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Mechanical Principles (Statics) help please

  1. Feb 17, 2015 #1
    hey, so I'm currently studying electronic engineering on year 2 of level 3 and having to study mechanical principles this year which I don't really know anything about and my tutor doesn't explain properly. I would appreciate some help on the method of calculating the questions. I've added clear images of the questions as there would be too much to type out

    known equations that I need:

    R=√(ΣFx^2+ΣFy^2)
    θ=tan^-1(ΣFy/ΣFx)

    σ=F/A
    ε=σ/E
    A=πr^2
    τ=Fp/Ap
    S=τ/Y
    Combined Factor of Safety=Sy/(Sx^2 + 3τxy^2)^1/2

    Attempt so far, not sure if it's correct but I've tried with limited knowledge about mech p.

    Task 1

    a) Σfx=20cos(30) + 10cos(150) -25
    =17.32-8.66-25
    =-16.34N

    Σfy=20sin(30) - 10sin(150) - (f=mg=2(10)=20N)
    =10-5-20
    = -25

    R=√(16.34^2+25^2)
    =√(267+625)
    =√892
    =29.87N

    θ=tan^-1(-25/-16.34)
    =56.83°

    ∴R=29.87N at an angle of 56.83° from the positive horizontal axis

    b) Equilibrant=29.87N 180+56.83=236.83°

    ∴E=29.87N at an angle of 236.83° from the positive horizontal axis

    C) I didn't really understand the question, I assumed it meant show the resultant on the space diagram but it says calculate so I'm not sure

    Task 2

    Total load due to U.D.L=6x2=12KN
    Centre load distance from Ra=6+3=9m
    Centre load distance from Rb=3m

    Taking moments of Rb:

    Rb x 12 - 10 x 3 - 5 x (3+3) - (2x6)(9)
    ∴ 12Rb-30-30-108=0
    ∴Rb=(30+30+108)/12
    =168/12
    =14KN

    Total load on beam = 10+5+(2x6) = 27KN

    Ra+14=27
    Ra=27-14
    Ra=13KN

    Task 3

    a) A=πr^2=π30^2=900π=2827.43mm^2
    σ=F/A=(30x10^3)N/(2.827x10^-3)m^2=10.61MPa

    b) for strain I got 50.52x10^-6 using ε=σ/E

    c) I got 0.126mm using dl=σlo/E

    d) for shear stress I got 381.97MPa using τ=Fp/Ap

    for shear strain I got 2.73 x 10^-3 using S=τ/y

    Couldn't do Task 4 at all, I know theres alot to read but would appreciate some help on abit of it lol cheers 961135_10204878600638237_1457999976_n.jpg 10952420_10204878600718239_603997395_n.jpg 10816139_10204878600838242_424724110_n.jpg 11004038_10204878601158250_1131994399_n.jpg

    961135_10204878600638237_1457999976_n.jpg 10952420_10204878600718239_603997395_n.jpg 10816139_10204878600838242_424724110_n.jpg 11004038_10204878601158250_1131994399_n.jpg
     
  2. jcsd
  3. Feb 17, 2015 #2

    CWatters

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    Look at the diagram again and check the term 10cos(150).

    ditto the term 10sin(150).
     
  4. Feb 17, 2015 #3

    CWatters

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    PS Might be better to create separate threads for Task 2, 3, 4.
     
  5. Feb 18, 2015 #4

    PhanthomJay

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    Gold Member

    task 2 looks good; Task 3, you do not have the correct shear force, the pin is in double shear, you are off by a multiple.
     
  6. Feb 20, 2015 #5
    thanks guys :) appreciate the help
     
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