# 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 mm

^{2}and that of brass bar is 3000 mm^{2}. These are subjected to a compressive load 100,000 N. If*E*= 0.2 MN/mm_{b}^{2}and*E*= 0.1 MN/mm_{b}^{2}, the stresses developed are:Discussion:

11 comments Page 1 of 2.
Imtiaz Hakim said:
5 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.

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:
5 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.

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.

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

Please anyone explain this.

Ankit Gautam said:
6 years ago

Please explain it clearly.

Tanvee said:
7 years ago

Not getting it. Please explain.

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.

You will get the answer.

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)

Suman Chapagain said:
8 years ago

The assumption made here is the formation made by two bars are equal.

Sanjuu said:
8 years ago

Explain in details.

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