chapter 4 stresses in beam equation of bending stress

Darshan Institute of Engineering & Technology Civil Engineering Department

Equation of bending stress (Flexure eqn)

(i) Before bending (ii) C/S of Beam Stress diagram (iii) Before bending

Consider a layer PQ at a distance Y from the neural axis.

after bending layer of PQ compressed to PQ’

Decrease in length of PQ layer

l PQ - P'Q'

l PQ - P'Q' Strain ( ) _______________(1)

original length PQ

Now from geometry of the fig two section OP’Q’ & OR’S’ are similar.

n

P'Q' P'Q' 1 1

R'S' R'S'

R'S'-P'Q' R'S' RS PQ Neutral layer

R'S'

PQ-P'Q' ______________________________(2)

PQ

from eq (1) & (2)

stre

R y R y

R R

R R Y Y

R R

Y

R

Ybending

R

ss ( ) E

E

____________________________________(3)Y

Y

R E

E

R

Now for fig. i.e. C/S of beam, in which PQ small layer is at a distance Y from NA.

Mechanics of Solids (3130608)

Chapter – 4 Stresses in beam


δa= Area of PQ layer

As we know that from qun (3)

Algebraic sum of all such moment about NA

Now, from qun (3)

Example 1 A simply supported beam of span 4.0 has a cross-section 200mm x 300mm. If the permissible

stress in the material of the beam is 20n/mm2, determine maximum udl it can carry.

Answer:

3 36 4

xx

2

200 300I 450 10

12 12

20 /

300150

2 2

bdmm

N mm

dy mm

nflexure eq M

I Y

RY

E

aa

NAabout force thisofMoment

a a layer PQ oflength in Decrease

2

R

EYY

R

EY

R

EY

a I

a

a M

2

2

2

YIR

E

YR

E

YR

E

R

Y

E

I

M




6620 450 10

M 600 10150

IN mm

Y

Now, for simply supported beam with U.D.L on entire span

26

max

2

6

M 600 108

4000 600 10

8

300

300

wl

w

w N mm

w kN m

Example 2 A cast iron test beam 30 mm square in cross-section 500 mm long is simply supported at end.

It fails at central point load at 4.32 KN. What load at free end will cause the failure of cantilever beam of

1 m span made of same material 30 mm x 60 mm in cross section?

Answer: (1) For simply supported beam

4 4 4

4

xx

3

n

2

30I 67500

12 12 12

3015

2 2

4.32 10 500540000

4 4

Mflexure eq

I

540000 120

67500 15

d bmm

dy mm

wlM N mm

Y

N mm

(2) For cantilever beam

3 33 4

xx

30 60I 540 10

12 12

3015

2 2

bdmm

dy mm




n

3

6

6

M M 120flexure eq

I 540 10 30

M 2.16 10

For cantilever M w

2.16 10 w 2160

1000

Y

N mm

l

N

w 2.16 KN

Example 3: A section of beam as shown in fig. is subjected to a bending stress of 10 kN.m about the major axis. Draw bending stress distribution across the section.

Answer:

(1) Center of a given beam

Sr.no Shape Area (mm2) Y (mm) AxY (mm

3)

1

A1=100x10

A1=1000

Y1=90+𝟏𝟎

𝟐

Y1=95

A1Y1=

95000

2

A2=90x10

A2=900

Y2=𝟗𝟎

𝟐

Y2=45

A1Y1=

405000

∑ A = 1900 ∑ AY = 135500

b

t

t

135500Y

1900

Y 71.32

Y 71.32

Y 100-71.32

Y 28.68

AY

A

mm

mm

mm

(2) Moment of inertia about centroidal horizontal axis

Sr.no. Area h (mm) Ah2 (mm4) IG or Iself (mm4) Ixx = IG+ Ah2(mm4)

1 A1=1000 h1=Yt- 𝒅𝟏

𝟐 =23.68 A1h1

2= 5607242.34 IG1=𝟏𝟎𝟎×𝟏𝟎𝟑

𝟏𝟐 = 8333.34 I1 = 569075.7

2 A2=900 h2=Yb- 𝒅𝟐

𝟐 = 26.32 A2h2

2= 623468.16 IG2=𝟏𝟎×𝟗𝟎𝟑

𝟏𝟐 = 607500 I2 = 1230968.16

C2 Y2

Y1

C1




(3) Bending stress

M = 10 KN = 10 X 103 X 10

3N·mm = 10 X 10

6N·mm

Ixx = 1.8 X 106 mm

4

Ytop = 100 – 71.32 = 28.36 mm

Ybottom = 71.32 mm Bending stress at top

M

I tY

6

6

10 10

1.8 10 28.68

2

top 159.34 N mm

Bending stress at bottom

M

I bY

6

6

10 10

1.8 10 71.32

2

bottom 392.22 N mm




Shear stress distribution across beam section (equation of shear stress)

Consider a small portion ABCD of length δx of abeam loaded with U.D.L as shown in fig (a).

