Neha Joshi

EduRev JEE

Neha Joshi
EduRev JEE
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Discussed Questions

Total strain energy stored in a simply supported beam of span, 'L' and flexural rigidity 'EI 'subjected to a concentrated load 'W' at the centre is equal to
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Neha Joshi answered  •  Oct 11, 2023
Strain energy
=
=
=
Alternative method: In a funny way you may use Castigliano’s theorem,
δ =
We know that δ = for simply supported beam in concentrated load at mid span.
Then δ = or
U = partially integrating
with respect to W we ge U =

A truss consists of horizontal members (AC, CD, DB  and EF) and vertical members (CE and DF) having length l each. The-members AE, DE and BF are inclined at 45° to the horizontal For the uniformly distributed load p per unit length on the members EF of the truss shown in figure given below, the force in the member CD is
[2003]
  • a)
    pl/2
  • b)
    pl
  • c)
    0
  • d)
    2pl/3
Correct answer is option 'A'. Can you explain this answer?

Neha Joshi answered  •  Oct 11, 2023
Total load on EF member = pl
where l = length of EF
For horizontal equilibrium
RA + RB = Pl

Taking moment about A, we have

Considering a point A.
For horizontal equilibrium FAC + FAE cos 45º = 0

For vertical equilibrium
RA – FAE sin 45º = 0


–ve sign shows that the force on member AC is opposite of assumed. Now considering a point C For horizontal equilibrium
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