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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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Answer:

a) 1/2

Explanation:

Overexertion accounted for more than  half of all events that resulted in a disabling condition.

Furthermore, 30% of all overexertion cases were reported in the services industry, on the other hand, 25% of injuries resulting from contact with objects and  equipment occurred in the manufacturing industry.

The above piece of information is taken from the bureau of labor statistics, Survey of Occupational Injuries and Illnesses

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3 years ago
A furniture manufacturer purchases a drill press machine enabled with 5G and edge computing capabilities to keep the machine ope
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A lot of manufacturer often uses 5G machines. How these capabilities could help improve safety of the operators is that it does includes an emergency switch for the operator so that one can manually shut off when needed.

<h3>Edge computing with 5G</h3>

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8 0
2 years ago
Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1
tia_tia [17]

Answer:

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of <em>n-th</em> packet is :  (n - 1) × L/R

where L :  packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate =  2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

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queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

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