9111-SRM MCET Mechanical Engineering
UNIT IV
DEFLECTION OF BEAMS
PART A
1. Write the advantages of Macaulay’s method over integration method.
(AU Apr/May 2023)
Use of Macaulay's technique is very convenient for cases of discontinuous and/or
discrete loading. Typically partial uniformly distributed loads (u.d.l.) and uniformly varying
loads (u.v.l.) over the span and a number of concentrated loads are conveniently handled
using this technique.
2. State moment area theorems. (AU Apr/May 2023, Nov/Dec 2023, Nov/Dec 2021)
Mohr’s theorem 1:
The change of slope between any two point is equal to the net area of the BM
diagram between these points divided by EI.
Mohr’s theorem 2:
The total deflection between any two point is equal to the moment of the area of the
BM diagram between these two point about the last point divided by EI.
3. What are the disadvantages of double integration method (AU Nov/Dec
2021)
The disadvantages of double integration method are
- Complex loading conditions
- Difficulties in applying boundary conditions
- Time-consuming
- Sensitivity to inaccuracies
- Limited to linear elastic materials
- Not suitable for beams with variable cross-sections
4. What is the maximum deflection at the free end of the cantilever beam carrying a
UDL of ‘w’/m over the entire span T (AU Nov/Dec 2020)
The maximum deflection at the free end of a cantilever beam carrying a uniformly
distributed load (UDL) of w/m over the entire span T is given by the equation:
4
wL
ymax = 8 EI
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5. A cantilever of span 1.8 m is carrying a point load at the free end. Find the deflection
at the free end, if the slope at the free end is 1o. (AU Nov/Dec 2015)
6. Distinguish between actual beam and conjugate beam (AU Apr/May 2023)
Actual Beam Conjugate Beam
Physical It is the real beam in the It is an imaginary beam
Reality structure. used for analysis.
Purpose To analyse forces, moments To find the deflections
and deflections in the beam. and rotations of the actual
beam.
Boundary Determined by the structure Based on the actual
Conditions (supports, loads) beam’s loading and
boundary conditions, but
with adjusted supports
Deflection Calculated using standard Calculated using the
Calculation formulas moment curvature
relationship applied to the
conjugate beam.
Support Determined by the problems Support conditions
Conditions physical setup (roller, hinge, depend on the actual
fixed) beam’s loading and
boundary conditions.
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7. Define Strain Energy (AU Apr/May 2018, Nov/Dec 2018)
When an elastic material is deformed due to application of external force, internal resistance
is developed in the material of the body, Due to deformation, some work is done by the
internal resistance developed in the body, which is stored in the form of energy. This energy
is known as strain energy. It is expressed in Nm.
8. Write about the procedures of Macaulay’s method to define deflection of beam
(AU Nov/Dec 2021)
a. Macaulay’s Method involves expressing the bending moment equation using Macaulay’s
brackets (step functions) to handle point loads and moments at specific locations on the
beam.
b. The deflection is calculated by integrating the bending moment equation twice, where the
first integration gives the slope and the second gives the deflection of the beam.
c. Boundary conditions are applied to solve for the constants of integration, providing the
final equation for deflection, which accounts for the beam's loading and support conditions.
9. What are the methods for finding out the slope and deflection at a section?
(AU Nov/Dec 2020, Nov/Dec 2016, Nov/Dec 2015)
The available methods to find the slope and deflection of beam are
- Double integration method
- Macaulay’s method
- Moment Area method
- Conjugate beam method
10. A cantilever beam is subjected to a point load W at the free end. What is the slope
and deflection at the free end? (AU Nov/Dec 2019)
Figure 4.1: A cantilever beam
2
WL
Slope at the free end θ =
2 EI
3
WL
Deflection at the free end y =
3 EI
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11. How the deflection & slope is calculated for the Cantilever beam by conjugate
beam method? (AU Nov/Dec 2018, Nov/Dec
2017)
12. What is the equation used in the case of double integration method?
(AU Nov/Dec 2015)
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13. Write down the equation for the maximum deflection of a cantilever beam carrying
a central point load ‘w’. (AU May / June
2017)
14. State the Maxwell’s reciprocal theorem.
(AU Nov/Dec 2019, Apr/May 2019, Nov/Dec 2017)
The Maxwell reciprocal theorem states that, “the work done by the first system of load due
to displacement caused by a second system of load equal the work done by the second
system of load due to displacement caused by the first system of load”.
