5 Bending Stress

8/14/2019 5 Bending Stress

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DMV 4343JAN ~ JUN `07

DEPARTMENT MANUFACTURING / PRODUCT DESIGN /MOULD / TOOL AND DIE

SEMESTER 4 / 6

COURSE MECHANICS OF MATERIALS DURATION 8 hrsCOURSE CODE DMV 4343 / DMV 5343 REF. NO.

VTOS NAME MISS AFZAN BINTI ROZALIMR RIDHWAN BIN RAMELI

PAGE 22

TOPICBENDING STRESS

SUB TOPIC5.1 Simple Bending Theory5.2 Non-Symmetry Bending5.3 Second Moment of Area5.4 Mohrs Circle5.5 Parallel Axes Theorem5.6 Stress and Deflection

Chapter 5 BENDING STRESS p1

INFORMATION SHEET


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5.1 Simple Bending Theory

In this chapter we continue our study of beams by determining how the stressresultants, the bending moment M(x) and the transverse shear force V(x), are

related to the normal stress and the shear stress at section x. Loads (transverse forces or

couple) applied to a beam cause it to deflect laterally, as illustrated in

Figure 5.1.

FIGURE 5.1 Transverse deflection of a beam

This lateral deflection, or bending, changes the initially straight longitudinal

axis of the beam into a curve that is called the deflection curve, shown dashed in

Figure 5.1. By relating the curvature of the deflection curve to the bending moment

M, we can determine the distribution of the normal stress x. You will discover that thisderivation includes all three of the fundamental types of equations: geometry of

deformation (in the strain-displacement analysis), material behavior (in the stress-

strain relations), and equilibrium (in the definition of stress resultants and in relating

stress resultants to the external loads and reactions). ,

Beam-Deformation Terminology.

To simplify this study of beams, we initially consider only straight beams that have a

longitudinal plane of symmetry (LPS), and for which the loading and support aresymmetric with respect to this plane, as illustrated in Figure 5.2.

FIGURE 5.2 Illustration of some beam-deformation terminology

Under these conditions, this longitudinal plane of symmetry is the plane of bending.


REF NO. :

PAGE :


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Pure Bending.

Let us begin our analysis of beams by examining the deformation of a uniform beam

segment subjected to pure bending, that is,

a segment whose material properties are constant along its length, and

for which M(x) is constant.

If equal couples MOare applied to the ends of an otherwise unloaded segment of beam, as

in Figure 5.3, the moment is constant along the segment and the segment is said to be in

pure bending.

(a) The cross

section before

deformation

(b) The undeformed beam (c) Plane section remain plane

FIGURE 5.3 A uniform beam segment undergoing pure bending

We can see that changes in length (strain, ), occurs at D*G* (compression) and A*E*

(tension). However, BF retains it original length; which is called neutral surface (NS).

Lines ABD and EFG in Figure 5.3b represent the edges of typical cross sections in the

undeformed beam; lines A*B*D* and E*F*G* in Figure 5.3c represent these same cross

sections after deformation, as seen from the front face of the beam.

From Figure 5.3c we can determine the following characteristics of a uniform beam

undergoing pure bending:

1. Since M(x) = MO = const, pure-bending deformation of a beam is uniform along the

length of the segment undergoing pure bending: so whatever happens at a typical

cross-sectionABD also happens at section EFG.

2. Pure bending has front-to-back symmetry. The only way that this can be possible is

for all cross sections like ABD and EFG, to remain plane and remain

perpendicular to the deflection curve.

3. In summary, when a beam undergoes pure bending, its deflection curve forms a



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circular arc, and its cross sections remain plane and remain perpendicular to

the deflection curve. Experiments show that this is, indeed, the way that beams

deform whensubjected to pure bending.

Strain-Displacement Analysis

Let us continue our analysis of the deformation of a uniform beam segment subjected to

pure bending, that is, a segment for which M(x) is constant.

Kinematic Assumptions of Bernoulli-Euler Beam Theory. The previous discussion of

pure bending can be summarized in the following four deformation assumptions of

Bernoulli-Euler beam theory:

1. The beam possesses a longitudinal plane of symmetry, and is loaded and supported

symmetrically with respect to this plane. This plane is called the plane of bending.

