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All questions of Deflection of Beams for Mechanical Engineering Exam
An overhang beam of uniform El is loaded as shown
The deflection at the free end will be- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
An overhang beam of uniform El is loaded as shown
The deflection at the free end will be
The deflection at the free end will be
a)
b)
c)
d)
 | Gate Funda answered |
The reaction RB at B is, ∑MC = 0
⇒
∴
∴ Total strain energy
So by Castigliano’s theorem, deflection at free end,
⇒
∴
∴ Total strain energy
So by Castigliano’s theorem, deflection at free end,
A 4-metre long beam, simply supported at its ends; carries a point load 'W at its centre. If the slope at the ends of beam is 1° then deflection at the centre of beam is - a)10.56 mm
- b)18.32 mm
- c)23.27 mm
- d)39.37 mm
Correct answer is option 'C'. Can you explain this answer?
A 4-metre long beam, simply supported at its ends; carries a point load 'W at its centre. If the slope at the ends of beam is 1° then deflection at the centre of beam is
a)
10.56 mm
b)
18.32 mm
c)
23.27 mm
d)
39.37 mm
 | Bhaskar Mukherjee answered |
Deflection at centre of beam 
Slope at ends of beam
Slope at ends of beam
A cantilever beam of span ‘l’ and uniform flexural rigidity ‘EI’ is loaded with an upward force ‘W’ at the mid point and downward force ‘P’ at the free end. The deflection at the free end will be zero if - a)W = 3P/2
- b)W = 2P
- c)W = 16P/5
- d)W = 5P
Correct answer is option 'C'. Can you explain this answer?
A cantilever beam of span ‘l’ and uniform flexural rigidity ‘EI’ is loaded with an upward force ‘W’ at the mid point and downward force ‘P’ at the free end. The deflection at the free end will be zero if
a)
W = 3P/2
b)
W = 2P
c)
W = 16P/5
d)
W = 5P
 | Ameya Deshpande answered |
Upward deflection at B due to load W,
Downward deflection at S due to load P,
For no deflection at B,
⇒ W = 16P/5
A simply supported beam of uniform flexural rigidity is loaded as shown in the given figure. The rotation at the end A is
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A simply supported beam of uniform flexural rigidity is loaded as shown in the given figure. The rotation at the end A is
a)
b)
c)
d)
| | Rajdeep Gupta answered |
The 
diagram of beam is,
diagram of beam is,
The slope at mid span is zero.
The difference between the slope at two points is 
area between these points.
area between these points.So slope at A,
Which of the foilowing statements are correct. Macaulay's method for calcuiation of slope and deflection in a beam is suitable for
1. Prismatic beams only.
2. Several concentrated loads and can be extended to uniformly distributed loads.
3. Both prismatic and Non-prismatic beams,- a)Only 1
- b)1 and 2
- c)Only 3
- d)2 and 3
Correct answer is option 'B'. Can you explain this answer?
Which of the foilowing statements are correct. Macaulay's method for calcuiation of slope and deflection in a beam is suitable for
1. Prismatic beams only.
2. Several concentrated loads and can be extended to uniformly distributed loads.
3. Both prismatic and Non-prismatic beams,
1. Prismatic beams only.
2. Several concentrated loads and can be extended to uniformly distributed loads.
3. Both prismatic and Non-prismatic beams,
a)
Only 1
b)
1 and 2
c)
Only 3
d)
2 and 3
| | Gauri Roy answered |
Macaulay's method for calculation of slope and deflection in a beam
Introduction
Macaulay's method is a widely used technique in structural engineering for calculating the slope and deflection of beams. It is a powerful tool that can be used to analyze both prismatic and non-prismatic beams subjected to various loading conditions. In this method, the beam is divided into smaller segments, and the bending moment equation is expressed in terms of these segments.
Statement 1: Macaulay's method is suitable for prismatic beams only
This statement is incorrect. Macaulay's method can be used to analyze both prismatic and non-prismatic beams. Prismatic beams are those that have a constant cross-section throughout their length, while non-prismatic beams have varying cross-sections along their length. Macaulay's method is applicable to both types of beams, making it a versatile technique.
Statement 2: Macaulay's method can be extended to uniformly distributed loads
This statement is correct. Macaulay's method can handle various loading conditions, including concentrated loads as well as uniformly distributed loads. By expressing the bending moment equation in terms of smaller segments, the method can effectively account for the distribution of loads along the beam's length. This allows for accurate calculation of the slope and deflection under uniformly distributed loads.
