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A coil chain of a crane required to carry a maximum load of 50 kN, is shown in the figure.

image.png

Find the diameter of the link stock, if the permissible tensile stress in the link material is not to exceed 75 MPa.

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The link is subjected to 3 types of stresses

1. Bending of the link due to curvature and eccentricity, 2. Direct tensile load on the link, 3. Double shear at the contact point, 4. crushing at the link at the point of contct (but the links are same material, hence this may be neglected)

of the above three, double shear corresponds to weakest portion, Now

Allowable shear strength => 75/2 = P/2A where P = applied load = 50,000N, 2A - double shear area; solving

A = 666 mm^2 => diameter = 29.14mm or dia = 30 mm

 

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the max limit of the stress is 75 Mpa. So in view we have the idea of what the FOS can be. (factor of safety). But since we are not sure what the material is, we can solve this in simple way 

permissible tensile stress= force/(area perpendicular to the force or load)

75 * 1e6 = 50 *1e3/(pi*(d_link/2)^2))

solving above equation leads to following diameter of chain: 0.0292 metres

which is 29.20 mm

Check the image below for reference..let me know what you think

solution.PNG

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7 hours ago, admin said:

A coil chain of a crane required to carry a maximum load of 50 kN, is shown in the figure.

image.png

Find the diameter of the link stock, if the permissible tensile stress in the link material is not to exceed 75 MPa.

All answers posted by members will be hidden so that others can post their answers without viewing the already posted answers.

 

 

29.14mm

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7 hours ago, admin said:

A coil chain of a crane required to carry a maximum load of 50 kN, is shown in the figure.

image.png

Find the diameter of the link stock, if the permissible tensile stress in the link material is not to exceed 75 MPa.

All answers posted by members will be hidden so that others can post their answers without viewing the already posted answers.

 

 

Given : P = 50 kN = 50 × 103 N ; σt
 = 75 MPa = 75 N/mm2
Let d = Diameter of the link stock in mm.
∴ Area, A = 4× d2 = 0.7854 d2
We know that the maximum load (P),
50 × 103 = σt
. A = 75 × 0.7854 d2 = 58.9 d2
∴ d 2 = 50 × 103 / 58.9 = 850 or d = 29.13 say 30 mm Ans

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On 5/17/2021 at 7:56 AM, admin said:

A coil chain of a crane required to carry a maximum load of 50 kN, is shown in the figure.

image.png

Find the diameter of the link stock, if the permissible tensile stress in the link material is not to exceed 75 MPa.

All answers posted by members will be hidden so that others can post their answers without viewing the already posted answers.

 

 

20.60mm

 

IMG20210517194257.jpg

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On 5/17/2021 at 4:19 AM, Balaji Nidaganti said:

Answer is, diameter (d)= 29mm.

Balaji, you have rounded down - ie to the "unsafe" side.  In fasteners of any kind, you would always round up...always "enhance" safety, not detract from it.

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Most of you here keep making the same mistake.
667mm². That's the surface over which our load is distributed. The surface of 2 circles. We need the diameter of a 333mm² circle.

Two people solved it correctly (~21mm). Some were on the right path, but ended up providing radius as the final result.

Edit:
How does the following work?

On 5/17/2021 at 4:26 AM, admin said:

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fa83d6dc-4327-4447-84d6-f2910cf74999_lg.

 

RX Marine was established in 1996 in Mumbai, India; as chemical manufacturing company catering exclusively to the needs of the marine industry. In a short span of 12 years the company has established itself as one of the leading wholesale suppliers of a wide range of chemicals for - Marine industry internationally - and other local industires and plants. Our client list bears testimony to this. The RXSOL policy has its foundations on two pillars of strength - a continuous investment in research and development to deliver premium quality products and a commitment to service.

Degreaser

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On 5/17/2021 at 9:32 AM, cadsanthanam said:

RX Marine was established in 1996 in Mumbai, India; as chemical manufacturing company catering exclusively to the needs of the marine industry. In a short span of 12 years the company has established itself as one of the leading wholesale suppliers of a wide range of chemicals for - Marine industry internationally - and other local industires and plants. Our client list bears testimony to this. The RXSOL policy has its foundations on two pillars of strength - a continuous investment in research and development to deliver premium quality products and a commitment to service.
Degreaser

 

 

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On 5/17/2021 at 5:26 AM, admin said:

A coil chain of a crane required to carry a maximum load of 50 kN, is shown in the figure.

image.png

Find the diameter of the link stock, if the permissible tensile stress in the link material is not to exceed 75 MPa.

All answers posted by members will be hidden so that others can post their answers without viewing the already posted answers.

 

 

29.13mm

 

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Okay...enough is enough...my turn!

Assuming the chain links form a cyclic and therefore have a natural load-spread between both legs and no stress concentration:

Stress = F/A

Stress = 75 000 000 N/m^2, F = 50 000 N

A = 50 000 / 75 000 000 = 0.000 667 m^2 for both legs

A = (pi r^2)/2 per leg  

so r = root (0.000 333 / Pi) = 0.010 301 m 

d = 2r = 0.020 601 m = 20.601 mm - always round to the safety side....so:

Therefore use 21mm dia stock minimum!

 

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