Mechanical and structural analysis miscellaneous
- The buckling load P = Pcr for the column AB in the figure, as KT approaches infinity, become
α π2EI L2
where α is equal to
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Since stiffness KT ≈ ∞,
it act as fixed end. At the top end there will be only axial deformation. so it also cuts as fixed end.PCϒ = 4. π2EI l2
∴ α = 4Correct Option: D
Since stiffness KT ≈ ∞,
it act as fixed end. At the top end there will be only axial deformation. so it also cuts as fixed end.PCϒ = 4. π2EI l2
∴ α = 4
- The bending moment diagram for a beam is given below: The shear force at sections aa' and bb' respectively are of the magnitude.
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Consider left of α – α'
Shear force = 0
(Since there is uniform bending moment) Consider right bb'
Shear force = change in bending moment
= (200 – 150) kNm = 50 kNmCorrect Option: C
Consider left of α – α'
Shear force = 0
(Since there is uniform bending moment) Consider right bb'
Shear force = change in bending moment
= (200 – 150) kNm = 50 kNm
- kN 75. A cantilever beam of length l width b and depth d is loaded with a concentrated vertical load at the tip. If yielding starts at a load P, the collapse load shall be
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Shape factor = MP ie, Plastic moment Me Elastic moment
= 1.5 for rectangular.
Ultimate load = yield point × shape factor yield point = P
∴ ultimate load = P ×1.5 = 1.5 PCorrect Option: B
Shape factor = MP ie, Plastic moment Me Elastic moment
= 1.5 for rectangular.
Ultimate load = yield point × shape factor yield point = P
∴ ultimate load = P ×1.5 = 1.5 P
- A three hinged parabolic arch ABC has a span of 20 m and a central rise of 4 m. The arch has hinges at the ends at the centre. A train of two point loads of 20 kN and 10 kN, 5 m apart, crosses this arch from left to right, with 20 kN load leading. The maximum thrust induced at the supports is
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Max thrust occurs when 20 kN coal reaches point B.
This is found out by drawing ILD.
Max Thrust = 20 × 1.25 + 10 × 1.25/2
= 31.25 kNCorrect Option: C
Max thrust occurs when 20 kN coal reaches point B.
This is found out by drawing ILD.
Max Thrust = 20 × 1.25 + 10 × 1.25/2
= 31.25 kN
- In a two dimensional stress analysis, the state of stress at a point is shown below.
If σ = 120 Mpa and τ = 70 MPa,
then σx and σy are respectively
AB = 4
BC = 3
AC = 5
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= 67.5 MPa
= 213.3 MPaCorrect Option: C
= 67.5 MPa
= 213.3 MPa