Bending Stress Ppt

7/27/2019 Bending Stress Ppt

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B ENDING STRESSES IN B EA MS

Beams are subjected to bending moment and shearing forces

which vary from section to section. To resist the bendingmoment and shearing force, the beam section developsstresses.

Bending is usually associated with shear. However, for simplicity we neglect effect of shear and consider momentalone ( this is true when the maximum bending moment isconsidered---- shear is ZERO) to find the stresses due tobending. Such a theory wherein stresses due to bending aloneis considered is known as PURE BENDING or SIMPLEBENDING theory.


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Example of pure bending

W W

SFD-

+

a aA B

VA= W VB= W

C D

BMD

Wa Wa

+Pure bending

between C & D


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BENDING ACTION:

Sagging

M

NEUTRAL AXIS

NEUTRAL LAYER

c

Neutral Axis t


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Hogging

c Neutral Axis

t

Neutral layer


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BENDING ACTION

Sagging -> Fibres below the neutral axis (NA) get stretched -> Fibres

are under tension

Fibres above the NA get compressed -> Fibres are in compression

Hogging -> Vice-versa

In between there is a fibre or layer which neither undergoes tensionnor compression. This layer is called Neutral Layer (stresses arezero).

The trace of this layer on the c/s is called the Neutral Axis.


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Assumptions made in Pure bending theory

1) The beam is initially straight and every layer is free toexpand or contract.

2) The material is homogenous and isotropic.

3) Young s modulus (E) is same in both tension andcompression.

4) Stresses are within the elastic limit.

5) The radius of curvature of the beam is very large incomparison to the depth of the beam.


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6) A transverse section of the beam which is plane before bendingwill remain plane even after bending.

7) Stress is purely longitudinal.


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Note:

homogeneous: of the same kind throughout

Isotropic: of equal elastic properties in all directions.

B S 7

Longitudinal axis

w1

A transverse section of the beam = the cross section of the beam


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DERIVATION OF PURE BENDING EQUATION

Relationship between bending stress and radius of curvature.

PART I:


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Consider the beam section of length dx subjected to pure

bending. After bending the fibre AB is shortened in length,whereas the fibre CD is increased in length.

In b/w there is a fibre (EF) which is neither shortened in lengthnor increased in length (Neutral Layer).

Let the radius of the fibre E'F be R . Let us select one more fibreGH at a distance of y from the fibre EF as shown in the fig.

EF = E'F = dx = R d

The initial length of fibre GH equals R d After bending the new length of GH equals

G'H= ( R+y) d

= R d + y d


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Change in length of fibre GH = ( R d + y d ) - R d = y d

Therefore the strain in fibre GH= change in length / original length= y d / R d

= y/R

If is the bending stress and E is the Young s modulus of the material,then strain

= /E

/E = y/R => = (E/R) y---------(1)

= (E/R) y => i.e. bending stress in any fibre is proportional to thedistance of the fibre (y) from the neutral axis and hence maximum

bending stress occurs at the farthest fibre from the neutral axis.


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Note: Neutral axis coincides with the horizontal centroidalaxis of the cross section

N A

c

t


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on one side of the neutral axis there are compressive stresses and onthe other there are tensile stresses. These stresses form a couple,whose moment must be equal to the external moment M. Themoment of this couple, which resists the external bending moment,is known as moment of resistance.

Moment of resistance

c

Neutral Axist


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Moment of resistance

Consider an elemental area da at a distance y from the neutral axis.

The force on this elemental area = da

= (E/R) y da {from (1)}

The moment of this resisting force about neutral axis =

(E/R) y da y = (E/R) y da

da y

N A


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Total moment of resistance offered by the beam section,

M'= (E/R) y da= E/R y da

y da =second moment of the area =moment of inertia about theneutral axis.

M'= (E/R) I NA

For equilibrium moment of resistance (M') should be equal to

applied moment Mi.e. M' = M

Hence. We get M = (E/R) I NA


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(E/R) = (M/I NA)--------(2)

From equation 1 & 2, (M/I NA )= (E/R) = ( /y) ----

BENDING EQUATION.

(Bernoulli-Euler bending equation)

Where E= Young s modulus, R= Radius of curvature,M= Bending moment at the section,

I NA= Moment of inertia about neutral axis, = Bending stress

y = distance of the fibre from the neutral axis


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(M/I)=( /y)

or = (M/I) y

Its shows maximum bending stress occurs at the greatest distancefrom the neutral axis.

Let y max = distance of the extreme fibre from the N.A.

(max) = maximum bending stress at distance y max

(max) = (M/I) y max

where M is the maximum moment carrying capacity of the section,

SECTION MODULUS:

M = (max) (I /y max)

M = (max) (I/y max) = (max) Z

Where Z= I/y max= section modulus (property of the section)

Unit ----- mm 3 , m3


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(1) Rectangular cross section

Z= INA/ ymax

=( bd 3/12) / d/2

=bd 2/6

section modulus

b

N A

Y max=d/2

d


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(2) Hollow rectangular section

Z= INA / ymax

=1/12(BD 3-bd 3) / (D/2)

=(BD 3-bd 3) / 6D

(3) Circular sectionZ= INA / ymax

=( d4/64) / (d/2)

= d3/ 32

B

bD/2

Ymax =D/2

d/2 D

N A

d

N AY max=d/2


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(4) Triangular section

b

hN A

Y max = 2h/3Z = I NA / Y max

=(bh 3 /36) / (2h/3)

=bh 2/24

h/3


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(1) Calculate the maximum UDL the beam shown in Fig. can carry

if the bending stress at failure is 50 MPa & factor of safety to begiven is 5.

