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madreJ [45]
3 years ago
6

Compare the tensile load capacity of a 5/16-18 UNC thread and a 5/16-24 UNF thread made of the same material.

Engineering
1 answer:
ehidna [41]3 years ago
6 0

Answer:

Explanation:

The material strength is given as 100,000 PSI  

To find the tensile load capacity of a threaded rod, we multiply the Strength of the material with the Stress area for the bolt.

A_{t} = 0.7854 (D -  {0.9743} / n))²

Where D is the Basic Major Diameter

n is the Number of threads per inch

At is the Tensile strength area of the bolt.

For 5/16 - 18 UNC thread

D = 0.3125

n = 18

Therefore the tensile load capacity is = 100000 X (0.7854 X (0.3125 - 0.9743/ 18)²

= 5243 lb.

Similarly for 5/16 - 24 UNF , only the n value changes to 24

we get the tensile load capacity = 5806.6 lb

Hence the fine thread bolt is stronger because it has more tensile load capacity.

For metric Bolts

We have to consider all values in SI units

Strength = 689 MPa

A_{t} = {4} / (D - 0.938194 X pitch)^{2}

We get for M8 -1.25

Tensile load capacity as = 689 X 36.6 = 25223 N = 5670 lb

For M8 -1

Tensile load capacity as = 689 X 39.167 = 26986 N = 6067 N

Strength of 5/16 - 14 Acme thread

We use the imperial formula as mentioned above,

The tensile capacity is 4634 lb.

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a ball is subject to two forces F1 and F2. The magnitudes of the two forces are 45.0 N and 70.0 N respectively. In the figure be
tino4ka555 [31]

Answer:

  • F1.x ≈ -28.93
  • F1.y ≈ 34.47
  • F2.x = 70
  • F2.y = 0
  • (F1+F2).x ≈ 41.07
  • (F1+F2).y ≈ 34.47
  • |F1+F2| ≈ 53.62
  • ∠(F1+F2) ≈ 40.0°

Explanation:

A suitable calculator can show you the vector components and their resultant in polar or rectangular format. (See attached.) 2D vectors are conveniently treated as complex numbers, which is why the y-component values are shown as imaginary.

(The 50° angle measured from the -x axis is equivalent to 130° measured from the +x axis, which is the reference we're using here.)

If you'd like to compute the vector components by hand, they are ...

  (x, y) = magnitude×(cos(angle), sin(angle))

This notation is sometimes abbreviated <em>magnitude cis angle</em>, a reference to the complex number form x+yi.

8 0
2 years ago
Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia
Mrrafil [7]

Answer:

The fin temperature in °C at a distance of 10 cm from the base = 33.78°C

Explanation:

The following assumptions will be made to solve this problem

- The heat transfer coefficient does not change with the time or distance.

- The temperature of the fins varies just in only one direction.

The temperature of the fin at x = 10 cm = 0.10 m from the base can be calculated from the temperature variation with distance formula for a very long fin.

(T - T∞) = (T₀ - T∞)e⁻ᵐˣ

T = T(x) = temperature at any point along the fin

T∞ = temperature at the tip of the fin = ambient temperature = 25°C

T₀ = temperature at the base of thw fin = 50°C

x = any distance along the length of the fin from the base of the fin = 0.1 m

m = √(hP/KA)

h = Heat transfer coefficient = 123 W/m².K

P = perimeter in contact with the base = πD = π × 0.03 = 0.0943 m

K = thermal conductivity = 150 W/m.K

A = surface area in contact with the base = πD²/4 = π(0.03)²/4 = 0.0007071 m²

m = √(123 × 0.0943)/(150 × 0.0007071)

m = 10.46

mx = 10.46 × 0.1 = 1.046

(T - 25) = (50 - 25) e⁻¹•⁰⁴⁶

T = 25 + 25 e⁻¹•⁰⁴⁶ = 25 + 8.78 = 33.78°C

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