time, I explained a simple method for determining the deflection of a beam.
This time, we will introduce methods for solving statically indeterminate problems in beams, similar to the statically indeterminate problems of trusses we discussed previously. Furthermore, we will also introduce the concept of combined beams made of different materials.
The method I'm introducing today is unlikely to be used to actually determine the amount of deflection in real-world simulations, given the current state of simulation technology.
However, it is important to understand this well, as it is useful for imagining the deformation of structures during the conceptual stage, such as in layouts and sketches.
Furthermore, regarding the composite beams in the latter half of the design, it is now very common to combine different materials to improve strength and rigidity. Therefore, it is important to be familiar with this concept. The use of composite materials is expected to increase exponentially in the future.
This can also be calculated through simulation, so you probably won't need to do the calculations yourself very often, but if you don't understand the meaning, the simulation won't work correctly.
In any case, even if you don't actually use it, there are many situations where not understanding it will cause problems, so please make sure to understand it.
The statically indeterminate problem of beams
I explained what statically determinate and statically indeterminate problems are in the context of trusses, but I'll explain them again.
Static determination means that stress and deformation can be determined solely by the balance of external forces (force, moment) and internal forces (stress, shear force).That is.
On the other handStatically indeterminate means that the stress and deformation cannot be determined solely from the balance of external and internal forces; further conditions, such as the geometric relationships between deformations, must be known to determine them..
As always, the name doesn't really matter, but the point is whether or not it can be determined solely by the balance of forces.
This time, we will be dealing with the problem of statically indeterminate beams.
The statically determinate and statically indeterminate problems were introduced on a previous page, so please take a look if you're interested.
The statically indeterminate problem of beams
Let's set up an example problem, as usual.
There is a beam of length l, with one end supported by a base A and the other end fixed to a wall. A uniformly distributed load q is applied to it. We will find the deflection and deflection angle of this beam.
The coordinate system uses the leftmost point as the origin.
Considering the balance of forces and moments as usual, we get the following (RA, RB, and MB are unknowns).
When considering the balance of forces
$ RA+RB=ql $
Next, let's consider the balance of moments around point A.
$ MB - RBl + \frac{ql^2}{2} = 0 $
Normally, we would find the unknowns RA, RB, and MB here, but since there are only two equations for three variables, we cannot find them. This is a statically indeterminate problem that cannot be solved by the equilibrium of external forces.
Since it's not possible to obtain the necessary information, we'll proceed with the unknowns remaining as RA, RB, and MB.
Consider the bending moment at an arbitrary coordinate x.
$M=\frac{q}{2}x^2+RAx$
Substitute the above equation into the differential equation for deflection.
$ \frac{d^2y}{dx^2}=\frac{1}{EI}(\frac{q}{x^2}-RAx) $
Solving this will
$ \frac{d^2y}{dx^2}=\frac{1}{EI}(\frac{q}{6}x^3-\frac{RA}{6}x2+C1) $
$ y=\frac{1}{EI}(\frac{q}{24}x^4-\frac{RA}{6}x^3+C1x+C2) $
Here, as usual, we find C1 and C2 from the boundary conditions of the beam. The boundary conditions are that the deflection is 0 at the left end of the beam (x=0) and the deflection angle is 0 at the right end (x=l).
$ C1=-\frac{ql^3}{6}+\frac{RAl^2}{2}, C2=0 $
Normally, we would be able to derive the equations for the beam's deflection angle and deflection amount here, but in this example, the unknowns RA, RB, and MB remain. As mentioned above, the equilibrium of forces and the equilibrium of moments are insufficient, so we can solve the problem by using the boundary condition that the deflection amount is 0 at the right end of the beam (x=l) (geometric conditions for solving statically indeterminate problems).
$ RA=\frac{3}{8}ql, RB=\frac{5}{8}ql, MB=\frac{1}{8}ql^2 $
Substitute this into the formulas for the angle of deflection and the amount of deflection.
$ たわみ角θ=\frac{q}{48EI}(8x^3-9lx^2+l^3)、たわみ量y=\frac{q}{48EI}(2x^4-3lx^3+l^3x) $
Since the equilibrium equation is insufficient for the external force variables, we use the deflection equation to impose geometric constraints and solve the problem.
However, in the case of statically indeterminate problems, the equations to be set up and the calculations to be performed increase compared to statically determinate problems, making the calculations very cumbersome. Therefore, I will introduce a simpler method.
Superposition method
Let's consider a beam that is exactly the same as the one in the previous example.
At this time, the external forces acting on the beam are a uniformly distributed load q and a reaction force RA from the base A.
In fact, you can treat each of these as a separate beam problem and then add up the results later. This is called the superposition method.
The needles are separated into q and RA.
Each of these can be broken down into a simple cantilever beam problem. Their deflections can be easily calculated using the multiple integral method or the area moment method, as previously introduced.
