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 2 of 4.

Ataa alkaby said:   6 years ago
The right answer is 0.0275mm.

Zia Ullah said:   6 years ago
The correct answer is 0.02275 mm.

Sudip Byapari said:   8 years ago
2. 274 is the correct answer.

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.

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.

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

Vikas said:   8 years ago
I agree @Situ Gupta.


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