document30

Problem set:[Professor Video1][Professor Video2] [Faculty Video][Professor

Note][Faculty Note]

1. A simply supported beam, 2 in wide by 4 in high and 12 ft long is subjected to a concentrated load

of 2000 lb at a point 3 ft from one of the supports. Determine the maximum fibre stress and the

stress in a fibre located 0.5 in from the top of the beam at mid span.[Ans: fmax = 10,125 psi, f =

5,062.5 psi]

2. A flat steel bar, 1 inch wide by ¼ inch thick and 40 inches long, is bent by couples applied at the

ends so that the midpoint deflection is 1.0 inch. Compute the stress in the bar and the magnitude of

the couples. Use E = 29 × 106 psi.[Ans: f = 18.1 ksi, M = 188.3 lb.in]

Strength of Materials/ Unit 8/ Module 3 Symmetric

Beam Bending

Subject/Unit Name/Module Name http://10.20.3.1:8080/engineering/I_YEAR/EM/M30...

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3. A 50-mm diameter bar is used as a simply supported beam 3 m long. Determine the largest

uniformly distributed load that can be applied over the right two-thirds of the beam if the flexural

stress is limited to 50 MPa. [Ans: w = 690.29 N/m]

4. A concrete floor slab is reinforced by 5/8-in-diameter steel rods placed 1.5 in. Above the lower

face of the slab and spaced 6 in. On centers. The modulus of elasticity is 3.6*106 psi for the

concrete used and 29*106 psi for the steel. Knowing that a bending moment of 40kip. In. Is

applied to each 1-ft width of the slab, determine (a) the maximum stress in the concrete, (b) the

stress in the steel. [Ans: (a) 1.306 ksi (b) 18.52 ksi]



Additional problems:

1. A high strength steel band saw, 20 mm wide by 0.80 mm thick, runs over pulleys 600 mm in

diameter. What maximum flexural stress is developed? What minimum diameter pulleys can be

used without exceeding a flexural stress of 400 MPa? Assume E = 200 GPa. [ Ans: fmax =

266.67 MPa, d = 400mm]

2. A simply supported rectangular beam, 2 in wide by 4 in deep, carries a uniformly distributed load

of 80 lb/ft over its entire length. What is the maximum length of the beam if the flexural stress is

limited to 3000 psi? [Ans: L=11.55 ft]



3. A beam with an S380 x 74 section carries a total uniformly distributed load of 3W and a

concentrated load W, as shown in Fig. Determine W if the flexural stress is limited to 120 MPa.

[Ans: 48.1 kN]

4. A steel bar (Ek = 210GPa) and an aluminum bar (Ea= 70GPa) are bonded together to form the

composite bar shown. Determine the maximum stress in (a) the aluminum, (b) the steel, when the

bar is bent about a horizontal axis, with M= 200N.m. [Ans: (a) 44.5 MPa (b) -80.1 MPa]

Example Problems:

1. A cantilever beam, 50 mm wide by 150 mm high and 6 m long, carries a load

that varies uniformly from zero at the free end to 1000 N/m at the wall. (a)

Compute the magnitude and location of the maximum flexural stress. (b)

Determine the type and magnitude of the stress in a fiber 20 mm from the top of

the beam at a section 2 m from the free end.



Solution:

Thus

(a) The maximum moment occurs at the support (the wall) or at x=6m.

=6000 N.m



(b) At a section 2m from the free and or at x=2m at fibre 20mm from the top of the

beam:

=2000/9 N.m

2. In a laboratory test of a beam loaded by end couples, the fibers at layer AB in Fig

are found to increase 60 × 10–3mm whereas those at CD decrease 100 × 10–3 mm

in the 200-mm-gage length. Using E = 70 GPa, determine the flexural stress in

the top and bottom fibres.



Solution:

x= 0.6(120-x)

x+0.6x = 0.6(120)

1.6x = 72

x= 45mm



From Hooke’s Law

=35 MPa tension

=70 MPa compression

3. A box beam is composed of four planks, each 2 inches by 8 inches, securely

spiked together to form the section shown in Fig. Show that INA = 981.3 in4 If

wo = 300 lb/ft, find P to cause a maximum flexural stress of 1400 psi.