n m
∑ ( P i ) A ( δ i )B = ∑ ( P j ) A ( δ j )B
i=1 j=1
15. Draw conjugate beam for a double side over hanging beam (AU May / June
2017)
Figure 4.2: Double side over hanging beam
16. State Castigliano’s first Theorem. (AU Nov/Dec 2018)
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The partial derivative of the strain energy with respect to a displacement equals the force
associated with that displacement.
∂U
δi =
∂ Fi
Where,
δ i=displacement of the point of application of the force F iin the direction of F i
U = strain energy
17. In a support beam of 3m span carrying uniformly distribution load throughout the
length the slope at the support is 1°. What is the max deflection in the beam?
(AU Apr/May 2019)
18. State Castigliano’s Second Theorem. (AU Apr/May
2018)
The partial derivative of the strain energy with respect to a force equals the displacement
associated with that force.
∂ Ui
= W1
∂ δ1
∂ Ui
= W 2 This is Castigliano's second theorem
∂ δ2
∂U
When rotation are to be determined, θ =
∂ Mi
19. Define the terms Resilience, Proof Resilience and Modulus of Resilience
(AU Apr/May 2019, Nov/Dec 2018, Apr/May 2017, Nov/Dec
2017)
Resilience: It is ability of a material to absorb energy under elastic deformation and to
recover this energy upon removal of load. Resilience is indicated by the area under the stress
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strain curving to the point of elastic limit. In a technical sense, resilience is the property of a
material that allow it return to its original shape after being de formed.
Proof Resilience: It is defined as the maximum energy that can be absorbed within the
elastic limit without creating a permanent distortion.
Modulus of Resilience: It is the ratio of the proof resilience to the volume of the body.
20. In a cantilever beam, the measured deflection at, free end was 8 mm when a
concentrated load of 12 kN was applied at it’s mid span. What will be the deflection at
mid – span when the same beam carries a concentrated load of 7 kN at the free end?
(AU Apr / May 2015)
Figure 4.3: Deflection at mid – span
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PART B
1. A simply supported beam of 12m span carries a concentrated load of 30kN at a
distance of 9m from the end A. Determine the deflection at the load point and the
slopes at the load point and at the two ends. Take 1=2x10 9mm4 and E = 205 GPa. Use
moment area method. (AU Nov/Dec 2023)
Figure 4.4: Simply supported beam
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2. Using the moment area method, determine the slope and deflection at free end of the
cantilever beam when it is subjected to uniformly distributed load over entire length
and point load at the free end. (AU Apr/May 2023)
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Figure 4.5: Slope and deflection at free end of the cantilever
beam
3. Difference between Conjugate beam method for computation of slopes and
deflections in determinate beam (AU Nov/Dec 2021)
The Conjugate Beam Method is a classical technique used to compute the slopes and
deflections in statically determinate beams. It has some key differences when compared to
other methods, such as Double Integration or Moment-Area Method.
Conjugate Beam Method Other Methods
Basic Concept The real beam is Double Integration Method
transformed into a uses the relationship
conjugate beam that has the between the bending
same length but different moment and curvature of
boundary conditions. the beam to derive a set of
The conjugate beam is equations, which are
loaded in a manner where integrated to find the
the load at any point is deflections and slopes.
proportional to the bending Moment-Area Method
moment at the involves the direct
corresponding point of the calculation of deflections
real beam divided by the by analyzing the area under
flexural rigidity (EI). the moment-curvature
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diagram for the beam.
Transformatio The original beam is There is no transformation
n of the replaced by a conjugate of the beam. The original
Problem beam with the same length beam is analyzed directly
but boundary conditions using its bending moment
modified according to the and flexural rigidity to
real beam's supports and compute the deflections and
loading conditions. slopes.
Boundary The boundary conditions of Double Integration directly
Conditions the conjugate beam are applies the boundary
derived from the boundary conditions of the real beam
conditions of the real beam. to the deflection and slope
equations without
transforming the beam into
another system.
