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)