This document summarizes the moment-area method for calculating deflections in beams. It discusses how the bending moment diagram can be divided into areas that correspond to rotations of the elastic curve. The sum of these areas multiplied by the distance to the centroid gives the tangential deviation, which can be used to determine the deflection. The method is applicable to statically indeterminate beams using superposition. Boundary conditions and how to handle different support types are also covered.

1. Chapter 10
Deflection of Beams
Mechanics of Solids

2. Part A- Deflection by Integration
Moment-Curvature Relation
• It is assumed that bending takes place only about one of the principal
axes of the cross section.
• consider Fig. 1, where the radius of curvature 𝜌 of the elastic curve
can change along the span.
• If distance 𝑦 from the neutral surface to the strained fibers is
measured in the usual manner as being positive upwards, the
deformation ∆𝑢 of any fiber,
∆𝑢 = −𝑦 ∆𝜃

3. Fig. 1: Deformation of a beam in bending
• The fibers lying in the curved neutral surface of the deformed beam,
characterized in Fig. 1 (d) by fiber 𝑎𝑏, are not strained at all.

4. • 𝑑𝑢 𝑑𝑠 is the normal strain in a beam fiber at a distance 𝑦 from the
neutral axis.
𝑑𝑢
𝑑𝑠
= 𝜀
• With the aid of Fig. 1(c), it is seen that, since ∆𝑠 = 𝜌 ∆𝜃,
lim
∆𝑠→0
∆𝜃
∆𝑠
=
𝑑𝜃
𝑑𝑠
=
1
𝜌
= 𝜅
which is the definition of curvature 𝜅.
• Fundamental relation between curvature of the elastic curve and the
normal strain,
1
𝜌
= 𝜅 = −
𝜀
𝑦

5. • For the elastic case, since 𝜀 = 𝜀 𝑥 = 𝜎𝑥 𝐸 and 𝜎𝑥 = − 𝑀𝑦 𝐼,
1
𝜌
=
𝑀
𝐸𝐼
• This equation relates bending moment 𝑀 at a given section of an
elastic beam having a moment of inertia 𝐼 around the neutral axis to
the curvature 1 𝜌 of the elastic curve.
Governing Differential Equation

6. • Since, the deflections tolerated in the vast majority of engineering
structures are very small, slope 𝑑𝑣 𝑑𝑥 of the elastic curve is also very
small.
1
𝜌
≈
𝑑2 𝑣
𝑑𝑥2
=
𝑀
𝐸𝐼
Fig. 2: Moment and its relation to curvature

7. • The positive direction of the 𝑣 axis is taken to have the same sense as
that of the positive 𝑦 axis and the positive direction of the applied
load 𝑞, Fig. 2.
• If the positive slope 𝑑𝑣 𝑑𝑥 of the elastic curve becomes larger as 𝑥
increases, curvature 1 𝜌 is positive.
• For the elastic curve, at the level of accuracy, one has 𝑑𝑠 = 𝑑𝑥.
• The square of the slope 𝑑𝑣 𝑑𝑥 is negligibly small compared with
unity,
• There is no horizontal displacement of the points lying on the neutral
surface, i.e., at 𝑦 = 0.

8. Alternative Forms of the Governing Equation
• For beams with constant flexural rigidity 𝐸𝐼,
𝐸𝐼
𝑑2 𝑣
𝑑𝑥2 = 𝑀 𝑥 𝐸𝐼
𝑑3 𝑣
𝑑𝑥3 = 𝑉 𝑥 𝐸𝐼
𝑑4 𝑣
𝑑𝑥4 = 𝑞 𝑥
• The choice of one of these equations for determining 𝑣 depends on
the ease with which an expression for load, shear, or moment can be
formulated.