2. There is a longitudinal plane perpendicular to the plane of bending that remains free

of strain (i.e., x = 0) as the beam deforms. This plane is called the neutral surface

(NS). The intersection of the neutral surface with a cross section is called the neutral

axis (NA) of the cross section. The intersection of the neutral surface with the plane

of bending is called the axis of the beam: it forms the deflection curve of the

deformed beam.

3. Cross sections, which are plane and are perpendicular to the axis of the

undeformed beam, remain plane and remain perpendicular to the deflection

curve of the deformed beam.

4. Deformation in the plane of a cross section (i.e., transverse strains y and z) may

be neglected in deriving an expressionforthe longitudinal strain x.

The third of the preceding assumptions is crucial to the development of the Bernoulli-Euler

beam theory; it leads to a practical theory of bending ofbeams that is comparable to the

theories of axial deformation and torsion covered previously.

Strain-Displacement Analysis; Longitudinal Strain.Because of Assumptions 1 and 4, the fibers in any plane parallel to the xy plane behave

identically to the corresponding fibers that lie in the xy plane (i.e., the plane of bending).

Therefore, bending deformation is independent of the coordinate z, so the drawing in

Figure 5.4 represents the deformation of any plane in the beam parallel to the xy plane.

Using Figure 5.4 and the preceding four deformation assumptions, we can develop an

expression for the extensional strain x, in a longitudinal fiberatcoordinates (x, y, z) in the

beam.



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(a) The undeformed beam segment (b) The deformed beam segmentFIGURE 5.4 The geometry of deformation of a beam segment, showing the plane of

bending.

In Figure 5.4a, points A and P lie in the cross-sectional plane at coordinate x in the

undeformed beam: similarly, points B and Q lie in lie in the cross-sectional plane at (x + x)

in the undeformed beam. Line segment PQ is parallel to the x axis and lies at distance +y

above the NS (xz plane). Therefore, in the undeformed beam the infinitesimal fibers AB and

PQ are both of Iength x. From Assumption 3, points A* and P* lie in a plane that is

perpendicular to the neutral surface of the deformed beam, and points B* and Q* lie in a

plane that also is perpendicular to the deformed neutral surface. According to Assumption 2,

however, the length ofA*B*, a fiber lying in the neutral surface, is unchanged; that is, the

length of A*B* is still x, as indicated in Figure 5.4b. Finally, by virtue of Assumption 4, A*P*

= AP = y, and B*Q* = BQ = y. Figure 5.4 therefore, embodies all four deformation

assumptions of Bernoulli-Euler beam theory, so we can use it in deriving an expression for

the extensional strain of a longitudinal fiber.

From the general definition of extensional strain, we can express the extensional strain in the

longitudinal fiber PQ as

x x (x, y, z) =lim

(P*Q* -

PQ)=

lim x* - x

Q P PQ x 0 x

Considering A*B* to be the arc of a circle of radius (x) subtending an angle *, we get

A*B* =

x= *

Similarly,

P*Q* = x* = ( y) *

Combining these two equations, we get



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I = A y2 dA :

Figure 5.6

Iz = A y2 dA

So,

=

h/2y2 bdy

-h/2

= by3 h/2

3 -h/2

Iz =bh3

12

From Figure 5.6;

dA = bdy

Area moment of inertia

Recall

M = - Ay

dA

and = - Ey /

M = - A y (- Ey / ) dA

I= E / - A y2 dA

M =EI

= EIk Where EI = flexural rigidity

To get rid of radius of curvature which requires longer calculation to find, we have tocombine these two equations:

M =EI

So,

E =M

I

x =- E y

= - M yI

x =- M y

Flexural FormulaI

S is elastic section modulus where S = I / c,

c = h / 2

then

S= bh2 / 6



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Then max at ymax = c

max =

- M

c =

- M

Flexural Formula

I S

Where

max =The maximum normal stress in the member, which occurs at a point

on the cross-sectional area farthest away from the neutral axis

M =

The resultant internal moment, determined from the method of

sections and the equations of equilibrium, and computed about the

neutral axis of the cross section

I =The moment of inertia of the cross sectional area computed about

the neutral axis

c =The perpendicular distance from the neutral axis to a point farthest

away from the neutral axis, where max acts

EXAMPLE 5.1

A simple cast iron (E = 175 GPa) beam of rectangularcross section carries a load of 5 kN/m. Determine:

(a) The maximum tensile and compressive stresses

at the midspan,

(b) The normal stress and strain at a point A, and

(c) The radius of curvature of the beam at B.

