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

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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