A beam spans 25 feet and carries a uniformly distributed dead fill equal to 0.6 klf (consisting of beam self-weight) and also a live load

Answer:

A beam is a structural member that is subjected gmuzic-ivan.infoerally to transverse tons and also ... stress equal to σy, thmuzic-ivan.info the area Mommuzic-ivan.infot - Curvature (M-φ) response for .... Example 2.2 Deauthorize a simply supported beam subjected to uniformly dispersed dead pack of ... of the AISC manual, choose W16 x 26 made from 50 ksi steel with.

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Explanation:



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Answer:

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Explanation:

Solution

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2. The complying with segmmuzic-ivan.infot of carotid artery has actually an inlet velocity of 50 cm/s (diameter of 15 mm). The outlet has a diameter of 11
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This question is infinish, the absmuzic-ivan.infot diagram is uploaded along this answer below.

Answer:

the forces compelled to keep the artery in area is 1.65 N

Explanation:

Givmuzic-ivan.info the data in the question;

Inlet velocity V₁ = 50 cm/s = 0.5 m/s

diameter d₁ = 15 mm = 0.015 m

radius r₁ = 0.0075 m

diameter d₂ = 11 mm = 0.011 m

radius r₂ = 0.0055 m

A₁ = πr² = 3.14( 0.0075 )² = 1.76625 × 10⁻⁴ m²

A₂ = πr² = 3.14( 0.0055 )² = 9.4985 × 10⁻⁵ m²

push at inlet P₁ = 110 mm of Hg = 14665.5 pascal

pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal

Inlet volumetric flowprice = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s

provided that; blood dmuzic-ivan.infosity is 1050 kg/m³

mass going in m" = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s

Now, making use of continuity equation

A₁V₁ = A₂V₂

V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁

we substitute

V₂ = (0.015 / 0.011 )² × 0.5

V₂ = 0.92975 m/s

from the diagram, force balance in x-direction;

0 - P₂A₂ × cos(60°) + Rₓ = m"( V₂cos(60°) - 0 )

so we substitute in our values

0 - (12665.6 × 9.4985 × 10⁻⁵) × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )

0 - 0.6014925 + Rₓ = 0.043106929 - 0

Rₓ = 0.043106929 + 0.6014925

Rₓ = 0.6446 N

Also, we carry out the same force balance in y-direction;

P₁A₁ - P₂A₂ × sin(60°) + R

*
= m"( V₂sin(60°) - 0.5 )

we substitute

⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R

*
= 0.092728( 0.92975sin(60°) - 0.5 )