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
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

Rosnee said:   8 years ago
Please explain the calculation.

Sanjuu said:   8 years ago
Explain in details.

Suman Chapagain said:   8 years ago
The assumption made here is the formation made by two bars are equal.

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.

You will get the answer.
(1)

Chhaya said:   7 years ago
Thanks @Suman.

Tanvee said:   7 years ago
Not getting it. Please explain.

Ankit Gautam said:   6 years ago
Please explain it clearly.

Monika Verma said:   6 years ago
Please anyone explain this.

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.

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= 70000÷3500= 20N/mm2.
Stress in bronze =30000÷3000= 10N/mm2.
(2)


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