Statement 3: Macaulay's method is suitable for both prismatic and non-prismatic beams
This statement is correct. As mentioned earlier, Macaulay's method can be used for both prismatic and non-prismatic beams. The method does not rely on the beam's cross-sectional properties remaining constant, making it applicable to a wide range of beam geometries. Whether the beam is prismatic or non-prismatic, Macaulay's method provides an effective approach to determine the slope and deflection.
Conclusion
In conclusion, Macaulay's method is a versatile technique for calculating the slope and deflection in beams. It can be applied to both prismatic and non-prismatic beams, making it suitable for a wide range of structural analysis problems. Additionally, the method can handle various loading conditions, including concentrated loads and uniformly distributed loads. Therefore, statement 1 is incorrect, statement 2 is correct, and statement 3 is correct. The correct option is 'B' - 1 and 2.
Introduction
Macaulay's method is a widely used technique in structural engineering for calculating the slope and deflection of beams. It is a powerful tool that can be used to analyze both prismatic and non-prismatic beams subjected to various loading conditions. In this method, the beam is divided into smaller segments, and the bending moment equation is expressed in terms of these segments.
Statement 1: Macaulay's method is suitable for prismatic beams only
This statement is incorrect. Macaulay's method can be used to analyze both prismatic and non-prismatic beams. Prismatic beams are those that have a constant cross-section throughout their length, while non-prismatic beams have varying cross-sections along their length. Macaulay's method is applicable to both types of beams, making it a versatile technique.
Statement 2: Macaulay's method can be extended to uniformly distributed loads
This statement is correct. Macaulay's method can handle various loading conditions, including concentrated loads as well as uniformly distributed loads. By expressing the bending moment equation in terms of smaller segments, the method can effectively account for the distribution of loads along the beam's length. This allows for accurate calculation of the slope and deflection under uniformly distributed loads.
Statement 3: Macaulay's method is suitable for both prismatic and non-prismatic beams
This statement is correct. As mentioned earlier, Macaulay's method can be used for both prismatic and non-prismatic beams. The method does not rely on the beam's cross-sectional properties remaining constant, making it applicable to a wide range of beam geometries. Whether the beam is prismatic or non-prismatic, Macaulay's method provides an effective approach to determine the slope and deflection.
Conclusion
In conclusion, Macaulay's method is a versatile technique for calculating the slope and deflection in beams. It can be applied to both prismatic and non-prismatic beams, making it suitable for a wide range of structural analysis problems. Additionally, the method can handle various loading conditions, including concentrated loads and uniformly distributed loads. Therefore, statement 1 is incorrect, statement 2 is correct, and statement 3 is correct. The correct option is 'B' - 1 and 2.
The maximum deflection of simply supported beam occurs at zero- a)bending moment location
- b)shear force location
- c)slope location
- d)shear force location and also zero bending moment location
Correct answer is option 'C'. Can you explain this answer?
The maximum deflection of simply supported beam occurs at zero
a)
bending moment location
b)
shear force location
c)
slope location
d)
shear force location and also zero bending moment location
| | Anmol Menon answered |
Explanation:
- Simply supported beam is a type of beam that is supported at two ends and free to rotate at each end.
- When a load is applied on the beam, it causes bending and deflection in the beam.
- The maximum deflection in a simply supported beam occurs at the mid-span of the beam.
- The slope of the beam at the mid-span is zero, which means that the beam is neither rising nor falling at that point.
- Therefore, the correct option is C, which states that the maximum deflection occurs at the slope location.
- The bending moment and shear force are related to the stress and strain in the beam, but they do not directly affect the deflection of the beam.
- However, the maximum bending moment and shear force occur at the supports of the beam, where the beam is fixed and cannot rotate.
- Therefore, the maximum bending moment and shear force occur at the supports, but the maximum deflection occurs at the mid-span of the beam.
- The maximum deflection in a simply supported beam can be calculated using the following formula:
δmax = (5WL^4)/(384EI)
where δmax is the maximum deflection, W is the load applied on the beam, L is the span of the beam, E is the modulus of elasticity of the material of the beam, and I is the moment of inertia of the beam.
- Simply supported beam is a type of beam that is supported at two ends and free to rotate at each end.
- When a load is applied on the beam, it causes bending and deflection in the beam.
- The maximum deflection in a simply supported beam occurs at the mid-span of the beam.
- The slope of the beam at the mid-span is zero, which means that the beam is neither rising nor falling at that point.
- Therefore, the correct option is C, which states that the maximum deflection occurs at the slope location.
- The bending moment and shear force are related to the stress and strain in the beam, but they do not directly affect the deflection of the beam.
