Deflection of Beam MCQ Quiz - Objective Question with Answer for Deflection of Beam - Download Free PDF
Last updated on May 12, 2025
Latest Deflection of Beam MCQ Objective Questions
Deflection of Beam Question 1:
Which of the following statements is true regarding a fixed beam?
Answer (Detailed Solution Below)
Deflection of Beam Question 1 Detailed Solution
Explanation:
- A fixed beam has its ends fixed, meaning it cannot rotate or displace at those ends. Due to the constraints:
-
The deflection at the fixed ends is zero because the beam is not allowed to move vertically at the supports.
-
The slope at the fixed ends is also zero because there is no rotation allowed at the fixed supports.
Additional Information
Simply Supported Beam
Definition: A simply supported beam is one that is supported at both ends, with one end typically resting on a roller and the other on a hinge or pin.
Key Characteristics:
-
The beam is free to rotate at the supports, meaning there is no moment resistance at the supports.
-
It experiences vertical reactions at the supports but no bending moments at the supports.
- The deflection is present along the length of the beam, and it is typically greatest at the midpoint.
- Typical Applications: Bridges, building floors, and roof structures.
Deflection of Beam Question 2:
The maximum deflection of a cantilever beam of length (L) with a point load (W) at the free end is
Answer (Detailed Solution Below)
Deflection of Beam Question 2 Detailed Solution
Explanation:
Deflection and slope of various beams are given by:
|
\({y_B} = \frac{{P{L^3}}}{{3EI}}\) |
\({\theta _B} = \frac{{P{L^2}}}{{2EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{8EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{6EI}}\) |
|
\({y_B} = \frac{{M{L^2}}}{{2EI}}\) |
\({\theta _B} = \frac{{ML}}{{EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{30EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{48EI}}\) |
\({\theta _B} = \frac{{w{L^2}}}{{16EI\;}}\) |
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\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = 0\) |
\({\theta _B} = \frac{{ML}}{{24EI}}\) |
|
\({y_c} = \frac{{M{L^2}}}{{8EI}}\) |
\({\theta _B} = \frac{{ML}}{{2EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{192EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
|
\({y_c} = \frac{{w{L^4}}}{{384EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
Where, y = Deflection of beam, θ = Slope of beam
Deflection of Beam Question 3:
The value of maximum deflection i.e., \({{W{L^4}} \over {8EI}}\) is true for which loading condition? (Where L is the length of span, El is the flexural rigidity)
Answer (Detailed Solution Below)
Deflection of Beam Question 3 Detailed Solution
Explanation:
The deflection value for different conditions are as tabulated below:
Deflection of Beam Question 4:
For a cantilever beam of length L carrying a point load P at the free end, what is the deflection at the free end?
Answer (Detailed Solution Below)
Deflection of Beam Question 4 Detailed Solution
Explanation:
Deflection and slope of various beams are given by:
|
\({y_B} = \frac{{P{L^3}}}{{3EI}}\) |
\({\theta _B} = \frac{{P{L^2}}}{{2EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{8EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{6EI}}\) |
|
\({y_B} = \frac{{M{L^2}}}{{2EI}}\) |
\({\theta _B} = \frac{{ML}}{{EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{30EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{48EI}}\) |
\({\theta _B} = \frac{{w{L^2}}}{{16EI\;}}\) |
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\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = 0\) |
\({\theta _B} = \frac{{ML}}{{24EI}}\) |
|
\({y_c} = \frac{{M{L^2}}}{{8EI}}\) |
\({\theta _B} = \frac{{ML}}{{2EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{192EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
|
\({y_c} = \frac{{w{L^4}}}{{384EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
Where, y = Deflection of beam, θ = Slope of beam
Deflection of Beam Question 5:
A cantilever beam having square cross section of side 'a' is subjected to an end load. If 'a' is increased by 19%, then the tip deflection decreases approximately -
Answer (Detailed Solution Below)
Deflection of Beam Question 5 Detailed Solution
Concept:
\(I = \frac{{{a^4}}}{{12}}\)
\({\delta _{free\;end}} = \frac{{P{L^3}}}{{3EI}},\;\delta \propto \frac{1}{I} \Rightarrow \delta \propto \frac{1}{{{a^4}}}\)
∵ Initial side = a
Final side = 1.19 a
\(\therefore {\delta _i} = \frac{k}{{{a^4}}},{\delta _f} = \frac{k}{{{{\left( {1.19a} \right)}^4}}} = \frac{k}{{2.005\;{a^4}}} = 0.498\frac{k}{{{a^4}}}\)
Where k → constant ⇒ δf = 0.498 δi
\(\frac{{{\delta _f} - {\delta _i}}}{{{\delta _i}}} = \left( {\frac{{{\delta _f}}}{{{\delta _i}}} - 1} \right) \times 100\% = \left( {0.498 - 1} \right) \times 100\% = \left( { - 0.501} \right) \times 100\)
= 50% decreases.
