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

Concept:

The conditions to be satisfied for the plastic methods of analysis are as follows:

1. Equilibrium Condition: 

• All the equilibrium conditions, i.e. summation of all the forces and moments should be equal to zero.

2. Mechanism Condition: 

• The structure at collapse must be capable of deforming as a mechanism due to the formation of plastic hinges, i.e., the ultimate load is reached when a mechanism forms. this is also called the continuity condition.

3. Yield Condition: 

• This is also called the plastic moment condition. The bending moment at any section must not exceed the full plastic moment of the section (M ≤ Mp).

Concept:

Shape Factor: It is defined as the ratio of moment carrying capacity of the plastic section over that of the elastic section when the yielding just starts.

\(Shape\;Factor = \frac{{{Z_P}{\sigma _y}}}{{{Z_e}{\sigma _y}}} = \frac{{{M_P}}}{{{M_y}}} = \frac{{{Z_P}}}{{{Z_e}}}\)

Shape factor values for different sections:

Calculation:

Plastic modulus of section ZP = 4.8 × 10-4 m3

Plastic moment Capacity MP = 120 kNm

Plastic moment capacity is given by,

\({M_P} = {Z_P}{\sigma _y}\)

Yield Stress is given by,

\({\sigma _y} = \frac{{{M_P}}}{{{Z_P}}} = \frac{{120 \times 1000}}{{4.8 \times {{10}^{ - 4}}}} = 250 \times {10^6}\;N/{m^2} = 250\;MPa\)

Concept:

Plastic section modulus is given by,

\(Z_p=\frac{A}{2}(\overline {{y_1}} + \overline {{y_2}} )\)

Elastic section modulus is given by, Z_e = I/y

Concept:

Plastic section modulus, \(Z_p=\frac{A}{2}(\overline {{y_1}} + \overline {{y_2}} )\)

⇒ \({Z_p} = \frac{{a \times 2a}}{2}\left( {\frac{a}{2} + \frac{a}{2}} \right)\)

Z_p = a³

Elastic section modulus, \({Z_e} = \frac{{\frac{{a \times {{\left( {2a} \right)}^3}}}{{12}}}}{{\frac{{2a}}{2}}}\)

Z_e = (2/3) a³

Ratio of Plastic section modulus to elastic section modulus,

\(\frac{{{Z_p}}}{{{Z_e}}} = \frac{{{a^3}}}{{\frac{2}{3} \times {a^3}}} = 1.5\)

Explanation:

Plastic design:

• Plastic design is a method by which elements are selected considering the system's overall ultimate capacity.
• It utilizes the full strength of steel and hence also called the ultimate load method
   

Advantages of plastic design:

•  Realization of uniform and realistic F.O.S for all parts of the structures
•  Simplified analytical procedure and readily of obtaining design moments since there is no need to satisfy elastic strain compatibility conditions.
• Saving of material over elastic methods resulting in lighter structures.
• No effect due to temperature changes, settlement of supports, imperfection, erection method, because their only effect is to change the amount of rotation required.
   

Disadvantages of plastic design:

• Obtaining collapse load is difficult if the structure is reasonably complicated.
• There is little saving in column design.
• Difficult to design for fatigue.
• Lateral bracing requirements are more than stringent than elastic design because of its lighter structure which may lead to dimensional instability of the stucture.

Explanation:

Shape factor

• It is defined as the ratio of the plastic moment to yield moment.
• It is a function of cross-section or shape and is represented by ' f '.
• \(f = \frac{{{M_P}}}{{{M_y}}} = \frac{{{f_y}{Z_{Pz}}}}{{{f_y}{Z_{ez}}}} = \frac{{{Z_{Pz}}}}{{{Z_{ez}}}}\)

Shape factor for different cross-sectional shapes:

Explanation:

Plastic hinge

• The deformation of a part of a beam wherever plastic bending happens or the greatest deformation (i.e. curvature).
• If an indeterminate structure has redundancy r, then insertion of r plastic hinges makes it statically determinate and any further insertion of hinges converts this statically determinate structure into a mechanism.
• Hence for collapse the number of plastic hinges required = r + 1

