Beam Deflection

Objectives In this course you will learn the following • Importance of computation of deflection. • Computation of deflection using moment area method. 4.1)

Introduction When a structure is subjected to the action of applied loads each member undergoes deformation due to which the axis of structure is deflected from its original position. The deflections also occur due to temperature variations and lack-of-fit of members. The deflections of structures are important for ensuring that the designed structure is not excessively flexible. The large deformations in the structures can cause damage or cracking of non-structural elements. The deflection in beams is dependent on the acting bending moments and its flexural stiffness. The computation of deflections in structures is also required for solving the statically indeterminate structures. In this chapter, several methods for computing deflection of structures are considered. 4.2 Moment Area Method

The moment-area method is one of the most effective methods for obtaining the bending displacement in beams and frames. In this method, the area of the bending moment diagrams is utilized for computing the slope and or deflections at particular points along the axis of the beam or frame. Two theorems known as the moment area theorems are utilized for calculation of the deflection. One theorem is used to calculate the change in the slope between two points on the elastic curve. The other theorem is used to compute the vertical distance (called tangential deviation) between a point on the elastic curve and a line tangent to the elastic curve at a second point. Consider Figure 4.1 showing the elastic curve of a loaded simple beam. On the elastic curve tangents are drawn on points A and B . Total angle between the two tangents is denoted as

. In order to find out

, consider the incremental change in angle

over

an infinitesimal segment located at a distance of from point B . The radius of curvature and bending moment for any section of the beam is given by the usual bending equation

where R is the radius of curvature; E is the modulus of elasticity; I is the moment of inertia; and M denotes the bending moment. The elementary length

and the change in angle

are related as,

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Substituting R from Eq. (4.2) in Eq. (4.1) (4.3) The total angle change which is expressed as

can be obtained by integrating Eq. (4.3) between points A and B

(4.4a) or,

Area of M / EI diagram between A and B

(4.4b)

The difference of slope between any two points on a continuous elastic curve of a beam is equal to the area under the M / EI curve between these points. The distance dt along the vertical line through point B is nearly equal to. (4.5)


Integration of dt between points A and B yield the vertical distance the tangent from point A on the elastic curve. Thus

between the point B and

(4.6) since the quantity M /EI represents an infinitesimal area under the M /EI diagram and distance from that area to point B, the integral on right hand side of Eq. (4.6) can be interpreted as moment of the area under the M/EI diagram between points A and B about point B . This is the second moment area theorem. If A and B are two points on the deflected shape of a beam, the vertical distance of point B from the tangent drawn to the elastic curve at point A is equal to the moment of bending moment diagram area between the points A and B about the vertical line from point B , divided by EI . Sign convention used here can be remembered keeping the simply supported beam of Figure 4.1 in mind. A sagging moment is the positive bending moment diagram and has positive area. Slopes are positive if measured in the anti-clockwise direction. Positive deviation that the point B lies above the tangent from the point A .

indicates

Example 4.1 Determine the end slope and deflection of the mid-point C in the beam shown below using moment area method .

Solution: The M / EI diagram of the beam is shown in Figure 4.2(a). The slope at A , obtained by computing the

can be

using the second moment area theorem i.e.


(clockwise direction) The slope at B can be obtained by using the first moment area theorem between points A and B i.e.

(anti-clockwise) (It is to be noted that the . The negative sign is because of the slope being in the clockwise direction. As per sign convention a positive slope is in the anti-clockwise direction) The deflection at the centre of the beam can be obtained with the help of the second moment area theorem between points A and C i.e.

(downward direction) Example 4.2 Using the moment area method, determine the slope at B and C and deflection at C of the cantilever beam as shown in Figure 4.3(a). The beam is subjected to uniformly distributed load over entire length and point load at the free end Solution: The moment curves produced by the concentrated load, W and the uniformly distributed load,w are plotted separately and divided by EI (refer Figures 4.3(b) and (c)). This results in the simple geometric shapes in which the area and locations of their centroids are known. Since the end A is fixed, therefore, points A and C

. Applying the first moment-area theorem between

( negative sign is due to hogging moment)

(clockwise direction) The slope at B can be obtained by applying the first moment area theorem between points B and C i.e.

–


(clockwise direction)

The deflection at C is equal to the tangential deviation of point C from the tangent to the elastic curve at A (see Figure 4.3(d)). = moment of areas under M / EI curves between A and C in Figures 4.3(b) and (c) about C

(downward direction)


Example 4.3 Determine the end-slopes and deflection at the center of a non-prismatic simply supported beam. The beam is subjected to a concentrated load at the center. Solution: The M/EI diagram of the beam is shown in Figure 4.4(b).

Applying second moment-area theorem between points A and B ,

(clockwise direction) Applying first moment area theorem between A and C .

(anti-clockwise direction) Applying second moment area theorem between A and C .


(downward direction) Example 4.4 Determine the slope and deflection at the hinge of the beam shown in the Figure 4.5 (a).

Solution: The bending moment diagram is shown in Figure 4.5(b).

Since the end A is fixed, therefore, points A and B (refer Figure 4.5(c))

. Applying the first moment-area theorem between

(clockwise direction)

Applying second moment area theorem between points A and B ,

(downward direction)


Applying second moment area theorem between points B and D ,

(anti-clockwise direction) From the first moment area theorem between points B and D

(clockwise direction) Example 4.5 Determine the vertical deflection and slope of point C of the rigid-jointed plane frame shown in the Figure 4.6(a).

Solution: The M/EI and deflected shape of the frame are shown in the Figures 4.6(a) and (b), respectively. As the point A is fixed implying that theorem between points A and B ,

. Applying first moment area

(looking from the left side)

(anti-clockwise direction) Applying second moment area theorem between points B and C

The vertical displacement of point C


(downward direction) Applying first moment area theorem between point B and C

(anti-clockwise direction Objectives In this course you will learn the following

Computation of deflection using conjugate beam method. Conjugate Beam Method The conjugate beam method is an extremely versatile method for computation of deflections in beams. The relationships between the loading, shear, and bending moments are given by

where M is the bending moment; V is the shear; and w ( x ) is the intensity of distributed laod. Similarly, we have the following

A comparison of two set of equations indicates that if M / EI is the loading on an imaginary beam, the resulting shear and moment in the beam are the slope and displacement of the real beam, respectively. The imaginary beam is called as the “ conjugate beam ” and has the same length as the original beam. There are two major steps in the conjugate beam method. The first step is to set up an additional beam, called "conjugate beam,” and the second step is to determine the “ shearing forces ” and “

bending moments ” in the conjugate beam. The loading diagram showing the elastic loads acting on the conjugate beam is simply the bending-moment diagram of the actual beam divided by the flexural rigidity EI of the actual beam. This elastic load is downward if the bending moment is sagging. For each existing support condition of the actual beam, there is a corresponding support condition for the conjugate beam. Table 4.1 shows the corresponding conjugate beam of different types of actual beams. The actual beam as well as the conjugate beam are always in static equilibrium condition . The slope of (the centerline of) the actual beam at any cross-section is equal to the “ shearing force ” at the corresponding cross-section of the conjugate beam. This slope is positive or anti-clockwise if the “ shearing force ” is positive — to rotate the beam element anti-clockwise — in beam convention . The deflection of (the centerline of) the actual beam at any point is equal to the “ bending moment ” of the conjugate beam at the corresponding point. This deflection is downward if the “ bending moment ” is positive — to cause top fiber in compression — in beam convention . The positive shearing force and bending moment are shown below in Figure 4.7.


Table 4.1 Real and Conjugate beams for different structures REAL SRUCTURE

CONJUGATE STRUCTURE


Example 4.6 Determine the slope and deflection of point A of the of a cantilever beam AB of length L and uniform flexural rigidity EI. A concentrated force P is applied at the free end of beam.


Solution: The conjugate beam of the actual beam is shown in Figure 4.8(b). A linearly varying distributed upward elastic load with intensity equal to zero at A and equal to PL/EI at B. The free-body diagram for the conjugate beam is shown in Figure 4.8(c). The reactions at A of the conjugate beam are given by


The slope at A ,

and the deflection

at the free end A of the actual beam in Figure

4.8(d) are respectively, equal to the “shearing force” and the “bending moment” the fixed end A of the conjugate beam in Figure 4.8(d).