Then the deflections y1 and y2 at each tip are given by the following equations.
$ y1=\frac{q}{24EI}(x^4-4l^3x+3l^4) $
$ y2=-\frac{RA}{6EI}(x^3-3l^2x+2l^3) $
Adding up the deflections of each simple beam gives the same amount of deflection as the original beam. This is called the superposition method.
Since RA is still unknown, the condition for solving the statically indeterminate problem is that when the left end of the beam is x=0, the deflection is 0.
$ y=y1+y2=\frac{ql^4}{8EI}-\frac{RAl^3}{3EI}=0 $
From this, $RA=\frac{3}{8}ql$, and the reaction forces RB and MB at the fixed end B are obtained from the balance of forces and the balance of moments at the tip A.
$ RA+RB=ql $
$ MB - RBl + \frac{ql^2}{2} = 0 $
From $RB=\frac{5}{8}ql$ and $MB=\frac{1}{8}ql^2$, the deflection can be calculated using the following formula.
$ y=y1+y2=\frac{q}{48EI}(2x^4-3lx^3+l^3x) $
Yes, it can be found quite easily.
When calculating the deflections of y1 and y2, the resulting bending moments are relatively simple, so I recommend using the area moment method rather than the multiple integral method.
Let me give you another example. This is a problem that troubled me when I was a student.
Problem 2 of the statically indeterminate beam
Let's start by setting up an example problem. Suppose two cantilevered beams of length l are arranged in an X shape, with a load P applied at the intersection.
To make this easier to calculate, we can name each base A, B, C, and D.
From here, in order to use the superposition method, we will consider beams AB and CD separately. Let the loads be P1 and P2 respectively (assuming that their second moments of area are different).
This makes a simple double-supported beam. Now let's find the maximum deflections y1 and y2 for each.
$ y1=\frac{P1l^3}{48EI1}、y2=\frac{P2l^3}{48EI2}（それぞれの最大たわみが発生する場所ははりの中央） $
Considering the superposition conditions, the load is $P=P1+P2$, and since the deflection is the same where beams AB and CD intersect, $y1=y2$. (Geometric constraints on a statically indeterminate problem)
From P1=P-P2 and y1=y2, we have \frac{P1}{I1}=\frac{P2}{I2}.
$ P1=\frac{I1}{I1+I2}P, P2=\frac{I2}{I1+I2}P$
Thus, the deflection angle, deflection amount, and reaction forces RA, RB, RC, and RD at each base can be easily determined.
This concludes my introduction to the statically indeterminate problem of beams.
Combination needles
Now let's consider beams when different materials are combined.
In actual design and product development, as far as I know, combinations of aluminum and iron are quite common. Other combinations include resin and copper, carbon fiber and resin (CFRP), and, more unusually, titanium and magnesium are sometimes used to create single parts or components through press-fitting or bonding.
Such materials are called composite materials.
When composite materials are created using a clever combination of materials, the drawbacks of each material can be negated, resulting in a very useful material (though expensive).
However, if the physical properties of the materials are too different (material properties, stress-strain diagram), the dissimilar components may separate or slip during deformation, rendering the process meaningless (adhesion, delamination).
While material manufacturers are certainly involved in this kind of research, surprisingly, Japanese universities and companies are also putting in a lot of effort and possess world-class capabilities.
While composite materials are excellent, their composition of dissimilar materials makes strength calculations extremely cumbersome, as it requires considering the elastic modulus and cross-section of each individual material.
Therefore, we treat the composite material as a single component and calculate its hypothetical second moment of area.
Determining the virtual second moment of area of a composite material.
Let's set up an example problem as usual.
Consider a beam with the dimensions shown in the diagram below, where the main members are aluminum and the bottom thickness t is iron. Let the elastic modulus of aluminum be Ea and the elastic modulus of iron be Ef. Of course, iron is stronger than aluminum, so note that Ef > Ea.
Let's consider a hypothetical second moment of area here.
To find the second moment of area, we will consider the bending stress and bending moment within the beam's cross-section, just as we did with the first and second moments of area explained previously for beginners.
First, let's consider the bending stress that occurs at an arbitrary distance y in the cross-section of the example problem.
If ρ is the radius of curvature when the beam is deflected, the bending stress can be expressed as follows.
Bending stress σa = \frac{Ea}{ρ}y within an aluminum component
$ Bending stress σf = \frac{Ef}{ρ}y within a steel member $
The force acting on an infinitesimal area dA = bdy is,
The load dPa applied to a small area within an aluminum component is \frac{Ea}{ρ}ybdy
$ Load dPf = \frac{Ef}{ρ}ybdy $
Becomes
Here, please remember that the sum of bending stresses within the same cross-section is zero, and that the bending stress moment is generated by the bending stress.
