Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 3 (Q.No. 9)
9.
A steel rod of 2 cm diameter and 5 metres long is subjected to an axial pull of 3000 kg. If E = 2.1 x 106, the elongation of the rod will be
2.275 mm
0.2275 mm
0.02275 mm
2.02275 mm.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
38 comments Page 4 of 4.

Dheeraj said:   6 years ago
Correct answer is C.

Chandan said:   5 years ago
0.2274 ans
P-3000*10 N.
A-314.16mm2 (D -20mm).
L-5000mm.
E-2.1*10^6.
L = 0.2274mm.

Pradeesh said:   5 years ago
C is the correct answer.
(1)

TOFIQUE AKHTAR said:   4 years ago
We know that,

Modulus of Elasticity (E) = Stress/Strain.
= (P/A) / (dL/L).
Or, E = PxL/AxdL.
Or, dL = PxL/AxE

Given, P = 3000 Kg. A = (π/4) x 202 mm2 L = 5000 mm E = 2.1 x 106 Kg/mm2
So, dL = 3000x5000/{(π/4)x20x20x2.1x106 } mm,
= 3000x5000x7x4x10 / 22x20x20x21x106 mm [ Put π = 22/7]
1/44 mm,
= 0.022727 mm.

Hardik Kumar said:   4 years ago
Force = 3000 kg.
Converting it into Newton,
Force = 3000*9.18 N,

dL = PL/AE.
dL = (3000 * 9.81 * 5000)/((π/4)*20*2.1*10^6).
dL = 0.222mm.

B.Singh said:   3 years ago
Dia = 20 mm
Lenght = 5000 mm
Load (P) = 3000 kg.

Note;

{E = 2.1x10^6 N/mm^2} (given)
Convert E into kg/mm^2
So, E = 0.21x10^6 kg/mm^2 (Approx).

Delta = PL/AE = (3000 x 5000 x 4) / (pie x 20 x 20 x 0.21 x 10^6)
Delta = 0.2273 mm.

Nitin said:   3 years ago
I agree @ Subham,

Because they given 'e' in kg/cm^2, So only we have to convert length in cm, after calculation we get the answer in cm, then multiplied by "10" that gives the answer in mm.

So finally option "A" right answer.
(1)

Deep Das said:   3 years ago
The right answer is 'C'.


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