Note that points downward because A (i.e. sagging moment) .

at

causes tension in bottom fiber of the beam at

Example 4.7 Determine the slope at A and deflection of B of the beam shown in Figure 4.9(a) using the conjugate beam method.

Solution: The vertical reaction at A in the real beam is given by The bending moment at any point X at a distance x from A is given by

The corresponding conjugate beam and loading acting on it are shown in Figure 4.9(b). The loading on the

beam varies parabolically with maximum value as

.


The slope at A ,

in the original beam will be equal to the shear force at A in the conjugate beam, thus,

(clockwise direction) The deflection of B in the real beam will be equal to the bending moment at B in conjugate beam i.e.

(downward direction

Example 4.8 Determine the deflection at the free end of the beam shown in Figure 4.10 using conjugate beam method and verify by moment area method.

Solution: (a) Conjugate beam method The corresponding conjugate beam and loading are shown in Figure 4.10(b). The loading is upward linearly distributed load with maximum value of

at

B.

Taking moment about point B, the vertical reaction at A in the conjugate beam is given by

The bending moment at C (by taking moment about C ) is given by


(sagging type) Hence, the deflection of point C will be equal to

in the downward direction

(b) Verification by moment-area method

Applying second moment area theorem between points A and B will give the slope at A i.e.

Further, applying moment area theorem between point A and C


(clockwise direction) Since

(downward direction) Objectives In this course you will learn the following

Computation of deflection using principle of virtual work ( PVW ). Application to pin-jointed structure. Application of PVW to beams and frames. Simplified PVW for beams and frames using multiplication of bending moment diagram. Principle of Virtual Work Consider a structural system subjected to a set of forces ( … referred as P force) under stable equilibrium condition as shown in Figure 4.11(a). Further, consider a small element within the structural system and stresses on the surfaces caused by the P forces are shown in Figure 4.11(b) and referred as

.

Let the body undergoes to a set of compatible virtual displacement . These displacements are imaginary and fictitious as shown by dotted line. While the body is displaced, the real forces acting on the body move through these displacements. These forces and virtual displacements must satisfy the principle of conservation of energy i.e. (4.8)

(4.9)


This is the principle of virtual work.

If a system in equilibrium under a system of forces undergoes a deformation, the work done by the external forces ( P ) equals the work done by the internal stresses due to those forces, (

).

In order to use the above principle for practical applications, we have to interchange the role of the forces and displacement. Let the structure acted upon by a virtual force is subjected to real displacements then the Eq. (4.9) can be written as

(4.10) This is the principle of complimentary virtual work and used for computing displacements. Consider a structure shown in Figure 4.12(a) and subjected to P force and it is required to find the displacement of point C in the direction specified. First apply a virtual force at C in the required direction. Next apply the external (real) loads acting on the structures as shown in Figure 4.12(a) with the virtual force remain in the position. The displacement of C in the required ditection be elements deform by an amount

and the internal

. Using Eq. (4.10) (4.11)

The left hand side of Eq. (4.11) denotes the external work done by the virtual force real dispolacement

. On the other hand, the right hand side of Eq. (4.11) represents the internal work

done by the virtual internal element forces d f moving through the displacement Since

moving through the

is arbitrary and for convenience let

.

=1 (i.e. unit load). The Eq. (4.11) can be re-written as

(4.12) where f denotes the internal force in the members due to virtual unit load. The right hand side of Eq. (4.12) will directly provide the displacement of point C due to applied external forces. This method is also known as unit load method. Similarly for finding out a rotation, place as

at any point of a loaded structure, the corresponding Eq. (4.12) will take (4.13)

where interested

denotes the internal force in the members due to virtual unit moment applied in the direction of .


4.4.1

Application to Pin-Jointed Structures

Consider a pin-jointed structure as shown in Figure 4.13 and subjected to external force P 1 , P 2 and P 3 . Let the vertical displacement of point C ,

is required. Under the action of the real external load, let the


axial force in typical member be and therefore, the deformation of the member AE are the length and axial rigidity of typical member). Apply a unit vertical load at C and substituting in Eq. (4.12) leads to

( L and

The basic steps to be followed for finding the displacements of the pin-jointed structure are

1.

