- #1

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- 4

- Homework Statement
- 1. In the formula D = Eh^3/(12(1-v^2)), h is given as 0,1 +- 0,002 and v as 0,3 +- 0,02. Express the approximate maximum error in D in terms of E.

2. The coefficient of rigidity (n) of a wire of length (L) and uniform diameter (d) is given by n = AL/d^4, where A is constant. If errors of +-0,25% and +-1% are possible in measuring L and d respectively, determine the maximum percentage error in the calculated value of n.

- Relevant Equations
- 1.

- D = Eh^3/(12(1-v^2))

- ΔD = ∂D/∂h * Δh + ∂D/∂v * Δv

2.

- n = AL/d^4

- Δn = ∂n/∂L * ΔL + ∂n/∂d * Δd

Attempt at question No. 1:

ΔD = ∂D/∂h * Δh + ∂D/∂v * Δv

∂D/∂h = 3Eh^2/(12(1-v^2))

∂D/∂v = 2Eh^3/(12(1-v^2)^2)

Δh = +- 0,002

Δv = 0,02

h = 0,1

v = 0,3

ΔD = 3Eh^2/(12(1-v^2)) * Δh + 2Eh^3/(12(1-v^2)^2) * Δv

Because the problem asked for maximum percentage error then I decided to use the positive value of Δh = 0,002, because the value of Δv is positive, is this right?

ΔD = 3E(0,1)^2/(12(1-(0,3)^2)) * (0,002) + 2E(0,1)^3/(12(1-(0,3^2)^2) * (0,02)

Finish the equation and then the result will be:

ΔD = 4,5746 * 10^-6 E.

Really skeptical about this answer though.

Attempt at question No. 2:

Δn = ∂n/∂L * ΔL + ∂n/∂d * Δd

∂n/∂L = A/d^4

∂n/∂d = -4AL/d^5

ΔL = +-0,25% L

Δd = +-1% d

Δn = A/d^4 * (0,25%*L) + -4AL/d^5 * (1%*d)

Δn = AL/d^4 * (0,0025) + -4AL/d^4 * (0,01)

Δn = 0,0025n - 0,04n = -0,0375n = -3,75%n

I'm not confident in this one though, I think the answer should be 4,25%, but I don't know how to achieve that value.