Solution:



Check if the shear at P is positive as assumed

-900 + 0.25P = -900 + 0.25(6680.63)

=770.16 lb



Thus, P = 6680.63 lb

4. A bar obtained by bonding together pieces of steel (Ek = 29*106 psi) and brass

(Eb= 15*106 psi) has the cross section shown fig. Determine the maximum stress

in the steel and in the brass when the bar is in pure bending with a bending

moment M= 40kip.in.

Solution:

The transformed section corresponding to an equivalent bar

made entirely of brass is shown in Fig.b. Since

The width of the central portion of brass, which replaces the

original steel portion, is obtained by multiplying the original

width by 1.933 we have

(0.75 in.)(1.933) = 1.45 in.

Note that this change in dimension occurs in a direction parallel to the neutral axis. The moment of

inertia of the transformed section about its centroid axis is

and the maximum distance from the neutral axis is c=1.5

in. We find the maximum stress in the transformed

section:



The value obtained also represents the maximum stress in the brass portion of the original

composite bar. The maximum stress in the steel portion, however, will be larger than the value

obtained for the transformed section, since the area of the central portion must be reduced by the

factor n = 1.933 when we return from the transformed section to the original one. We thus

conclude that

5. Two steel plates have been welded together to form a beam in the shape of the T

that has been strengthened by securely bolting to it the two oak timbers shown.

The modulus of elasticity is 12.5 GPa for the wood and 200 GPa for the steel.

Knowing that a bending moment M = 50kN.m is applied to the composite beam,

determine (a) the maximum stress in the wood, (b) the stress in the steel along the

top edge.



Solution:

Transformed Section: we first compute the ratio

Multiplying the horizontal dimensions of the steel

portion of the section by n = 16, we obtain a

transformed section made entirely of wood.

Neutral Axis: The neutral axis passes through the

centroid of the transformed section. Since the

section consists of two rectangles, we have

Centroid Moment of Inertia: Using the parallel-axis theorem:



a. Maximum Stress in Wood: The wood farthest from the neutral axis is

located along the bottom edge, where c2 = 0.200 m.

b. Stress in Steel: Along the top edge c1 = 0.120m. From the transformed

section we obtain an equivalent stress in wood, which must be multiplied

n to obtain the stress in steel.

Faculty Note

1. The Basic Kinematic Assumption

In the simplified engineering theory of bending, to establish the relation among

the applied bending moment, the cross-sectional properties of a member, and the

internal stresses and deformations, the approach applied earlier in the torsion

problem is again employed. This requires, first, that a plausible deformation



assumption reduce the internally statically in-determinate problem to a determinate

one; second, that the deformations causing strains be related to stresses through the

appropriate stress-strain relations; and, finally, that the equilibrium requirements of

external and internal forces be met. The key kinematic assumption for the deformation

of a beam as used in the simplified theory is discussed in this section. A generalization

of this assumption forms the basis for the theories of plates and shells.

For present purposes, consider a horizontal prismatic beam having a cross

section with a vertical axis of symmetry; see Fig. 1(a). A horizontal line through the

centroid of the cross section will be referred to as the axis of a beam. Next, consider a

typical element of the beam between two planes perpendicular to the beam axis. In

side view, such an element is identified in the figure as abcd. When such a beam is

subjected to equal end moments Mz acting around the z axis, Fig. 1(b), this beam

bends in the plane of symmetry and the planes initially perpendicular to the beam axis

slightly tilt. Nevertheless, the lines such as ad and bc becoming a’d’ and b' c' remain

straight. This observation forms the basis for the fundamental hypothesis 2 of the

flexure theory. It may be stated thus: plane sections through a beam taken normal to its

axis remain plane after the beam is subjected to bending.



As demonstrated in texts on the theory of elasticity, this assumption is

completely true for elastic, rectangular members in pure bending; if shears also exist,

a small error is introduced. Practically, however, this assumption is generally

applicable with a high degree of accuracy whether the material behaves elastically or

plastically, providing the depth of the beam is small in relation to its span. In this

chapter, the stress analysis of all beams is based on this assumption.

