Why ultimate load is less than yield load?

[FK123]

i have made a beam with both ends fixed. when I apply load at center then the ultimate load will be 8Mp/L=8fyz/L which came to be 440KN

My question is how to calculate the load at which beam starts yielding?

I have used E=200000N/mm2 and initial yield strength=220N/mm2

 

[SteamKing]

i have made a beam with both ends fixed. when I apply load at center then the ultimate load will be 8Mp/L=8fyz/L which came to be 440KN

My question is how to calculate the load at which beam starts yielding?

I have used E=200000N/mm2 and initial yield strength=220N/mm2

We'll need a lot more information than you've given us to figure out what you've done.

First, tell us the length of the beam.

Second, tell us what you mean when you write "8Mp/L=8fyz/L". About all I can guess is that L = length of the beam. Everything else, who knows?

For a fixed-fixed beam with a central load P, the max. bending moment is M_MAX = PL / 8, which occurs at the ends of the beam and in the center.

While the bending stress remains below the yield stress, then the bending stress σ is calculated by

σ = M y / I,

where M = bending moment,
y = distance from the neutral axis of the beam
I = second moment of area of the beam cross section.

shear stress τ is calculated by

τ = VQ / (I * t)

where V = shear force,
I = second moment of area of the beam cross section (same value used above)
Q = first moment of area
t = span of the cross section in way of the shear stress location.

As you can see, you need to know something about the beam cross section in order to calculate the stresses created when the beam is loaded.

Because of the nature in which a fixed-fixed beam is supported, there will be additional shear stresses created at the locations where the maximum bending moment occurs, so these shear stresses must be calculated and combined with the bending stress to give the proper stress value at those locations, so that one can determine the load at which yielding of the material in the beam occurs.

Once the stress in the outer fibers of beam exceeds the yield stress of the material, the formula above no longer applies and a plastic analysis must be performed.

 

[FK123]

Thankyou
Length of beam=1000mm Width of beam=100mm height of beam=100mm
Ultimate load = 8* Plastic Moment/ length
where Plastic moment = Yield stress * plastic section modulus

I was supposed to do non linear analysis of beam. so i have used the von mises model to define the beam material property. where E=200000N/mm2 and initial yield strength=220N/mm2 and poison ratio=0.28
my professor told me to calculate manually the load at which plastic hinge will be created and the load at which material starts yielding. Therefore i have used Ultimate load = 8* Plastic Moment/ length to calculate load at which plastic hinge created.

But i didnt know how to calculate load at yield?

 

[SteamKing]

Thankyou
Length of beam=1000mm Width of beam=100mm height of beam=100mm
Ultimate load = 8* Plastic Moment/ length
where Plastic moment = Yield stress * plastic section modulus

I was supposed to do non linear analysis of beam. so i have used the von mises model to define the beam material property. where E=200000N/mm2 and initial yield strength=220N/mm2 and poison ratio=0.28
my professor told me to calculate manually the load at which plastic hinge will be created and the load at which material starts yielding. Therefore i have used Ultimate load = 8* Plastic Moment/ length to calculate load at which plastic hinge created.

But i didnt know how to calculate load at yield?

Finding the load at yield is simple. When the beam bends elastically, the outer fibers see the highest bending stresses. Find the load P such that the bending stress in the outer fiber of the beam reaches the yield stress of the material.

Assuming that the cross section of this beam is rectangular, the maximum bending stress at the outer fibers will be

σ = M y / I = 220 N/mm², where y = depth of the beam / 2 and I = bh³/12. Solve for M, which is also M = P*L/8.

 

[FK123]

thankyou

 

[FK123]

I have calculated the load at plastic hinge i.e 440KN
and the load at yield is 293.3KN

when i do nonlinear analysis it shows the maximum load of 306KN. why does not the software runs till 440KN since 440 is ultimate load where plastic hinge will be created? the graph below shows the maximum load 306KK. the graph is between load and displacement