Structural Analysis MCQ Quiz in मराठी - Objective Question with Answer for Structural Analysis - मोफत PDF डाउनलोड करा

Concept:

Static Indeterminacy: If the equilibrium equations are enough to analyze for unknown reactions, the structure is said to be statically indeterminate.

Ds = Dse + Dsi

Dse = r – S

Dsi = 3 × Number of closed loop (For portal frame)

Dsi = m- 2j+ 3 (For truss structure)

Kinematic Indeterminacy: It is the total number of possible degree of freedom of all the joints.

Dk = 3J - r + h (For beam & portal frame)

Dk = 2J - r + h (For truss structure)

where, Dse = External Indeterminacy, Dsi = Internal Indeterminacy, Dk = Kinematic Indeterminacy, r = No. of unknown reactions, S = No. of equilibrium equation, m = No. of mombers, h = No. of plastic hinges & J = No. of joints

Calculation:

Here, r = 7 and J = 6

D_k = 3 × 6 – 7 + 2

∴ D_k = 13

Concept:

Initially the horizontal thrust is towards right at A.

When support B sinks.

∑ M_B = 0

V_A × 1 – H_A × δ  = 0

\(\Rightarrow {H_A} = \frac{{{V_A} \cdot l}}{\delta }\)

So, H_A is acting towards left

So overall horizontal thrust will decrease.

Mistake Points When the support sinks/settles the horizontal thrust acts in the opposite direction, hence the horizontal thrust gets decreased. 

Alternate Method:

The equation for horizontal thrust in a 2-hinged parabolic arch is

\(H\; = \;\frac{{\smallint \frac{{{M_x}ydx}}{{EI}}\; + \;\alpha tl}}{{\smallint \frac{{{y^2}dx}}{{E{I_C}}}\; + \;\frac{l}{{AE}}\; + \;k}}\)

Where A = Horizontal thrust

E = Modulus of elasticity of the arch material

I = moment of inertia of Arch cross-section

A = Area of cross-section of the arch

α = Coefficient of thermal expansion

t = Difference of temperature

l = length of arch, k = yielding of supports

Here there is no clear mention of temperature Change, (t = 0),

so the horizontal thrust will decrease as the support sinks, i.e k has some value.

And H inversely proportional to k

D_s = D_se + D_si - R

D_se = (2 + 1) - 3 = 0

D_si = m - (2j - 3) = 10 - (2 × 5 - 3) = 3

Where m = no of members

J = no of pin joints

R = 0

D_s = 3

Explanation:

As we know that 

Fixed end moments correspond to reactions at support such that the slope at supports will be zero 

As conjugate beam method 

The slope at supports is equal to the reaction at supports of the conjugate beam.

So we can predict that the reaction ay support of the conjugate beam is inversely proportional to the moment of inertia (MOI)

Greater MOI  of the beam, lesser will be support reaction of conjugate beam hence lesser will be the slope.

So slope decreased with an increase in MOI the moment required to make slope zero will be lesser means the fixed end moment will increase. 

Given:

Maximum bending moment at C:

Maximum bending moment at C will occur when average loading just to the left of C is equal to average loading just to right of C.

\(\frac{x}{{5 - x}} = \frac{4}{6}\)

So, x = 2 m

Maximum bending moment = Area of ILD below loading (rectangle + triangle) × w

= [(1.2 × 5) + 0.5 × (5 × 1.2)] × 10

= 90 KN-m

Concept:

When in addition to equilibrium equations, compatibility equations are used to evaluate the unknown reactions and internal forces in any structure, such analysis is called indeterminate analysis, and such structure is called indeterminate structure.

We have two distinct methods of analysis for such an indeterminate structure:

a. Force method of analysis

b. Displacement method of analysis

Difference between force method and Displacement method:

Concept:

Total Indeterminacy is given by,

Total Indeterminacy = External indeterminacy (D_se) + Internal indeterminacy (D_si)

External Indeterminacy (D_se) = R - 3

Where, R = Number of external Reaction

Internal indeterminacy (D_si) = 3C

Where, C = Number of closed loop

Calculation:

Given,

R = 6, C = 3

External Indeterminacy (D_se) = R - 3 = 6 - 3 = 3

Internal Indeterminacy (D_si) = 3C = 3 × 3 = 9

Total Indeterminacy = D_se + D_si = 3 + 9 = 12

Concept:

The force in the member AB will be determined by method of joints, i.e. by analyzing each joints.

For this we will be fixing some rule, they are

• for every joint we will consider the unknown forces are going away from the joint
• and the  forces are considered positive.
• So after determination, if one forces comes positive then, it will mean that it is tensile and if it comes negative then it would mean that it is compressive.

Calculations:

Taking moment about A, we get - 

R_C × 6 = 5 × 6 sin 60

⇒ R_C = 5 sin 60

⇒  RC = \(\frac{{5\sqrt 3 }}{2}\)

Analyzing joint C:

R_C + R_CB sin 60 = 0

⇒ RCB sin 60 = - RC 

⇒ RCB × \(\frac{{\sqrt 3 }}{2}\) = - \(\frac{{5\sqrt 3 }}{2}\)

⇒ RCB = - 5 kN [-ve denotes that the force will be toward the joint, hence it will be compressive in nature]

Concept:-

The given figure shows the fixed end moments under given loading:-

The fixed end moment at end A

\(M_A=-{wL^2\over 30}\)

\(M_B={wL^2\over 20}\)

\(\frac{{{M_A}}}{{{M_B}}}\; = {{wL^2\over 30}\over{wL^2\over 20}} \;\frac{2}{3}\)

Hence, the ratio of the moment will be 2/3.

Concept:

For two hinged semi-circular arches:

Horizontal thrust is given by:

\({{\rm{H}}_{\rm{A}}} = {{\rm{H}}_{\rm{B}}} = \frac{{\smallint \frac{{{{\rm{M}}_{\rm{x}}}{\rm{ydx}}}}{{{\rm{E}}{{\rm{I}}_{\rm{c}}}}}}}{{\smallint \frac{{{{\rm{y}}^2}{\rm{dx}}}}{{{\rm{E}}{{\rm{I}}_{\rm{c}}}}}}}\)

Where,

Mx = Moment at A – A’, y = vertical distance,

E = Modulus of elasticity, and

 I = Moment of inertia of c/s of the arch.

When  W load acts at the crown, then

Horizontal thrust (H) = w/π

∴ H is independent of the radius of the arch.

So, the ratio will be 1: 1: 1.

Note

Important Points:

 For a Two Hinged Semicircular arch subjected to concentrated load (W) at any other point which makes an angle α with the horizontal, then the horizontal thrust,

\(H = \left( {\frac{W}{\pi }} \right){\sin ^2}\left( \alpha \right)\;\)