Shear force (S.F) & Bending moment (B.M) will change at every point along the length of the

beam.

Let, M = B.M at AB

M+ δM = B.M at CD

F = S.F at AB

F+ δF = S.F at CD

I = M.I. of the section about the neutral axis.

Now consider an element strip EF at distance Y from the neutral axis.

a = cross section area of strip EF.

Σ = bending stress at AB.

Now, using

Y

Y

I

M

I

M eq flexure n

similarly, Y across CD section

M

I

acting across CD acting across AB

M Mbending stress

I

Y

Force Force

a

____ i a

stress area

M MY a

I

M

Y a _____ iiI

M

A C

E F

B D

Y Y N.A

M+δM

δX

d

b Fig.(a)

Fig.(b)




Net unbalanced force on the strip

Now, total unbalance force above neutral axis may be found out by integration above equation

between 0 &2

d

Ya

dYYadYYa

dYaY

I

M F

I

M

I

M

I

MF

2

d

0

2

d

0

2

d

0

Where, A = area of beam above N.A

Y = distance between the centroid of the area & N.A

bI

bI

YA

x

M

Area

forceshear total stressshear ,

YAF

bx

YAM

M

Now

FSF

ngsubstituti

.x

M

aY

aYaY

I

M

I

M -

I

MM




Prove that the maximum shear stress in a rectangular section of beam is 1.5 times of average

shear stress.

Consider rectangular beam as shown in fig.

b = width, d = depth

(i) Average shear stress:-

We know that average shear stress is given as

τavg = 𝐹

𝑏×𝑑

(ii) Maximum shear stress:-

Maximum shear stress occurs at neutral axis which lies at 𝑑

2 form top & bottom.

avg

db

F

db

F

bd

F

bbd

ddbF

bddY

bddhear

Ib

YFA

5.1

5.12

32

3

2

3

12

42

12 I ,

2

2

2

d A ,

max

avg

max

3max

3

max

d

𝑑

2

b

𝑌 =𝑑

4

𝑑

2




Prove that the maximum shear stress in a circular section of beam is 4/3 times of average

shear stress.

Consider a plane circular lamina having diameter‘d’

(i) Average shear stress:-

τavg = 𝐹

𝐴=

𝐹𝜋

4𝑑2=

4𝐹

𝜋𝑑2

(ii) Maximum shear stress:-

avg

d

Fd

F

d

F

dd

ddF

bd

ddhear

Ib

YFA

3

4

33.13

4

43

16

Now,

3

16

64

3

2

8

db , 12

I

3

2d

3

4rY ,

8 )

4(

2

1 A ,

max

2

2

avg

max

24

2

max

3

22

max

Maximum shear stress lies at neutral axis which

passes through its centroid.

The centroid of circle lies at 𝑑

2 form top & bottom.

A

A

d = 2r = dia

𝑌 =4𝑟

3𝜋

N




Example . A section of beam as shown in fig. is subjected to a shear force of 20KN. Draw

shear stress distribution across the section.

Ans:-

Shear force on beam = 20 KN

(i) shear stress at junction of flange & web.

Ytop = 28.68 mm, Ybottom = 71.32 mm

A Y = 100 10 × 28.68−10

2 = 23680 mm

3

(ii) Shear stress at C.G.

mm 100 b

1.8 I ,bI

YFA

2

2

6

3

N/mm 26.3 10

1002.63

stress.shear in the increasesudden is therejunction, the

/ 63.2100108.1

236801020

bI

At

mmNYFA

3 712.25424

71.174423680

2

68.181068.1855.2810100

mm

YA

2

6

3

max / 25.28100108.1

712.254241020

bImmN

YFA

100 mm

100 mm

10 mm

10 mm

τ = 0 τ = 2.63

N/mm2 26.3 N/mm

2

τmax = 28.25

N/mm2

Yt=

28.68

mm

Yb=

71.32

mm

10

mm

100 mm

28.68

mm

N A

shear stress diagram




Ex- Fig. shows a beam cross section subjected to a shear force of 200 KN. Determine the

shearing stress at neutral axis & sketch the shear stress distribution diagram.

Ans. (1) Center of a given section

Sr.no Shape Area (mm2) Y (mm) A*Y (mm3)

1

A1=10*120 A1=1200

Y1=

10+120+𝟏𝟎

𝟐

Y1= 135

A1Y1=

162000

2

A2=10*120 A2=1200

Y2 =10+𝟏𝟐𝟎

𝟐

Y2 =70

A1Y1= 84000

3

A2=10*120 A2=1200

Y2 =𝟏𝟎

𝟐

Y2 =5

A1Y1= 6000

∑ A = 3600 ∑ AY = 252000

(2) Moment of inertia about centroidal horizontal axis.

