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 force-equilibrium 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 one-moment 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:
Dk = 3J – re + rr
J = No. of joints
re = External support Reactions
rr = Released Reactions
re = 7
J = 6
rr = m – 1
m = No. of members meeting at internal hinge
rr = 2 – 1 = 1
rr = 2 – 1 = 1
Dk = 3 × 6 – 7 + 2
Dk = 13
The total degree of indeterminacy (external + internal) for the bridge truss is :
Dsi = m – (2j – 3)
m = 20
j = 10
Dsi = 20 – (2 × 10 – 3) = 3
Dse = re – 3 = 4 – 3 = 1
Total indeterminacy = 3 + 1 = 4
A two hinged semi-circular arch of radius R carries a concentrated load W at the crown. The horizontal thrust is
H = W/π sin2α
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 left-hand abutment. The horizontal trust on the abutment is:
VA = VB = wL/2
⇒ 4 (h – h1) = h
3h = 4h1
aking moment about hinge at D
Static indeterminacy of the frame shown
Ds = Dse + Dsi - 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
Dse = (2 + 1) - 3 = 0
Dsi = 3C = 3× 7 = 21
R = 1 (for hinge support) + 2 (roller support)
∴ Ds = 0 + 21 - 3 = 18
In a two-hinged 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
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.
Y1 = 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