Engineering Deflection and Slope of a Simply Supported Beam

[Tomarama24]

Homework Statement

College assignment help for computer software modelling.

Relevant Equations

M= R_A x-ω(x-a) ((x-a))/2+ω(x-b) ((x-b))/2

M/I=stress/y=E/R

Hello,

I have a mechanical college assignment for calculating the slope and deflection of a simply supported beam. I have solved for both deflection (y) and slope (dy/dx) but I am not sure if they are correct as I cannot confirm it with simulation software online.

The assignment requires me to review a suitable universal beam size from appropriate data tables to given design specifications for slope and deflection. Then to critique and justify the choice of beam with computer software to model the application. I do not have access to any paid software for this and I am stuck. I am unsure how to proceed in this area.

I have attached the initial question and the drawing of the beam. Using Macaulay's method I calculated at x=2.5m:
Deflection (y) = -2.61667µm
Slope (dy/dx) = 256.668 x10^-9 rads. (This seems to small)

Does anyone have any recommendations what software I could sue to verify this? I tried to use SkyCiv, but I don't know the moment of inertia or the Young's modulus only that EI = 50MN/m^2.

Thank you.

 

[pbuk]

If you want us to give any help you must post your workings, preferably in $\LaTeX$. Having said that,

(This seems to small)

Perhaps: when this happens it is usually because of using mm instead of m, g instead of kg etc. so check your units.

I don't know the moment of inertia or the Young's modulus only that EI = 50MN/m^2.

I think you have misunderstood what the question is asking you to do: as I read it you are supposed to select a beam design and material then size it so that $EI = 50 \ \mathrm{MNm^{-2}}$.

 

[Tomarama24]

I don't know how to use the Latex on this site. Also attached this in a word file if it's easier to read.

My workings:
R1+R2=15kN+25kN+(10kN×2m)=60kN
(15kN×15m)+(10kN×2m×2.5m)+(25kN×3.5m)=R2×5m
R2=(160 kNm)/5m=32 kN
R1=60kN-32kN=28kN
M= R_A x-ω(x-a) ((x-a))/2+ω(x-b) ((x-b))/2
M=28x-15(x-1.5)-25(x-3.5)-(10(x-1.5)^2)/2+(10(x-3.5)^2)/2
EI (d^2 y)/(dx^2 )=(28x^2)/2-(15(x-1.5)^2)/2-(25(x-3.5)^2)/2-(10(x-1.5)^3)/6+(10(x-3.5)^3)/6+C
EI y=(28x^3)/6-(15(x-1.5)^3)/6-(25(x-3.5)^3)/6-(10(x-1.5)^4)/24+(10(x-3.5)^4)/24+Cx+D
Using R1 for the boundary condition where x=0 & y=0
EI×0=(28〖×0〗^3)/6-(15(0-1.5)^3)/6-(25(0-3.5)^3)/6-(10(0-1.5)^4)/24+(10(0-3.5)^4)/24+(C×0)+D
∴0=0-0-0-0+0+0+D
D=0

EI×0=(28〖×5〗^3)/6-(15(5-1.5)^3)/6-(25(5-3.5)^3)/6-(10(5-1.5)^4)/24+(10(5-3.5)^4)/24+C×5+0
C=(-4.01667×〖10〗^2)/5=-80.3334
With x=2.5m and EI=50MN/m^2
50×〖10〗^6×y=(28〖×2.5〗^3)/6-(15(2.5-1.5)^3)/6-(25(2.5-3.5)^3)/6-(10(2.5-1.5)^4)/24+(10(2.5-3.5)^4)/24+(-80.3334×2.5)+0

Eliminating negative bracket fractions as they do not cause deflection.
∴50×〖10〗^6×y=(28〖×2.5〗^3)/6-(15(2.5-1.5)^3)/6-(10(2.5-1.5)^4)/24+(-80.3334×2.5)+0

∴y=(-130.8335)/(50×〖10〗^6 )=-2.61667×〖10〗^(-6)

EI dy/dx=(28(2.5)^2)/2-(15(2.5-1.5)^2)/2-(25(2.5-3.5)^2)/2-(10(2.5-1.5)^3)/6+(10(2.5-3.5)^3)/6+(-80.3334)
∴EI dy/dx=87.5-7.5-12.5-1.667+1.667+(-80.3334)
∴EI dy/dx=-12.8334
dy/dx=(-12.8334)/(50×〖10〗^6 )=-256.668×〖10〗^(-9) rads

This is as far as I have gotten, I am not sure how to translate this to a universal data table for a beam selection.

 

[erobz]

What makes you think the slope is too small at $\frac{L}{2}$? We expect the slope to be $\frac{dy}{dx} = 0$ somewhere just to the right of $\frac{L}{2}$ because of the slight asymmetrical loading. There could be a units error- its a lot of tabulation to check - which I haven't done, but there is a logical reason for the "smallness" of the slope in that region in my opinion.

P.S. see LaTeX Guide

 

[pbuk]

I don't know how to use the Latex on this site.

This is as far as I have gotten, I am not sure how to translate this to a universal data table for a beam selection.

No, of course not, that is looking at the problem from the wrong end. UB selection tables show elastic modulus $E$ and second moment of area $I$ and you are given the value of $EI$. Find one that (approximately) matches, model it under the specified conditions and see if the deflection (approximately) matches your calculations.

 

[Lnewqban]

Welcome @Tomarama24 !

You may find some good ideas here:

https://civilengineeronline.com/problemsolver3.htm

https://www.engineersedge.com/beam_bending/double_integration_method_example_4_15560.htm

https://mathalino.com/reviewer/mech...roblem-663-deflections-simply-supported-beams

 

[erobz]

Ok, I got motivated enough to check your work. Your solution methodology is fine, and your calculations are correct up to the evaluation of the derivative on the last step. Check to see if values inside the parenthesis are negative before you include them in the sum, and your $EI$ is also converted incorrectly. You are looking for units of $\frac{\text{kN}}{\text{m}^2}$, you have used $\frac{\text{N}}{\text{m}^2}$.

You should be getting something in the vicinity (magnitude)of $2 \cdot 10^{-3} \text{rad}$...so you were correct...slope was too small.

 

[Tomarama24]

Ok, I got motivated enough to check your work. Your solution methodology is fine, and your calculations are correct up to the evaluation of the derivative on the last step. Check to see if values inside the parenthesis are negative before you include them in the sum, and your $EI$ is also converted incorrectly. You are looking for units of $\frac{\text{kN}}{\text{m}^2}$, you have used $\frac{\text{N}}{\text{m}^2}$.

View attachment 356505

You should be getting something in the vicinity (magnitude)of $2 \cdot 10^{-3} \text{rad}$...so you were correct...slope was too small.

Hello erobz,

Thank you for the clarification. While I did eventually manage to factor in the kN part of this equation into my final calculations, I did not exclude those two parts from the final equation.

My final answer was $$-0.000256668 rads$$

I have now submitted this assignment and cannot make any changes.

 

[erobz]

Hello erobz,

Thank you for the clarification. While I did eventually manage to factor in the kN part of this equation into my final calculations, I did not exclude those two parts from the final equation.

My final answer was $$-0.000256668 rads$$

I have now submitted this assignment and cannot make any changes

bummer.