Solve this problem

geonana · 2009-11-09T18:37:37+00:00

Hi, Todd, hope this question qualifies, as I have no clue how to solve it I've provided you a drawing to the right to assist you in coming up with the answers. You need the following information to solve for the Northerly (R1) and Westerly (R2) .... Point Load (P) = 167.4571 Length (L) between (R1) and (R2) is 1000 Length (b) between (P) and (R2) is 319.442 And that's all you should need to know to solve for (R1) and (R2).

Solve this problem

geonana
@geonana-i9IU5M

Updated: Oct 24, 2024

Views: 1.1K

Hi, Todd, hope this question qualifies, as I have no clue how to solve it

I've provided you a drawing to the right to assist you in coming up with the answers. You need the following information to solve for the Northerly (R1) and Westerly (R2) ....

Point Load (P) = 167.4571
Length (L) between (R1) and (R2) is 1000
Length (b) between (P) and (R2) is 319.442

And that's all you should need to know to solve for (R1) and (R2).

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Kaustubh Katdare

Administrator • Nov 9, 2009

Re: Mechanical engineering: Solve this problem

@Geonana: Todd runs a funny video blog. I'm moving your question to appropriate section.

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CIVILPRINCESS

Member • Nov 9, 2009

geonana
Hi, Todd, hope this question qualifies, as I have no clue how to solve it

I've provided you a drawing to the right to assist you in coming up with the answers. You need the following information to solve for the Northerly (R1) and Westerly (R2) ....

Point Load (P) = 167.4571
Length (L) between (R1) and (R2) is 1000
Length (b) between (P) and (R2) is 319.442

And that's all you should need to know to solve for (R1) and (R2).
I think i have the solution to your problem...

algebric sum of vertical forces=0
i.e. R1+R2=167.4571----(1)

The moment about any point in the beam is =0
lets take the moment about the point where R2 acts
R1x1000-167.4571x319.442=0
=>R1=53.4928
And substitute this in eq.(1)

R1=53.4928
R2=113.9643

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geonana

Member • Nov 10, 2009

Ok, something is not quite right because the solution needs to be close to these coords: (within 3 km)N 53° 29.000 W 113° 58.000

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heba66

Member • Nov 11, 2009

Hi geonana ,
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CIVILPRINCESS

Member • Nov 11, 2009

geonana
Ok, something is not quite right because the solution needs to be close to these coords: (within 3 km)N 53° 29.000 W 113° 58.000
But why is the solution in degrees.😕
if you apply degree to my solution you are getting the answer..
i.e.When you convert to degree 53.4928=>53degree29minutes
if you know the reason please let me know....:smile:

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geonana

Member • Nov 12, 2009

THX, just needed to convert it...it's going to work! THX again 😀😁

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Guttu

Member • Nov 12, 2009

CIVILPRINCESS
But why is the solution in degrees.😕
if you apply degree to my solution you are getting the answer..
i.e.When you convert to degree 53.4928=>53degree29minutes
if you know the reason please let me know....:smile:
that person is geocaching puzzler. It's part of a puzzle.

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mannucivil

Member • Nov 13, 2009

geonana
Hi, Todd, hope this question qualifies, as I have no clue how to solve it

I've provided you a drawing to the right to assist you in coming up with the answers. You need the following information to solve for the Northerly (R1) and Westerly (R2) ....

Point Load (P) = 167.4571
Length (L) between (R1) and (R2) is 1000
Length (b) between (P) and (R2) is 319.442

And that's all you should need to know to solve for (R1) and (R2).
this is determinate beam ,solv it by using equilibrium condition

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shubhrajeet tiwa

Member • Nov 18, 2009

let us apply newton equilibrium eq we have
1)net force in y dir=0
therefore
r1+r2=p---(1)
net moment about r1 =0
i.e
r2*1000=p*(1000-319.442)---(2)
solve simultaneously get ans
how ever to find difflection
generate an poerator <f(x)> =0;{f(x)=0 or (-)eve}
then bending moment eq is: r1*<l-x>-p*<b-x>----(3)
now since
d^sy/dx^2=eq(3)/(e*i);were e is young modulus and i is moment of inertia
integrate twise will give you defliction having to constat of integration
to find there value
put y=0 at point r1 and r2
were y is difliction

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