Bending Stress.pptx

JJ310- STRENGTH OF MATERIALS

BENDING STRESSES

BENDING STRESS

1. To know about convention symbols for

bending stress

2. Derive the equation of bending

stresses

3. Understanding types of section

standard

4. Calculate the neutral axis (NA) for standard of

section

5. Calculate the second moment of

area (I) for the standard section

6. Use the bending equations to solve problems involving

strength and bending for the

simply supported and cantilever beam

1. CONVENTION SYMBOLSSYMBOLS DESCRIPTION UNIT

E

F

I

Ic

IG

IPN

Ixx

M

R

Young’s Modulus

Concentrated Load

Second Moment Area

Second Moment Area around centroid axis

Second moment area around the center of gravity

Second moment area around the neutral axis

Second moment area around x – x axis

Bending Moment

Radius of Curvature

(

N/

m2

)

( N )

( m4 )

( m4 )

( m4 )

( m4 )

( m4 )

( Nm )

( m )

STRESSES IN BEAMS

Forces and couples acting on the beam cause bending (flexural stresses) and shearing stresses on any cross section of the beam and deflection perpendicular to the longitudinal axis of the beam.

If couples are applied to the ends of the beam and no forces act on it, the bending is said to be pure bending. If forces produce the bending, the bending is called ordinary bending.

ASSUMPTIONS

In using the following formulas for flexural and shearing stresses, it is assumed that:-

a plane section of the beam normal to its longitudinal axis prior to loading remains plane after the forces and couples have been applied

the beam is initially straight and of uniform cross section and that the modulus of elasticity in tension and compression are equal.

ALLOCATION OF BENDING STRESS

y

M

M

P.N.

R

O

A’

B’E’

P’F’

Q’C’

D’

E

C D

QP

F

BA

MM

y

Rajah 9.2 (a)

Rajah 9.2 (b)

Referring on beside figure, the original position of the beam is shown partially in Figure 9.2 (a)

When the beam is curved, the part will be show as in Figure 9.2 (b).

For such a curvature, the longitudinal layer of the beam along the top surface (AB) will be in compression, while the lower surface (CD) will be in tension.

Between the layers AB and CD, the length of one layer does not change when the bent beam.

y

M

M

P.N.

R

O

A’

B’E’

P’F’

Q’C’

D’

E

C D

QP

F

BA

MM

y

Rajah 9.2 (a)

Rajah 9.2 (b)

The layers is called the neutral axis (N.A) or neutral plane.

The bending stress () at N.A = 0.

y

M

M

P.N.

R

O

A’

B’E’

P’F’

Q’C’

D’

E

C D

QP

F

BA

MM

y

Rajah 9.2 (a)

Rajah 9.2 (b)

Distance of neutral axis from the center of curvature is called the radius of curvature (R).

For any layer of PQ (a distance y from the N.A) is curved and elongated and it become to P'Q'.

The layers along N.A become to E'F.

EF = E’F’ = R original length, PQ = EF = R

Final length, P’Q’ = (R+y)

y

M

M

P.N.

R

O

A’

B’E’

P’F’

Q’C’

D’

E

C D

QP

F

BA

MM

y

Rajah 9.2 (a)

Rajah 9.2 (b)

length Original

PQ of elongation The Strain PQ

R

R - y) (R

R

y

R

y E PQ,at Stress

R

E

y

.......................eq. (1)

y

M

M

P.N.

R

O

A’

B’E’

P’F’

Q’C’

D’

E

C D

QP

F

BA

MM

y

Rajah 9.2 (a)

Rajah 9.2 (b)

Rajah 9.2: Allocation of bending stress

N

AP

y

Based on Figure 9.2 (c), the cross section area for shaded strip is = A.

Suppose that the longitudinal stress on the line = .

The total moments of beam cross section:-

yA x y.R

E N.A aroundMoment

R

yE. σ proven,been Have

A.yR

E strip on the force theSo,

A.yR

E 2

A .yR

E M 2 I

R

E M

Note;

But to balance the bending beam, the moment of resistance must be equal to the applied moments.

