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 the 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, we can use the following steps:
1. Start with the equation for the bending deflection of a simply supported beam under a point load P:
δ = -P * l_AB^3 / (3 * E * I)
Here, δ is the bending deflection, P is the applied load, l_AB is the span between A and B, E is the Young's modulus of ash, and I is the moment of inertia of the cross-section.
Now, let's move on to the second part of your question:
To estimate the deflection δ of the bat at B just before it breaks, as a function of the Young's modulus E, bending strength σ_b of ash wood, span l_AB, and bat radius r, we need to consider the maximum normal stress in the bat when it reaches the bending strength.
2. The equation for maximum normal stress in the beam due to bending is given by:
σ_max = M * c / I
Here, σ_max is the maximum normal stress, M is the bending moment, c is the distance from the neutral axis to the outermost fiber, and I is the moment of inertia of the cross-section.
3. The bending moment M can be expressed in terms of the applied load P, span l_AB, and distance x from point A to the location of interest (point B in this case):
M = P * (l_AB - x)
4. The distance c from the neutral axis to the outermost fiber can be approximated as the radius of the bat, r.
5. Substitute the expressions for M and c into the equation for maximum normal stress:
σ_max = P * (l_AB - x) * r / I
6. The bat breaks when the maximum normal stress reaches the bending strength σ_b. So we set σ_max equal to σ_b:
σ_b = P * (l_AB - x) * r / I
7. Solve this equation for x, which represents the location along the span where the maximum stress occurs:
x = (P * l_AB * r) / (σ_b * I) - l_AB
8. Finally, substitute the value of x into the equation for the bending deflection derived in part 1:
δ = -P * x^2 * (3 * l_AB - 2 * x) / (6 * E * I)
Now, with the obtained expression, you can substitute the values of σ_max, l_AB, E, and r to find the deflection δ of the bat at B just before it breaks.