In pure bending of a prismatic beam, the beam axis deforms into a part of a

circle of radius , (rho) as shown in Fig. 1(b). For an element defined by an

infinitesimal angle , the fiber length ef of the beam axis is given as .

Hence,

where the reciprocal of defines the axis curvature k (kappa). In pure bending

of prismatic beams, both and k are constant.



The fiber length gh located on a radius - y can be found similarly. Therefore,

the difference between fiber lengths gh and ef identified here as can be

expressed as follows

By dividing by ds and using Eq. 1, the last term becomes k. Moreover, since the

deflection and rotations of the beam axes are very small, the cosines of the angles

involved in making the projections of and ds onto the horizontal axis are very

nearly unity. Therefore, in the development of the simplified beam theory, it is possible

to replace by du, the axial fibber deformation, and ds by dx. Hence, by dividing

Eq. 2 by ds and approximating by du/dx, which according to Eq. 2 is the normal

strain one has

This equation establishes the expression for the basic kinematic hypothesis for

the flexure theory. However, although it is clear that the strain in a bent beam varies

along the beam depth linearly with y, information is lacking for locating the origin of the

y axis. With the aid of Hcoke's law and an equation of equilibrium, this problem is

resolved in the next section.

2. The Elastic Flexure Formula

By using Hooke's law, the expression for the normal strain given by Eq. 3can be

recast into a relation for the normal longitudinal stress

In this equation, the variable y can assume both positive and negative values.

Two nontrivial equations of equilibrium are available to solve the beam flexure



problem. One of these determines the origin for y: the second completes the solution

for the flexure formula. Using the first one of these equations, requiring that in pure

bending, the sum of all forces at a section in the x direction must vanish, one has

where the subscript A indicates that the summation of the infinitesimal forces

must be carded out over the entire cross-sectional area A of the beam. This equation

with the aid of Eq. 4 can be rewritten as

where the

constants E and k

are taken outside

the second

integral. By

definition, this

integral.

,

where is the

distance from the

origin to the

centroid of an

area A. Since

here this integral

equals zero and

area A is not zero, distance y must be set .equal to zero. Therefore, the z axis must

pass through the centroid of a section'. According to Eqs.3 and 4, this means that

along the z axis so chosen, both the normal strain , and the normal stress , are



zero. In bending theory, this axis is referred to as the neutral axis of a beam. The

neutral axis for any elastic beam of homogeneous material can be easily determined

by finding the centroid of a cross-sectional area.

Based on this result, linear variation in strain is schematically shown in Fig. 1(c).

The corresponding elastic stress distribution in accordance with Eq. 4 is shown in Fig.

1(d). Both the absolute maximum strain and the absolute maximum stress

occur at the largest value of y.

Alternative representations of the elastic bending stress distribution in a beam

are illustrated in Fig. 2. Note the need for awareness that the problem is three-

dimensional, although for simplicity, two-dimensional representations are generally

used. The locus of a neutral axis along a length of a beam defines the neutral surface,

as noted in Fig. 3.

To complete the derivation of the elastic flexure formula, the second relevant

equation of equilibrium must be brought in: the sum of the externally applied and the

internal resisting moments must vanish, i.e., be in equilibrium. For the beam segment

in Fig. 4(a), this yields

A negative sign in front of the

integral is necessary because the

compressive stresses , develop a

counter clockwise moment around the z

axis. The tensile stresses below the

neutral axis, where y's have a negative

sign, contribute to this moment in the

same manner. This sign also follows

directly from Eq. 4. From a slightly

different point of view, Eq. 7 states that the



clockwise external moment Mz is balanced by the counter clockwise moment

developed by the internal stresses at a section. Recasting Eq. 7 into this form, and

recognizing that E and k are constants,

In mechanics, the last integral, depending only on the geometrical properties of

a cross-sectional area, is called the rectangular moment of inertia or second moment

of the area A and will be designated in this text by I. It must be found with respect to the

cross section's neutral (centroid) axis. Since I must always be determined with respect

to a particular axis, it is often meaningful to identify it with a subscript corresponding to

such an axis. For the case considered, this subscript is z, i.e.,

With this notation, Eq. 6-8 yields the following result:

This is the basic relation giving the curvature of an elastic beam subjected to a

specified moment.



