Methods of Structural Analysis MCQ Quiz - Objective Question with Answer for Methods of Structural Analysis - Download Free PDF

Explanation:

• In the conjugate beam method, the bending moment diagram of the real beam is treated as the load on the conjugate beam.

• The support conditions in the conjugate beam change according to the boundary conditions of the real beam:

  • A fixed support in the real beam becomes a hinged support in the conjugate beam.

  • This is because a fixed end in the real beam (zero displacement and rotation) corresponds to a point in the conjugate beam where the slope is zero but rotation is allowed, like a hinge.

• This transformation helps analyze beam deflections and rotations using simpler static analysis of the conjugate beam.

 Additional InformationTable depicting real beam and its corresponding conjugate beam: 

Explanation:

K_CB = 4EI/L (Far end is fixed)

KCA = 3EI/L (Far end is pinned)

K₁₁ = KCB + KCA 

      = 4EI/L + 3EI/L  = 7EI/L

Explanation:

This is a case of pure transverse displacement of a fixed-fixed prismatic beam with no rotation — typically referred to as "sway" in structural analysis.

• Beam AB is fixed at both ends.

• One end is displaced transversely by Δ \Delta" id="MathJax-Element-27-Frame" role="presentation" style="position: relative;" tabindex="0">Δ$\Delta$ $\Delta$ (without rotation).

• This introduces equal and opposite end moments to maintain equilibrium.

For such a case, the induced moment at each end (A or B) is:\(6EI \Delta / L^2\)

Additional Information

• Transverse displacement refers to the movement of a structural member perpendicular to its longitudinal axis.

• For a horizontal beam, transverse displacement means vertical movement.

• For a vertical column, transverse displacement means horizontal movement (also called sway).

• A beam fixed at both ends is displaced sideways (laterally) at one end without rotation.

• This displacement is parallel to the direction of the applied shear or lateral force, not along the length of the beam.

• To maintain fixity (zero rotation), internal moments develop at both supports.

Explanation:

• Stiffness (K) measures a structure’s resistance to deformation under load.

• Flexibility (F) measures how much a structure deforms under a given load.

• For a linear elastic element, these two are reciprocals of each other.

• Therefore, if stiffness increases, flexibility decreases, and vice versa.

Additional Information

• This inverse relationship is fundamental in structural analysis and design.

• Stiffness depends on material properties (like Young’s modulus) and geometry.

• Flexibility matrices and stiffness matrices are key concepts in methods like the Flexibility Method and Stiffness Method of structural analysis.

Explanation:

Stiffness (S): It is defined as the force per unit deflection or moment required per unit rotation.

Case 1:

For the far end being Hinged Supported or freely supported:

If a unit rotation is to be caused at an end A for the far end being hinged support. Moment of 3EI/L is to be applied at end and hence stiffness for the member is said to be 3EI/L

Case 2:

For the far end being Fixed Supported:

If a unit rotation is to be caused at an end A for the far end being fixed supported. Moment of 4EI/L is to be applied at end and hence stiffness for the member is said to be 4EI/L

Case 3: when the far end is free

Beam offers no resistance to the moment.

S = 0

Mohr’s Theorem I:

The angle between the two tangents drawn on the elastic line is equal to the area of the Bending Moment Diagram between those two points divided by flexural rigidity.

\(\theta = \frac{{\left[ {Area\;of\;bending\;moment\;diagram} \right]}}{{EI}}\)

Mohr’s Theorem II:

The deviation of a point away from the tangent drawn from the other point is given by the moment of area of bending moment diagram about the first point divided by flexural rigidity.

\(\delta = BB' = \frac{{\left[ {Area\;of\;bending\;moment\;diagram} \right] \times \bar x}}{{EI}}\)

I^st condition:

For a cantilever beam subjected to load W at distance of L/3 from free end, the deflection is given by:

\({\rm{y}}{{\rm{c}}_1} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{2{\rm{L}}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{2{\rm{L}}}}{3}} \right)^2} \times \frac{{\rm{L}}}{3} = {\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}^3}\)

II^nd condition:

For a cantilever beam subjected to load W at distance of 2L/3 from free end:

\({\rm{y}}{{\rm{c}}_2} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{\rm{L}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{\rm{L}}}{3}} \right)^2} \times \frac{{2{\rm{L}}}}{3} = \frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}\)

\({\rm{Ratio\;}}\left( {\rm{r}} \right) = \frac{{{\rm{y}}{{\rm{c}}_1}}}{{{\rm{y}}{{\rm{c}}_2}}} = \frac{{{{\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}}^3}}}{{\frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}}} = \frac{{28}}{8} = \frac{7}{2}\)

\(\therefore \frac{{{\rm{y}}{{\rm{c}}_2}}}{{{\rm{y}}{{\rm{c}}_1}}} = \frac{2}{7}\)

Explanation:

• The column analogous method is the force method used to analyze the indeterminate structure.
• In the method of column analogous, the actual structure is considered under the action of applied loads and the redundant acting simultaneously.The load on the top of the analogous column is usually the B.M.D. due to applied loads on simple spans and therefore the reaction to this applied load is the B.M.D. due to redundant on simple spans.
• It is based on the analogy between moments at ends and pressure at the edges of the short column. If both ends of the beam are fixed then the end moments are called fixed end moments.

• An analogous column will be a short column with cross-section dimensions L and 1/EI.​

so, area of analogous column = L/EI

Concept:

• The moment distribution method is best suited for the rigid 2D frame as for rigid 2D frame because only one kind of moment (Mz) is acting at every joint of the structure the moment developed at the joint is distributed to all connected members at that particular joint.
• The moment distribution method only takes into account the moment effect not the axial force effect.
• As trusses are designed to carry axial force, so it is not desired to do moment distribution for pin-jointed truss or trussed beam.
• For space frame, there are three kinds of the moment (Mx, My & Mz) acting at every joint. So, it is very difficult and inconvenient to do moment distribution for spaced frame.

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. 

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:-

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.

\({M_\theta }\; = \frac{W}{{16}} \times \left( {{R^2}\left( {3 + \mu } \right) - {r^2}\left( {1 + 3\mu } \right)} \right)\)

Where M_θ = Circumferential moment

R = radius of slab

μ = Poisson’s ratio

r = any section at a distance r from the centre of the slab

W = load on circular slab

For the maximum circumferential moment at centre

r = 0 and μ = 0

Explanation:

fixed end moments

Fixed end

\(M_{FAB}= \frac{{Pab(L + b)}}{{2{L^2}}}\)

So \(M_{FAB} = \frac{-{Pab(L + b)}}{{2{L^2}}}\)

Negative sign as anticlockwise rotation is taken

Concept:

The FEM’s when a beam is subjected to point load and clear loading such as uDL is given below:

Calculation:

Given:

a = 2m

b = 3m

L = a + b = 5m

P = 50 kN

\(M_{AB}^F = - \frac{{P{b^2}a}}{{{L^2}}} = - \frac{{50\; \times\; {{\left( 3 \right)}^2}\; \times \;2}}{{{{\left( 5 \right)}^2}}} = - 36\;kNm\;\left( \uparrow \right)\)

(-ve sign means that \(M_{AB}^F\) is in Anticlockwises

\(M_{AB}^F = + \frac{{P{a^2}b}}{{{L^2}}} = \frac{{50\; \times \;{{\left( 2 \right)}^2}\;\times\; 3}}{{{{\left( 5 \right)}^2}}} = + 24\;kN\;M\left( \downarrow \right)\)