Test: Structural Analysis- 2 - Civil Engineering (CE) MCQ

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.

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

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 

H = W/π sin²α

If α = 90^∘, H = W/π

V_A = V_B = wL/2


 

⇒ 4 (h – h₁) = h

3h = 4h₁
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

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. 

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

For maximum positive shear force at C:

Using Muller Breslau Principle.

y13=356y13=356

Y₁ = 0.3

Maximum positive shear force at  (S.F)_max1 =108kN

For maximum negative shear force at C:

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