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

• FEAnalyst
FEAnalyst
TL;DR Summary
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:

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 ?

Last edited:
I think this should help you figure out how to do it:

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: $$\delta_{s}=\frac{5qL^{4}}{384EI+8kL^{3}}$$

Lnewqban and jack action
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.

Last edited:
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

Last edited:

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