Stress and Elongation of bar due to Self Weight | Tie Bar of Uniform Strength | Strength of Materials | Solid Mechanics | By Akhand Dutta

Stress and Elongation of bar due to Self Weight | Tie Bar of Uniform Strength | Strength of Materials | Solid Mechanics | By Akhand Dutta

    Stress and Elongation in a bar due to self-weight

    or

    Derive the expression for the stress produced in a bar due to self-weight.

    or

    Derive the expression for the elongation produced in a bar due to self-weight.

    or

    Find the expression for the stress and elongation produced in a bar due to self-weight.



    Stress and Elongation of the bar due to Self Weight




    Length of bar = 'l' metres
    Area of bar = 'A' m^2
    Density = P kg/m^2

    Weight of bar NPTS = AyP
    (Ay = Volume)

    Stress, at section NP:

    S = Force at NP / Area of cross-section of bar

    S = AyPg / A = 9.81Py N/m^2

    S = 9.81Py                   ................................................................................. (1)

    (1) shows that stress due to self-weight is directly proportional to length 'y'.

    => Stress at lower end = 0.

    Smax = 9.81Pl


    If we assume dy to be very small, then thickness in LM and NP are equal.

    Then, strain in length dy = S / E

    = 9.81Py / E.

    Extension in length dy = 9.81Py.dy / E

    So, Total extension of bar: Integrating within 0 to l,

    dl = ∫0⟶l (9.81 Py)dy / E

    dl = 9.81Pl^2 / 2E



    Tie Bar of Uniform Strength

    or

    Derive the relation between the areas of the cross-section of a tie bar of uniform strength.

    or

    Find out the relation between the areas of the cross-section of a tie bar of uniform strength.


    Tie Bar of Uniform Strength

    Load Applied = 'F' newtons

    Stress = 'S' N/m^2 
    *Stress is same everywhere as the bar is of uniform strength.

    P = Density in kg/m^3
    A = Area of the cross-section at QQ.
    A+dA = Area of the cross-section at NN.

    If A varies from A1 to A2 from RR to MM:

    For Section RR: 

    S = F / A1 ................................................................... (1)
    => A1S = F

    For Section QQ:

    AS = F + (mass of QR)*g .................................................... (2)

    For Section NN:

    (A + dA)S = F + (mass of QR)*g + (mass of NQ)*g
    (A + dA)S = AS + (mass of NQ)*g ......................................... (From 2)

    AS + dA*S = AS + (mass of NQ)
    S*dA = PAyg dy

    dA / A = Pgdy / S

    Integrating within limits:

    ∫ A1 ⟶ A (dA / A) = (Pg / S) ∫ y ⟶ 0 (y dy)

    ln (A / A1) = Pgy / S

    A / A1 = e^(Pgy / S)

    A = A1 e^(Pgy / S)

    On putting y = l, A = A2

    A2 = A1 e^(Pgl / S)
    (g = 9.81)




      

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    "Meet Mr. Akhand Dutta, the visionary Founder and Owner of CEWA (Civil Engineering With Akhand Dutta). Currently, The Placement Coordinator for Structural Engineering at the prestigious National Institute of Technology, Kurukshetra. With a lif…

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