Top Deflection of Beam MCQ Objective Questions
In the case of a beam simply supported at both ends, if the same load instead of being concentrated at centre is distributed uniformly throughout the length, then deflection at centre will get reduced by
Answer (Detailed Solution Below)
Deflection of Beam Question 6 Detailed Solution
Download Solution PDFConcept:
Deflection at centre of a simply supported beam due to point load (W):
\(δ_1=\frac{WL^3}{48EI}\)
Deflection at centre of a simply supported beam due to uniformly distribute load (W = wL):
\(δ_2=\frac{5wL^4}{384EI}=\frac{5WL^3}{384EI}\;\;\;(\because w=\frac{W}{L})\)
Calculation:
Given:
\(\frac{\delta_2}{\delta_1}=\frac{5WL^3}{384EI}\times \frac{48EI}{WL^3}=\frac{5}{8}\)
Hence,
The deflection will be reduced by
\(={\delta _1} - {\delta _2} = \left( {1 - \frac{5}{8}} \right){\delta _1} = \frac{3}{8}{\delta _1}\)
A free end of a cantilever beam rotates by 0.001 radians under a point load 10 kN. Then deflection at the free end due to a moment of 100 KN - m is:
Answer (Detailed Solution Below)
Deflection of Beam Question 7 Detailed Solution
Download Solution PDFConcept:
Slope for cantilever beam with point load,
\(\theta = \frac{{P{L^2}}}{{2EI}} = 0.001\)
∴ \(\frac{{{L^2}}}{{2EI}} = \frac{{0.001}}{P}\)
Now, deflection for cantilever beam with moment,
\({\rm{\Delta }} = \frac{{M{L^2}}}{{2EI}} = M \times \frac{{0.001}}{P} = 100 \times \frac{{0.001}}{{10}} = 0.01\;m = 10\;mm\)
A cantilever beam of length, L, with uniform cross-section and flexural rigidity, EI, is loaded uniformly by a vertical load, w per unit length. The maximum vertical deflection of the beam is given by
Answer (Detailed Solution Below)
Deflection of Beam Question 8 Detailed Solution
Download Solution PDFConcept:
Area moment method
Theorem 1: The difference of slope of any two points of a beam is equal to the area of\(\frac{M}{{EI}}\) diagram between those points.
θB – θA = Area of \(\frac{M}{{EI}}\) diagram between B and A.
Theorem 2: The difference of deflection of two points of a beam is equal to the moment of area of\(\frac{M}{{EI}}\)diagram between those points.
YB – YA = Moment of area of\(\frac{M}{{EI}}\)diagram between B and A.
YB – YA = (Ax̅)
Calculation:
δmax = Ax̅
\({δ _{max}} = \frac{1}{3} \times L \times \frac{{W{L^2}}}{{2EI}} \times \frac{3}{4}L\)
\(\Rightarrow {δ _{max}} = \frac{{W{L^4}}}{{8EI}}\)
Which of the following represents the bending at a section of the beam?