Compute the axial force in various members (i.e.

) due to applied external forces.

2.

Compute the axial force in various members (i.e. required displacement of the point.

3.

Compute the product

4. 5.

The summation will provide the desired displacement. The axial force shall be taken as positive if tensile and negative if compressive.

) due to unit load applied in the direction of

for all members.

The positive implies that the desired displacement is in the direction of applied unit load and negative quantity will indicate that the desired displacement is in the opposite direction of the applied unit load.

Example 4.9 Find the horizontal and vertical deflection at joint C of the pin-jointed frame shown in Figure 4.14. AE is constant for all members.

Solution: Calculate forces i.e. force in various members of the truss due to the applied loading. These can be obtained by considering the equilibrium of various joints as marked in Figure 4.14(b).


Table 4.2 Member

For

Length

For L

AB BC CD DA

L L L L

-P 0 -P 0

0 0 -1 0

0 0 PL 0

AC

The computation of

L 0 0 -1 0

0 0 PL 0

0

0

for two desired displacements of pin-jointed frame are shown in Table 4.2.

Horizontal displacement of joint C ,

Vertical displacement of joint C ,


Example 4.10 For the pin-jointed structure shown in the Figure 4.15, find the horizontal and vertical and E =200,000

displacement of the joint D . The area of cross-section, A =500 members.

Solution: The axial rigidity of the members, desired displacements is presented in Table 4.3 Table 4.3

kN. The computation of the

For

Member Length L(m)

for all the

For L

L

AB

2

1920

0

0

BC

2

480

0

0

CD

2

480

0

0

DE

2

120

1

-

EF

2

1080

1

-

CE

2

480

0

0

BF

2

480

0

0

-

The horizontal deflection of

The vertical deflection of

-

= 55.2 mm

= - 8.31 mm

= 8.31 mm

Application to beams and frames In order to find out the vertical displacement of C of the beam shown in Figure 4.16(a), apply a unit load as shown in Figure 4.16(b).


The internal virtual work is considered mainly due to bending and caused due to internal moments under going the rotation due to the applied loading. (internal virtual work done by shearing forces and axial forces is small in comparison to the bending moments and hence ignored). Since the where will take a shape of

is the moment due to applied loading, the Eq. (4.12) for the displacement of C

(4.15) The basic steps to be followed for finding the displacement or slope of a beams and frames are summarized as

1.

Compute the bending moment (i.e.

2.

Compute the bending moment (i.e. displacement or slope.

3.

Compute the integral over the entire members of the beam or frame which will provide the desired displacement. The bending moment shall be taken as positive if sagging and negative if hogging (in case of beams).

4.

) due to applied external forces. ) due to unit load applied in the direction of required

The positive implies that the desired displacement is in the direction of applied unit load and negative quantity will indicate that the desired displacement is in the opposite direction of the applied unit load.

Example 4.11 Determine the slope and deflection of point A of the cantilever beam AB with length L and constant flexural rigidity EI.


Solution: Deflection under the Load - Apply a vertical unit load at point A of the beam as shown in Figure 4.17(b). Consider any point X at a distance of x from A ,

The vertical deflection of point A is given by

Slope at the free end: Apply a unit couple at point A of the beam as shown in Figure 4.17(c). Consider any point X at a distance of x from A .

The slope at A is given by

Example 4.12. Determine mid-span deflection and end slopes of a simply supported beam of span L carrying a udl w per unit length.


Solution: Mid-span deflection : Apply a unit load at mid span as shown in Figure 4.18(b). Consider any point X at a distance of x from A

(0< x < L ) (0< x < L/2 )

( L /2< x < L ) The vertical deflection of point C is given by

End slopes : Applying a unit couple at A as shown in Figure 4.18(c). Consider any point X at a distance of x from A

(0< x < L ) (0< x < L/2 ) The slope at A is given by


Due to symmetry

(anti-clockwise direction)

Example 4.13 Determine vertical deflection and rotation of point B of the beam shown in Figure 4.19(a). The beam is subjected to a couple

at C .

