Chapter 6 Stresses in Beams.(SOM-201)

8/8/2019 Chapter 6 Stresses in Beams.(SOM-201)

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Vijay Gupta

MEC201

Mechanics of Solids



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Stresses in Beams



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Pure Bending



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Equilibrium: Resistive Moment

Is this positive or negative?



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Origin of Resisting Moment

Compression near topExtension near bottom



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Origin of Resisting Moment...

Extension near top

Compression near bottom



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Beam of symmetrical section

x

M b

y

z



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Longitudinal Strains

1

2

3

4

A BC D y

A BC D

2 4

1 3

d

y

- y

CDfinal = (

CDoriginal =

=



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Longitudinal Stresses

x = -y/

x = -Ey/

-ve

+ve

Compression

Tension



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Neutral plane

Neutral plane

y



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Neutral axis



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Equilibrium

y



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Equilibrium...

y

Maximum tensile stress occurs on extreme fibres



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Rectangular Section

z

y

b

hdy

Location y c g of centroid =



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Rectangular Section

z

y

b

h/2

dy

With y measured from

centroidal axis:



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Example: Cantilevered Beam

P P PL

x

BMD

- PL

bh



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P P PL

x

BMD

- PL

bh



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Cantilevered Beam

bh

bh



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Section

orientation

Factor

b 10 mm 30 mm 3h 30 mm 10 mm 1/3

I zz

2.25×108 m4 0.25×108 m4 1/32

xx

200 MPa 600 MPa 3

Radius of curvature (at

the point of

maximum BM

15 m 1.67 m 1/32

Cantilevered Beam



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Parallel Axis Theorem

bh

N N

P P y 0

y 1

y

I zz

about NA is the minimum



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Introduction to Beams

Common shapes are

I Angle Channel

Common materials are steel and wood



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Introduction to Beams

The parallel portions on an I-beam or H-

beam are referred to as the flanges. The

portion that connects the flanges is referredto as the web.

FlangesWeb

Flanges

Web



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10

50

10

50

100 N 200 N250 N

50 N

x1 m 2 m 3 m

A beam with angle section



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10

50

10

50

Angle section

x

y

(16.1, 16.1)



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Angle section

x

y 10

50

10

50

(16.1, 16.1)

B

A



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SFD

100 N.m

50 N.mBMD

100 N 200 N250 N

50 N

x1 m 2 m 3 m


150 N

100 N50 N

10

50

10

(16.1, 16.1)50



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x-location 1 m 2 m

Bending

moment, Mb

100 N.m +50 N.m

Stressdistribution

Maximum

tensile stress

At top plane

(y = 0.0339 m)

11.3 MPa

At bottom plane

(y = 0.0161 m)

2.7 MPa

Maximum

compressive

stress

At bottom plane

(y = 0.0161 m)

5.4 MPa

At top plane

(y = 0.0339 m)

5.6 MPa




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T-Section

z

y Area A = 20

y c g =(5×12 + 2×8)/20

= 76/20 = 3.8

1

2

To find I zz about the NA, we first find the I of eachof the two areas about their own centroids, and

the use parallel axis theorem to shift to the actual

NA.



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T- Section...

For +ve M B: Max tensile

stress is much more

than the maxcompressive stress



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I-Section

z

y Area A = 28

y c g = 0

To find I zz

about the NA, we first find the I of each

of the three areas about their own centroids, and

the use parallel axis theorem to shift to the actual

NA.

1

2

3



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I-Section

I of Section 1 about its own C A

= 6×23/12 = 4

This transferred to NA (at a distance

of (5 ± 3) = 2 is = 4 + 12×(2)2 = 52

Similarly, I of Section 2 about its own C A (which is

also the NA) = 2×23

/12 = 1.3.

I for the section is 2×52 + 1.3= 105.3 units

z

y

1

23



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Comparison of three sections of the

same area

60

30 54

18

2020

20

38.6

13



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Section

Location of NA from the

base

30 mm 38 mm 30 mm

Izz

5.40×107 m4 5.21×107 m4 6.77×107 m4

Stress distribution

y for max tensile stress 30 mm 38 mm 30 mm

y for max comp. stress +30 mm +22 mm +30 mm

max tensile stress 5.56×104Mb Pa 7.29×104Mb Pa 4.43×104Mb Pa

Max comp. stress 5.56×104Mb Pa 4.22×104Mb Pa 4.43×104Mb Pa

Comparison of three sections of

the same area



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Example

100100 200

Max stresses occur

at the second

support. The beamwill tend to break

here.



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Example:- Stresses in Beam

W eb : y ma x = [387/2-33.3] = 160.2 mm

= 82.4 MPa



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P roblem:- The beam ABCD with a rectangular cross section carries

the loading shown in figure. Determine the magnitude and location of the

maximum bending stress in the beam .



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R A=1kN R D=3kN

1kN

3kN

-1kN.m

4kN.m

3kN.m

I = 2.133 x 10 -6 m4

(4000)(0.04)

2.133 x 10 -6

ma x = 75 M P a



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P roblem:- An T section is used as a simply supported beam to carry

the uniformly distributed load of magnitude of 3W and a concentrated

load w. What is the maximum allowable value of W if the working stress

in bending is 120 Mpa?

A

d = 48.75 mm INN= 11.918 X 106 mm4



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A

R A=2.3 W R D=1.7 W

0.7W

2.3W

1.7W

M AB = 2.3Wx ± 0.5 Wx 2

M BC = 4.5W ± 0.7 Wx

M CD = 8.5W ± 1.7 Wx

2.64W 2.4W

1.7W

M ma x = 2.64W ma x = 120 M P a

ma x = M y/ I NN y = 150 ± 48.75

= 101.25 mm

W=(120 x 10 6 x 11.918 x10 -6 )/(2.64 x .10125)

W= 5.35 kN



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Composite beam

z

y



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Composite beam

Neutral

axis



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Composite beam

Determination of neutral plane:

Neutral

axis



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Composite beam

y y'

Neutral axis



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Reinforced concrete beam

400

50

200

y

NA

205



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Reinforced concrete beam

NA

245

205

155



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Shear stresses in beam



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P

P/ 2 P/ 2

SFD

BMD

x

x

Neutral axis



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x

y

z

P

P/ 2 P/ 2

SFD

BMD

x

x

y

x

z

xy

yx



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P

P/ 2 P/ 2

SFD

BMD

x

x

Neutral axis y

x

d x

M M+d M



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y

b

A¶

y



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b

h

P P PL

x

SFD- P

x BMD

- PL

y

Q at y = 0 is bh2/8; at y = h/ 2 is 0.



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An Example...

Parabolic distribution



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Another Example

xy = V Q/I

zz b

Q xy



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An I beam

15 10 15

10

20

10

375 NSFD 375 N

375 N 375 N

0.5 m 0.5 m

750 N 750 N500 N/m

2 m

93.75 N.m

BMD

93.75 N.m



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An I beam

y y

15 10 15

10

20

10

A

C

B



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An I beam

15 10 15

10

20

10

A

C

B

Flanges resist bending

moments, webs resist

shear forces



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Built-up beam

20 20

20

20

120

50100

z

y y

F s = 6.28 kN



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Built-up beam

20 20

20

20

120

50

100

z

y



Built-up beam

y

s = 0.12 m

Chapter 6 Stresses in Beams.(SOM-201)

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Transcript of Chapter 6 Stresses in Beams.(SOM-201)