Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 2 (Q.No. 9)
9.
A steel rod of sectional area 250 sq. mm connects two parallel walls 5 m apart. The nuts at the ends were tightened when the rod was heated to 100°C. If αsteel = 0.000012/C°, Esteel = 0.2 MN/mm2, the tensile force developed at a temperature of 50°C, is
Discussion:
7 comments Page 1 of 1.
Karn chaurasiya said:
6 years ago
Stress = Eat =0.2*10^6 *0.000012 * 50 = 120N/mm^2.
(1)
Shivank said:
6 years ago
Stress= αET = 120 n/mm2.
NARAHARI said:
7 years ago
STRESS = E*change in strain.
Change in starin=α*change in temp.
10^6N/mm^2*0.00012*(100-50)=120N/mm^2.
Change in starin=α*change in temp.
10^6N/mm^2*0.00012*(100-50)=120N/mm^2.
Aamir said:
7 years ago
In question its actually tensile stress, otherwise tensile force =30000N.
(1)
Omar said:
9 years ago
$ = change in length.
$ = 5 * 0.000012 * 50 = 0.003,
£= $/L=0.003/5 = 0.0006,
s = E* £ = 0.2 * 0.0006 = 0.00012 MN,
s = 0.00012 * 1000000 =120 N/mm2.
$ = 5 * 0.000012 * 50 = 0.003,
£= $/L=0.003/5 = 0.0006,
s = E* £ = 0.2 * 0.0006 = 0.00012 MN,
s = 0.00012 * 1000000 =120 N/mm2.
Mahesh said:
9 years ago
6 = excxt = 2 x 1.2 x 50 = 120n/mm2.
James said:
9 years ago
How did you get the answer?
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