By substituting Eq. 10 into Eq. 4, the elastic flexure formula 5 for beams is

obtained.

The derivation of this formula was carried out with the coordinate axes shown in

Fig. 5(a). If the derivation for a member having a doubly symmetric cross section were

done with the coordinates shown in Fig.5 (b), the expression for the longitudinal

stress would read

The sign reversal in relation to Eq. 11 is necessary because a positive My

causes tensile stresses for positive z's.

Application of these equations to biaxial bending as well as an extension of the

bending theory for beams with unsymmetrical cross sections is considered in Sections

11 and 14. In this part of the chapter, attention is confined to beams having symmetric

cross sections bent in the plane of symmetry. For such applications, it is customary to

recast the flexure formula to give the maximum normal stress directly and to



designate the value of by c. It is also common practice to dispense with the

sign as in Eq. 11 as well as with subscripts on M and I. Since the normal stresses must

develop a couple statically equivalent to the internal bending moment, their sense can

be determined by inspection. On this basis, the flexure formula becomes

In conformity with the above practice, in dealing with bending of symmetric

beam sections, the simplified notation of leaving out z subscripts in Eq. 11 on M and I

will be employed often in this text.

The flexure formula and its variations discussed before are of unusually great

importance in applications to structural and machine design. In applying these

formulas, the internal bending moment can be expressed in newton-meters [N.m] or

inch-pounds [in-lb], c in meters [m] or inches [in], and I in m 4 or in 4. The use of

consistent units as indicated makes the units of or: [N.m][m]/[m 4] = N/m

2 = Pa, or [in-lb]

[in]/[in 4] =[lb/in

2] = psi, as to be expected.

It should be noted that as given by Eqs. 11 or 12 is the only stress that

results from pure bending of a beam. Therefore, in the matrix representation of the

stress tensor, one has

In concluding this discussion, it is interesting

to note that due to Poisson’s ratio, the compressed

zone of a beam expands laterally; 6 the tensile



zone contracts. The strains in the y and z directions

are , where is

given by Eq. 11. This is in complete agreement with

the rigorous solution. Poisson's effect, as may be

shown by the methods of elasticity, deforms the

neutral axis into a curve of large radius; and the

neutral surface becomes curved in two opposite

directions; see Fig. 6. In the previous treatment, the

neutral surface was assumed to be curved in one

direction only. These interesting details are not significant in most practical problems.

Procedure Summary and Extensions

The same three basic concepts of engineering mechanics of solids that were

used in developing the theories for axially loaded bars and circular shafts in torsion

are used in the preceding derivation of flexure formulas. These may be summarized

as follows:

1. Equilibrium conditions (statics) are used for determining the internal

resisting bending moment at a section.

2. Geometry of deformation (kinematics) is used by assuming that plane

sections through a beam remain plane after deformation. This leads to

the conclusion that normal strains along a beam section vary linearly

from the neutral axis.

3. Properties of materials (constitutive relations) in the form of Hooke's law

are assumed to apply to the longitudinal normal strains. Poisson effect of

transverse contraction and expansion is neglected.



The first two of the enumerated concepts remain fully applicable. Only the third,

dealing with the mechanical properties of materials must be modified. As an example

of a change necessary for such cases consider the beam having the cross section

shown in Fig. 7(a). This beam is made up of two materials, 1 and 2, bonded together at

their interface. The elastic modules for the two materials are E1 and E

2, where

the-subscripts is identify the material. For the purposes of discussion assume that E2

>E1.

When such a composite beam is bent, as for a beam of one material the strains

vary linearly, as shown in Fig. 7(b). However, the longitudinal stresses depend on the

elastic moduli and are as shown in Fig. 7(c). At the interface between the two

materials, whereas the strain for both materials is the same, the stresses are different,

and depend on the magnitudes of E1 and E

2. The remaining issue in such problems

consists of locating the neutral axis or surface. This can be easily done for beams

having cross sections with symmetry around the vertical axes.