9. Boundary Conditions
• Several types of homogeneous boundary conditions are as follows:
Fig. 3: Homogeneous boundary conditions for beams with constant 𝐸𝐼
(a) both conditions are kinematic; in (c) both are static;
In (b) and (d), conditions are mixed

10. 1. Clamped or fixed support: In this case, the displacement 𝑣 and the
slope 𝑑𝑣 𝑑𝑥 must vanish.
𝑣 𝑎 = 0 and 𝑣′ 𝑎 = 0
2. Roller or Pinned support: At the end considered, no deflection 𝑣
nor moment 𝑀 can exist.
𝑣 𝑎 = 0 𝑀 𝑎 = 𝐸𝐼𝑣′′ 𝑎 = 0
3. Free end: Such an end is free of moment and shear. Hence,
𝑀 𝑎 = 𝐸𝐼𝑣′′ 𝑎 = 0 𝑉 𝑎 = 𝐸𝐼𝑣′′ 𝑥=𝑎
′ = 0
4. Guided support: In this case, free vertical movement is permitted,
but the rotation of the end is prevented. The support is not able
capable of resisting any shear.
𝑣′
𝑎 = 0 𝑉 𝑎 = 𝐸𝐼𝑣′′ 𝑥=𝑎
′
= 0

11. • Consider the three cases shown in Fig. 4.
Fig. 4: Beams having
different loading
• The beam in Fig. 4(a) is indeterminate to the first degree, as any one of
the reactions can be removed and the beam will remain stable. The
boundary conditions shown in Fig. 3(a) apply for end 𝐴, and those in
Fig. 3(b), for end 𝐵.

12. • In some problems, discontinuities in the mathematical functions for
either load or member stiffness arise along a given span length.
• Such discontinuities, occur at concentrated forces or moments and at
abrupt changes in cross-sectional areas affecting 𝐸𝐼.
• At any juncture of the two zones of a beam where a discontinuity
occurs, the deflection and the tangent to the elastic curve must be the
same regardless of the direction from which the common point is
approached, Fig. 5.
Fig. 5: Unacceptable geometric conditions in a continuous elastic curve

13. Direct Integration Solutions
• As, 𝐸𝐼𝑣iv
= 𝑞 𝑥 . By successively integrating this expression four
time, the formal solution for 𝑣 is obtained. Thus,
𝑑2 𝑣
𝑑𝑥2 =
𝑑
𝑑𝑥
𝑑𝑣
𝑑𝑥
=
𝑑𝑣′
𝑑𝑥
=
𝑀
𝐸𝐼

14. • Starting with, 𝐸𝐼𝑣′′ = 𝑀 𝑥
• In these equations, constants 𝐶1, 𝐶2, 𝐶3, and 𝐶4 must be determined
from the conditions at the boundaries.
• In above equation, constants 𝐶1, 𝐶2, 𝐶3 𝐸𝐼, and 𝐶4 𝐸𝐼 are usually the
initial values of 𝑉, 𝑀, 𝜃, and 𝑣 at the origin.

15. Procedure Summary
• Three basic concepts:
1. Equilibrium conditions (statics) are used for a beam element to
establish the relationship between the applied load and shear, as
well as between the shear and bending moment.
2. Geometry of deformations (kinematics) is used by assuming that
plane sections through a beam element remain plane after
deformation. such planes intersect and define beam strains and the
radius of curvature for an element.
3. Properties of materials (constitutive relations) in the form of
Hooke’s law, are assumed to apply only to longitudinal normal
stresses and strains.

16. Deflections by Superposition
• The deflections of beams subjected to several or complicated loading
conditions are usually synthesized from the simpler loadings, using
the principle of superposition.
• The problem in Fig. 6 can be separated into three different cases as
shown. The algebraic sum of the three separate solutions gives the
total deflection.
Fig. 6: Resolution of a complex problem into several simpler problems

17. • The superposition procedure for determining elastic deflection of
beams can be extended to structural systems consisting of several
flexural members.
Fig. 7: A method of analyzing deflection of frames
• By multiplying angle 𝜃 𝐵 by length 𝑎 of the vertical member, the
deflection of point 𝐶 due to rotation of joint 𝐵 is determined.

