Deflection of Beam MCQ Quiz in తెలుగు - Objective Question with Answer for Deflection of Beam - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

From the Double integration method, we know,

\(EI \times \frac{{{d^2}y}}{{d{x^2}}} = {M_x}\)    -- (i)

On integrating (i) we get slope (dy/dx) at a given point

\(EI \times \frac{{dy}}{{dx}} = {M_x} \times x + {C_1}\) -- (ii)

On integrating (ii), we get deflection (y) at a given point

\(EI \times \left( y \right) = {M_x} \times \frac{{{x^2}}}{2} + {C_1}x + {C_2}\)

Where EI is flexural rigidity

Mx is the moment at section x – x

Calculation

Given,

For cantilever beam at \(x = 0,\frac{{dy}}{{dx}} = 0\;\& \;x = 0,~y = 0\) (x taken from fixed-end).

From this we get C1 = C2 = 0,  Hence

\(y = \frac{{M{x^2}}}{{2EI}}\)  

Now at x = L/2, the deflection will be 

\(y = \frac{{M{L^2}}}{{8EI}}\)

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:

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

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

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

Here load 9 kN is causing anticlockwise moment. Assume a Moment ‘M’ acting at the free end in the clockwise direction.

Deflection at the free end due to load P (downward):

\({\delta _P} = \frac{{P{l^3}}}{{3EI}}\)

Deflection at the free end due to Moment M (upward):

\({\delta _M} = \frac{{M{l^2}}}{{2EI}}\)

For zero deflection at the free end:

\(\frac{{P{l^3}}}{{3EI}} = \frac{{M{l^2}}}{{2EI}} \Rightarrow M = \frac{2}{3}Pl = \frac{2}{3} \times 9 \times 2 = 12\;kN.m\;\left( {Clockwise} \right)\)

Explanation:

• The maximum deflection of a simply supported beam of span ‘L’ carrying a lateral or transverse load at the centre of the span, flexural rigidity being EI, is given by:

​\(δ = \frac{{W{L^3}}}{{48EI}}\).

• From the above formula, it can be seen that, deflection (δ) is inversely proportional to the area moment of inertia (I), modulus of elasticity (E) and directly proportional to the transverse load and span of the beam.
• So, to reduce the deflection in the given case, the area moment of inertia should be increased.​

Additional InformationDeflection and slopes of various beams are given by:

Where, y = Deflection of the beam, θ = Slope of beam

Concept:

Area-Moment Method:

• It is also known as Mohr's method.
• This method establishes a procedure that utilizes the area of the moment diagrams [actually, M/EI diagram] to evaluate the slope or deflection at selected points along the axis of a beam.
• This method is applicable to both prismatic and non-prismatic beams. In this method continuity of slope is assumed. Hence in this method is not applicable where there is a sudden break in the continuity of slope such as an internal hinge or internal links are present.

Theorem-1:

• The change in slope from any point 'A' to 'B' is equal to the area of the M/EI diagram between 'A' and B.
• Here the area is taken positive if the M/EI diagram is positive i.e. sagging and vice-versa.

Theorem-2:

Deflection at any point 'A' with respect to the tangent at any point 'B' is equal to the moment of area of M/EI diagram between 'A' and 'B" about the point 'A'.

Concept:

Deflection of beam:

• The deformation of a beam is usually expressed in terms of its deflection from its original unloaded position.
• The deflection is measured from the original neutral surface of the beam to the neutral surface of the deformed beam.
• The configuration assumed by the deformed neutral surface is known as the elastic curve of the beam.

Methods of Determining Beam Deflections:

Explanation:

The moment distribution method is a structural analysis method for statically indeterminate beams and frames developed by Hardy Cross.

Concept:

The maximum deflection of the simply supported beam is given by (span l, central load W)

\(\Delta =\frac{WL^3}{48EI}\)

where, W = Central load, L = Length of the beam, E = Modulus of elasticity, I = Moment of Inertia

Calculation:

Given: 

W = 9 kN = 9000 N, and L = 4 m = 4000 mm, E = 200 GPa = 200 × 10³ MPa, I = 12 × 106 mm4

\(\Delta =\frac{WL^3}{48EI}=\frac{9000\times(4\times10^3)^3}{48\times200\times 10^3\times12\times10^6}=5~mm\)

Additional Information 

Slope and deflection of some important beam-load conditions are given below:

Concept

Slope at ends when a simply supported beam of the span (L) is  point load ‘W’ at the centre  is given by 

\({\theta _B} = \frac{{W{l^2}}}{{16EI\;}}\)

\({y_c} = \frac{{W{l^3}}}{{48EI}}\)

Important Points

Deflection and slope of various beams are given by:

Concept:

The deflection of the beam at the free end due to load P is \(\Delta = \frac{{P{l^3}}}{{3EI}}\)

Since the deflection in beam and spring are equal, therefore stiffness of both i.e. beam and spring are in parallel connection.

For parallel combination:

k_eq = k₁ + k₂

where k₁ = stiffness of spring and k₂ = stiffness of beam.

Calculation:

Given:

Deflection of the beam due to point load P if spring is not there \(\Delta = \frac{{P{l^3}}}{{3EI}}\)

Stiffness \({k_2} = \frac{P}{\Delta }\)

\({k_2} = \frac{P}{{\frac{{P{l^3}}}{{3EI}}}} = \frac{{3EI}}{{{l^3}}}\)

\(\therefore \;{k_{eq}} = k + \frac{{3EI}}{{{L^3}}} = \frac{{k{L^3}\;+\;3EI}}{{{L^3}}}\)

∴ Transverse deflection under the load

\(\Rightarrow \Delta_{net}= \frac{P}{{{K_{eq}}}} = \frac{{P{L^3}}}{{K{L^3} + 3EI}}\)

\( \Rightarrow \Delta_{net} = \frac{{P{L^3}}}{{3EI}}\;\left( {\frac{1}{{\frac{{K{L^3}}}{{3EI}}\;+\;1}}} \right)\)

\(\Rightarrow \Delta = \frac{{P{L^3}}}{{3EI}}\;\left( {\frac{{3EI}}{{K{L^3}\;+\;3EI}}} \right)\)

Shortcut Method:

Deflection of beam = Deflection of spring

\(\frac{{\left( {P\;-\;R} \right){L^3}}}{{3EI}} = \frac{R}{K}\)

\( \Rightarrow \left( {\frac{1}{K} + \frac{{{L^3}}}{{3EI}}} \right)R = \frac{{P{L^3}}}{{3EI}}\)

\(R = \frac{{P{L^3}}}{{3EI}} \times \frac{{3EIK}}{{3EI\;+\;{L^3}}}\)

\(\Delta_{net} = \frac{R}{K} = \frac{{P{L^3}}}{{3EI}}\;\left( {\frac{{3EI}}{{3EI\;+\;K{L^3}}}} \right)\)

Explanation:

\({\left( {B.M} \right)_x} = \frac{{W{x^2}}}{2}\)

At \(x = L\;;\;B.M = \frac{{W{L^2}}}{2}\)

[M/EI diagram]

As per the moment area theorem,

\({y_B} - {y_A} = \frac{{A\; \times \;\bar x}}{2}\)

\({y_B} - 0 = \frac{1}{3} \times \frac{{W{L^2}}}{{2EI}} \times L \times \frac{{3L}}{4}\)  (∵ yA = 0)

\({y_B} = \frac{{W{L^4}}}{{8EI}}\) = δ_max

Additional Information

SFD and BMD of various type of load in Cantilever Beam: