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Which one of the following structures is statically determinate and stable?
A structure will be statically determinate if the external reactions can be determined from forceequilibrium equations. A structure is stable when the whole or part of the structure is prevented from large displacements on account of loading.
In option '1', the structure is stable and there are three reaction components which can be determined by two force equilibrium conditions and onemoment equilibrium condition.
The structure in option '2' is stable but statically indeterminate to the second degree.
The structure is shown in option '3' has both reaction components coinciding with each other, so the moment equilibrium condition will never be satisfied and the structure will not be under equilibrium.
In option '4' the structure is not stable as a little movement in the horizontal direction leads to the large displacement and there is no resistance offered by the structure in the horizontal direction.
Determine the kinematic indeterminacy of the beam shown below:
D_{k} = 3J – r_{e} + r_{r}
J = No. of joints
r_{e} = External support Reactions
r_{r} = Released Reactions
r_{e} = 7
J = 6
r_{r} = m – 1
m = No. of members meeting at internal hinge
At B:
r_{r} = 2 – 1 = 1
At D:
r_{r} = 2 – 1 = 1
Therefore,
D_{k} = 3 × 6 – 7 + 2
D_{k} = 13
The total degree of indeterminacy (external + internal) for the bridge truss is :
D_{si} = m – (2j – 3)
m = 20
j = 10
D_{si }= 20 – (2 × 10 – 3) = 3
D_{se} = r_{e} – 3 = 4 – 3 = 1
Total indeterminacy = 3 + 1 = 4
A two hinged semicircular arch of radius R carries a concentrated load W at the crown. The horizontal thrust is
H = W/π sin^{2}α
If α = 90^{∘}, H = W/π
A three – hinged parabolic arch rid of span L and crown rise ‘h’ carries a uniformly distributed superimposed load of intensity ‘w’ per unit length. The hinges are located on two abutments at the same level and the third hinge at a quarter span location from the lefthand abutment. The horizontal trust on the abutment is:
V_{A} = V_{B} = wL/2
⇒ 4 (h – h_{1}) = h
3h = 4h_{1}
aking moment about hinge at D
D_{s} = D_{se} + D_{si}  R
Dse = External Indeterminacy
Dsi = Internal Indeterminacy
= 3 X No. of cuts required to make stable cantilever (C)
R = No. of Reactions added unnecessarily to make stable cantilever
Calculations:
D_{se} = (2 + 1)  3 = 0
D_{si} = 3C = 3× 7 = 21
R = 1 (for hinge support) + 2 (roller support)
∴ D_{s} = 0 + 21  3 = 18
In a twohinged parabolic arch (consider arch to be initially unloaded) an increase in temperature induces 
As loading is UDL and the shape of the arch is parabolic so there will be no moment in the arch rib.
But as the temperature increases in a two hinged arch (degree of indeterminacy one), the horizontal thrust will increase.
Moment due to horizontal thrust is – Py.
So maximum bending moment will be at crown as crown has highest value of y.
Find the maximum tension in the cable in KN if a unit load of 10 KN can move in either direction in the beam AB?
Let the 10 KN load is at a distance x from support A
Taking moment about A
T sin θ × 8 = 10x
⇒ T=
So ‘T’ will be maximum when ‘x’ becomes maximum i.e; x = 8 m
A uniformly distributed load of 80 kN per meter run of length 3 m moves on a simply support girder of span 10 m. The magnitude of the ratio of Maximum negative shear force to the maximum positive shear force at 4 m from the left end will be _______ (up to two decimal places).
For maximum positive shear force at C:
Using Muller Breslau Principle.
y13=356y13=356
Y_{1} = 0.3
Maximum positive shear force at (S.F)_{max1 }=108kN
For maximum negative shear force at C:
Find the maximum reaction developed at B when a udl of 3KN/m of span 5m is moving towards right in the beam shown below:
The influence line diagram for the reaction at B is shown below:
To get the maximum reaction at B, avg. load on AD = avg. load on DC
x/12 = 5−x/4
x = 3.75m
for 4m → 1.5
(4−1.25)m = 2.75m→1.5/4 × 2.75 = 1.0312
So the maximum reaction at B will be
= 1/2×(3.75) × (1.03+1.5) × 3+1/2 × (1.25) × (1.03+1.5) × 3
= 18.975 KN
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