If we put this into an equation,
The sum of bending stresses is \int_{-e1}^{e2-t} dPa+\int_{e2-t}^{e2} dPf=0
The moment due to bending stress is \int_{-e1}^{e2-t} ydPa＋\int_{e2-t}^{e2}ydPw=M
We could simply calculate the above formula by separating the aluminum and iron parts, but that would be cumbersome, so let's try a slightly different approach. We'll use mathematical techniques to virtually represent the two separate materials as one of them.
This time, we'll stick to iron (either is fine). We'll try to express the formula for the load on a small cross-sectional area within an aluminum component using the elastic modulus of iron by manipulating the width and the elastic modulus of aluminum.
$ Load dPa on a small cross-sectional area within an aluminum member = \frac{Ea}{ρ}(\frac{Ef}{Ef})ybdy = \frac{Ef}{ρ}(\frac{Ea}{Ef}b)ydy $
If we consider $ \frac{Ea}{Ef}b$ in the above equation as the width of a hypothetical cross-section, then the aluminum part can be thought of as iron with a width of $ \frac{Ea}{Ef}b$.
Using this property, we can rewrite the formula for the sum of bending stresses.
$ \int dPa+\int dPf=\int_{-e1}^{e2-t} \frac{Ea}{ρ}ybdy+\int_{e2-t}^{e2} \frac{Ef}{ρ}ybdy=\frac{Ef}{ρ}(\int_{-e1}^{e2-t} \frac{Ea}{Ef}bydy+\int_{e2-t}^{e2}bydy) $
The expression $ \int_{-e1}^{e2-t} \frac{Ea}{Ef}bydy+\int_{e2-t}^{e2}bydy$ in this equation can be expressed as the first moment of area when the cross-section is considered to be uniform iron.
Similarly, if we rewrite the equation for the moment due to bending stress,
$ \int_{-e1}^{e2-t} ydPa＋\int_{e2-t}^{e2}ydPw=\int_{-e1}^{e2-t} \frac{Ea}{ρ}y^2bdy+\int_{e2-t}^{e2} \frac{Ef}{ρ}y^2bdy=\frac{Ef}{ρ}(\int_{-e1}^{e2-t} \frac{Ea}{Ef}by^2dy+\int_{e2-t}^{e2} by^2dy)=M $
The expression $ \int_{-e1}^{e2-t} \frac{Ea}{Ef}by^2dy+\int_{e2-t}^{e2} by^2dy$ in this equation can be expressed as the second moment of area when the cross-section is considered to be uniform iron.
This can be illustrated as follows:
In other words, a component made of aluminum and iron can be treated as a uniform iron component by keeping the height the same and setting the width of the aluminum part to $ \frac{Ea}{Ef}b $.
This transformed second moment of area is called the equivalent cross-section.
By setting the second moment of area of the equivalent cross-section as I and using the properties of the second moment of area, the stress at the upper end and the stress at the lower end σ² of the member can be expressed by the following equations.
Stress at the upper end σ1 = -\frac{M}{I}e1 (aluminum part)
$ Stress at the lower end σ² = \frac{M}{I}e² (iron part) $
It's important to note that while the deformation and strain of the beam can be treated as equivalent because the difference in elastic modulus is absorbed by a virtual width, the bending stress does not accurately represent that of the original member.
Therefore, the stress at the upper end (aluminum part) cannot be expressed by the above formula, so we will recalculate it using the equivalent strain.
$ \frac{σ1a}{Ea}=\frac{σ1}{Ef} (the strain of the aluminum part is equal to the strain of the equivalent cross-section), σ1a=-\frac{Ea}{Ef}\frac{M}{I}e2 $
This method is convenient because it allows you to combine the calculations of the second moment of area for each component in a composite material, which can be cumbersome. Furthermore, since this method can be used with multiple types of components, I believe it's an ideal technique for components with multiple layers of laminate.
Summary
This isn't something you'll use frequently in a design setting, but I'd like you to understand the concept.
It's useful when conceptualizing layouts and sketches.
When solving statically indeterminate beam problems, in addition to the balance of external and internal forces, be sure to also consider the geometric relationship between the beam's deflection angle and deflection amount.
When multiple types of loads are applied to a beam, each load can be considered separately and then added together.
- Components made up of multiple types of materials are replaced with a single virtual cross-section.
Understanding this much should be enough.
Composite materials, in particular, are sure to see increased adoption in the future, so it's helpful to be able to visualize their potential.
I've finally managed to explain most of the needles up to this point.
Next time, as a summary of beams, we will introduce a collection of data on the second moment of area of typical cross-sections and the deflection of typical beams.
To those who found this article helpful in understanding design:
While there is basically no textbook covering this content, and it is my own original work, I will introduce the textbook that I have been using since I was a student.
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