The document discusses two problems involving determining stress in mechanical linkages: 1) A crane coil chain carrying a 50 kN load must have a link diameter of at least 30 mm to ensure the stress does not exceed 75 MPa. 2) A cast iron link transmitting a 45 kN load produces tensile stresses of 50 MPa at section A-A and 64.3 MPa at the narrower section B-B, as calculated using the respective rectangular areas.

1. 22565 EMD : 1.9 : Fundamental of Design
1) A coil chain of a crane required to carry a maximum load of 50 kN, is shown
in Figure. Find the diameter of the link stock, if the permissible tensile stress in
the link material is not to exceed 75 MPa.
Data : Load = 50KN , Permissible tensile stress = 75Mpa, d=?
Stress = Force / Area
Area =  / 4 x d2
= F / A
75 = 50 x 1000 / ( / 4 x d2 )
d = 29.13  30 mm

2. 22565 EMD : 1.9 : Fundamental of Design
2) A cast iron link, as shown in Figure, is required to transmit a steady tensile
load of 45 kN. Find the tensile stress induced in the link material at sections
A-A and B-B.
Data : Force = 45KN , Area = Section A-A and Section B-B
Find : Tensile stress at A-A and at B-B
Consider first section A-A
Rectangular Area = 20 x 45 = 900 mm2
Stress = Force / Area = 45 x 1000 N/ 900 mm2 = 50 N/ mm2

3. 22565 EMD : 1.9 : Fundamental of Design
Consider first section B-B
Rectangular Area = (75 x 20) – (40 x 20) = 700 mm2
Stress = Force / Area = 45 x 1000 N/ 700 mm2 = 64.30 N/ mm2