Solution: Vertical deflection of B : Apply a unit load at B as shown in Figure 4.19(b). Consider any point X at a distance of x from C

(0< x < a + b ) (0< x < b ) =

(b < x < a + b )

The vertical deflection of point B is given by

(i.e. in the upward direction) Rotation of B : Apply a unit couple at B as shown in Figure 4.19(c). Consider any point X at a distance of x from C

(0< x < a + b ) (0< x < a + b )

The rotation of point B is given by


Example 4.14. Determine horizontal deflection of C and slope at A of a rigid-jointed plane frame as shown in Figure 4.20(a). Both members of the frame have same flexural rigidity, EI .

Solution: Horizontal deflection Consider AB : ( x measured A )

of C : Apply a unit load C as shown in Figure 4.20(b).

Consider BC : ( x measured C )


The horizontal deflection of point C is given by

Rotation at A : Applying a couple at A as shown in Figure 4.20(d). Consider AB : ( x measured A )

Consider BC : ( x measured C )


The slope at A is given by

Moment diagrams multiplication method for beams and frames Recall the Eq. (4.15) in which the bending deflection of the beams and frames are obtained by the and ) over a length of the members. integration of the two bending moments variations (i.e. However, for a uniform beam section (i.e. EI is constant) such integrals can be readily derived depending

upon the various shapes of the bending moment diagrams. The computation of integral is given in the Table 4.A1. The various steps for this method for finding deflections of the beams and frame are: 1.

Draw the bending moment diagram of given beam or frame due to applied external loading (i.e. diagram).


2.

Draw the corresponding bending moment diagram due to unit load applied in the direction of interested deflection (i.e.

diagram).

with the help of results shown in Table

Compute the desired deflection by computing the 4.A1.

Example 4.15 Determine the deflection under the load and point D of a simply supported beam with overhang as shown in Figure 4.21

Figure 4.21 Solution: Bending moment diagram (i.e. diagram) due to concentrated load W is shown in Figure 4.21(b). Deflection under the Load : Apply a vertical unit load in place of W . The bending moment diagram due to this load is shown in Figure 4.21(c). The vertical deflection under the load is obtained by multiplying the bending moment diagrams of Figure 4.21(b) and (c) and is given by

(refer Table 4.A1)

Deflection of the free end : Apply a unit vertical load acting upward at point D of the beam. The bending moment diagram due to this load is shown in Figure 4.21(d). The vertical deflection under the load is obtained by multiplying the bending moments diagrams of Figure 4.21(b) and (d) and is given by


(refer Table 4.A1)

Example 4.16 Using the diagram multiplication method, determine the deflection under the load and end slopes of a non-prismatic simply supported beam.

Solution: Bending moment (B.M.) diagram (i.e. shown in Figure 4.22(b).

diagram) due to concentrated load W on the beam is

Mid-span deflection : Apply a unit load in the downward direction at C . Deflection at C is given by multiplying the diagrams of Figure 4.22 (b) and (c) as follows

Slope at A : Apply a unit couple at A acting in the clockwise direction and plot the bending moment diagram of the beam as shown in Figure 4.22(d). The slope at A is given by multiplying the diagrams of Figure 4.22 (b) and (d) as follows


(clockwise direction) Slope at B : Apply a unit couple at B acting in the anti-clockwise direction and plot the bending moment diagram of the beam as shown in Figure 4.22(e). The slope at B is given by multiplying the diagrams of Figure 4.22 (b) and (e) as follows

Example 4.17 Using the diagram multiplication method, determine the horizontal displacement and rotaion of point C of the rigid-jointed plane frame shown in Figure 4.23. Both the members of the frame have same EI value.

Solution: The free-body and bending moment diagram (B.M.D.) of the frame due to applied loading are shown in Figures 4.23(b) and (c), respectively.


