Using a Table provided in Lecture 2 determine longitudinal sound speed in aluminum. Assume Young's modulus E= 10*10^6 lbf/in^2 , Poisson ratio v=0.33 and density of rho=5.13 slug/ft^3. Use imperial units ft/s in your answer.

Question

Using a Table provided in Lecture 2 determine longitudinal sound speed in aluminum. Assume Young's modulus E= 10*10^6  lbf/in^2 , Poisson ratio v=0.33 and density of rho=5.13 slug/ft^3.  Use imperial units ft/s in your answer.

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We need to determine the longitudinal wave speed in a homogeneous isotropic aluminum material using the given material constants, and express the result in ft/s in imperial units. The standard relation for the longitudinal (compressional) wave speed in an isotropic solid is:
c_l = sqrt( (K + 4/3 μ) / ρ )
where K is the bulk modulus, μ is the shear modulus, and ρ is the density.
For an isotropic solid, K and μ can be expressed in terms of Young's modulus E and Poisson's ratio ν as:
μ = E / (2(1 + ν))
K = E / (3(1 − 2ν))
Plugging these into the expression gives an equivalent compact form:
c_l = sqrt( E * [ 1/(3(1 − 2ν)) + 2/(3(1 + ν)) ] / ρ )
This is often also written as c_l = sqrt( E(1 − ν) / [ (1 + ν)(1 − 2ν) ρ ] ), which is algebraically the same after simplification.
Now insert the provided numbers and pay careful attention to units:
- E is given as 10 × 10^6 lbf/in^2. This is 10^7 psi. To use ρ in slug/ft^3, convert E to lbf/ft^2 by multiplying by 144 (since 1 ft^2 = 144 in^2): E = 1.44 × 10^9 lbf/ft^2.
- ν = 0.33.
- ρ = 5.13 slug/ft^3.
Compute the factor F that multiplies E/ρ inside the square root. Using the form c_l = sqrt( E(1 − ν) / [ (1 + ν)(1 − 2ν) ρ ] ), we evaluate:
1 − ν = 0.67
1 + ν = 1.33
1 − 2ν = 1 − 0.66 = 0.34
Thus the denominator factor is (1 + ν)(1 − 2ν) = 1.33 × 0.34 ≈ 0.4522.
So the inside of the square root becomes: E × (1 − ν) / [ (1 + ν)(1 − 2ν) × ρ ] = (1.44 × 10^9) × 0.67 / (0.4522 × 5.13).
Compute the numerator: 1.44 × 10^9 × 0.67 ≈ 9.648 × 10^8.
Compute the denominator: 0.4522 × 5.13 ≈ 2.321 × 10^0 ≈ 2.321.
Divide: 9.648 × 10^8 / 2.321 ≈ 4.160 × 10^8.
Take the square root to obtain c_l: c_l ≈ sqrt(4.160 × 10^8) ≈ 2.04 × 10^4 ft/s, i.e., about 20,400 ft/s.
Alternative formulation (c_l = sqrt( E [ 1/(3(1 − 2ν)) + 2/(3(1 + ν)) ] / ρ )) yields a numerically consistent result when using the same E, ν, and ρ after unit conversion; evaluating that form with the given values gives a similar magnitude on the order of tens of thousands of ft/s, typically around 20,000 ft/s for aluminum with these parameters.
What if your target or provided answer is 16,750 ft/s? A few possibilities could explain the discrepancy:
- Different unit treatment: if E were not properly converted to ft-based units (lbf/ft^2) or if ρ used a different basis (e.g., lbm/ft^3 with g_c adjustments), the numeric result would shift.
- Slightly different ν or E: small changes in ν (0.33 vs 0.34) or E (e.g., 10^7 vs 10^7.1) can noticeably affect c_l because the relationship is under a square root.
- Using an approximate or alternative form of the formula could yield a somewhat different value, especially if certain conversion factors are approximated.
In any case, the key steps are consistent: convert E to consistent units with ρ, compute μ and K (or use the single-form c_l = sqrt( E(1 − ν) / [ (1 + ν)(1 − 2ν) ρ ] )), and then take the square root to obtain the longitudinal wave speed in ft/s. The computed result with the provided numbers and proper unit handling lands near 20,000 ft/s, not 16,750 ft/s, suggesting a possible mismatch in the given target value or in unit conventions used for that number.

MENG-4089-06/MENG-5072-01/MENG-5072-01D HW 1

Using a Table provided in Lecture 2 determine longitudinal sound speed in aluminum. Assume Young's modulus E= 10*10^6 lbf/in^2 , Poisson ratio v=0.33 and density of rho=5.13 slug/ft^3. Use imperial units ft/s in your answer.

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