This part I have:
1)Derive an equation for the bending deflection δ of the bat at B, where the ball hits the bat, in terms of the applied load, P, the span, lAB, the Young's modulus of ash, E and the moment of inertia of the cross section, I.
Answer:
-P*l_AB^3/(3*E*I)
2) Estimate the deflection δ of the bat at B, where it hits the ball, just before it breaks, as a function of the Young's modulus, E, and the bending strength, σb, of the ash wood, and of the span of the bat between A and B, lAB, and the bat radius, r. You can assume that the bat breaks when the maximum normal stress reaches the bending strength of ash.
Express your answer in terms of σmax,lAB,E, and r.
|δ|=?
To derive an equation for the bending deflection δ of the bat at point B, where the ball hits the bat, we can use the principles of beam bending. Here's how you can approach it:
Step 1: Start with the governing equation for beam bending, which is based on Euler-Bernoulli beam theory:
M = -EI(d^2y/dx^2)
where M is the bending moment, y is the deflection, x is the distance along the beam, E is the Young's modulus, and I is the moment of inertia of the cross-section.
Step 2: Apply the equilibrium equation for bending moment at point B, where the ball hits the bat:
-M = P(l_AB - x)
where P is the applied load and l_AB is the span of the bat between points A and B.
Step 3: Integrate the bending moment equation twice with respect to x to obtain the deflection equation:
d^2y/dx^2 = -P(l_AB - x)/EI
Integrating twice will give us a second-order differential equation, and solving this equation will give us the bending deflection δ in terms of the given parameters.
Step 4: Assuming that the deflection is zero at point A (where the bat is clamped), we can consider x = 0 at point A. This allows us to simplify the equation further.
Step 5: Solve the differential equation to obtain the bending deflection equation by integrating twice and applying the boundary condition:
d^2y/dx^2 = -P(l_AB - x)/EI
Integrating once: dy/dx = -P(l_ABx - (x^2)/2)/EI + C1, where C1 is the constant of integration.
Integrating for the second time: y = -P(l_ABx^2/2 - (x^3)/6)/EI + C1x + C2, where C2 is the constant of integration.
Step 6: Apply the boundary condition to find C1 and C2. Since the deflection is zero at point A (x = 0), we have y(0) = 0. Substituting these values, we obtain C2 = 0.
Step 7: Simplify the equation to get the final expression for the bending deflection δ at point B:
y = -P(l_ABx^2/2 - (x^3)/6)/EI + C1x
Simplifying further and substituting x = l_AB, where the ball hits the bat:
δ = -Pl_AB^3/(6EI) + C1*l_AB
Considering the deflection at point B, we can assume that the deflection due to applied load P is zero. Therefore, the deflection equation becomes:
δ = C1*l_AB
Since the deflection at point B varies with the applied load, the constant of integration C1 represents the deflection under pure bending moment. When there is no pure bending moment, C1 = 0, and the final expression for the bending deflection at point B is:
δ = -Pl_AB^3/(6EI)
Now, let's move on to estimating the deflection of the bat just before it breaks.
To estimate the deflection δ of the bat at point B just before it breaks, we need to consider maximum bending stress and its relationship with the bending strength of the bat material ash.
Step 1: Express the maximum bending stress σ_max in terms of parameters E, I, and δ. The maximum bending stress occurs at the outermost fiber of the bat:
σ_max = (My)/I = (EδI)/I = Eδ
Step 2: The bat breaks when the maximum normal stress reaches the bending strength σ_b of ash. Therefore, we can equate the maximum bending stress σ_max with the bending strength σ_b:
σ_b = Eδ
Step 3: Solve the equation for δ to obtain the deflection at point B just before the bat breaks:
δ = σ_b/E
Since the deflection at point B just before the bat breaks depends on the bending strength σ_b, Young's modulus E, the span of the bat l_AB, and the bat radius r, the final expression for |δ| in terms of these parameters is:
|δ| = |σ_b|/(E*l_AB*r)