Deflection of Beam MCQ Quiz in తెలుగు - Objective Question with Answer for Deflection of Beam - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Mar 12, 2025
Latest Deflection of Beam MCQ Objective Questions
Top Deflection of Beam MCQ Objective Questions
Deflection of Beam Question 1:
A cantilever beam of length, L, and flexural rigidity, EI, is subjected to an end moment, M, as shown in the figure. The deflection of the beam at x = L/2
Answer (Detailed Solution Below)
Deflection of Beam Question 1 Detailed Solution
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
\(y = \frac{{M{L^2}}}{{8EI}}\)
Deflection of Beam Question 2:
The ratio of the deflections of the free end of a cantilever due to an isolated load at 1/3rd and 2/3rd of the span is
Answer (Detailed Solution Below)
Deflection of Beam Question 2 Detailed Solution
Ist condition:
For a cantilever beam subjected to load W at distance of L/3 from free end, the deflection is given by:
\({\rm{y}}{{\rm{c}}_1} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{2{\rm{L}}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{2{\rm{L}}}}{3}} \right)^2} \times \frac{{\rm{L}}}{3} = {\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}^3}\)
IInd condition:
For a cantilever beam subjected to load W at distance of 2L/3 from free end:
\({\rm{y}}{{\rm{c}}_2} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{\rm{L}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{\rm{L}}}{3}} \right)^2} \times \frac{{2{\rm{L}}}}{3} = \frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}\)
\({\rm{Ratio\;}}\left( {\rm{r}} \right) = \frac{{{\rm{y}}{{\rm{c}}_1}}}{{{\rm{y}}{{\rm{c}}_2}}} = \frac{{{{\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}}^3}}}{{\frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}}} = \frac{{28}}{8} = \frac{7}{2}\)
\(\therefore \frac{{{\rm{y}}{{\rm{c}}_2}}}{{{\rm{y}}{{\rm{c}}_1}}} = \frac{2}{7}\)
Deflection of Beam Question 3:
A cantilever beam is shown in the figure. Find the magnitude and direction of a moment to be applied at free end for zero vertical deflection.
Answer (Detailed Solution Below)
Deflection of Beam Question 3 Detailed Solution
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:
\(\frac{{P{l^3}}}{{3EI}} = \frac{{M{l^2}}}{{2EI}} \Rightarrow M = \frac{2}{3}Pl = \frac{2}{3} \times 9 \times 2 = 12\;kN.m\;\left( {Clockwise} \right)\)
Deflection of Beam Question 4:
A simply supported laterally loaded beam was found to deflect more than a specified value. Which of the following measures will reduce deflection?
Answer (Detailed Solution Below)
Deflection of Beam Question 4 Detailed Solution
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:
|
\({y_B} = \frac{{P{L^3}}}{{3EI}}\) |
\({\theta _B} = \frac{{P{L^2}}}{{2EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{8EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{6EI}}\) |
|
\({y_B} = \frac{{M{L^2}}}{{2EI}}\) |
\({\theta _B} = \frac{{ML}}{{EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{30EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{48EI}}\) |
\({\theta _B} = \frac{{w{L^2}}}{{16EI\;}}\) |
|
\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = 0\) |
\({\theta _B} = \frac{{ML}}{{24EI}}\) |
|
\({y_c} = \frac{{M{L^2}}}{{8EI}}\) |
\({\theta _B} = \frac{{ML}}{{2EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{192EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
|
\({y_c} = \frac{{w{L^4}}}{{384EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
Where, y = Deflection of the beam, θ = Slope of beam
Deflection of Beam Question 5:
In moment area method, the deflection of a point A from a tangent at 'B' is equal to the
Answer (Detailed Solution Below)
Deflection of Beam Question 5 Detailed Solution
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'.
Deflection of Beam Question 6:
Which of the following methods is NOT used for finding deflection of beam?
Answer (Detailed Solution Below)
Deflection of Beam Question 6 Detailed Solution
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:
Method | Details |
Double-integration method | This method is best when there is continuity in the applied loading. |
Moment-Area Method | The method is especially suitable when the deflection or angle of rotation at only one point of the beam is desired. |
Strain-energy method (Castigliano's Theorem) |
Castigliano's Theorem lets us use strain energies at the locations of forces to determine the deflections. The Theorem also allows for the determining of deflections for objects with changing cross-sectional areas. |
Conjugate-beam method |
This method is based on the construction of a conjugate beam, defined as an imaginary beam of length equal to that of the original beam and loaded with an elastic weight M/EI, where M is the BM of the actual beam. This method is especially useful for simply supported and cantilever beams with varying flexural rigidities. |
Method of Superposition | The slope and deflection of the beam caused by several different loads acting simultaneously can be found by superimposing the slopes and deflections caused by the loads acting separately. |
Explanation:
The moment distribution method is a structural analysis method for statically indeterminate beams and frames developed by Hardy Cross.
