Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 3 (Q.No. 11)
11.
If the width b and depth d of a beam simply supported with a central load are interchanged, the deflection at the centre of the beam will be changed in the ratio of
Discussion:
14 comments Page 1 of 2.
Lokesh said:
3 years ago
In the case of simply supported, deflection =pl^3/48EI.
So, from this, we can conclude deflection is inversely proportional to I,
So simply, X2/X1=db^3/bd^3= (b/d) ^2.
So, from this, we can conclude deflection is inversely proportional to I,
So simply, X2/X1=db^3/bd^3= (b/d) ^2.
(3)
Leka said:
5 years ago
Option D is correct.
Wl^2/16EI(old)÷Wl^2/16EI(new),
=1/I(old) ÷1/I(new),
=I(new) ÷ I(old),
=(b^3 * d)/(b*d^3),
=(b/d)^2.
Wl^2/16EI(old)÷Wl^2/16EI(new),
=1/I(old) ÷1/I(new),
=I(new) ÷ I(old),
=(b^3 * d)/(b*d^3),
=(b/d)^2.
Himzz the great said:
9 years ago
According to me option C is right, as first we always take I = bd^3/12 and then db^3/12 which gives us the ratio (d/b)^2.
Deepak Gupta said:
7 years ago
Option D is correct. Because deflection is inversely proportional to I. So, the value is getting after reversing.
Abhash poudel said:
10 years ago
pl^3/48EI = (pl^3.12)/(48E.bd^3).
pl^3/48EI = (pl^3.12)/(48E.db^3).
db^3/bd^3 = (b/d)^2.
pl^3/48EI = (pl^3.12)/(48E.db^3).
db^3/bd^3 = (b/d)^2.
GURUDAS said:
8 years ago
I old /I new.
= db^3/bd^3,
= (b/d)^2.
Because I is at the bottom and it will revers.
= db^3/bd^3,
= (b/d)^2.
Because I is at the bottom and it will revers.
Rishi said:
7 years ago
Option c is correct. I new/I old should be done because to compute the change.
Bhaskar Manna said:
8 years ago
I agree @Himzz. Option C is the right answer.
(1)
B.Singh said:
3 years ago
@All.
We have to always consider Change/Old.
We have to always consider Change/Old.
Rishabh said:
10 years ago
Ratio of old/new will give I new/I old.
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