Horizontal deflection of C : Apply a horizontal force at C as shown in Figure 4.23(d) and plot the bending moment diagram as shown in Figure 4.23(e). The horizontal deflection at C is given by multiplying the diagrams of Figure 4.22 (c) and (e) as follows

Rotation of C : Apply a unit couple at C as shown in Figure 4.23(f) and plot the bending moment diagram as shown in Figure 4.23(g). The slope at C is given by multiplying the diagrams of Figure 4.22 (c) and (g) as follows

(anti-clockwise direction)


Objectives In this course you will learn the following

Deflection by strain energy method Evaluation of strain energy in member under different loading. Application of strain energy method for different types of structure. Deflection by Strain Energy Method The concepts of strain, strain-displacement relationships are very useful in computing energy-related quantities such as work and strain energy. These can then be used in the computation of deflections. In the special case, when the structure is linear elastic and the deformations are caused by external forces only, (the complementary energy U * is equal to the strain energy U ) the displacement of structure in the direction of force expressed by

is

(4.16) This equation is known as Castigliano's theorem. It must be remembered that its use is limited to the calculation of displacement in linear elastic structures caused by applied loads. The use of this theorem is equivalent to the virtual work transformation by the unit-load theorem.

1 Calculation of Strain Energy When external loads are applied on an elastic body they deform. The work done is transformed into elastic strain energy U that is stored in the body. We will develop expressions for the strain energy for different types of loads. Axial Force : Consider a member of length L and axial rigidity AE subjected to an axial force P applied gradually as shown in the Figure 4.24. The strain energy stored in the member will be equal to the external work done by the axial force i.e

(4.17)

Figure 4.24 Member subjected to axial force Bending Moment: Consider a beam of length L and flexural rigidity EI subjected to a general loading as shown in Figure 4.25. Consider a small differential element of length, dx . The energy stored in the small element is given by

(4.18) The total strain energy in the entire beam will be


(4.19)

Figure 4.25 Member under bending Shear Force: The strain energy stored in the member due to shearing force is expressed by

(4.20) where V is the shearing force; and is the shearing rigidity of the member. Twisting Moment: The strain energy stored in the member due to twisting moment is expressed by

(4.21) where T is the twisting moment; and GJ is the torsional rigidity of the member.

Example 4.18 Find the horizontal deflection at joint C of the pin-jointed frame as shown in Figure 4.26(a). AE is constant for all members.

Solution: The force in various members of the frame is shown in Figure 4.26(b). Calculation of strain energy of the frame is shown in Table 4.4.


Table 4.4 Member

Length ( L )

Force ( P )

AB BC BD CD

L L

P P

L

0

0

Horizontal displacement of joint C ,

Example 4.19 A bar of uniform cross-section is bent into a quadrant of circle of radius R . One end of the bent is fixed and other is free. At the free end it carries a vertical load W . Determine the vertical and horizontal deflection at A . Solution:


Vertical displacement of A : The vertical displacement of A is given by

For evaluation of the total strain energy in the system, consider a small element Figure. The bending moment at this element,

as shown in the

. Thus,

Since there is no horizontal force acting at point A , apply a horizontal force, F at A as shown in Figure 4.27(b). From the Castigliano's theorem, the horizontal displacement of A due to applied external load W is given by

The bending moment at the small element horizontal displacement of A

is

. Thus, the


(i.e. deflection is in

direction)

Example 4.20 Determine the deflection of the end A of the beam as shown in Figure 4.28. The flexibility of the spring is

.

Solution: Reactions at support B and C are (upward) and

(downward)

Force in the spring = Reaction, Deflection under the load is given by

where

is the total strain energy stored in the system;

member AB ; is the energy stored in the member BC ; and Strain energy in the spring is given by

is the energy stored in the

= strain energy in the spring.

Consider member AB : ( x measured from A )

Consider member BC : ( x measured from C )


Thus,

The deflection of point A,

Objectives In this course you will learn the following

Bending deflection of beams due to temperature variation. Bending Deflection due to Temperature Variation Consider a beam member (refer Figure 4.29) subjected to temperature gradient the depth of beam such that

over

(4.22) where

= temperature at the top of the beam; and

= temperature at the bottom of the beam.

The deflection of the beam due to temperature variation is shown in Figure 4.29(b). It is assumed that temperature varies linearly through the depth, d and is the coefficient of thermal expansion of the material.


