A 30-mm-diameter copper rod is 1 m long with a yield strength of 70 MPa. Determine the axial force necessary to cause the diamet
1 answer:
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S = 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S = 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S > σ
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Answer:
(a) Electric field at 0.250 m is zero.
(b) Electric field at 2.90 m is zero.
(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.
(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.
Explanation:
Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.
Electric field inside and outside the metal sphere is :
E = 0 for r ≤ R ( inside )
= for r > R ( outside )
Here K is electric constant and r is the distance from the center of the metal sphere.
(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.
(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.
(c) Electric field at 3.10 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.
E = -
E = - 5.15 x 10³ V/m
(d) Electric field at 8.00 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.
E = -
E = - 0.77 x 10³ V/m