Simply supported beam with a spring support in the middle - deflection

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The discussion focuses on calculating the maximum deflection of a simply supported beam with a spring support in the middle under a uniformly distributed load (UDL). The initial derivation using a known formula for maximum deflection yielded a result of 6.1928 mm, which differed from the finite element analysis (FEA) result of 5.1796 mm. The user questions the validity of their conversion of load and whether the simplified approach used for concentrated forces can be applied to UDLs. They explore alternative calculations and derive a new formula that aligns more closely with the FEA results, indicating that accounting for the spring's force proportional to deflection is crucial. The conversation emphasizes the importance of correctly treating UDLs in deflection calculations for statically indeterminate beams.
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How to calculate the maximum deflection of a simply supported beam with spring support in the middle when it's subjected to uniformly distributed load ?
Hi,

I'd like to calculate the maximum deflection of a simply supported beam with spring support in the middle and UDL (uniformly distributed load) acting on the whole beam:

beam scheme.JPG


Here's my derivation, starting from the known formula for maximum deflection of a simply supported beam with UDL and no spring: $$\delta_{b}=\frac{5 \left( \frac{F}{L} \right) L^{4}}{384EI}$$ $$k_{b}=\frac{F}{\delta_{b}}=\frac{384EI}{5L^{3}}$$ $$k_{t}=k_{b}+k_{s}=\frac{384EI}{5L^{3}}+k_{s}$$ $$\delta_{t}=\frac{F}{k_{t}}=\frac{5FL^{3}}{384EI+5kL^{3}}$$

The problem is that when I substitute the data: ##F=2000 \ N##, ##L=500 \ mm##, ##E=210 \ GPa##, ##I=\frac{a^{4}}{12}=\frac{12^{4}}{12}=1728 \ mm^{4}##, ##k=100 \ \frac{N}{mm}##

the result is: $$\delta_{t}=\frac{5 \cdot 2000 \cdot 500^{3}}{384 \cdot 210000 \cdot 1728+5 \cdot 100 \cdot 500^{3}}=6.1928 \ mm$$

while from FEA, I get: ##5.1796 \ mm## and I believe this result is correct. What's wrong with my formula ?

Interestingly, I got a very good agreement for the same case with a concentrated force in the middle instead of UDL (using the same approach, just a different base formula for the deflection of the beam). Does it mean that the ##q=\frac{F}{L}## conversion is incorrect here ? How should I treat it then ?
 
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Well, I guess that I could use one of the standard methods to calculate the deflection of this statically indeterminate beam from scratch but I don’t know how to treat the spring. Also, the aforementioned simple approach based on the known formula for the deflection of a beam without a spring works well for the case with concentrated force in the middle so it seems that it’s just a matter of properly accounting for the UDL. Unless this simplified approach won’t work with UDL but I hope it’s not the case.
 
Lnewqban said:
Would this approach to a different situation be useful to yours?

https://mathalino.com/reviewer/strength-materials/problem-709-propped-beam-spring-support
Thank you very much. Using this approach, I get: $$\delta=\frac{5qL^{4}}{384EI}=8.9705 \ mm$$ $$\delta - \delta_{s}=\delta_{R}$$ $$8.9705- \frac{R}{k}=\frac{RL^{3}}{48EI}$$ $$8.9705- \frac{R}{100}=\frac{R \cdot 500^{3}}{48 \cdot 210000 \cdot 1728}$$ $$R=522.26 \ N$$ $$\delta_{s}=\frac{R}{k}=\frac{522.26}{100}=5.2226 \ mm$$ so pretty close to the simulation result. I also substituted everything and rearranged the equation to get a single formula: $$\delta_{s}=\frac{5qL^{4}}{384EI+8kL^{3}}$$
 
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You should have 4 equations to work with:

$$ \uparrow^+ \sum F = R_A - wL -k y|_{x=L/2}+R_B = 0 $$

$$ \circlearrowright^+ \sum_{A} M = k y|_{x=L/2} \frac{L}{2}+ wL \frac{L}{2} - R_BL = 0 $$

Elastic Eqn's:
$$ EI \theta = \frac{R_A x^2 }{2} -\frac{w x^3}{6} + C_{\theta} \tag{slope}$$

$$ EI y = \frac{R_A x^3 }{6} -\frac{w x^4}{24} + C_{\theta}x + C_y \tag{deflection}$$

with conditions:

##y|_{x=0} = 0##

##\theta|_{x=L/2} = 0 ##

by evaluating ## y|_{x=L/2}## I believe you can combine all these to solve the system.
 
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When I solve the system I get:

$$y_c=−\frac{5}{8}\left( \frac{wL^4}{ 48EI+kL^3} \right) \approx−5.22~ \rm{mm} $$

$$ R_A=R_B=\frac{1}{2}(ky_c+wL) \approx 738.9 ~\rm{N} $$

$$ Fc=−ky_c \approx 522.2 ~\rm{N} $$

EDIT: I found an algebra error. I see this agrees with the result in #5
 
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