Consider a small element of length dx . The strain at top and bottom of the small elements are

(4.23a)

(4.23b) The curvature of the beam is given by

(4.24) The equation (4.24) can be used for finding out the bending deflection in beams due to temperature variation. If the beam is restrained from rotation, the moment induced in the beam will be given by

(4.25)

The equation (4.25) is obtained by equating the right hand side of equation (4.24) to bending theory.

from the simple

Temperature deflections of a cantilever beam:

Consider a cantilever beam as shown in Figure 4.30 subjected to temperature gradient over the depth. Integrating the equation (4.24)

(4.26)

(4.27)

Boundary conditions: At x = 0, and v = 0 will give the values of arbitrary constants as .The slope and deflection of the free end of the cantilever beam are


(4.28a)

(4.28a) Objectives In this course you will learn the following

Maxwell-Betti Law of reciprocal deflection. Illustrative examples for proving law of reciprocal deflection Maxwell-Betti Law of Reciprocal Deflections Maxwell-Betti Law of real work is a basic theorem in the structural analysis. Using this theorem, it will be established that the flexibility coefficients in compatibility equations, formulated to solve indeterminate structures by the flexibility method, form a symmetric matrix and this will reduce the number of deflection computations. The Maxwell-Betti law also has applications in the construction of influence lines diagrams for statically indeterminate structures. The Maxwell-Betti law, which applies to any stable elastic structure (a beam, truss, or frame, for example) on unyielding supports and at constant temperature, states: The deflection of point A in direction 1 due to unit load at point B in direction 2 is equal in the magnitude to the deflection of point B in direction 2 produced by a unit load applied at A in direction 1. The Figure 4.31 explains the Maxwell-Betti Law of reciprocal displacements in which, the displacement

is equal to the displacement

.


In order to prove the reciprocal theorem, consider the simple beams shown in Figure 4.32. Let a vertical force at point B produces a vertical deflection at point A and point B as shown in Figure 4.32(a). Similarly, in Figure 4.32(b) the application of a vertical force

at point A produces a vertical deflections

and

respectively. Let us evaluate the total work done by the two forces are applied in different order to the zero to their final value. Case 1:

at points A and B , and

when they

applied and followed by

…. (a) Work done when

is gradually applied

s …. (b) Work done when

is gradually applied with

at

in place


Total work done by the two forces for case 1 is

(4.29) Case2:

applied and followed by

(c) Work done when

…. (d) Work done when

is gradually applied

is gradually applied with

in place

Total work done by the two forces for case 2 is

(4.30) Since the final deflected position of the beam produced by the two cases of loads is the same regardless of the order in which the loads are applied. The total work done by the forces is also the same regardless of the order in which the loads are applied. Thus, equating the total work of Cases 1 and 2 give

(4.31) If

, the equation (4.31) depicts the statement of the Maxwell-Betti law i.e.

The Maxwell-Betti theorem also holds for rotations as well as rotation and linear displacement in beams and frames. Example 4.21 Verify Maxwell-Betti law of reciprocal displacement for the direction 1 and 2 of the pin-jointed structure shown in Figure 4.33(a).

Solution: Apply the forces and in the direction 1 and 2, respectively. The calculation of total strain energy in the system is given in Table 4.5.


Table 4.5

Member

Length

AB

L

AC

Since

Force P -(

L

) P1

, hence the Maxwell-Betti law of reciprocal displacement is proved.

Example 4.22 Verify Maxwell-Betti law of reciprocal displacement for the cantilever beam shown in Figure 4.34(a).


Solution: Apply the forces calculated below.

and

in the directions 1 and 2, respectively. The total strain energy stored is

Consider any point X at a distance x from B .

Since

, the Maxwekk-Betti law of reciprical displacement is proved.

Example 4.23 Verify Maxwell-Betti law of reciprocal displacement for the rigid-jointed plane frame with reference to marked direction as shown in Figure 4.35(a). EI is same for both members.


Solution: Apply the forces

and

in the directions 1 and 2, respectively as shown in Figure 4.35(b).

Consider AB : ( x measured from A )

Consider BC : ( x measured from B )

Thus


The displacement in the direction 1 due to unit load applied in 2 is

The displacement in the direction 2 due to unit load applied in 1 is

Since

, proves the Maxwell-Betti law of reciprocal displacements.


Beam Deflection

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