- However, the maximum bending moment and shear force occur at the supports of the beam, where the beam is fixed and cannot rotate.
- Therefore, the maximum bending moment and shear force occur at the supports, but the maximum deflection occurs at the mid-span of the beam.
- The maximum deflection in a simply supported beam can be calculated using the following formula:
δmax = (5WL^4)/(384EI)
where δmax is the maximum deflection, W is the load applied on the beam, L is the span of the beam, E is the modulus of elasticity of the material of the beam, and I is the moment of inertia of the beam.
For cantilever beam shown in the figure, the deflection at C due to a couple M applied at Bis equal to
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
For cantilever beam shown in the figure, the deflection at C due to a couple M applied at Bis equal to
a)
b)
c)
d)
| | Pallabi Bajaj answered |
Slope and deflection at B,
[As there is no moment between B and C so BC remains linear at θB slope]
A cantilever beam of length ’L’ is subjected to a concentrated load P at free end what is the deflection at the centre of beam?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
A cantilever beam of length ’L’ is subjected to a concentrated load P at free end what is the deflection at the centre of beam?
a)
b)
c)
d)
| | Debolina Chavan answered |
Using Maxwell’s reciprocal theorem,
Deflection at C when load is applied at B = Deflection at B when load is applied at C
A horizontal beam of length l of uniform cross-section is pinned at ends A and B. At end A there is an anticlockwise moment M and at end B there is clockwise moment 2M. The slope at end A and end B are respectively.- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
A horizontal beam of length l of uniform cross-section is pinned at ends A and B. At end A there is an anticlockwise moment M and at end B there is clockwise moment 2M. The slope at end A and end B are respectively.
a)
b)
c)
d)
| | Ashish Chakraborty answered |
Two identical cantileversare loaded as shown in figure. If slope at the free end of the cantilever in figure (a) is θ, the slope at the free end of the cantilever in figure (b) will be.
- a)θ/3
- b)θ/2
- c)2θ/3
- d)θ
Correct answer is option 'D'. Can you explain this answer?
Two identical cantileversare loaded as shown in figure. If slope at the free end of the cantilever in figure (a) is θ, the slope at the free end of the cantilever in figure (b) will be.
a)
θ/3
b)
θ/2
c)
2θ/3
d)
θ
| | Madhurima Banerjee answered |
Slope in figure (a) is given by,
Where
∴
Slope in figure (b) is
∴ θb = θ
Where
∴
Slope in figure (b) is
∴ θb = θ
The given figure shows a cantilever of span L subjected to a concentrated load P and moment M at the free end. Deflection at the free end is given by
- a)PL2/2EI + ML2/3EI
- b)PL3/3EI + ML2/2 El
- c)PL3/2EI + ML2/3EI
- d)PL3/48El + ML2/48EI
Correct answer is option 'B'. Can you explain this answer?
The given figure shows a cantilever of span L subjected to a concentrated load P and moment M at the free end. Deflection at the free end is given by
a)
PL2/2EI + ML2/3EI
b)
PL3/3EI + ML2/2 El
c)
PL3/2EI + ML2/3EI
d)
PL3/48El + ML2/48EI
| | Rahul Chauhan answered |
Deflection due to concentrated load acting at the end of cantilever= Pl3/3EI
Deflection due to moment acting at the end of cantilever= Ml2/2EI
Therefore, from principle of superposition, the deflection due to both moment and concentrated load acting at the end of cantilever will be Pl3/3EI + Ml2/2EI
Deflection due to moment acting at the end of cantilever= Ml2/2EI
Therefore, from principle of superposition, the deflection due to both moment and concentrated load acting at the end of cantilever will be Pl3/3EI + Ml2/2EI
Calculate the reaction at the roller support for the cantilever beam shown in the figure?
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Calculate the reaction at the roller support for the cantilever beam shown in the figure?
a)
b)
c)
d)
| | Tanishq Menon answered |
Let P be the roller reaction.
Upward deflection of beam at A due to P.
Upward deflection of beam at A due to P.
Downward deflection of beam at A due to w.
A cantilever beam of length ‘l’ carries a UDL ‘w’ from fixed end till mid span. The deflection at the free end is given by- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A cantilever beam of length ‘l’ carries a UDL ‘w’ from fixed end till mid span. The deflection at the free end is given by
a)
b)
c)
d)
| | Kavya Mehta answered |
The slope at mid span
∴ Deflection at free end,
A simply supported beam AB of span ‘l’ has a uniform cross-section throughout. It carries a central concentrated load W and another load which is uniformly distributed over the entire span, its total magnitude being W. The maximum deflection in the beam is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A simply supported beam AB of span ‘l’ has a uniform cross-section throughout. It carries a central concentrated load W and another load which is uniformly distributed over the entire span, its total magnitude being W. The maximum deflection in the beam is
a)
b)
c)
d)
| | Navya Saha answered |
Maximum deflection will occur at centre
Consider the following statements.
