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scoundrel [369]
4 years ago
9

For an isotropic material, E and ν are often chosen as the two independent engineering constants. There are other elastic consta

nts: the shear modulus G, the bulk modulus K, and the Lames’ constants, μ and λ.
Suppose that G = E/[2(1+v)].
Prove that μ and λ, and K, satisfy the following relations:

μ = G = 3K(1-2v)/[2(1+v)], λ = Ev/ [(1+v)(1-2v)] = 2vG/ (1-2v),
K = E/ 3(1-2v) = λ+(2/3)μ

To prove it, apply pure shear stress τ, and connect it to pure shear strain γ by G, ie. τ = Gγ. Then in the 45° orientation, consider its normal stresses σ, and σ: are related to its normal strains ε, and e, through E and v in Hooke's law. We can easily establish this relation.
Engineering
1 answer:
pav-90 [236]4 years ago
8 0

Answer:

khgy

Explanation:

nbv

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3 years ago
Construct a Mohr circle for the stress element at A in problem 2. Using ruler and compass, draw the Mohr circle to the scale. Dr
serious [3.7K]

Answer:

hello your question is incomplete attached below is the missing diagram to the question and the detailed solution

Answer : principal stresses : 0.82 MPa, -33.492 MPa

              shear stress = 17.157 MPa

              ∅ = 9.09 ≈ 10°

Explanation:

The principal stress ( б1 ) = 0.82 MPa

                                 ( б2 ) = -33.492 MPa

The shear stress = 17.157 MPa

∅ = 9.09 ≈ 10°

attached below is the detailed solution and the Mohr's circle

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3 years ago
An experiment compares the initial speed of bullets fired from two handguns: a 9 mm and a 0.44 caliber. The guns are fired into
olasank [31]

Answer:

1.176

Explanation:

When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.

v2 * (M + mb) = v1 * mb

Where

v1: muzzle velocity of the bullet

M: mass of the bob

mb: mass of the bullet

v2: mass of the bob with the bullet after being hit

v2 = v1 * mb / (M + mb)

Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.

Ek = 1/2 (M + mb) * v2^2

Ep = (M + mb) * g * h

Ek = Ep

1/2 (M + mb) * v2^2 = (M + mb) * g * h

1/2 * (v1 * mb / (M + mb))^2 = g * h

1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h

v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)

v2 = \sqrt{\frac{g *h * (M+ mb)^2}{\frac{1}{2} * mb^2}}

The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.

h = L * (1 - cos(a))

v2 = \sqrt{\frac{g * L * (1 - cos(a)) * (M+ mb)^2}{\frac{1}{2} * mb^2}}

For the 9 mm:

v2 = \sqrt{\frac{9.81 * L * (1 - cos(4.3)) * (10+ 0.006)^2}{\frac{1}{2} * 0.006^2}} = \sqrt{L} * 391

For the 0.44 caliber:

v2 = \sqrt{\frac{9.81 * L * (1 - cos(10.1)) * (10+ 0.012)^2}{\frac{1}{2} * 0.012^2}} = \sqrt{L} * 460

The ratio is 460 / 391 = 1.176

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Answer:

Detailed solution is given below

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