For beams of several different materials, the elastic moduli for each material

must be identified. Let Ei be such an elastic modulus for the i

th material in a composite

cross section. Then Eq. 4 can be generalized to read

Where from Fig. 7(a), In this relation yb is arbitrarily measured

from the bottom of the section, and locates the neutral axis as shown.



Since for pure bending the force Fx at a section in the x direction must vanish,

following the same procedure as before, and substituting Eq. 14 into Eq. 5.

The last expression differs from Eq. 6 only by not placing Ei outside of the

integral. By substituting into Eq. 15, and recognizing that is a

constant,

and

Where the integration must be carried out with appropriate Ei’s, for each

material. This equation defines the modulus-weighted centroid and locates the neutral

axis.

Essentially the same process is used for inelastic bending analysis of beams by

changing the stress-strain relations. The first two of the enumerated basic concepts

remain applicable.

The developed theory for elastic beams of one material is in complete

agreement with the mathematically exact solution 7 based on the theory of elasticity

for pure bending of an elastic rectangular bar. However, even for this limited case, the

boundary conditions at the ends require the surface stresses to be distributed over

the ends as given by Eq. 11. For this case plane sections through a beam remain

precisely plane after bending. However, in usual applications, per Saint-Venant's

principle, it is generally assumed that the stresses, at a distance about equal to the



depth of a member away from the applied moment, are essentially uniform and are

given by Eq. 11. The local stresses at points of force application or change in cross

section are calculated using stress concentration factors. In applications the theory

discussed is routinely applied to any kind of cross section, whether a material is elastic

or plastic.

In conclusion it should be noted that, in all cases in pure bending, the stresses

acting on the area above the neutral axis develop a force of one sense, whereas those

below the neutral axis develop force acting in the opposite direction. An example is

shown in Fig. 7(d) where the tension T is equal to the compression C, and the T - C

couple is equal to the moment Mz. This method of reducing stresses to forces and a

couple can be used to advantage in some problems.

3. Computation of the Moment of Inertia

In applying the flexure formula, the rectangular moment of inertia I of the cross-

sectional area about the neutral axis must be determined. Its value is defined by the

integral of y2 dA over the entire cross-sectional area of a member, and it must be

emphasized that for the flexure formula, the moment of inertia must be computed

around the neutral axis. This axis passes through the centroid of the cross-sectional

area. It is shown in Sections 6-15 and 6-16 that for symmetric cross sections, the

neutral axis is perpendicular to the axis of symmetry. The moment of inertia around

such an axis is either a maximum or a minimum, and for that reason, this axis is one of

the principal axes for an area. The procedures for determining centroid and moments

of inertia of areas are generally thoroughly discussed in texts on statics. 9 However,

for completeness, they are reviewed in what follows.

The first step in evaluating I for an area is to find its centroid. An integration of y2

dA is then performed with respect to the horizontal axis passing through the area's

centroid. In applications of the flexure formula the actual integration over areas is

necessary for only a few elementary shapes, such as rectangles, triangles, etc. Values

of moments of inertia for some simple shapes may be found in texts on statics as well

as. in any standard civil or mechanical engineering handbook (also see Table 2 of the



Appendix). Most cross-sectional areas used may be divided into a combination of

these simple shapes. To find I for an area composed of several simple shapes, the

parallel-axis theorem (sometimes called the transfer formula) is necessary; its

development follows.

Consider that the area A shown in Fig. 8 is a part of a complex area of a cross

section of a beam in flexure. The centroid axis zc for this area is at a distance dz from

the centroid z axis for the whole cross sectional area. Then, by definition, the moment

of inertia Izc of the area A around its zc axis is

On the other hand, the moment of inertia It of the same area A around the z axis

is

By squaring the quantities in the parentheses and placing the constants outside

the integrals,

Here the first integral according to Eq.

17 is equal to Izc, the second integral vanishes

as ye passes through the centroid of A, and the

last integral reduces to A . Hence,



This is the parallel-axis theorem. It can be stated as follows: the moment of

inertia of an area around any axis is equal to the moment of inertia of the same area

around a parallel axis passing through the area's centroid, plus the product of the

same area and the square of the distance between the two axes.

