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:
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= 70000÷3500= 20N/mm2.
Stress in bronze =30000÷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= 70000÷3500= 20N/mm2.
Stress in bronze =30000÷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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