Answer (Detailed Solution Below)
Deflection of Beam Question 9 Detailed Solution
Download Solution PDFConcept: -
We know,
The radius of curvature is given by –
\(R = \frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}\)
Also, from bending equation,
\(R = \frac{{EI}}{M}\)
Here, EI – flexural rigidity and M – moment
Assuming small deflection of beam, such that the slope of the elastic curve dy / dx is very small. Then the term (dy / dx)2 can be neglected.
Hence,
\(R = \frac{1}{{\frac{{{d^2}y}}{{d{x^2}}}}} = \frac{{EI}}{M}\)
\(\Rightarrow M = EI\frac{{{d^2}y}}{{d{x^2}}}\)
A cantilever of beam of span l is fixed at one end, the other end resting freely on the middle of a simply supported cross-beam of the same span and section. If the cantilever beam is now loaded with a uniform load of w per unit length, find the reaction at the free end offered by the cross beam.
Answer (Detailed Solution Below)
Deflection of Beam Question 10 Detailed Solution
Download Solution PDFConcept:
Deflection of cantilever beam due to point load = \(\frac{Pl^3}{3EI}\)
Deflection of cantilever beam due to udl = \(\frac{wl^4}{8EI}\)
Deflection of SSB due to point load at mid point = \(\frac{Pl^3}{48EI}\)
Solution:
Deflection at the end of cantilever beam is given by,
\(=\frac{wl^4}{8EI}-\frac{Rl^3}{3EI}\)
Deflection at midpoint of SSB.
=\(\frac{Rl^3}{48EI}\)
Equating both the cases,
\(\frac{wl^4}{8EI} -\frac{Rl^3}{3EI}= \frac{Rl^3}{48EI}\)
\(\frac{wl^4}{8EI} = \frac{Rl^3}{48EI} + \frac{Rl^3}{3EI}\)
\(\frac{wl^4}{8EI}=(\frac{1+16}{48})\frac{Rl^3}{EI}\)
\(R=\frac{6wl}{17}\)
A cantilever beam of length L has flexural rigidity EI up to length L/2 from the fixed end and EI/2 for the rest. It carries a moment M at the free end. The slope at the free end is given by-
Answer (Detailed Solution Below)
Deflection of Beam Question 11 Detailed Solution
Download Solution PDFConcept:
The slope at the free end of a cantilever beam with varying flexural rigidity is determined using the moment-area theorem.
Given:
- Flexural rigidity up to length L/2 from the fixed end: \( EI \)
- Flexural rigidity for the remaining length: \( EI/2 \)
- Moment applied at the free end: \( M \)
Calculation:
Using the moment-area theorem, the total slope at the free end is given by the sum of contributions from both segments.
For the first segment (0 to L/2) with flexural rigidity \( EI \):
\( \theta_1 = \frac{M (L/2)}{EI} \)
For the second segment (L/2 to L) with flexural rigidity \( EI/2 \):
\( \theta_2 = \frac{M (L/2)}{(EI/2)} = \frac{2M (L/2)}{EI} = \frac{ML}{EI} \)
Total slope at the free end:
\( \theta = \theta_1 + \theta_2 = \frac{ML}{2EI} + \frac{ML}{EI} \)
\( \theta = \frac{3ML}{2EI} \)
Final Answer: \( \frac{3ML}{2EI} \)
The reaction of the prop of a propped cantilever beam of span I with UDL W kN/m is
Answer (Detailed Solution Below)
Deflection of Beam Question 12 Detailed Solution
Download Solution PDFCalculation:
Deflection of end B = 0
(Downward deflection due to uniformly distributed load) – (upward deflection due to RB) = 0
\(\frac{{w{l^4}}}{{8EI}} - \frac{{{R_B}{l^3}}}{{3EI}} = 0\)
\(\begin{array}{l} {R_B} = \frac{3}{8}wl\\ {R_A} = wl - \frac{3}{8}wl = \frac{5}{8}wl \end{array}\)
Here RB is the reaction of the prop.
