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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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aleksandr82 [10.1K]

Answer:

int()

Explanation:

float() is using decimals, so that can't be it, like float(input( "how much does this cost?"))

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By elimination, int() is correct.

Hope this helps!

7 0
2 years ago
A 6V battery is connected in series with two bulbs. What would the voltage drop be across each bulb?
Vitek1552 [10]

Answer:

3V

Explanation:

6 0
3 years ago
A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and lo
Zinaida [17]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, e_{min} = 0.51

Void ratio in the loosest state, e_{max} = 0.87

Now,

Dry density, \gamma_d=\frac{\gamma_t}{1+w}

=\frac{18}{1+0.05}

= 17.14 kN/m³

Also,

\gamma_d=\frac{G\gamma_w}{1+e}

here, G = Specific gravity = 2.7 for sand

17.14=\frac{2.7\times9.81}{1+e}

or

e = 0.545

Relative density = \frac{e_{max}-e}{e_{max}-e_{min}}

= \frac{0.87-0.545}{0.87-0.51}

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

or

S = 0.2477

or

S = 0.2477 × 100% = 24.77%

7 0
3 years ago
Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1
babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A     ⇒    P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E)    ⇒  ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒  ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0
3 years ago
Por que no puedo rasterizar una capa en Photoshop?
Jobisdone [24]

Answer:

¿Qué aplicación estás usando? podría ser porque te equivocaste un paso

8 0
3 years ago
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