Deflection of Beam Question 7:
A Simply supported beam of span 4 m and I = 12 × 106 mm4 is subjected to a central load of 9 kN. The deflection of the beam when E = 200 GPa is
Answer (Detailed Solution Below)
Deflection of Beam Question 7 Detailed Solution
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:
Beam - condition |
Slope |
Deflection |
|
\(\frac{{{\rm{P}}{{\rm{l}}^2}}}{{2{\rm{EI}}}}\) |
\(\frac{{{\rm{P}}{{\rm{l}}^3}}}{{3{\rm{EI}}}}\) |
|
\(\frac{{{\rm{P}}{{\rm{L}}^3}}}{{6{\rm{EI}}}}\) |
\(\frac{{{\rm{P}}{{\rm{l}}^4}}}{{8{\rm{EI}}}}\) |
|
\(\frac{{{\rm{Ml}}}}{{{\rm{EI}}}}\) |
\(\frac{{{\rm{M}}{{\rm{l}}^2}}}{{2{\rm{EI}}}}\) |
|
\(\frac{{{\rm{P}}{{\rm{L}}^2}}}{{16{\rm{EI}}}}\) |
\(\frac{{{\rm{P}}{{\rm{l}}^3}}}{{48{\rm{EI}}}}\) |
|
\(\frac{{{\rm{P}}{{\rm{L}}^3}}}{{24{\rm{EI}}}}\) |
\(\frac{{5{\rm{P}}{{\rm{L}}^4}}}{{384{\rm{EI\;}}}}\) |
|
0 |
\(\frac{{{\rm{P}}{{\rm{L}}^3}}}{{192{\rm{EI}}}}\) |
|
0 |
\(\frac{{{\rm{P}}{{\rm{L}}^4}}}{{384{\rm{EI\;}}}}\) |
Deflection of Beam Question 8:
A simply supported beam of span ‘l’ carries a point load ‘W’ at the centre, the slope at the support will be
Answer (Detailed Solution Below)
Deflection of Beam Question 8 Detailed Solution
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:
|
\({y_B} = \frac{{P{L^3}}}{{3EI}}\) |
\({\theta _B} = \frac{{P{L^2}}}{{2EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{8EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{6EI}}\) |
|
\({y_B} = \frac{{M{L^2}}}{{2EI}}\) |
\({\theta _B} = \frac{{ML}}{{EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{30EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{48EI}}\) |
\({\theta _B} = \frac{{P{L^2}}}{{16EI\;}}\) |
|
\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = 0\) |
\({\theta _B} = \frac{{ML}}{{24EI}}\) |
Deflection of Beam Question 9:
A cantilever of length l, and flexural rigidity El, stiffened by a spring of stiffness k, is loaded by a transverse force P, as shown.
Answer (Detailed Solution Below)
Deflection of Beam Question 9 Detailed Solution
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:
keq = 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
\(\frac{{\left( {P\;-\;R} \right){L^3}}}{{3EI}} = \frac{R}{K}\)
\( \Rightarrow \left( {\frac{1}{K} + \frac{{{L^3}}}{{3EI}}} \right)R = \frac{{P{L^3}}}{{3EI}}\)
\(R = \frac{{P{L^3}}}{{3EI}} \times \frac{{3EIK}}{{3EI\;+\;{L^3}}}\)
\(\Delta_{net} = \frac{R}{K} = \frac{{P{L^3}}}{{3EI}}\;\left( {\frac{{3EI}}{{3EI\;+\;K{L^3}}}} \right)\)
Deflection of Beam Question 10:
For uniformly distributed load (w) acting on the beam with one end fixed and one end free if length of beam is L, distance from left hand is x, young’s modulus and moment of inertia are E and I then find the maximum deflection (δmax) of beam:
Answer (Detailed Solution Below)
Deflection of Beam Question 10 Detailed Solution
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)
\({y_B} = \frac{{W{L^4}}}{{8EI}}\) = δmax
Additional Information
SFD and BMD of various type of load in Cantilever Beam: