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
Discussion:
38 comments Page 1 of 4.
Deep Das said:
3 years ago
The right answer is 'C'.
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.
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)
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.
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.
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.
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.
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.
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.
Pradeesh said:
5 years ago
C is the correct answer.
(1)
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.
P-3000*10 N.
A-314.16mm2 (D -20mm).
L-5000mm.
E-2.1*10^6.
L = 0.2274mm.
Dheeraj said:
6 years ago
Correct answer is C.
Amjad Ali said:
6 years ago
Right answer is 0.02274mm.
Zia Ullah said:
6 years ago
The correct answer is 0.02275 mm.
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