In calculations, Eq. 18 must be applied to each part into which a cross-sectional

area has been subdivided and the results summed to obtain Iz for the whole section,

i.e.,

After this process is completed, the z subscript may be dropped in treating

bending of symmetric cross sections.

4. Applications of the Flexure Formula

The largest stress at a section of a beam is given by Eq. 13, and

in most practical problems, it is this maximum stress that has to be determined.

Therefore, it is desirable to make the process of determining as simple as

possible. This can be accomplished by noting that both I and c are constants for a

given section of a beam. Hence, i/c is a constant. Moreover, since this ratio is only a

function of the cross- sectional dimensions of a beam, it can be uniquely determined

for any cross-sectional area. This ratio is called the elastic section modulus of a

section and will be designated by S. With this notation, Eq. 13 becomes

Or stated otherwise

maximum bending stress = bending moment /elastic section modulus



If the moment of inertia I is measured in in4 (or m

4) and c in m or S is measured

in in3 (or m

3). Likewise, if M is measured in in-lb (or N-m), the units of stress, as before,

become psi (or N/m2). It bears repeating that the distance c as used here is measured

from the neutral axis to the most remote fiber of the beam. This makes I/c = S a

minimum, and consequently M/S gives the maximum stress. The efficient sections for

resisting elastic bending have as large an S as possible for a given amount of

material. This is accomplished by locating as much of the material as possible far from

the neutral axis.

The use of the elastic section modulus in Eq. 21 corresponds somewhat to the

use of the area term A in Eq. 13 (or = P/A). However, only the maximum flexural stress

on a section is obtained from Eq. 21, whereas the stress computed from Eq. 13 holds

true across the whole section of a member.

Equation 21 is widely used in practice because of its simplicity. To facilitate its

use, section moduli for many manufactured cross sections are tabulated in

handbooks. Equation 21 is particularly convenient for the design of beams. Once the

maximum bending moment for a beam is determined and an allowable stress is

decided upon, Eq. 21 may be solved for the required section modulus. This

information is sufficient to select a beam. This is necessary inasmuch as a shear force

which in turn causes stresses, usually also acts at a beam section. The interaction of

the various kinds of stresses must be considered first to gain complete insight into the

problem.

The following two examples illustrate calculations for bending stresses at

specified sections, where, in addition to bending moments, shears are also required

for equilibrium. As shown in the next chapter, the presence of small or moderate

shears does not significantly affect the bending stresses in slender beams. Both

moment and shear frequently occur at the same section simultaneously.

5. Beams of Composite Cross Section

Important uses of beams made of different materials occur in practice. Wooden

beams are sometimes reinforced by metal straps, plastics are reinforced with fibres,



and reinforced concrete is concrete with steel reinforcing bars. The elastic bending

theory discussed before can be readily extended to include such beams of composite

cross section.

Consider an elastic beam of several materials bonded together with a vertical

axis of symmetry as shown in Fig. 9(a). The elastic moduli Ei for the different materials

are given. As for a homogeneous material, the longitudinal extensional strains are

assumed to vary linearly as shown.

in Fig. 9(b). The neutral axis for this section, passing through the modulus-

weighted centroid, is located by the distance yb and can be calculated using Eq. 16.

The stresses shown in Fig. 9(c) follow from Eq. 14. At the interfaces between two

materials, depending on the relative values of their Ei's, a sharp discontinuity in stress

magnitudes arises.

Following the same procedure as in Eq. 7, the resisting bending moment

where the curvature K, being constant for the section, is taken outside the

integral, and defines symbolically the value of the integral in the middle

expression. Hence



and by substituting this relation into Eqs. 3 and 14,

where the last expression is an analogue to Eq. 11, and can be immediately

specialized for a homogeneous beam.

In calculations of bending of composite cross sections, sometimes it is useful to

introduce the concept of an equivalent or transformed cross-sectional area in one

material. This requires arbitrary selection of a reference Ei , defined here as E

ref. Using

this notation the integral in Eq.15, for constant curvature k, can be recast as follows:

Where . Therefore a beam of composite cross section can be

considered to have the mechanical properties of the reference material, provided the

differential areas dA are multiplied by ni, the ratio of Ei to Eref. After transforming a

cross section in this manner, conventional elastic analysis is applicable. In

transformed sections the stresses vary linearly from the neural axis in all materials.