\(R_B=\frac{3}{8}{\rm{Wl\;}}\left( {{\rm{kN}}} \right)\)
For a simply supported subjected to uniformly distributed load, if the length of the beam is doubled, deflection becomes ______ times.
Answer (Detailed Solution Below)
Deflection of Beam Question 13 Detailed Solution
Download Solution PDFConcept:
The standard deflection and slope formulas of simply supported beam under uniformly distributed load is given below:
Calculation:
Given:
L1 = L, L2 = 2L
Deflection under UDL at centre,
\(y_c=\frac{5}{{384}}\frac{{w{L^4}}}{{EI}} \)
⇒ \(y_c \propto L^4\)
\(\frac{y_{c1}}{y_{c2}}=(\frac{L_1}{L_2})^4\)
\(\frac{y_{c1}}{y_{c2}}=(\frac{L}{2L})^4\)
\(\frac{y_{c1}}{y_{c2}}=\frac{1}{16}\)
yc2 = 16 yc1
Important Points
Deflection and slope of various beams are given by:
|
\({y_B} = \frac{{P{L^3}}}{{3EI}}\) |
\({\theta _B} = \frac{{P{L^2}}}{{2EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{8EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{6EI}}\) |
|
\({y_B} = \frac{{M{L^2}}}{{2EI}}\) |
\({\theta _B} = \frac{{ML}}{{EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{30EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{48EI}}\) |
\({\theta _B} = \frac{{w{L^2}}}{{16EI\;}}\) |
|
\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = 0\) |
\({\theta _B} = \frac{{ML}}{{24EI}}\) |
|
\({y_c} = \frac{{M{L^2}}}{{8EI}}\) |
\({\theta _B} = \frac{{ML}}{{2EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{192EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
|
\({y_c} = \frac{{w{L^4}}}{{384EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
Where, y = Deflection of beam, θ = Slope of beam
Bending moment at any section in a conjugate beam gives _______ in the actual beam.
Answer (Detailed Solution Below)
Deflection of Beam Question 14 Detailed Solution
Download Solution PDFExplanation:
Real Beam |
Conjugate beam |
Free end |
Fixed end |
Internal hinge |
Internal pin or roller support |
End pin or roller connection |
Remains same |
M/EI diagram of the real beam due to the top applied load |
Loading on Conjugate beam |
Slope at any point in the real beam |
Shear force at that point or section in Conjugate beam |
Deflection at any point in the real beam |
Bending moment at that point or section in Conjugate beam |
Two identical simply supported beams in which beam ‘A’ carries a central load of W and beam ‘B’ carries a uniformly distributed load such that wl = W, where w = uniformly distributed load and I = span of the identical beams, then the ratio of maximum deflection between B and A
Answer (Detailed Solution Below)
Deflection of Beam Question 15 Detailed Solution
Download Solution PDFExplanation:
Deflection and slope of various beams are given by:
|
\({y_B} = \frac{{P{L^3}}}{{3EI}}\) |
\({\theta _B} = \frac{{P{L^2}}}{{2EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{8EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{6EI}}\) |
|
\({y_B} = \frac{{M{L^2}}}{{2EI}}\) |
\({\theta _B} = \frac{{ML}}{{EI}}\) |
|
\({y_B} = \frac{{w{L^4}}}{{30EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{48EI}}\) |
\({\theta _B} = \frac{{w{L^2}}}{{16EI\;}}\) |
|
\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\) |
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) |
|
\({y_c} = 0\) |
\({\theta _B} = \frac{{ML}}{{24EI}}\) |
|
\({y_c} = \frac{{M{L^2}}}{{8EI}}\) |
\({\theta _B} = \frac{{ML}}{{2EI}}\) |
|
\({y_c} = \frac{{P{L^3}}}{{192EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
|
\({y_c} = \frac{{w{L^4}}}{{384EI}}\) |
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) |
Where, y = Deflection of beam, θ = Slope of beam
∴ Ratio of maximum deflection between B and A = \(\frac{ \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}}{\frac{{P{L^3}}}{{48EI}}}\) = \(\frac{5}{8}\)