If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.
<h3>What is strain?</h3>
Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.
- L = 20 cm d x 1 = 0.21 cm
- dx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3
- (assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333
- b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333
- = 1561.84 MPa.
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Answer:
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Explanation:
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Answer:
SPCA factor
Single payment compound amount factor.
Total amount pay A = $24,230.95 (Approx)
Interest paid = $4,230.95 (Approx)
Explanation:
Given:
P = $20,000
n = 6 year
r = 3.25%
Find:
Total amount pay A
Computation:
A=p(1+r)ⁿ
A=20,000[1+3.5%]⁶
A=20,000[(1.0325)⁶]
Total amount pay A = $24,230.95 (Approx)
Interest paid = $24,230.95 - 20,000
Interest paid = $4,230.95 (Approx)
Answer:
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0 it is a incompressible flow.
c) ap = (16/3) i + (32/3) j + (16/3) k
Explanation:
Given
V = xy² i − (1/3) y³ j + xy k
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0
then
∇V = ∂(xy²)/∂x + ∂(− (1/3) y³)/∂y + ∂(xy)/∂z
⇒ ∇V = y² - y² + 0 = 0 it is a incompressible flow.
c) ap = xy²*∂(V)/∂x − (1/3) y³*∂(V)/∂y + xy*∂(V)/∂z
⇒ ap = xy²*(y² i + y k) - (1/3) y³*(2xy i − y² j + x k) + xy*(0)
⇒ ap = (xy⁴ - (2/3) xy⁴) i + (1/3) y⁵ j + (xy³ - (1/3) xy³) k
⇒ ap = (1/3) xy⁴ i + (1/3) y⁵ j + (2/3) xy³ k
At point (1, 2, 3)
⇒ ap = (1/3) (1*2⁴) i + (1/3) (2)⁵ j + (2/3) (1*2³) k
⇒ ap = (16/3) i + (32/3) j + (16/3) k
Answer:The distinction between a permanent magnet and an electromagnet is essentially one in how the field is created, not the properties of the field afterwards. So electromagnets still have two poles, still attract ferromagnetic materials, and still have poles that repel other like poles and attract unlike poles.
Explanation:
