A strut 2.5m long is 60mm in diameter. One end of the strut is fixed while its other end is hinged. Find the safe compressive load with FOS=3.5. Take E=2.1x105N/mm2

 A strut 2.5m long is 60mm in diameter. One end of the strut is fixed while its other end is hinged. Find the safe compressive load with FOS=3.5. Take E=2.1x10⁵N/mm²

 Solution

Step 1: Data

Length of the column = 2500mm

Diameter of the column = 60mm

Condition = one end hinged and other end fixed

Crippling load =??

FOS=3.5

E=2.1x10⁵N/mm²

Step 2: Calculation of moment of inertia

I = π d ⁴ / 64

I= π (60)⁴ / 64

I = 0.636X10⁶mm⁴

Step 3: Calculation of  crippling load

Condition = one end hinged and other end fixed

P = 2 Π ²E I/ L²

P = 2 Π ²⁽2.1x10⁵) (0.636X10⁶) / (2500)²

P = 421.81 KN

Step 4: Calculation of safe load

Safe load = crippling load /FOS

Safe load = 421.81/3.5

Safe load = 120.52KN