The television coverage of the World Series showed several remarkable slow-motion views of the deflection of the bat as it hit the ball. Major league wood baseball bats are made of ash. A typical major league bat has a length of 0.95m and a tapering circular cross section. The ball typically hits the bat 150mm from the free end (see figure below). Occasionally, the bat breaks as it hits the ball.
For this problem, you can assume that the bat is a cylinder of constant radius and that the batter holds the end of the bat rigidly, so that it is loaded as a cantilever beam.
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.
δ =
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| =
Calculate the value for the deflection δ of the bat at B, given that ash has a Young's modulus, E=10GPa and a bending strength, σb=100MPa, and that lAB=0.8m and the (assumed constant) radius of the bat is 22mm.
the first one I have:
sigma=-(P*l_AB^3)/(3*E*I)
somebody can help me?
To derive the equation for the bending deflection δ of the bat at B, we can use the principles of bending of a cantilever beam.
1. Start by considering the bending moment M at point B where the ball hits the bat. The bending moment can be calculated as the product of the applied load P and the distance from the point where the load is applied to B, which is lAB - 150mm.
M = P * (lAB - 150mm)
2. The bending moment is related to the bending deflection δ at B through the equation:
M = (E * I) * (d^2δ / dx^2)
Where E is the Young's modulus of ash and I is the moment of inertia of the circular cross-section of the bat.
3. Since the bat is a cantilever beam, the deflection at the free end (B) is zero, and the slope of the deflection at the free end is also zero.
δ(B) = 0
(dδ / dx)(B) = 0
4. By integrating the equation from step 2 twice, we can solve for the bending deflection δ:
δ(x) = (P / (6 * E * I)) * ((lAB - 150mm)^2 - x^2) * (x - lAB + 150mm)
Where x is the distance from the fixed end of the bat to any point along its length.
Now let's calculate the estimate of the deflection |δ| of the bat at B just before it breaks.
1. The maximum normal stress σmax in the bat occurs at the point where the deflection is the greatest. In this case, it is at the point B where the ball hits the bat. Using the formula for the maximum normal stress in a cantilever beam:
σmax = (M * y) / I
where y is the distance from the neutral axis of the cross-section to the point B, which is the radius r of the bat.
2. The bat breaks when the maximum normal stress reaches the bending strength σb of the ash wood, so we can set σmax = σb and solve for the deflection δ:
δ = (σb * I) / (E * r)
Now let's calculate the value for the deflection δ of the bat at B using the given values:
E = 10 GPa = 10^10 Pa
σb = 100 MPa = 10^8 Pa
lAB = 0.8 m
r = 22 mm = 0.022 m
Substituting these values into the equation for δ, we get:
δ = (10^8 Pa * π * (0.022 m)^4) / (10^10 Pa * 0.022 m)
Calculating this expression, we find:
δ = 0.001 π m ≈ 0.00314 m