1. Conjugate beam can be used to determine slopes and deflection in a non-prismatic beam.
2. Conjugate beam may be statically indeterminate.
3. Conjugate beam method gives absolute slope and deflection.The correct answer is- a)1 and 3
- b)1 and 2
- c)2 and 3
- d)1, 2 and 3
Correct answer is option 'A'. Can you explain this answer?
Consider the following statements.
1. Conjugate beam can be used to determine slopes and deflection in a non-prismatic beam.
2. Conjugate beam may be statically indeterminate.
3. Conjugate beam method gives absolute slope and deflection.
1. Conjugate beam can be used to determine slopes and deflection in a non-prismatic beam.
2. Conjugate beam may be statically indeterminate.
3. Conjugate beam method gives absolute slope and deflection.
The correct answer is
a)
1 and 3
b)
1 and 2
c)
2 and 3
d)
1, 2 and 3
| | Lakshmi Datta answered |
Conjugate Beam Method in Structural Analysis
The conjugate beam method is a technique used in structural analysis to determine the slope and deflection of a beam. It involves replacing the real beam with a hypothetical beam known as the conjugate beam, which has the same length, load, and support conditions as the real beam. However, the conjugate beam is free from internal forces and moments, making it easier to analyze.
Statements and their Explanation
1. Conjugate beam can be used to determine slopes and deflection in a non-prismatic beam.
A non-prismatic beam is a beam whose cross-section changes along its length. The conjugate beam method can be used to determine the slope and deflection of such beams. The method involves finding the conjugate beam of the non-prismatic beam and analyzing it using the standard formulae for a prismatic beam.
2. Conjugate beam may be statically indeterminate.
A statically indeterminate beam is a beam whose support reactions cannot be determined by equilibrium equations alone. The conjugate beam method may result in a statically indeterminate system, depending on the support conditions of the real beam. In such cases, additional equations or compatibility conditions may be required to solve the problem.
3. Conjugate beam method gives absolute slope and deflection.
The conjugate beam method gives the slope and deflection of the beam at a particular point. However, it does not give the absolute slope and deflection, which depend on the boundary conditions of the real beam. Therefore, the results obtained from the conjugate beam method should be corrected for the actual boundary conditions of the real beam.
Conclusion
In summary, the conjugate beam method is a useful technique for determining the slope and deflection of beams, including non-prismatic beams. However, it may result in a statically indeterminate system, and the results obtained should be corrected for the actual boundary conditions of the real beam.
The conjugate beam method is a technique used in structural analysis to determine the slope and deflection of a beam. It involves replacing the real beam with a hypothetical beam known as the conjugate beam, which has the same length, load, and support conditions as the real beam. However, the conjugate beam is free from internal forces and moments, making it easier to analyze.
Statements and their Explanation
1. Conjugate beam can be used to determine slopes and deflection in a non-prismatic beam.
A non-prismatic beam is a beam whose cross-section changes along its length. The conjugate beam method can be used to determine the slope and deflection of such beams. The method involves finding the conjugate beam of the non-prismatic beam and analyzing it using the standard formulae for a prismatic beam.
2. Conjugate beam may be statically indeterminate.
A statically indeterminate beam is a beam whose support reactions cannot be determined by equilibrium equations alone. The conjugate beam method may result in a statically indeterminate system, depending on the support conditions of the real beam. In such cases, additional equations or compatibility conditions may be required to solve the problem.
3. Conjugate beam method gives absolute slope and deflection.
The conjugate beam method gives the slope and deflection of the beam at a particular point. However, it does not give the absolute slope and deflection, which depend on the boundary conditions of the real beam. Therefore, the results obtained from the conjugate beam method should be corrected for the actual boundary conditions of the real beam.
Conclusion
In summary, the conjugate beam method is a useful technique for determining the slope and deflection of beams, including non-prismatic beams. However, it may result in a statically indeterminate system, and the results obtained should be corrected for the actual boundary conditions of the real beam.