Therefore:-

When combined with equation 1, then:-

axis neutral around area secondmoment isA . I 2 y

R

E

I

M or I.

R

E M

R

E

y

σ

I

M This equation is used to solve

the problem of the bending stresses

BENDING MOMENT OF BEAM (WITH UNIFORMLY DISTRIBUTED LOAD)

For Simply Supported Beam:-

8 wLM

2

max

The maximum bending moment can be determined using the formula below:

For Cantilever Beam

2 wLM

2

max

The maximum bending moment can be determined using the formula below:

BENDING MOMENT OF BEAM (WITH UNIFORMLY DISTRIBUTED LOAD)

WE WILL CONTINUE……….AFTER TAKE 5….

EXERCISE:-

1. What is the meaning of symbol (I) in the general bending stress below

A. Second moment of inertiaB. CentroidC. Radius of curvatureD. Bending moment

R

E

y

σ

I

M

EXERCISE:-

2. What is the meaning of symbol (M) in the general bending stress below


R

E

y

σ

I

M

EXERCISE:-

3. What is the meaning of symbol (y) in the general bending stress below


R

E

y

σ

I

M

EXERCISE:-

4.The formula bending moment for For Simply Supported Beam is:-

A.

B.

C.

2 wLM

2

max

8 wLM

2

max

16 wLM

2

max

EXERCISE:-

5.The formula bending moment Cantilever Beam:-

A.

B.

C.

2 wLM

2

max

8 wLM

2

max

16 wLM

2

max

ANSWER:-

1. A2. D3. B4. B5. A

INTRODUCTION – SECOND MOMENT OF AREA

In this unit, the bending equation will be used to determine the bending stress of standard forms.

To obtain the bending stress, the position of neutral axis (N.A) and second moment of area (or moment of inertia) for standard forms (I) should be calculated.

Rectangular: Second moment of area is defined as

Thus, for a rectangular section, second moment of area in PN, is

SECOND MOMENT OF AREA– RECTANGULAR SECTION

b

A B

dA

dy

y

DC

P.N.d

dA y I 2

12

bd

3

y b

dy y b

dy y I

3

2/

2/

3

2d/2

d/2-

22/

2/P.N.

d

d

d

d

Figure 10.1: Rectangular cross-section beam

Second moment of area at the N.A axis

A

B

b bC

Dd

FE

P.N.

Rajah 10.2: Rasuk Berkeratan Rentas I

IN.A = IACEF – Ib.b ;

b.b = part of shaded area

12

)(bd 2 -

12

BD

33

To obtain the second

moment of area by using the cut-off

SECOND MOMENT OF AREA– RECTANGULAR I-SECTION

Figure 10.3 shows a circle of radius r. The shaded elements shown in the figure has an area dA and therefore the following equation is formed: -

dA = rd

SECOND MOMENT OF AREA– CIRCULAR SECTION

dro

r

dr

Rajah 10.3: Rasuk Berkeratan Rentas Bulat

From the polar coordinate system:-y = r sin

Second moment of area at N.A for a circular section is given by:-

SECOND MOMENT OF AREA– CIRCULAR SECTION

4

πr

θdθ sin 4

r

4

r dθ sinθ

θdr rd θ sin r

dA y I

4

22π

0

40

r

0

42π

0

22r

0

2π

0

2P.N.

0

o

dro

rdr

Rajah 10.3: The Circular cross section

Parallel axis theorem states that the second moment of area on any axis parallel to the NA (the axis X - X) is equal to the second moment of area around the axis through the centroid of the (NA) plus the product of cross-sectional area and square of the distance between the axis parallel to the PN

THEOREM OF PARALLEL AXIS

P

y

N

h

dA

xx

y’

dA y I 2xx

Rajah 10.4: Rectangular section beam

P

y

N

h

dA

xx

y’

Rajah 10.4: Rectangular section beam

If the NA line is drawn parallel to the line x-x, then the formula above can be described as follows: -

dAh dA y'2h dA )(y'

dAh h 2y' ) (y'

)h y' ( I

h y' y

22

22

2xx

THEOREM OF PARALLEL AXIS

The first integral is the second moment of area about an axis through the center of the form.

The second integral is the first moment of area about an axis through the center of the form, so it is equal to zero.

The last integral is to the total cross-sectional area. Next the equation above can be written as:-

dAy'

Ixx = IP.N. + Ah2NOTE: h = distance from the midpoint of

part of section in parallel with the neural axis.

STANDARD SECTION SCHEDULE

From the equation that has been made, we can be brief as the table below: -

STANDARD SECTION SCHEDULE

CENTROID - A COMBINATION FORM

The form of a diagram can be produced by combining several basic form or cut the original diagram.

The form in Figure 10.5 (a) is produced by combining the rectangular ABCD with semicircular ADP.

D

CB

A

P

CB

D

CB

DA A

P

C

p

D

y2

y1

A D

For Figure 10.5 (a) a high centroid of every basic form of BC is not the same that y1 y2

CENTROID - A COMBINATION FORM

y

Therefore,

Rajah 10.5(b)

)A (A

)yA y(A

A

Ay y

2 1

2211

CENTROID - TRUNCATED FORM

In Figure 10.6 (a) part of square DEFG is cut and removed from the original rectangular form of ABCH.

CB

D

H

G

A

E

F

Rajah 10.6(a)

Figure 10.6 (a) a high centroid for each basic shape of the line BC is equal (Figure 10.6 (b)). So: -

y

CB

D

H

G

A

E

F

y2y1 D

G

E

F

CB

HA

Rajah 10.6(b)

Referring to Figure 10.7, form of L can be produced by combining two rectangular or separated and cut methods of rectangular EDGF from ABCG.

y

y2

y1

E D

CB

FA

D

G

E

F

CB

GA

If using the cut and split, use the following formula:

)A (A

)yA y(A

A

Ay y

2 1

2211

CENTROID - TRUNCATED FORM

If the for every layer of the upper surface to the bottom surface of the beam is determined, the values can be plotted on a graph as follows. The graph shows the distribution of the bending stress.

Look at the value of is not dependent on cross-sectional width of a strip. In layer N.A, = 0.

THE STRESS DISTRIBUTION

P.N.

5 cm

7 cm

- tensile

= 0

- compressive

THANK YOU…

Q & A

EXERCISE:-

1. The equation of the second moment of area for rectangular shape is:-

A.

B.

C.

D.

4

d2

12

d3

12

d4

4

d3

2. The equation of the second moment of area for circular section is:-A.

B.

C.

D.

EXERCISE:-

4

r4

8

r4

4

r2

12

r2

3. This below formula is state of :-

A. Second moment of area

B. Position of neutral axis

C. Theorem of parallel axis

D. Modulus of section

dA y I 2xx

EXERCISE:-

4. Which is the formula is to determine the distance of centroid?A.

B.

C.

D.

EXERCISE:-

A

Ay

A

Ah

A

Ay 2

Ah

Ay

5. Which if the unit of second moment of area?A. m4

B. m2

C. Nm2

D. N/m2

EXERCISE:-

6. Based the figure on right side, the label of x and y to determine second moment of area for rectangular shape is:- A. X = h and Y = dB. X = d and Y = bC. X = b and Y = dD. X = d and Y = h

X

Y

EXERCISE:-

EXAMPLE 1

A T bar with a length 6 meters incur of a concentrated load. Each load is 16 kN at a

distance of 1 m from both ends of the beam. Bar is simply supported on both ends as in

Figure 10.3 (a). The cross section bar is shown in Figure 10.3 (b). Calculate the following:

-

a. Distance of neutral axis from the bottom of the beam.

b. Second moment of area around the neutral axis.

c. The radius of curvature at the midspan of the beam.

d. Maximum the compressive and the tensile bending stress in the beam

Given E= 200 GN / m2

80 mm

20 mm

15 mm

60 mm

16 kN 16 kN

1 m1 m

6 m

SOLUTION EXAMPLE 1:-

i. Calculate the distance of neutral axis from the bottom of the beam.

Step 1:

Divide of those sections into two parts. Calculate the centroid of the area and distance for each part of the sectional base.


Step 2:

Determine the distance on the bottom of neutral axis cross section

mm 65

1200) 1200(

) 40 x (1200 ) 90 x (1200

AA

yA yA y

21

2211

Since we using the

method of cut sections, use this formula.

Second moment of area at the neutral axis is

IP.N. = ( IC1 + A1h12 ) + ( IC2 + A2h2

2 )

= ( 40,000 + ( 1200 x 252 ) ) + ( 640,000 + ( 1200 x 252 ) )

= 2.18 x 106 mm4

= 2.18 x 10-6 m4

SOLUTION EXAMPLE 1:-ii. Calculate the second moment of area around the neutral axis.

Step 3:

Calculate the second moment of area and distance h for each section.

iii. The radius of curvature at the midspan of the beam.

Step 4:

Use formula:-

From the loading shown, we find that the order of loading is symmetrical, so the reactions:-

R1 = R2 = 16 kN

From B.M.D, the bending moment of the midspan beam is:-


16 kN16 kN

1 m1 m

6 m

R1 R2

( + )

(-)

G.D.R

( + )

G.M.L.

M

EI R

R

E

y

I

M

M = 16 kNm ( hogging )


m 27.25

10 x 16

10 x 2.18 x 10 x 200 R

3

-69

The radius of curvature at the midspan of the beam.

M

EI R

R

E

y

I

M


iv. Maximum the compressive and the tensile bending stress in the

beam.

Step 5:

Refer to the diagram of the beam section, we find that:-

ylower > yupper

The maximum produced at the lower surface is :

ymax = 65 mm = 0.065 m

) tensile( N/mm 10 x 477

10 x 18.2

0.065 x 10 x 16

I

yM σ

26

6-

3maxmax

The maximum produced at the upper surface is ,

ymax = 35 mm = 0.035 m


) ecompressiv ( N/m 10 x 256.8

10 x 18.2

0.035 x 10 x 16

I

yM σ

26

6-

3maxmax

A 1 meter cantilever beam loads uniformly distributed along the span of the beam. Beam cross section is shown in below figure, where EE is the upper surface of the beam. Determine the the following: -

i. the neutral axis of cross section.

ii. Second moment of area around the neutral axis.

iii. Maximum tensile and compressive stress in the beam results from sagging.

80 mm

40 mm

60 mm

120 mm

E E

1 m

20 kN/m

EXAMPLE 2:

y

Step 1:

Divide of those sections into two parts. Get wide and centroid of distance each part of the sectional base.

SOLUTION EXAMPLE 2:

i. Distance of neutral axis on the bottom of the cross section.

SOLUTION EXAMPLE 2:

mm 70

4800) 4800(

) 40 x 4800 ( ) 100 x (4800

A A

21

2211

AA

yyy


SOLUTION EXAMPLE 2:

Second moment of area around the neutral axis.

IP.N. = ( IC1 + A1h12 ) + ( IC2 + A2h2

2 )

= ( 640 x 103 + ( 4800 x 302 ) ) + ( 2560 x 103 + ( 4800 x 302 ) )

= 11.84 x 106 mm4

= 1.184 x 10-5 m4

SOLUTION EXAMPLE 2:

iii. Maximum bending moment occurs at the bar met the wall of :

Mmax = w L2 = (20 x 103) (12 ) = 10,000 Nm = 10 kNm

2 2

Since the beam is hogging, the top surface will have the lower surface tension and compression

SOLUTION EXAMPLE 2:

ylower maximum = 70 mm

yupper maximum = 120 – 70 = 50 mm

SOLUTION EXAMPLE 2:

MN/m 59.12 10 x 1.184

10 x 70 x 10,000 ecompressiv σ

MN/m 42.23 10 x 1.184

10 x 50 x 10,000 tensileσ

I

yM σ

y

σ

I

M

25-

3-

max

25-

3-

max

maxmaxmax

EXAMPLE 3:a. A cross section of a beam as shown in below figure.

Calculate :-i. The neutral axis for the beamii. second moment of area around the neutral axis.

b. If the beam is simply supported at both ends carrying a uniform load of 30 kN / m on the entire spans with a length 3m, calculate the bending stresses in the beam on: -

iii. the upper surfaceiv. the lower surface

80 mm

20 mm

20 mm10 mm

100 mm

10 mm

100 mm

P.N.

40 mm

a. The section can be considered a T-shaped grafting yield from second and first part. While the center is divided punched a hole as rectangular shape (part 3).

Get area for each of divisions involved :-

Part 1 A1 = 80 x 20 = 1600 mm2

Part 2 A2 = 40 x 100 = 4000 mm2

Part 3 A3 = 20 x 100 = 2000 mm2

Calculate the distance y from the base of the T section.

y1 = 100 + 10 = 110 mm ; y2 = 50 mm;

y3 = 10 + 50 = 60 mm

SOLUTION EXAMPLE 3:

P.N.

1

2

3

80 mm20 mm

20 mm

10 mm

100 mm

10 mm

100 mm

P.N.

40 mm

i. Dapatkan nilai

SOLUTION EXAMPLE 3:

P.N.

1

2

3

A

Ay y formulan menggunakadengan y

mm 71.1

2000) - 4000 1600(

60) x (2000 - 50) x (4000 110) x (1600

A A A

y A - y A y A y

321

332211

ii. Get a second moment of area around the neutral axis.

h1 = y1 - = 110 – 65.3 = 44.7 mm

h2 = - y2 = 65.3 – 50 = 15.3 mm

h3 = - y3 = 65.3 – 60 = 5.3 mm

Get Ah2 value of each part.

A1h12 = 1600 x ( 44.7 )2 = 3.2 x 106 mm4

A2h22 = 4000 x ( 15.3 )2 = 936 x 103 mm4

A3h32 = 2000 x ( 5.30 )2 = 56 x 103 mm4

SOLUTION EXAMPLE 3:

_

y _

y _

y

Use Ic = formula to determine second moment of

area for each section.

IC1 =

IC2 =

IC3 =

IP.N. = ( IC1 + A1h12 ) + ( IC2 + A2h2

2 ) - ( IC3 + A3h32 )

= 5.8 x 106 mm4

= 5.8 x 10-6 m4

SOLUTION EXAMPLE 3:

12

bd3

433

mm 10 x 53 12

20 x 80

433

mm 10 x 3.33 12

100 x 40

463

mm 10 x 1.67 12

100 x 20

b. Get the reaction force at both ends A and B.

Total force on A = Total force on B RA = RB

Therefore, RA = RB =

Mmax will occur on midspan of the beam. Therefore,

Mmax = wL2 / 8

= 30 kN/m (32) / 8 = 33.75 kNm

SOLUTION EXAMPLE 3:

3 m

30 kN/m

RA RB

kN 45kN 2

330

SOLUTION EXAMPLE 3:

I

y M σ

y

σ

I

M maxmaxmax

26-

3

upper MN/m 318 x108.5

0.0547 x 10 x 33.75 σ

26-

3

lower MN/m 380 x108.5

0.0653 x 10 x 33.75 σ

We obtained that :

ylower = 65.3 mm Therefore, yupper = 120 – 65.3 = 54.7 mm

Thus;

EXERCISE A 5 m cantilever beam loads uniformly distributed along

the span of the beam. Beam cross section is shown in the figure below, where EE is the upper surface of the beam. Determine the the following: -

i. the neutral axis of cross section.


iii. Maximum tensile and compressive stress in the beam results from sagging.

100 mm

20 mm

60 mm

120 mm

E E

5m

40 kN/m

y

Q & A

Thank You….

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Bending Stress.pptx