Deflection in a Simply supported beam

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The discussion centers on the deflection of a simply supported beam subjected to a central load of 150 kN. The finite element (FE) solver indicated a maximum displacement of approximately 4 mm, while theoretical calculations using the formula δ = P * l³ / (48 * E * I) predicted a deflection of about 11 mm. The discrepancy arises due to the beam's short and deep geometry, which leads to significant shear deflection not accounted for in standard beam formulas. Additionally, the calculated bending stress of 11.25 GPa exceeds the yield strength of typical carbon steels, indicating potential structural failure.

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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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