18. • The portion of a beam between the supports, as 𝐴𝐵 in Fig. 8, is
isolated and rotation of the tangent at 𝐵 is found.
Fig. 8: A method of analyzing deflections of an overhang
• In Fig. 9(a), a simple beam on a rigid support at one end and on a
yielding support with a spring constant 𝑘 at the other end.
• If 𝑅 𝐵 is the reaction at 𝐵, support 𝐵 settles ∆= 𝑅 𝐵 𝑘, Fig. 9(b).
• The deflection of a point such as 𝑏 is very nearly 𝜃1 𝑥 + 𝑐𝑏

19. Fig. 9: Deflections in a composite structure
• Deflection of beams in situations where hinges are introduced, Fig.
9(c), are treated similarly. The tangent to the adjoining elastic curve is
not continuous across a hinge.

20. • Method of superposition for determining deflections or reactions for
statically indeterminate beams.
• Consider the continuous beam analyzed. By removing the support 𝑅 𝐵,
the beam would deflect in the middle, Fig. 10(a).
• By applying force 𝑅 𝐵 in an upward direction, the required condition of
no displacement at 𝐵 can be restored. The respective expressions for
these deflections can be equated to find 𝑅 𝐵.
Fig. 10: Superposition of two solutions for determining reactions

21. Deflections in Unsymmetrical Bending
• If unsymmetrical bending takes place, deflections are calculated in
each of the principal planes and deflections so found are added
vectorially.
Fig. 11: Deflection of a beam subjected to unsymmetrical bending

22. • In Fig. 11, the 𝑦 and 𝑧 axes are the principal axes passing through the
centroid as well as the shear center of the cross section.
• A positive deflection 𝑣1 is shown for the beam deflection taking place
in the 𝑥𝑦 plane, similarly 𝑤1 corresponds to the deflection in the 𝑥𝑧
plane. Their vectorial sum 𝐴𝐴′ is the total beam deflection.
• If applied forces don’t act through the shear center for the cross
section, torsional stresses and deformations must also be considered.

23. Energy Methods for Deflections and Impact
• In beams, one can include, both the bending and the shear strain
energies.
• The procedure based on equating the internal strain energy 𝑈 to the
external work 𝑊𝑒 remains same as before.
• This method permits an assessment of deflections caused by bending
in relation to that caused by shear.
𝑈bending = 0
𝐿 𝑀2 𝑑𝑥
2𝐸𝐼
𝑈shear = vol
𝜏2 𝑑𝑉
2𝐺
• If force is gradually applied, 𝑊𝑒 = 1
2
𝑃∆ = 𝑈 = 𝑈bending + 𝑈shear

24. Inelastic Deflection of Beams
• Hooke’s law and superposition does not apply to inelastic problems,
since deflections are not linearly related to the applied forces.
• In some cases, piecewise linear solutions for small load or
displacement increments are made until the desired level of load or
displacement is reached. Such stepwise linear calculations are made
with the aid of a computer.
• Alternatively, time-consuming trial-and-error solutions are used to
calculate deflections in indeterminate beams.
• To develop simple solutions for ultimate strength of statically
determinate and indeterminate beams and frames assuming ideal
plastic behavior of material, a relationship between the bending
moment and curvature at a section of a beam must be developed.

25. Part B- Deflections by the Moment-Area Method
• It only determines the deflection due to the flexure of the beam;
deflection due to shear is neglected.
• The necessary theorems are based on the geometry of the elastic
curve and the associated 𝑀 𝐸𝐼 diagrams.
𝑑2 𝑣
𝑑𝑥2
=
𝑑
𝑑𝑥
𝑑𝑣
𝑑𝑥
=
𝑑𝜃
𝑑𝑥
=
𝑀
𝐸𝐼
Fig. 12: Interpretation of
a small angle change in
in an element

26. • From Fig. 12(a), quantity 𝑀 𝐸𝐼 𝑑𝑥 corresponds to an infinitesimal
area of the 𝑀 𝐸𝐼 diagram. This area is equal to the change in angle
between two adjoining tangents.
• If the small angle change 𝑑θ for an element is multiplied by a distance 𝑥
from an arbitrary origin to the same element, a vertical distance 𝑑𝑡 is
obtained, Fig. 12(b).
𝑑𝑡 = 𝑥 𝑑θ =
𝑀
𝐸𝐼
𝑥 𝑑𝑥
𝐴
𝐵
𝑑θ = θ 𝐵 − θ 𝐴 = ∆θ 𝐵 𝐴 = 𝐴
𝐵 𝑀
𝐸𝐼
𝑑𝑥
θ 𝐵 = θ 𝐴 + ∆θ 𝐵 𝐴
• In performing the summation, areas corresponding to the positive
bending moments are taken positive and vice versa.

