Undergrad Finding yield strength with a load/displacement curve

Click For Summary
To determine the yield strength from a load/displacement curve of a simply supported steel beam under a uniformly distributed load, one must identify the point where the curve deviates from linearity. Unlike tensile tests, where yielding is evident across the entire cross-section, bending leads to yield initiation at the outer fibers, making detection subtler. The maximum strain occurs at the beam's center, and strain can be calculated using the deflection equation and Hooke's Law. The yield stress can be estimated by calculating the normal stress using the maximum moment at the midpoint, derived from the load at the point of deviation. This approach allows for accurate assessment of yield strength in bending scenarios.
Ferdiss
Messages
2
Reaction score
0
Hi! I have done a simulation where I got out the load/displacement curve of a steel beam that is simply supported with an equally distributed load across the whole length (10 m). I want to field the point at which the steel yields. I am used to use the 0,2% value on a stress/strain curve, but how do I do it at a load/displacement curve? I have the units N on y-axis and m on x-axis.

Is it correct to say that strain is displacement(in z-direction, in the lateral direction of the beam) / l0(10 meters) ? Even though the displacement is in 'another direction' than the beam length?

So strain = 0,002 * 10 meters = 0,02 m?

Thank you!
 
Last edited:
Physics news on Phys.org
Welcome, @Ferdiss ! :cool:

Sorry, I can't understand your description of the task at hand.
Do you have any specification of material and type of that beam?
 
Hi, it is a steel beam S355. A simply supported beam with a uniformly distributed load :)

1713699870143.png
 
Hi @Ferdiss

No, you cannot compute strain that way. Unlike a bar in pure tension under axial load, this is a beam bending problem where strain is maximum at the outer fibers of the beam at the beam center location. If you have a plot of load vs displacement z, you can get the 0.002 z yield by drawing ne parallel from .002 z to the linear portion of curve and find the yield load and hence z. load. To convert to strain you need to compute z by the defection equation of the simple support beam (proportional to L^3/EI) and computes tress from the maximum moment (Mc/I) where c is rhe distance neutral axis to outer fiber. Then you compute strain using Hooke Law ( stress = E x strain) and find strain as proportional to zc/L^2 . This max strain at beam center.
 
Last edited by a moderator:
In a tensile test, the onset of yielding will show on practically the entire cross-section. So, spotting the point at which the force-displacement curve deviates from linear is relatively easy. For bending, the situation is different. In bending, yield begins at the portions of the beam furthest from the neutral axis and spreads to the entire cross section as the load increases. Thus, the deviation of the force-displacement curve from linearity is more subtle.

To find an estimate of the yield strength from a force-displacement curve, where the force here is the value of the distributed load q and the displacement is that of the beam at L/2, look for the first sign of deviation from linear. At that load, you can calculate the stress by using the normal stress in bending equation.

$$\sigma = \frac{Mc}{I}$$

and the maximum moment on the midpoint cross-section is

$$M = \frac{qL^2}{8}$$.

Plugging in the expression for M into the stress equation, with q equal to the load at deviation from linearity, is your estimate of the yield stress.
 
I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

Similar threads

Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 69 ·
3
Replies
69
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
7
Views
8K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
3
Views
2K