# Civil Engineering - Theory of Structures - Discussion

Discussion Forum : Theory of Structures - Section 2 (Q.No. 22)
22.
A compound bar consists of two bars of equal length. Steel bar cross-section is 3500 mm2 and that of brass bar is 3000 mm2. These are subjected to a compressive load 100,000 N. If Eb = 0.2 MN/mm2 and Eb = 0.1 MN/mm2, the stresses developed are:
σb = 10 N/mm2, σs = 20 N/mm2
σb = 8 N/mm2, σs = 16 N/mm2
σb = 6 N/mm2, σs = 12 N/mm2
σb = 5 N/mm2, σs = 10 N/mm2
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

Imtiaz Hakim said:   4 years ago
As it is a compound bar strain produced will be the same in both bars...

PsL/AsEs = PbL/AbL.

L will be cancelled out.
So
Ps/3500*2 = Pb/3000*.1 , Ps/700 = Pb/300.
Or Pb = 0.428ps.
Now P = PB +Ps = 0.428ps +ps = 1.428Ps.
100000=1.428Ps or Ps= 70000N and Pb = 30000N.
Stress = load/ area. stress in steel = 70000/3500 =20N.
Stress in brass = 30000/3000 =10N.
(3)

Imtiaz hakim said:   4 years ago
As it is a compound bar so the strain produce will be the same in both.

Strain PsL/AsEs is equal to PbL/ AbEb.
As length is the same it will be canceled out.
So Ps/3500*.2 = Pb/3000*.1.
Ps/700 = Pb/300.
.427Ps = Pb.
P= pb + ps.

P= .427ps + ps= 1.427ps.
100000= 1.427ps.
Ps = 70000.
100000-70000=Pb.
30000=pb.
Now stress in steel brass= 70000Ã·3500= 20N/mm2.
Stress in bronze =30000Ã·3000= 10N/mm2.
(2)

Arun said:   6 years ago
Stress = PL/AE.
6s = Ps L/As Es
6b=Pb L/Ab Eb.

Length of both bars is same hence L will be cancelled,
By putting values of the respective element.
6s= 20N/mm2 6b= 10 N/mm2.

Monika Verma said:   6 years ago

Ankit Gautam said:   6 years ago

Tanvee said:   7 years ago

Chhaya said:   7 years ago
Thanks @Suman.

Suman Chapagain said:   8 years ago
Deformation of steel = deformation of brass.

Ps.Ls/As.Es = Pb.Lb/Ab.Eb.
Here Ls and Lb cancels as lengths are equal.
You get from here is Ps =7Pb/3.
P = Ps + Pb.

Substitution here you get, Ps = 70,000 and Pb = 30,000.
Then stress = force/area.