Mechanical Engineering
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# Mechanical Numerical of the day

## Question

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

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

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2.06 cm

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30mm should be the link chain diameter.

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Known Data

Max.Tensile stress = 75MPa

Force = 50kN

Solution

Surface = Force/Max.Tensile stress = (50x10^3 N) / (75x10^6 Pa) = 0,667x10^-3 m^2

Surface = 2 x pi/4 x d^2

0,667x10^-3 m^2 = pi/2 x d^2

d = 21 mm

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It comes to about 3 cms dia.

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

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

29.14mm

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

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

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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TENSILE STRENGTH= FORCE/AREA

75000=50/A

A= COIN AREA CROSS SECTION= 3.13*D*D/4

D=COIN DIAMETER IN METER

SO , D=0.029METERS

SO THE COIN DIAMETER SHOULD BE GREATER THAN 0.029 METERS

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Bit rusty but here's my attempt. (Note: Used mathcha.io for math text editing! It's useful!)

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After solving I got d=29.1 mm

After solving d= 29.1 mm or d= 30 mm

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given data

`P=50KN=50000N

tensile stress= f= 75 MPa

DIAMETER WILL BE EQUAL TO = ROOT ((4*50000)/(3.14*75))

d= 29.13mm=30mm

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Where I get solution of this question?

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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.

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

20.60mm

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29,1 mm

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diameter will be 0.0103 m

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d=30 mm

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Tensile Stress = Load / Area

75 = 50*10^3 / (Pi D^2 / 4)

D^2*75 = 50000 / 0.7854

D^2 = 848.82

d = 29.13 mm.

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

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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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.

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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.

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

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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