Deflection in a Simply supported beam

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Discussion Overview

The discussion revolves around the deflection of a simply supported beam subjected to a central load of 150 kN. Participants explore the discrepancy between finite element (FE) analysis results and theoretical calculations of deflection, considering factors such as beam dimensions, material properties, and assumptions inherent in standard beam deflection formulas.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant reports a maximum displacement of ~4mm from FE analysis, while theoretical calculations suggest ~11mm using the formula delta = P* l^3 / (48*E*I).
  • Another participant notes that standard beam deflection formulas may not apply to short, deep beams where shear deflection could be significant.
  • A participant provides the cross-sectional dimensions of the beam (thickness = 2mm, height = 50mm) and reiterates the load and material properties.
  • Concerns are raised regarding the beam being overloaded, with calculated bending stress exceeding the elastic range for the material, suggesting that the deflection calculations may be incorrect.
  • Mathematical calculations are presented to support the theoretical deflection and bending stress values, indicating a significant discrepancy with the FE results.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of standard deflection formulas to the specific beam configuration, and there is no consensus on the validity of the FE analysis results versus theoretical predictions.

Contextual Notes

Limitations include potential assumptions about beam slenderness and material behavior under load, as well as the need for further details about the beam's cross-section to fully assess the situation.

pukb
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Hi All

As shown in the attached image, I have a simply supported beam with a load of 150kN acting at the center of the beam span.
E = 210 GPa
rho = 7800 kgm-3
span = 250 mm

After solving this in FE solver, I got maximum displacement to be ~4mm (at node 52)
But, on using the theoretical relation of delta = P* l^3 / (48*E*I), it is about ~11mm.

Can somebody please help me on this ? how to co-relate this.

Some observations:
1. There is no hourglassing.
2. Beam looks compressed by about 0.8mm at the center at the end of solution.
 

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Just eyeballing your problem, it appears you have a short, deep beam with a central load applied. The standard beam formulas for deflection are generally applicable to long slender beams where shear deflection doesn't need to be taken into account. The standard formulas are also approximations which assume that the slope of the beam in the deflected condition is very small. Without knowing more about the cross section of you beam, I can't provide any more insight into your dilemma.
 
Cross section:
Thickness = 2mm
height = 50 mm
 
pukb said:
Hi All

As shown in the attached image, I have a simply supported beam with a load of 150kN acting at the center of the beam span.
E = 210 GPa
rho = 7800 kgm-3
span = 250 mm

After solving this in FE solver, I got maximum displacement to be ~4mm (at node 52)
But, on using the theoretical relation of delta = P* l^3 / (48*E*I), it is about ~11mm.

Can somebody please help me on this ? how to co-relate this.

Some observations:
1. There is no hourglassing.
2. Beam looks compressed by about 0.8mm at the center at the end of solution.

pukb said:
Cross section:
Thickness = 2mm
height = 50 mm

Doing some simple calculations, your simply supported beam is wildly overloaded, with the calculated bending stress lying far outside the elastic range for the material used to construct it. Therefore, any deflections you are calculated using the deflection formula for a simply supported beam are certainly incorrect. The deflection your FE solver is giving you is suspect as well: this beam should probably have snapped in two.

Code:
Beam test:

L = 250 mm

depth = 50 mm
width =   2 mm

E = 210 GPa

P = 150 kN @ L/2

Simple supports

M = PL/4 = 150 kN * 0.25/4 = 9375 N-m

d = PL^3/(48EI)

I = (1/12)*0.002*0.05^3
I = 2.083E-8 m^4

d = 150000(0.25)^3/(48*210*10^9*2.083*10^-8)

d = 0.0111 m = 11.1 mm deflection

sigma = My/I = 9375 * 0.025 / 2.083*10^-8

sigma = 1.125*10^10 Pa = 11.25 GPa bending stress

yield strength = 186-758 MPa for steel

1 MPa = 145 psi

yield strength = 26970 - 109910 psi for carbon steels
 

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