Solution

The neutral axis z passes through the centroid C and

I =bh3

=(0.08m)(0.12m)3

= 11.52 x 10-6 m412 12

The section modulus of the cross sectional area

S =bh2

=(0.08m)(0.12m)2

= 192 x 10-6 m3

6 6



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a) At the midspan, the bending moment is M = 5 (4)2 / 8 = 10 kN.m. Because M is

positive, the maximum tensile and compressive stresses occur at the bottom and top

of fibers, respectively:

max =My

=(10 kN.m)(0.06m) = 52.1 MPa

I 11.52 x 10-6 m4

Or

These stresses act on infinitesimal elements at D and E

b) At a section through point A, the bending moment in M = 10(1) 5(1)2 / 2 = 7.5 kN.m,

and we have

A =- MyA =

- (7.5 kN.m) (-

0.02m) = 13 MPa

I 11.52 x 10-6 m4

The normal strain at point A is thus

A =A =

(13 x 106 N/m2)= 74.3

E 175 x 109 N/m2

c) The radius of curvature,

= - yA =

-(-0.02

m) = 269 mA 74.3


max =M

=(10 kN.m)

= 52.1 MPaS 192 x 10-6 m4


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5.2 Non-Symmetry Bending

In previous section, we have been considering flexural stress and strain in beams whose

cross-sectional shape and whose loading and support conditions produce bending that is

confined to a longitudinal plane of symmetry (LPS) of the beam. Where we assume:

a) the deflection of [be beam can be characterized by a deflection curve in the LPS

b) there is no tendency of the beam to twist

However, we also need to be able to analyze the behavior of beams that are not loaded and

supported in this simple manner.

(a) Components of load in two planes

of symmetry

(b) Bending moments due to an inclined load

(positive My and positive Mz shown)FIGURE 5.7 A doubly symmetric beam with inclined loading

If a beam is subjected to a non symmetry loadings such as in Figure 5.7, then the stress at

the defined center C should be contributed by the total of stress in the other two axes on the

plane which the load acting. Where, stress at x, x

x =Myz

-Mzy Flexural Formula

(non-symmetry bending)Iy Iz

This is illustrated in Figure 5.8 below.

FIGURE 5.8 Flexural stress due to inclined loading of a doubly symmetric beam.

It is noted that in Figure 5.8c that the stress at center C is zero. So;x = Myz* - Mzy* = 0



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EXAMPLE 5.2

A 1600 lb.in couple is applied to a wooden beam, of

rectangular cross section 1.5 x 3.5 in., in a plane forming

an angle of 30 with the vertical. Determine:

(a) The maximum stress in the beam

(b) The angle that the neutral surface forms with the

horizontal plane

Solution

a) The maximum stress in the beam, max

max = 1 + 2

=Mzy +

MyzIz Iy

We need to find My and Mz

Mz = M cos = 1.6 kNm (cos 30)= 1.386 kNm

My = M sin = 1.6 kNm (sin 30)= 0.8 kNm

Maximum stress

1 = Mz y =

(1.386 kNm)

(0.175m) = 452.6 kPa

Iz 535.9 x 10-6 m4

2 =My z =

(0.8 kNm)(0.075m)= 609.5 kPa

Iy 98.44 x 10-6 m4

max = 1 + 2= 452.6 kPa + 609.5 kPa= 1.062 Mpa

The distribution of the stresses

across the section.


Iz =bh3

=(0.15m)(0.35m)3

= 535.9 x 10-6 m412 12

Iy =bh3

=(0.35m)(0.15m)3

= 98.44 x 10-6 m412 12


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c) Angle of Neutral Surface with the horizontal plane.

tan

= (

Iz) tan

Iy

= 535.9 x 10-6 m4 tan

3098.44 x 10-6 m4

= 3.143

= tan -1 3.143= 72.4



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5.3 Second Moment of Area

The moment-area method provides a semigraphical technique for finding the slope and

displacement at specific points on the elastic curve of a beam or shaft.