Calculation:

Given,

Degree of redundancy, r = 2

The minimum number of plastic hinges required for the complete collapse of the structure, n = 2 +1 = 3.​

For the given beam:

Number of possible plastic hinges: 2 (B, A)

Case 1: Plastic hinge at point B, where cross-section changes

External work done (W_e­) = Work Done × Displacement = 2W × Δ

Internal work done (W_i) = Moment × Rotation

\(\therefore {{\text{W}}_{\text{i}}}={{\text{M}}_{\text{p}}}\times \text{ }\!\!\theta\!\!\text{ }={{\text{M}}_{\text{p}}}\times \frac{\text{ }\!\!\Delta\!\!\text{ }}{\text{L}/2}=\frac{2{{\text{M}}_{\text{p}}}\text{ }\!\!\Delta\!\!\text{ }}{\text{L}}\)

Equating external work done to external work done ⇒ W_e = W_i

\(\therefore {{\text{W}}_{\text{c}}}=\frac{{{\text{M}}_{\text{p}}}}{\text{L}}\)

Case 2: Plastic hinge at the support A

External work done (W_e) = 2W × Δ

Internal work done (W_i) = 1.5 M_p × θ

∴ W_e = W_i

\(\Rightarrow {{\text{W}}_{\text{c}}}=\frac{0.75}{\text{L}}{{\text{M}}_{\text{P}}}\)

Collapse load is the lower value of the above two cases:

∴ Collapse load (W_c) \(=\frac{0.75\text{ }\!\!~\!\!\text{ }{{\text{M}}_{\text{p}}}}{\text{L}}\)

Explanation:

Slender section:

The cross-section in which the elements buckle locally even before the attainment of yield stress are called slender sections.

Plastic section:

Cross-sections, which can develop a plastic hinge and have the rotation capacity required for the failure of the structure by the formation of a plastic mechanism are called plastic sections.

Compact section:

Cross-section which can develop the plastic moment of resistance but have inadequate plastic hinge rotation capacity for formation of a plastic mechanism before bucking are called compact section.

Semi-compact section:

The cross-section in which the extreme fibre in compression can reach yield stress but cannot develop the plastic moment of resistance due to local buckling are called semi-compact or a non-compact section.

Calculation: 

Step 1: Degree of static indeterminacy, D_s = 2

∴ Number of plastic hinges required for mechanism or collapse = Ds + 1 = 3

Step 2: Determine the location of plastic hinges

Plastic hinge will for, at the fixed end and under concentrated load

Step 3: Sketch possible mechanism

Determine the ultimate load by solving the equilibrium equations using principle of virtual work 

L/3 × θ = 2L/3 × α 

θ = 2α 

Internal work dne = M_Pθ + M_P(θ + α) + M_Pα 

External work done = W × L/3 × θ 

∴ From principal of virtual work

Internal work done = External work done 

MPθ + MP(θ + α) + MPα = W × L/3 × θ 

M_P 2α + M_P (2α + α) + MPα = W × L/3 × θ 

W = 9M_P/L

Number of possible plastic hinges (N) = 2

Degree of indeterminacy = Ds = 3 - 2 = 1

Plastic hinge required to form mechanism (n) = Ds + 1 = 1 + 1 = 2

By the principle of virtual work done,

External work done = Intenal workdone

\({\rm{W}} \times \left( {\frac{{\rm{L}}}{2} \times {\rm{\theta }}} \right) = {{\rm{M}}_{\rm{p}}}\left( {\rm{\theta }} \right) + {{\rm{M}}_{\rm{p}}}\left( {{\rm{\theta }} + {\rm{\theta }}} \right)\)

\(\therefore \;{{\rm{W}}_{\rm{u}}} = \frac{{6{{\rm{M}}_{\rm{u}}}}}{{\rm{L}}}\)