# BENDING STRESSES CLASS NOTES FOR MECHANICAL ENGINEERING

**BENDING STRESSES CLASS **

**NOTES FOR MECHANICAL **

**ENGINEERING**

### Bending and hence bending stresses

### are caused by a load perpendicular

### to the axis of the beam. This

### perpendicular load is called the

### transverse load or the lateral load.

### Bending is found in all types of beams,

### chimneys and dams. Bending is best

### resisted by moment of inertia of the beam.

### Bending theory is applicable for beam

### of one material. Composite beam is

### of two or more materials. Convert

### composite beam into a beam of

### one material. Stresses in a composite

### beam are found in one material.

### Find stresses in the other material

### using the same strain at the

### common surface.

**PURE BENDING EQUATION**

**M/I = σ/y = E/R**

**Where M is the bending moment applied ( Maximum bending moment from BMD), Nm**

**I = moment of inertia of the area about the centroidal axis, m**^{4}

^{4}

**σ is tensile or compressive stress at distance ‘y’ from the neural axis, MN/m**^{2}

^{2}

**y is the distance of a fiber from the neutral axis, m**

**E = Young’s modulus MN/m**^{2}

^{2}

**R is the radius of curvature, m**

**BENDING STRESSES **

**When both tensile and compressive stresses occur at the same time. ****Both of these are variable in magnitude. Maximum stresses occur on the extreme fibers. Zero stress occurs in the neutral fiber.**

**VARIATION OF BENDING STRESS AND BENDING STRAIN**

**Linear variation. Stress and strain are zero at the neutral axis**

**Stress and strain are maximum at the outermost fibers.**

**Bending stresses are in the direction of length.**

**Bending strains are in the direction of length.**

**Stress and strain are zero at the two extremities of the horizontal diameter of a round beam.**

**Stress and strain are maximum at the two extremities of the vertical diameter of a round beam.**

**Loads are perpendicular to length. Such loads are Lateral Loads or Transverse loads.**

**To resist maximum bending, depth of a rectangular beam is greater than the width. It gives higher value of the moment of inertia. Moment of inertia is a resistance to bending.**

### Fig. Variation of bending stress and bending strain in beams of symmetrical & non-symmetrical cross-sections

**SECTION MODULUS ‘Z’**

#### Section modulus is the ratio of moment of inertia to the maximum distance of outermost fiber from the neural axis. Its symbol is ‘z’. Its units are mm ^{3}.

#### Z = I/y_{max}

#### Value of y_{max} is same for uppermost & lowermost fibers in a symmetrical section. Symmetrical sections are rectangular, circular, square & I-sections. But value of y_{max }will be different for non-symmetrical area of cross sections. Non symmetrical sections are angle, channel, T and Triangular section. Bending moment equation use section modulus.

**M = σ Z**

#### Maximum stress is obtained using section modulus. It is applicable for the outermost fiber of the beam.

#### σ_{max} = M/(I/y_{max}) = M y_{max}/I = M/Z

#### The units of Z are mm^{3} and for a circular beam Z=πd^{3}/32

**MOMENT OF RESISTANCE ’M**_{R}’

_{R}’

#### It is the bending moment the beam can bear. Bending equation gives its value using **σ**_{allow}.

_{allow}

#### M_{R}/I **= σ**_{allow}/y = E/R,

_{allow}/y = E/R,

**BENDING MOMENT M**

#### Whereas ‘M’ is the actual BM acting on the beam. It is due to the actual loading on the beam.

#### M_{R} > M

**BENING STIFFNESS, EI**

#### Bending stiffness stands for minimum bending or for minimum bending stress

#### σ = M/EI, EI should be maximum to keep stress minimum.

#### EI is bending stiffness

**DEFINITION OF BENDING**

#### Bending is a part of a circular deformation. Initially straight member becomes like a bow or an arc.

** ASSUMPTIONS USED IN DERIVATION OF BENDING EQUATION**

#### (i) Beam is initially straight.

#### (ii) Beam is homogeneous i.e. composition is same throughout.

#### (iii) Material is continuous i.e. no voids or no foreign material is present.

#### (iv) Beam is isotropic material i.e. properties in x, y, z directions are same.

#### (v) A transverse plane (perpendicular to the axis) remains a transverse plane before and after the bending moment is applied.

#### (vi) A **constant bending moment** is applied on the beam.

#### (vii) No shear force is acting on the beam

**DIFFERENCE BETWEEN BENDING STRESSES AND SIMPLE STRESSES**

#### Bending stresses are simultaneously tensile as well as compressive stresses and these are of varying magnitude. Some fibers are in tension while others are in compression. There will be one fiber which will be under no stress and it is called the NEUTRAL FIBER. Stress in each fiber is different. Maximum stresses are in the outermost fibers. These maximum stresses are also called SKIN STRESSES because these are on the skin of the beam.

#### Simple stresses are either tensile OR compressive or shear and are of same constant magnitude in all the fibers. It is due to the reason that only one type of load is supposed to act at one time.

** Use of a beam with area with flanges and web.**

#### It has been found by analysis that Sections with web and flanges are much more efficient ( can resist more load in bending) than compact sections such as rectangular/square/circular sections having the same area of cross section. It is due to large value of moment of inertia for web and flange sections. Sections having larger moment of inertia are highly resistant to bending. Thus I -section beams are frequently used as beams in applications where large bending moment is present**.**

**Definition of neutral axis**

#### Neutral axis is a line in the cross section of the beam where neutral plane cuts the cross section. At this axis, there will be no stress. It is at the center of a symmetrical cross section. Neural axis is Not in the center of a T-section (non symmetrical section) beam. Neutral plane is a fiber of zero stress.

** ****Difference between Neutral layer and neutral axis.**

#### A.

**Neutral Layer**

##### It is a plane (area) under zero stress and under zero strain. It passes through the neutral axis of the beam.

**Neutral axis**

##### It is a line where neutral fiber cuts the cross section of the beam. It has zero stress or zero strain.

**Define a Flitched beam.**

##### Flitched beam is made of more than one material. In these beams, normally there are two materials. One of these is main material. The other material in small quantity is flitched material. Material in smaller quantity may be in two/three pieces. Each piece is a flitch. Example of a flitched beam is a timber been having a steel strip only at the top. It can have only at the bottom. Steel strip is on top and bottom. Such a beam is a flitched beam.

**BENDING STRESSES-****COMPOSITE BEAMS**

**DEFINITION OF A COMPOSITE BEAM**

#### A beam of two or more materials is a composite beam.

**STRESSES AND STRAINS IN COMPOSITE BEAMS**

#### Derivation of bending equation assumes the material is homogeneous and isotropic. A composite beam does not satisfy this assumption. Convert composite beam into a beam of one material. Convert a beam of wood and steel into a beam of wood.

#### MODULAR RATIO ‘m’

#### It is the ratio of the modulus of different materials that make up the beam. Modular ratio is greater than 1. Modular ratio converts dimensions of cross-section of composite beam into a beam of single material. Covert only the width. Obtain stress in the other material by multiplying by modular ratio. Thus, there will be discontinuity of stress in the common fiber(s). Stress in one material will be widely different from the stress in the other material at the joint fiber. Find stresses in the composite beam on the basis of the following assumption.

#### (i) A composite beam has same strain in the common fiber because of rigid connection assumed.

#### σ_{s}/E_{s} = σ_{w}/E_{w}

#### (ii) Stresses and strains will vary linearly in each of the two materials w.r.t ‘y’ from the neutral axis.

### Fig. Variation of bending stress and bending strain in a composite beam

#### An example makes the concept clear.

**NUMERICAL PROBLEM ON COMPOSITE BEAM**

**Fig. shows** the cross-section of a composite beam. A steel plate 20 mm x 120 mm reinforces the timber beam . Attach it at the top of the beam. A point load of 20 k N acts at the center of the span of 4m. Calculate the maximum stresses in timber and steel. Given E_{t }=10 GN/m^{2}, and E_{s} = 200 GN/m^{2}.

**SOLUTION**

#### Fig. Variation of bending stress and strain in a composite beam of Timber & Steel

#### (i) Draw the equivalent beam in timber as shown. Only steel portion will change.

#### (ii) Find the Neutral axis location by taking moments about the top surface.

#### y _{NA} =(2400 x 20 x 10 + 160 x 80 x 100)/(2400 x 20 + 160 x 80) = 29.3 mm

#### (iii) Find the moment of inertia of the entire beam of timber.

#### I = (1/12) 2400 x 20^{3} + 2400 x 20 x (19.3)^{2} + (1/12) 160 x (160)^{3} + 160 x 160 x (70.7)^{2}

#### = 12540 x 10^{4} mm^{4}

**(iv) **Find maximum bending moment M = WL/4 = (20 x3) /4 = 15 k N m

#### (v) Find σ in timber (which is the top fiber of steel)

#### Firstly σ_{top in converted timber} /y _{top} = M/I

#### Secondly σ_{top in converted timber} /29.3 =15 x 1000^{2}/(12540 x 10^{4})

**We get σ**_{top in converted timber } = 3.6 N /mm^{2 }

_{top in converted timber }= 3.6 N /mm

^{2 }

#### Now stress in steel

#### σ_{top in steel } /E_{s} = σ_{top in timber}/E_{t}

** ****σ**_{top in steel } = 3.6 x 20 = 72.0 N/mm^{2}

_{top in steel }= 3.6 x 20 = 72.0 N/mm

^{2}

** ****(vi) ****Find σ**_{max in timber} (which is at the bottom of steel)

_{max in timber}(which is at the bottom of steel)

#### M x 150.7/I = 15 x 1000 x 1000 x 150.7/(12540 x 10^{4})

** σ**_{max in timber} = 18.23 N/mm^{2 ANS}

_{max in timber}= 18.23 N/mm

^{2 ANS}

#### Stress in steel at the top level of timber =18.23 x 20 = 360.46 **N/mm**^{2}

^{2}

**PROBLEM**

** A beam of square section with side ‘a’ is used with its diagonal horizontal. Find the section modulus of the beam.**

#### A.

##### There will be one triangle above and another below the horizontal diagonal. The moment of inertia of one triangle about the base is bd^{3}/12= a^{4}/12. Therefore the total moment of inertia will be 2 x a^{4}/12.

##### Length of the diagonal =

##### Height of the triangle = [a^{2} – (a/ )^{2}]^{0.5} =a/

##### Therefore section modulus , Z = I/y_{max} = (2 x a^{4}/12) / a/ = a^{3}/3

**Q12. Find the ratio of second moment of area about the centroid axis to about the base of a triangle. Given base=200 mm and altitude = 600 mm.**

#### A.

##### Second moment of area is moment of inertia ‘I’

For a rectangle

##### Firstly I _{centroid axis } = bd^{3}/36, say= I_{1}

##### Secondly I _{base} = bd^{3}/12, say = I_{2}

##### Thirdly I_{1}/I_{2} = (bd^{3}/36)/( bd^{3}/36) = 1/3

https://www.mesubjects.net/wp-admin/post.php?post=4179&action=edit MCQ Bending stresses in beams