# 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
80 N/mm2
100 N/mm2
120 N/mm2
150 N/mm2
Explanation:
No answer description is available. Let's discuss.
Discussion:
7 comments Page 1 of 1.

Karn chaurasiya said:   4 years ago
Stress = Eat =0.2*10^6 *0.000012 * 50 = 120N/mm^2.

Shivank said:   5 years ago
Stress= αET = 120 n/mm2.

NARAHARI said:   5 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.

Aamir said:   6 years ago
In question its actually tensile stress, otherwise tensile force =30000N.

Omar said:   7 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.

Mahesh said:   7 years ago
6 = excxt = 2 x 1.2 x 50 = 120n/mm2.

James said:   8 years ago
How did you get the answer?