27. Fig. 13: Relationship between the 𝑀 𝐸𝐼 diagram and the elastic curve
• If the sum of the areas between any two points such as 𝐴 and 𝐵 is
positive, the tangent on the right rotates in the counterclockwise
direction; if negative the tangent on the right rotates in a clockwise
direction; Fig. 13(b).

28. • The quantity 𝑑𝑡 in Fig. 13(b) is due to the effect of curvature of an
element. By summing this effect for all elements from 𝐴 to 𝐵, vertical
distance 𝐴𝐹 is obtained.
• It is termed as the tangential deviation of a point 𝐴 from a tangent at
𝐵 and will be designated 𝑡 𝐴 𝐵.
• Second moment area theorem:
𝑡 𝐴 𝐵 = 𝐴
𝐵
𝑑θ 𝑥 = 𝐴
𝐵 𝑀
𝐸𝐼
𝑥 𝑑𝑥
𝑡 𝐴 𝐵 = Φ 𝑥
• Where, Φ is the total area of the 𝑀 𝐸𝐼 diagram between the two
points considered and 𝑥 is the horizontal distance to the centroid of
this area from 𝐴.

29. Fig. 14: Interpretation of signs for tangential deviations
• By analogous reasoning, the deviation of a point 𝐵 from a tangent at
𝐴 is
𝑡 𝐵 𝐴 = Φ𝑥1
𝑥1 is measured from the vertical line through point 𝐵; Fig. 14(b).

30. • Since no deflection is possible at a pinned or a roller support, the
elastic curve is drawn passing through such supports.
• At a fixed support, neither displacement nor rotation of the tangent to
the elastic curve is permitted, so the elastic curve must be drawn
tangent to the direction of the unloaded axis of the beam.
Statically Indeterminate Beams
• Statically indeterminate beams can readily be solved for unknown
reactions using the moment-area method by employing
superposition.

31. • It is recognized that restrained and continuous beams differ from
simply supported beams mainly by the presence of redundant
moments at the supports.
• Bending moment diagrams for these beams may be considered to
consists of two independent parts- one part for the moment caused
by all of the applied loading on a beam assumed to be simply
supported, and other part for the redundant end moments.
• The effect of redundant end moments is superposed on a beam
assumed to be simply supported.
• Application of a redundant moment at an end of a simple beam
results in a triangular-shaped moment diagram, with a maximum at
the applied moment, and a zero ordinate at the other end.

32. • Likewise, when end moments are present at both ends of a simple
beam, two triangular moment diagrams superpose into a trapezoidal-
shaped diagram.
• The known and the unknown parts of the bending-moment diagram
together give a complete bending-moment diagram. This whole
diagram can then be used in applying the moment-area theorems to
the continuous elastic curve of a beam.
• The geometric conditions of a problem, such as the continuity of the
elastic curve at the support or the tangents at built-in ends that
cannot rotate, permit a rapid formulations of equations for the
unknown values of the redundant moments at the supports.

33. • An alternative method of determining the redundant reaction
employs a procedure of plotting the bending-moment diagrams by
parts.
• In applying this method, only one of the existing fixed support is left in
place, creating a cantilever.
• Then separate bending diagrams for each one of the applied forces as
well as for the unknown reactions at the unsupported beam end are
drawn.
• The sum of all these moment diagrams for the cantilever make up the
complete bending-moment diagram.
• In either method, for beams of variable flexural rigidity, the moment
diagrams must be divided by the corresponding 𝐸𝐼’s.