Moment-Area Method also
applies the original
boundary conditions, but it
uses the areas under the
moment-curvature diagram
to determine deflections.
Calculation of The slope at any point in Double Integration requires
Slopes and the real beam is equal to the performing integrations of
Deflections deflection at the the bending moment
corresponding point in the equation to find expressions
conjugate beam. for slopes and deflections.
The deflection at any It is more algebraic.
point in the real beam is Moment-Area Method
equal to the slope at the calculates deflections and
corresponding point in the slopes using geometric
conjugate beam. properties of the moment
The method allows a diagram, which can be
relatively straightforward quick but requires careful
calculation by transforming handling of the areas under
the real beam into a new the curve.
structure.
Complexity The method is often simpler Double Integration can be
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to apply for beams with more complex as it involves
simple support conditions, solving differential
as the boundary conditions equations and applying
of the conjugate beam are boundary conditions step by
relatively easy to step.
determine. Moment-Area Method
It is particularly useful requires drawing accurate
for beams with standard moment diagrams and
loading calculating areas, which can
conditions(uniformly be time-consuming for
distributed loads, complex loading scenarios.
concentrated loads, etc.).
Applications Best suited for statically Double Integration is more
determinate beams with general and can be applied
simple boundary to both statically
conditions. determinate and
It is often used for indeterminate beams,
beams with simple loading though solving
conditions, where the indeterminate beams may
conjugate beam can be require additional steps.
easily derived. Moment-Area Method is
also widely applicable,
particularly for beams
subjected to varying
moment distributions.
4. A simply supported beam of uniform cross section carries a UD load ‘w’kN/m
throughout its length. Calculate the slope and deflection at the mid point by moment
area method. (AU Nov/Dec 2020, Nov/Dec 2019)
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Figure 4.6: Slope and deflection at the mid point
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5. A beam of length 5 m and of uniform rectangular section is supported at its ends and
carries uniformly distributed load 9 kN/m over the entire length. Calculate the width
and depth of the section if the maximum permissible bending stress is 7 N/mm² and the
central deflection is not to exceed 10 mm. E=1x104 N/mm2.
(AU Apr/May 2019, Nov/Dec 2018)
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6. A steel girder of uniform section 14 metres long is simply supported at the ends. It
carries concentrated loads of 90 kN and 60 kN at two points 3 metres and 4.5 metres
from the two ends respectively. Calculate: the deflection of the girder at the points
under the two loads and the maximum deflection. Take E = 210 x 10 6 kN/m 2 and I =
64 x 104 m². (AU Nov/Dec 2021, April/May 2017)
Figure 4.7: Simply supported beam
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7. A beam AB of span 4 m is simply supported at its ends. The beam carries a
concentrated load of 20 kN at 1m from the support A and uniformly distributed load of
10 kN/m in the right half span. Assume flexural rigidity EI = 4000 kN.m². Using
Macaulay's method, determine i) deflection in mid span, ii) Maximum deflection, iii)
Slope at A. (AU April / May 2018)
Figure 4.8: Simply supported beam
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8. A cantilever of length 2 m carries a point load of 20 kN at the free end and another
load of 20 kN at its centre. If E = 10 5 N/mm² and I = 108 mm 4 for the cantilever then
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determine by moment area method, the slope and deflection of the cantilever at the
free end. (AU Nov/Dec 2018)
Figure 4.9: Cantilever beam
Figure 4.10: Slope of the cantilever at the free end.
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9. Using conjugate beam method, obtain the slope and deflection at A, B, C and D of
the beam shown in figure. E=200 GPa and I=2x10² m². (AU Nov/Dec 2021)
Figure 4.11: Simply supported beam
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Figure 4.12: Slope and deflection at A, B, C and D of the conjugate beam
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10. A cantilever of length 2m carries a uniformly distributed load of 2.5kN/m run for a
length of 1.25m from the fixed end and a point load of 1kN at the free end. Find the
deflection at the free end, if the section is rectangular 12cm wide and 24cm deep and
E=1x104 N/mm2 (AU Apr/May 2018)
Figure 4.13: A cantilever beam
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