The actual stresses are obtained for the reference material, whereas the stresses in

the other materials must be multiplied by ni.

Professor Note

A member subjected to B.M or S.F undergoes bending.



The resistance to bending is obtained by bending

stresses in the cross section.

The resistance to shear force is obtained by shear

stresses in the cross sections.

In section AB there is no shear force acting only

bending moment. Such is the pure bending load and

stresses under these pure bending or simple bending

are analysed now.

In the section under bending

Shear force is absent, pure bending is causing reaction stresses and strains in the

cross-section.

(1)

For radius of bent R, stress intensity f is proportional to the distance from theneutral axis ‘y’.

About Neutral Axis:

Stress intensity in elemental area

Thrust force

Total moment resistance from beam section

But Moment of inertia of the section, INN about neutral axis

(2)



From (1) and (2) for a beam under bending

This relationship is used in solving many problems in Bending.

Flexural Loading: Shear and Moments in beams

Structures can have loads causing bending stresses

A beam of neutral axis NN’ is subjected to a bending moment M

Following assumptions are made

1. Material of beam is perfectly homogeneous & isotropic

2. Material obeys Hooke’s law

3. Transverse sections AB & CD before bending and after bending are plane

4. Each layer can independently bend, expand or contract, independent of the



layers (unaffected) above and below it

5. The young’s modulus (E) is the same in tension & compression

6. Loads are applied in the plane of bending

Neutral Surface: For the beam under moment load, the upper layer is under

compression and the lower layer is under tension expansion. Somewhere in between

is a layer, the fibers in the layer are neither in compression nor in tension such fibers

form the neutral surface.

Bending Equation Load, Shear & Moment Relationship

M = Bending moment acting at given section

σ = Bending stress

I = Moment of inertia of the section about neutral axis

y = Distance of fiber from the neutral axis

E = Young’s modulus of the beam material

R = Radius of curvature of the beam

Notes:-

1. Neutral axis always passes through it’s centroid



2. In case of symmetric sections, the neutral axis passes through it’s centre ofgravity, geometric centre, y is same on either side.

3. For non symmetric sections, first centroid is found than Ymin and Ymax to

extreme fibers is found. Select Ymax to find bending stress.

Example 1: A pump lever rocking shaft is shown in figure. Loads 25KN and 35KN

are acting at 150mm from the left and right hand bearing respectively. If maximum



allowable stress is 100 Mpa find the diameter of the shaft at the central portion.

RA + RB = 60KN

RB x 950 – 35 x 750 – 25 x 150 = 0

RB = 31.58 KN

RA = 28.42 KN

This load will cause moment on the beam member let the we have the relationof bending moment.

S at point D, MD = RB x 200 = 35.58 x 200 = 6432KN

Example 2:An axle 1m long has a 30KN fly wheel at the center. If the stress due



to bending is not to exceed 60 Mpa, find the diameter of the axle.

By symmetry

Flywheel at the center = 500mm

Moment at center

D = 108.3 = 110mm

Example 3 A beam of uniform rectangular cross section is fixed at one end. Atdistance 300mm from the fixed end an electric motor weighing 400N is fixed. Ifh=2b find the dimensions of the cross section if σallowable is 30 Mpa. (1Mpa= 1

N/mm2)



From bending moment equation

Example 4 A cast iron pulley transmits 10kw at 400rpm. The pulley diameter is1.2m and the cross section of the arm is elliptical, major axis twice the miner axis.If maximum permissible stress is 150Mpa find the dimensions of elliptical cross-section.

10kw at 400rmp be the torque transmitted ;

10



The tangential force on each arm F = 99.2 N

Maximum bending moment of the arm

Sectional modulus of elliptical section

Bending stress

b = 10.8 mm and a = 21.6 mm

Major axis = 2a = 43.2 mm

Minor axis = 2b = 21.6 mm

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