A simply supported beam with width ‘b' and depth ‘d’ carries a central load W and undergoes deflection δ at the centre. If the width and depth are interchanged, the deflection at the centre of the beam would attain the value- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A simply supported beam with width ‘b' and depth ‘d’ carries a central load W and undergoes deflection δ at the centre. If the width and depth are interchanged, the deflection at the centre of the beam would attain the value
a)
b)
c)
d)
| | Arshiya Roy answered |
For a simply supported beam deflection, δ is given by,
∴
Let δ1 be the deflection when the width and depth are interchanged,
∴
∴
∴
Let δ1 be the deflection when the width and depth are interchanged,
∴
∴
A simply supported beam of length CL' carries two equal unlike couples. M at the two ends. The central deflection of the beam is given by- a)ML2/4EI
- b)ML2/16EI
- c)ML2/64EI
- d)ML2/8EI
Correct answer is option 'D'. Can you explain this answer?
A simply supported beam of length CL' carries two equal unlike couples. M at the two ends. The central deflection of the beam is given by
a)
ML2/4EI
b)
ML2/16EI
c)
ML2/64EI
d)
ML2/8EI
 | Devanshi Iyer answered |
Central Deflection of a Simply Supported Beam Carrying Two Equal Unlike Couples
To find the central deflection of a simply supported beam carrying two equal unlike couples, we can use the principles of structural analysis and beam deflection.
Let's break down the problem and solution into the following steps:
1. Determine the reactions at the supports:
Since the beam is simply supported, the reactions at the supports will be equal and opposite. In this case, each support will have a reaction of M.
2. Calculate the bending moment distribution along the beam:
The bending moment at any point along the beam can be determined by considering the effect of the applied couples. Since the couples are equal and unlike, the bending moment will be constant throughout the length of the beam and will be equal to M.
3. Determine the equation for deflection:
The equation for deflection of a simply supported beam due to a uniformly distributed load can be given as follows:
δ = (5ML^4)/(384EI)
Where:
δ = deflection at the center of the beam
M = bending moment at any point on the beam (constant value of M in this case)
L = length of the beam
E = modulus of elasticity of the material
I = moment of inertia of the beam's cross-section
4. Substitute the values into the equation:
In this case, the length of the beam is CL. We can substitute the values into the equation to get the central deflection:
δ = (5M(CL)^4)/(384EI)
= (5MC^4L^4)/(384EI)
5. Simplify the equation:
We can simplify the equation further by canceling out common terms:
δ = (5C^4ML^4)/(384EI)
= (5ML^2C^4)/(384EI)
6. Compare the simplified equation with the given options:
Comparing the simplified equation with the given options, we find that the correct answer is option 'D':
δ = ML^2/(8EI)
Therefore, the central deflection of the simply supported beam carrying two equal unlike couples is given by ML^2/(8EI).
To find the central deflection of a simply supported beam carrying two equal unlike couples, we can use the principles of structural analysis and beam deflection.
Let's break down the problem and solution into the following steps:
1. Determine the reactions at the supports:
Since the beam is simply supported, the reactions at the supports will be equal and opposite. In this case, each support will have a reaction of M.
2. Calculate the bending moment distribution along the beam:
The bending moment at any point along the beam can be determined by considering the effect of the applied couples. Since the couples are equal and unlike, the bending moment will be constant throughout the length of the beam and will be equal to M.
3. Determine the equation for deflection:
The equation for deflection of a simply supported beam due to a uniformly distributed load can be given as follows:
δ = (5ML^4)/(384EI)
Where:
δ = deflection at the center of the beam
M = bending moment at any point on the beam (constant value of M in this case)
L = length of the beam
E = modulus of elasticity of the material
I = moment of inertia of the beam's cross-section
4. Substitute the values into the equation:
In this case, the length of the beam is CL. We can substitute the values into the equation to get the central deflection:
δ = (5M(CL)^4)/(384EI)
= (5MC^4L^4)/(384EI)
5. Simplify the equation:
We can simplify the equation further by canceling out common terms:
δ = (5C^4ML^4)/(384EI)
= (5ML^2C^4)/(384EI)
6. Compare the simplified equation with the given options:
Comparing the simplified equation with the given options, we find that the correct answer is option 'D':
δ = ML^2/(8EI)
Therefore, the central deflection of the simply supported beam carrying two equal unlike couples is given by ML^2/(8EI).
The stepped cantilever- is subjected to moments, M as shown in the figure below. The vertical deflection at the free end ( neglecting self weight) is 
- a)
- b)
- c)
- d)zero
Correct answer is option 'C'. Can you explain this answer?
The stepped cantilever- is subjected to moments, M as shown in the figure below. The vertical deflection at the free end ( neglecting self weight) is
a)
b)
c)
d)
zero
| | Shraddha Datta answered |
Using moment area method,
Deflection at B, wrt A
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