Sr.no. Area h (mm) Ah2 (mm4) IG or Iself (mm4) Ixx = IG + Ah2 (mm4)

1 A1=1200 h1=Yt- 𝒅𝟏

𝟐 =65 A1h1

2=5.07X106 IG1=

𝒃𝟏×𝒅𝟏𝟑

𝟏𝟐=1X104

I1 = 5.08X106

2 A2=1200 h2=Yt- Y2 = 0 A2h22=0

IG2=𝒃𝟐×𝒅𝟐

𝟑

𝟏𝟐=1.44X106

I2 = 1.44X106

3 A3=1200 H3=Yb- 𝒅𝟑

𝟐 = 65 A3h3

2=5.07X106 IG2=𝟏𝟎×𝟗𝟎𝟑

𝟏𝟐=1X104 I2 = 5.08X106

Ixx= I1 + I2+ I3=11.8 x 106 mm4

mm

A

AY

70

2

Y Y Y Y

3600

252000 Y

tb

120 mm

120 mm

10 mm

10 mm

120 mm

C2 Y2

Y1

C1

Y3 C3




(3) Shear stress

Shear stress at the junction of flange & web

Now, Shear stress at C point.

Shear stress at neutral axis.

From geometry of a given section is symmetric about centroidal axis (Horizontal &Vertical)

2

3

3

1

46

3

N/mm 8125.0

120106.11

78000105.14

Ib

YFA

mm 120b ,106.11

780002

1010120YA

Ib

YFA KN, 14.5F

B

B

xx

t

mmI

mmY

2

2

12

3

3

2

2

463

N/mm 75.910

120

b

b N/mm 75.9

elyAlternativ 10106.11

78000105.14

Ib

YFA

changes only width in mm 10bb

106.11 ,78000YA KN, 14.5F

BBCB

B

B

xx mmImm

2

3

3

3

1

221

11

2

46

N/mm 12 10106.11

96000105.14

mm 96000YA

2

60

2

10120

2

107010120

2YA

mm 10bb ,106.11 KN, 14.5F

D

xx

YA

YA

mmI

EcFBGA , ,

Yt=

70mm

A

C B

D

E

G

F

𝑌

1

2

3

𝑌1′

Yt=

70mm

C1

D

C2 𝑌2′

A

C B

D

E

G

F

τ = 0

0.815 N/mm2

9.75 N/mm2

12 N/mm2

τ = 0

0.815 N/mm2

9.75 N/mm2

shear stress distributiondiagram




Example As shows figure a cross section subjected to a shearing force of 200 kN. Determine the

shearing stress at neutral axis& sketch the shear stress distribution diagram across the section.

Ans. Given section is symmetric about Horizontal & Vertical axis

Moment of inertia about centroidal horizontal axis.

Now, Ixx = I1 + I2 + I3= 129.167 x 106 mm4

→ Stress at point B (shear)

→Stress at point C

S

n Area h (mm) Ah

2 (mm

4) IG or Iself (mm

4) Ixx = IG + Ah

2

1 A1=

50X100

=5000

h1=Yt- 𝟏𝟎𝟎

𝟐

= 100

A1h12=

5X107

IG1=𝒃𝟏×𝒅𝟏

𝟑

𝟏𝟐

=4.167X106

I1 = 5.4167x107

2 A2=

100X250

=25000

h2=0 A2h22= 0 IG2=

𝒃𝟐×𝒅𝟐𝟑

𝟏𝟐

=2.0834X107

I2 = 2.0834x107

3 A3= 5000

H3=Yb- 𝟏𝟎𝟎

𝟐

= 100

A3h32=

5X107

IG2=

4.167X106

I2 = 5.4167x107

mm 150 Y Y Y bottomtop

2

6

3

11

2

N/mm 48.15

5010167.129

100500010200

50bb ,1002

Y

KN 200F ,mm 500010050

B

B

t

Ib

YFA

mmmmd

Y

A

2N/mm 09.3250

50 BC

2

6

3

1

2

22

11

222

2111

N/mm 03.525010167.129

81250010200

50bb

mm 812500'2

'

42b

22db

Ib

YFA

mm

YA

YA

ddddA

G

𝑑

2

100mm

100mm

100mm

100

mm

100

mm

50

mm

b1= 50mm, d1= 100mm

b2= 250mm, d2= 100mm b3= 50mm, d3= 100mm

C

D E

A

B

G

F

Yt YB

𝑌1′

𝑌2′

C

A

B

G

C

A

B

G

τ = 0

3.09 N/mm2 15.48 N/mm

2

5.03 N/mm2

3.09 N/mm2

15.48 N/mm2




Draw shear stress distribution diagram for standard section

(1) Rectangular section (2) Solid circular section

(3) Triangular section (4) I - section

(5) T - section (6) T - section

(7) Hollow rectangular section (8) Solid square with diagonal horizontal

(Rhombus)

τNA =4

3τavg

τmax = 1.5 τavg τmax =

4

3τavg

τmax = 1.5 τavg

A N

τmax

τmax

τ= 0

τmax

h

4

3h

τmax = 1.125 τavg

C C



chapter 4 stresses in beam equation of bending stress

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