Question

18.35 Compute the required diameter of a steel push-rod

subjected to an axial compressive load of 10 kips.
The rod is to be made of AISI 1020 cold-drawn steel
(yield stress = 50 ksi). The length is 24 in. and the ends
are pinned. Use the Euler-Johnson formulas with a factor
of safety of 3.0.

Answer 1

Given:

Axial compressive load = 10 kips = 10000 lbs

Yield stress of AISI 1020 cold-drawn steel = 50 ksi

Length of the rod (L) = 24 in

Factor of safety (FOS) = 3

We need to find the diameter of the rod (d).

The Euler's critical load formula for a column with both ends pinned is given by:

Pcr = (pi^2 * E * I) / L^2

where,

Pcr = critical buckling load

E = Modulus of elasticity

I = Moment of inertia

L = Length of the column

The moment of inertia for a solid circular rod is given by:

I = (pi * d^4) / 64

The maximum compressive stress that the rod can withstand without buckling is given by the Euler-Johnson formula:

Pallow = (FOS * pi^2 * E * I) / L^2

where,

Pallow = Allowable compressive load

FOS = Factor of safety

E = Modulus of elasticity

I = Moment of inertia

L = Length of the column

The maximum load that the rod can withstand is equal to the yield load. Hence, we can write:

10,000 = (FOS * pi^2 * E * I) / L^2

Solving for the moment of inertia (I), we get:

I = (10,000 * L^2) / (FOS * pi^2 * E)

Substituting the values, we get:

I = (10,000 * 24^2) / (3 * pi^2 * 29 * 10^6)

I = 0.0112 in^4

Substituting this value of I in the moment of inertia equation, we get:

0.0112 = (pi * d^4) / 64

Solving for d, we get:

d = 0.524 in

Therefore, the required diameter of the steel push-rod is 0.524 inches.

Step-by-step explanation: