Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 1 (Q.No. 29)
29.
A steel rod 1 metre long having square cross section is pulled under a tensile load of 8 tonnes. The extension in the rod was 1 mm only. If Esteel = 2 x 106 kg/cm2, the side of the rod, is
Discussion:
4 comments Page 1 of 1.
Abdullah said:
4 years ago
Nice, Thanks @Deepak.
Deepak singroul said:
8 years ago
e = PL/AE.
So
0.1= (8000 * 100)/(A * 2000000).
A = (8000 *100) / ( 0.1*2000000).
A = 4 squre cm.
So
Side = √4 = 2 cm. Ans.
So
0.1= (8000 * 100)/(A * 2000000).
A = (8000 *100) / ( 0.1*2000000).
A = 4 squre cm.
So
Side = √4 = 2 cm. Ans.
(1)
SKS Pandey said:
8 years ago
Great @Sam nice explanation.
(1)
Sam said:
10 years ago
e = PL/AE.
Where,
e = extension in rod(0.1cm).
P = tensile load(8,000kg).
L = length of rod(100cm).
A = area of the x-section of rod(cm^2).
E = modulus of elasticity(2,000,000kg/cm^2).
Here x-section is a square so A=x^2.
Now x = square root(PL/eE).
x = [(8,000*100)/(.1*2,000,000)]^1/2.
x = 2cm.
Where,
e = extension in rod(0.1cm).
P = tensile load(8,000kg).
L = length of rod(100cm).
A = area of the x-section of rod(cm^2).
E = modulus of elasticity(2,000,000kg/cm^2).
Here x-section is a square so A=x^2.
Now x = square root(PL/eE).
x = [(8,000*100)/(.1*2,000,000)]^1/2.
x = 2cm.
(2)
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