--Original Question:
A hollow cylinder with outer radius r_0 and inner radius r_i is loaded by a uniformly distributed load q across its length L. The beam is supported by a roller at its left end and a pin L/5 from its right end.
What is the maximum bending moment in the beam? Express your answer in terms of q and L.
--Worked out solution:
www.jiskha.com/questions/1822768/a-hollow-cylinder-with-outer-radius-r-0-and-inner-radius-r-i-is-loaded-by-a-uniformly
--I'm just trying to figure out the steps: Why does the derivative of x become x? and how does a 2 show up in the denominator?
Thanks in advance for any further guidance someone can provide me.
Ha! I must say, discussing cylinder loads is quite a "circular" topic, but I'm happy to help with some "hilarious" guidance.
To understand why the derivative of x becomes x, we need to take a step back and put on our mathematical caps. Remember that the bending moment (M) in a beam is given by the equation M = -E * I * (d^2y/dx^2), where E is the Young's modulus and I is the moment of inertia.
In this problem, we have a hollow cylinder, so we need to calculate the moment of inertia (I) using the formula I = π/4 * (r_o^4 - r_i^4). Now, when we take the second derivative of y with respect to x, we get d^2y/dx^2 = d/dx(d/dx(y)). But wait, what is y? In this case, y represents the displacement of the beam at a given point, or in simpler terms, how much the beam bends.
Since we're dealing with a simple beam supported by a roller and a pin, the displacement equation for the beam is given by y = (1/2) * q * x^2, where q is the load and x is the distance from the roller support. Taking the derivative of y with respect to x gives us dy/dx = q * x, and taking the derivative once more gives us the second derivative, d^2y/dx^2 = q.
So, you see, the derivative of x doesn't become x itself, rather it becomes a constant value of q, because the second derivative of the displacement equation only depends on the load, not the distance x.
Now, let's address the curious appearance of the 2 in the denominator. In the equation for the bending moment M = -E * I * (d^2y/dx^2), the bending moment is proportional to the second derivative of the displacement. However, for a uniformly distributed load, the bending moment varies linearly with distance x, meaning it increases as x increases. To compensate for this, the bending moment formula is divided by a factor of 2 to account for the increase in moment along the length of the beam.
So there you have it! The derivative of x becomes a constant value of q because of the displacement equation, and the appearance of the 2 in the denominator is to compensate for the linear increase in bending moment. I hope this explanation brings a smile to your face, and if you have any more "circus-worthy" questions, feel free to ask!
In the derivation provided in the link you shared, the derivative of x does not become x. The equation that is being referred to is:
M = q * x * (L/5 - x) * (2/3)
Let's break down this equation step by step:
1. q is the uniformly distributed load applied across the length of the hollow cylinder.
2. x represents the distance measured from the left end of the cylinder to the section where we want to find the bending moment.
3. L/5 - x represents the distance between the pin support and the section where we want to find the bending moment. This represents the length of the remaining portion of the hollow cylinder.
4. (L/5 - x) * x represents the product of these two distances. It represents the area of the section where we want to find the bending moment. Note that when we take the derivative with respect to x, this term will help us account for how the bending moment changes as we move from one section of the beam to another.
5. The (2/3) term is a constant that arises in the derivation process and is specific to this particular beam loading and support configuration. It's not related to taking any derivative.
To find the maximum bending moment, you would need to take the derivative of the equation with respect to x, set it equal to zero, and solve for x. This will give you the value of x where the bending moment is maximum.
To understand why the derivative of x becomes x and why a 2 shows up in the denominator, let's break down the steps in the solution.
First, we need to find the shear force (V) as a function of x. The shear force is the sum of all the forces acting on the beam to the left or right of a particular point. Since the load is uniformly distributed, we can express it as q times the length of the beam (L). Thus, we have:
V(x) = q * (L - x)
Next, we need to find the bending moment (M) as a function of x. The bending moment is the integral of the shear force with respect to x. Integrating V(x) with respect to x, we get:
M(x) = ∫ V(x) dx
= ∫ (q * (L - x)) dx
= q * ∫ (L - x) dx
Now, let's evaluate this integral step by step:
∫ (L - x) dx = ∫ L dx - ∫ x dx
The integral of a constant (L in this case) with respect to x is simply the constant times x:
∫ L dx = Lx
The integral of x with respect to x can be calculated using the power rule for integration:
∫ x dx = (x^2)/2
Putting it all together, we have:
M(x) = q * (Lx - (x^2)/2)
Now, to find the maximum bending moment, we need to find the value of x that maximizes M(x). Since M(x) is a quadratic function with a negative coefficient for the x^2 term, its graph opens downwards and has a maximum value. The vertex of the parabola represents this maximum. The x-coordinate of the vertex can be found using the formula x = -b/(2a), where a is the coefficient of the x^2 term and b is the coefficient of the x term.
In our case, a = -1/2 and b = L. Plugging these values into the formula, we have:
x = -L / (2 * (-1/2))
= L
Thus, the maximum bending moment occurs at x = L. Substituting this into the expression for M(x), we get:
M(max) = q * (L(L) - (L^2)/2)
= q * (L^2 - L^2/2)
= q * (L^2/2)
So, the maximum bending moment in the beam is q * (L^2/2).
Regarding your question about the derivative of x becoming x and the 2 in the denominator, it is crucial to remember how integration works. When we integrate (L - x) with respect to x, -x becomes -x^2/2, and L becomes Lx. This occurs because the derivative of x with respect to x is 1, and the derivative of L with respect to x is 0 (since L is a constant). These derivatives are reversed when integrating. Additionally, the 2 in the denominator arises from the power rule for integration, specifically when integrating x to get (x^2)/2.