1. Elongation of Uniformly tapering Circular rod
Let us consider the uniformly tapering circular rod as shown in figure, length of the uniformly tapering circular rod of length L and larger diameter of the rod is D1 at one end and the diameter of other end is D2. The bar is subjected with an axial tensile load P
Let us consider a section at distance of x form lager end and take one infinitesimal smaller element of length dx.
The diameter of infinitesimal smaller element is Dx
Dx = D1-[(D1-D2)/L] X
Dx = D1- KX
Where we have assumed that K= (D1-D2)/L
The area of cross section of circular bar at a distance x from its larger diameter end is Ax
Ax = (П/4) (Dx)2
Ax = (П/4) (D1- KX) 2
Stress
The stress induced in circular bar at a distance x from its larger diameter end is σx and we will be given by
σx = P/ Ax
σx = P/ [(П/4) (D1- KX) 2]
σx = 4P/ [П (D1- KX) 2]
Strain
The strain induced in circular bar at a distance x from its larger diameter end is Ԑx and we will be given by
Strain = Stress / Young’s modulus of elasticity
Ԑx = σx /E
Ԑx = 4P/ [П E (D1- KX) 2]
Elongation of the bar
Let we assume dl is the Change in length of infinitesimal smaller element, it will be determined by recalling the concept of strain.
dl = Ԑx. dx
Where, Ԑx = 4P/ [П E (D1- KX) 2]
Now we will determine the total change in length of the uniformly tapering circular rod by integrating the above equation from 0 to L.
Check: If the bar is uniform diameter (d1=d2 = d) the elongation in the bar is = PL/AE
Example 1:
Find extension of tapering circular bar under axial pull for the following data: d1 = 20mm, d2 = 40mm, L = 600mm, E = 200 GPa. P = 40kN
Solution :
The elongation in tapered bar
dL = 4PL/(π E d1 d2)
Substitute the values in the above equation,
dL = 4x40,000x600 / (πx 200,000x20x40)
= 0.38mm. Ans.
2.Elongation of a Tapering bar of uniform thickness t and width varies from d to D
Take a element of length dx at a distance of x form end d as shown in figure.
Width of bar at a distance X form one end
bx = d + [(D - d) / L] . x
bx = d + k.x
Where k = (D - d) / L
Aera of the element dA = (d + kx).t
The elongation in the element of length dx is due to axial load P is
dL = Pdx / [Et(d + kx)],
The total elongation in the bar is obtained by integrating the above equation by applying the limits 0 to L
δL = ∫ Pdx / [Et(d + kx)],
= P/Et ∫{ dx /(d + kx)} ,
δL = [PL /{Et(D – d)}]. loge(D/d)
or if D= b2 and d= b1 this will write as
Example:
In the diagram shown below, a tensile load of 50 kN is applied axially. What is the increase in length of the rectangular tapered plate of thickness 10 mm. (Assume E = 220 Gpa)
Solution:
The elongation in the tapering bar of uniform thickness is = δL = [PL/{Et(D – d)}] .loge(D/d) /
The given data is
P= 50 kN =50x1000 N
D=b2= 100mm, d= b1 = 50mm
t = 10mm , L = 400mm
E= 220 GPa = 220000 N/mm2
Substitute the given data in the equation.
δL =0.126 mm Ans.
3.Elongation of a bar due to its own weight
Consider a bar of circular cross section and uniform diameter throughout. Consider the bar to be suspended from a rigid support and its top end, such that it is in a hanging in a vertical position as shown in the figure.
Let,
A = Uniform cross sectional area of the bar
E = Young’s modulus for the bar
L = Length of the bar
ρ = Weight of the bar per unit length of the material of the bar
Consider an element of length ‘dy’ at a distance of ‘y’ from the bottom of the bar.
The element being elongated due to the force ‘P’, at section x-x, as shown in the figure.
Weight of the portion below x-x = P = ρ × A × y
Change in the length of the element dy is
dL = PL/AE = (ρ×A×y)×dy/AE
= [ρA×y×dy] /AE
For total change in the length of the bar, we need to integrate along the length 0 to L
Total change in length = ∫[ρA×y×dy] /AE
On integrating, we get,
δL=(ρAL.L/2)/AE
If the total weight of the bar W = ρAL
This is the expression for the elongation of a uniform bar under self weight will becomes
δL = W L / 2AE
Example:
A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of steel is 7850 kg/m3 and E = 200 GPa, find the total elongation of the rod.
Solution:
The total Elongation in the bar is elongation due to its own weight and elongation due to external applied load
Total elongation= δ=δ1+δ2
The elongation due to its own weight
δ1=WL/2AE
Where:
W =ρAL = 7850(9.81/1000)3x[300(150x1000)]
δ1=WL/2AE
Where:
W =ρAL = 7850(9.81/1000)3x[300(150x1000)]
W = 3465.3825 N
L = 150(1000) = 150 000 mm
A = 300 mm2
E = 200 000 MPa
Thus,
δ1=(3465.3825x150000)/ (2x300x200000)
L = 150(1000) = 150 000 mm
A = 300 mm2
E = 200 000 MPa
Thus,
δ1=(3465.3825x150000)/ (2x300x200000)
δ1=4.33 mm
Elongation due to applied load is,
δ2=PL/AE
Where:
P = 20 kN = 20 000 N
L = 150 m = 150 000 mm
A = 300 mm2
E = 200 000 MPa
Thus,
δ2=(20000x150000)/(300x200000)
δ2=PL/AE
Where:
P = 20 kN = 20 000 N
L = 150 m = 150 000 mm
A = 300 mm2
E = 200 000 MPa
Thus,
δ2=(20000x150000)/(300x200000)
δ2=50 mm
Total elongation= δ=δ1+δ2
δ=4.33+50 = 54.33 mm Ans.
Total elongation= δ=δ1+δ2
δ=4.33+50 = 54.33 mm Ans.
Assignment:
Q1. A circular rod 0.2m long, tapers form 20mm diameter at one end to 10mm at other end. On applying an axial pull of 6KN it was found to extend by 0.068mm. Determine the Young’s Modulus of the rod material.
Q2.Determine the elongation of the tapered cylindrical aluminum bar as shown in figure caused by 30 kN axial load. Use elastic modulus of E=72 GPa
Q.3 A steel rod having a cross-sectional area of 400 mm2 and a length of 50 m is suspended vertically from one end. It supports a tensile load of 40 kN at the lower end. If the unit mass of steel is 7850 kg/m3 and E = 200 GPa, find the total elongation of the rod.
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