We have to assume:

1) The beam is initially straight,

2) It is elastically deformed by the loads,

3) Slope and deflection of the elastic curve is very small,

4) Deformation are caused by bending.

Moment area method is based on 2 theorems

- Theorem 1 : First Moment Area

- Theorem 2 : Second moment of Area

The vertical deviation of the tangent at a point (A) on the elastic curve with respect to the

tangent extended from another point (B) equals the moment of area under the M/EI diagram

between these two points (A and B). This moment is computed about point (A) where the

vertical deviation (tA/B) is to be determined.

Consider the following beam:

(a)

(c) dt is the vertical deviation of the tangent

on each side of the differential element dx.(b)

FIGURE 5.9 A beam experiencing distributed load

From Figure 5.9, we are ought to find tA/B as shown in (c). Mathematically we know that, s

= r;

So, ds = x d = dt

From the 3rd

assumption, we assume the slope and deflection of the elastic curve is verysmall, which mean ds = dt = x d



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From Theorem 1, which is not covered in this lecture:

d =M

dxEI

Hence

dt = d =M

dxEI

t = xM

dxEI

So,

tA/B = B

xM

dxA EI

Since the centroid of an area is found from x dA = x dA, and M/EI dx represent the areaunder the M/EI diagram, (Figure 5.9 b), we can also write:

tA/B = x B M

dxA EI

Where:

X is the distance from A to the centroid of the area under the M/EI diagram between A

and B.

Note that:

tA/B = tB/A



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EXAMPLE 5.3

Determine the displacement of points B and C of

the beam shown. EI is constant.

Solution:Construct the M/EI diagram

B = tB/A

C = tC/A

Moment Area Theorem

Consider area from A to B:

B = tB/A = (L

) [( -MO )(

L)] = -

MOL2

4 EI 2 8EIConsider area from A to C:

C = tC/A = (L

) [( -MO )( L )] = -

MOL2

2 EI 2EI

Both tB/A and tC/A is negative, showing that point B and C is below tangent at A.



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5.4 Mohrs Circle

It is proven that:

R = (Ix - Iy )2 + Ixy22

Procedure to construct Mohrs Circle

1) Compute Ix, Iy and Ixy

2) Construct the circle,

- abscissa represents the moment of inertia I

- ordinate - represents the product of inertia Ixy

3) Determine the coordinate of center of the circle, C from origin where

C =(Ix + Iy)

2

4) Plot reference point A having coordinate A (Ix, Ixy)

- Ix is always positive

- Ixy will be either positive or negative

5) Connect reference point A to the center C, where AC = radius of circle, R

6) Determine AC by trigonometry

7) Determine Imin and Imax : points that intersect abscissa (product of inertia Ixy = 0)

8) Determine 2p1

FIGURE 5.10 Mohrs Circle showing the important point.

EXAMPLE 5.4



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Use Mohrs circle to determine the principal moments

of inertia for the beams cross-sectional area shown

below, with respect to axes passing through the

centroid.

Solution

Compute Ix, Iy and Ixy

Using parallel axes theorem (Section 5.5), we get

Ix = 2.90 x 109 mm4

Iy = 5.60 x 109 mm4

Ixy= -3.00 x 109 mm4

C =(Ix + Iy) =

(2.90 + 5.60)= 4.25

2 2

A (Ix, Ixy) ; A (2.90, -3.00)

(a)

(b)

Imax = C + R = 4.25 + 3.29 = 7.54 x 109 mm4

Imin = C - R = 4.25 - 3.29 = 0.960 x 109 mm4

From figure (b):

2p1 = 180 - tan

-1

(|BA|/|BC|)= 180 - tan-1 (|3.00|/|1.35|)

= 114.2

p1 = 57.1

The major principal axis (for Imax = 7.54 x 109 mm4) is

therefore oriented at an angle p1 = 57.1, measured

counterclockwise, from the positive x axis. The minor

axis is perpendicular to this axis. The results are shown

in Figure (a).

