The document contains two examples of calculating the diameter of rods under compressive loads. The first example calculates that a wrought iron rod with a 350 kN compressive load needs a diameter of 95 mm to not exceed the permissible stress of 52.5 MPa. The second example calculates that a piston rod supporting a piston with 0.7 MPa steam pressure and 300 mm diameter needs a diameter of 35 mm to not exceed the maximum compressive stress of 40 MPa for the material.

1. 22565 EMD : 1.9 : Fundamental of Design
1) A wrought iron rod is under a compressive load of 350 kN. If
the permissible stress for the material is 52.5 N/mm2, calculate
the diameter of the rod.
Data : Load = 350KN ,
Permissible Compressive stress = 52.5Mpa, d=?
Compressive Stress = Force / Area
Area =  / 4 x d2
= F / A
52.5 = 350 x 1000 / ( / 4 x d2 )
d = 92.15  95 mm

2. 22565 EMD : 1.9 : Fundamental of Design
2) The diameter of a piston of the steam engine is 300 mm and
the maximum steam pressure is 0.7 N/mm2. If the maximum
permissible compressive stress for the piston rod material is
40 N/mm2, find the size of the piston rod.
Data : Pressure = 0.7 N/mm2 , D=300mm, Permissible
compressive stress = 40Mpa, Force = ?, d=?
Pressure = Force / Area
Force = Pressure X Area = 0.7 X  / 4 x D2
Force = 0.7 X (3.14/4 x 3002 ) = 49455 N
D
d
Pressure

3. 22565 EMD : 1.9 : Fundamental of Design
Force = 0.7 X (3.14/4 x 3002 ) = 49455 N
Area =  / 4 x d2
= F / A
40 MPa= 49455 N / ( / 4 x d2 ) mm2
d = 31.15  35 mm