PLASTIC MOMENT

I just want to correct the solution of Hashem Al Marimi.

In structural steel design, most steel beams are designed inaccordance with the allowable stress design method, wherein anactual bending stress fb inducedby the load man not exceed itsallowable Fy. Hence this method is kept within its elasticrange.

On the other hand, the plastic method takes advantage of thesubstantial reserve strength of a steel beam that exist after theyield stress has been reached.Therefore, the plastic theoryutilizes the stress strain relationships through the plastic rangeup to the start of strain hardening.. With this,equationsfor elastic theory is NO LONGER applicable.

Here is the solution to the problem:

Mmax based on shear and moment diagram = 3P

This Mmax = 3P wil also be the Mp or the plastic moment

Again since this is plastic design, is not applicable,instead we will beusing the following formulas for plastic design:

1.
2.

where:
Fy = yield strength of steel (MPa)
Mp = plastic moment (kN-m)
Z = plastic section modulus (m³)


Our Working Equation:


So,




STEP 1 - Get the total area:

Atotal = 200(300) - 100(200) = 40,000 mm²


STEP 2 - The Plastic Neutral Axis (PNA)will divide the section intotwo equal area.
Area above = Area below

50(100) + 2 (50) (x) = 20,000
x = 150mm (this is obvious since the section is symmetrical)


STEP 3 - Solve for Z

Z = ΣAy = [50(100(125) + 50 (100) (75) (2)] x 2
Z = 2,750,000 mm³ or 2.75 x 10^-3m³

STEP 4 - Substitute the values





or

-Maximum bending moment in the beam:kN.m

-Locate the c.g. : from symmetrymmm

- Moment of inertia for the crosssection:


- Since :

- Solving for P we get:
P = 212.78 kN ANS..