(c)


R = CA = (Ix - Iy )2 + Ixy22

= (1.35)2 + (-3.00)2= 3.29


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5.5 Parallel Axes Theorem

From Figure 5.10 below, it is known that the moment of inertia is

Ix = A y2 dA

Iy = A x2 dA

FIGURE 5.10

However, if the moment of inertia for an area is known about a centroidal axis we can

determine the moment of inertia of the area about a corresponding parallel axis using the

parallel axis theorem. Consider the following Figure 5.11.

FIGURE 5.11

In this case, a differential element dA is located at an arbitrary distance y from the centroidal

x axis, whereas the fixed distance between the parallel x and x axes is defined as d y. Since

the moment of inertia of dA about the x axis is dIx = (y + dy)2 dA, then for the entire area,

Ix = A (y + dy)2 dA = A y2 dA + 2dy A ydA + dy2 A dA= Ix = 0

Finally we get

Ix = Ix + A dy2

Iy = Iy + A dx2



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FIGURE 5.12

The product of inertia

Ixy = A xy dA

FIGURE 5.13

Consider the shaded area shown in Figure 5.13, where x and y represents a set of

centroidal axes, and x and y represent a corresponding set of parallel axes. Since the

product of inertia of dA with respect to the x and y axes is dIxy = (x + dx)(y + dy) dA, then for

the entire area,

Ixy = A (x + dx)(y + dy)dA = A xy dA + dx A ydA + dxdy A dA

The first term on the right represents the product of inertia of the area with respect to the

centroidal axis, Ixy. The second and third terms are zero since the moments of area are

taken about the centroidal axis. Realizing that the fourth integral represent the total area A,

we therefore have the final result

Ixy = Ixy + A dxdy



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EXAMPLE 5.5

A T beam with dimension shown is given moment at the cross section, M = 4 kMn.

Determine

a) The location of the neutral axis of the cross section

b) The moment of inertia with respect to the neutral axis, and

c) The maximum tensile stress and the maximum compressive stress on the

cross section

Solution

a) Locate the neutral axis

We can use first moment of area method where

y A = y1 A1 + y2 A2 = (0.55m)(0.5m)(0.1m) + (0.25m)(0.1m)(0.5m)= 0.04m3

Where

A = A1 + A2 = (0.5m)(0.1m) + (0.1m)(0.5m) = 0.1m2

Then

y =yA

=0.04m3

= 0.4mA 0.1m2

b) The moment of inertia with respect to the neutral axisThe moment of inertia of a rectangle about an axis through its own centroid is



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Ix =bh3

12

And, from parallel axes theorem, the moment of inertia about an axis through C

parallel to the axis through the centroid C is

I = Ix1 + A dy12 + Ix2 + A dy22

=(0.1m)(0.5m)3

+ (0.5m)(0.1m)(0.15m)2 +(0.5m)(0.1m)3

+ (0.1m)(0.5m)(0.15m)212 12

= 3.33 x 10-3 m4

c) The maximum tensile stress and the maximum compressive stress on the cross

section

x =- My

I

(max)C =My

=-(4kNm)(0.2m)

= -240 kPaI 3.33 x 10-3 m4

(max)T =My

=-(4kNm)(-0.4m)

= 480 ksiI 3.33 x 10-3 m4



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5.6 Stress and Deflection

The fundamental assumptions of the technical theory for slender beams are based

upon the geometry of deformation. We can state them as follows:

1. The deflection of the beam axis is small compared with the span of the beam. The

slope of the deflection curve is therefore very small and the square of the slope is

negligible in comparison with unity. If the beam is slightly curved initially, the

curvature is in the plane of the bending, and the radius of curvature is large in

relation to its depth (p > 10/z).

2. Plane sections initially normal to the beam axis remain plane and normal to that

axis after bending (for example, a-a). This means that the shearing strains y is

negligible. The deflection of the beam is thus associatedprincipallywith the axial or

bending strains ex. The transverse normal strains ey and the remaining strains

(z, xz,yz) may also be ignored.

3. The effect of the shearing stresses xy on the distribution of the axial or bending

stress x is neglected. The stresses normal to the neutral surface, y, are small

compared with xand may also be omitted. This supposition becomes unreliable in

the vicinity of highly concentrated transverse loads.

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