NUM ERI CAL PROBL EM S

w / unit r un

5 m

200 mm

300 mm

Maximum stress = 50 N/mm

Allowable (permissible) stress = 50/5 =10 N/mm 2


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We have to consider section of the beam where the BM is max, and

stress should be calculated at the farthest fibre from the neutral axis.E/R=M/I NA = b/y

M/I NA= b/y =>

I NA = bd/12= (200 300)/12= 45 107 mm 4

Ymax= d/2=300/2= 150 mm

BM max =wl/8= w (5000) /8

(w 5000/8) / 45 107 = 10/150

w= 9.6 N/mm = 9.6 kN/m


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(2) For the beam shown in Fig. design a rectangular section making thedepth twice the width. Max permissible bending stress = 8 N/mm .Also

calculate the stress values at a depth of 50mm from the top & bottom atthe section of maximum BM.

b

d=2b 2.5 m 3.5 m

9 KN

12 KN/m

A B


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MA=0

(126 3) + (9 2.5) -V B 6 = 0

VB= 238.5/6 =39.75 kN

Fy = 0

VA + V B=(12 6)+ 9

VA= 41.25 kN


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9 KN

39.75 kN

41.25 kN

11.25 kN

2.25 kN+

-

2.5 m 3.5 m

12 KN/m

A B

C

Max. bending moment will occur at the section where the shear forceis zero. The SFD shows that the section having zero shear force isavailable in the portion BC. Let that section be X-X, considered at adistance x from support B as shown below. The shear force at thatsection can be calculated as


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2.5 m 3.5 m

12 KN/m

A B

X

x X

-VB+12 x =0 i.e. -39.75+12x=0x = 39.75/12 =3.312 m.BM is max @ 3.312m from B.

BM@xx = 39.75 3.31 - 123.31(3.31)/2= 65.84 kN-m = 65.84 10 6 N mm


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Now M/I NA = b/y

65.8410 6/(b(2b) 3/12) = 8/b

b= 1.58.2310 6

b= 231.11 mm , d= 2b= 462.22 mm


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231.11mm

231.11

231.11

462.22mm

8 N/mm2

8 N/mm 2

50 mm

50 mm

c

t

From similar triangles,

8/ 231.11 = c/(231.11-50) = t / (231.11-50)

c = 6.27 N/ mm 2( compressive) & t = 6.27 N/ mm 2(tensile)

N A


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(3)A Rolled Steel Joist (RSJ) of 200mm 450 mm of 4m span issimply supported on its ends. The flanges are strengthened by two300mm 20mm plates one riveted to each flange. The second

moment of the area of the RSJ equals 35060 104

mm4. Calculate the

load the beam can carry for the following cases, if the bending stressin the plates is not to exceed 120 MPa, (a) greatest centralconcentrated load (b) maximum UDL throughout the span

300

450

20

20

200

4mRSJ


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I NA=I NA (RSJ)+MI due to plates about NA

= (35060 104 )

+2 [(300 (20)/12+300 20 (235)]=1.01 109 mm 4

300

450

20

20

200

N A

245 mm

245 mm


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(a) M/I NA= b/Y [M max=PL/4]Ymax = (450+2 20) /2= 245mm

=120N/mm 2

(P 4000) / 4 (1.01 109)=120/245P = 4.95 10 5 N

( b) M/I NA= b/Y [M max= wl 2/8]Ymax = (450+2 20) /2= 245mm

w = 247.35 N/ mm = 247.35 KN/m


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(4) An I section beam has 200 mm wide flanges and an overall depth o500 mm. Each flange is 25 mm thick and the web is 20 mm thick. At acertain section the BM is M. Find what percentage of M is resisted byflanges and the web.

200

500

25

20


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From similar triangle principle

max / =250/y =( max y) /250

Area of the element =200 dy

Force on the element = stress area

P= ( max y/250) ( 200 dy)

The moment of resistance of this about the NA equals

= (max y/250) ( 200 dy) y

=(4/5) y max dy


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Therefore moment of resistance of top flange =

Total moment of resistance of both the flanges

=2.26x10 6 max

dy ydy y M F

250

225

2

max

250

225max

2

5

42

5

42

250

225

max

2

5

4dy y

% moment resisted by flanges =(M F/M) 100

=(2.16 106 max )/(2.86 106 max) 100 = 79.02%

Therefore % moment resisted by the web= 20.98%


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(5)Locate & calculate the position and magnitude of

maximum bending stress for the beam shown.

10mm 5mm

500 N

80 mm

x

X

X

Let us consider a section X- X at a distance of x from the free end.


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Bending stress is not maximum at left end (10 mm dia end) because at that end bending moment may be maximum but I xx is also maximum.

Y I

M


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Diameter at X-X , D x =5 + x/16

Dx=5 + 0.0625 x

Therefore I xx= Dx4/64 = (5 + 0.0625x) 4 /64

M/I= b/y

Mxx= 500x,

y= y max @ section x-x = D x/2

b(x-x) = (M xx y max) / I xx

Dx/2

b(x-x) = (500 x D x) / (2 I xx)


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b(x-x) = (500 x D x )/ (2 Dx4/64 )

= (5092.96 x) / D x3

=(5092.96 x) / (5+0.0625x)3

= (5092.96 x ) (5+ 0.0625x) -3

096.5092)0625.05()}0625.0)0625.05(3()96.5092{( 3

4

x x x

96.5092)0625.05(

)0625.0)0625.05(3()96.5092(3

4

x

x x

Now, to have maximum bending stress, d b(xx) /dx = 0

5+ 0.0625x =0.1875x

x = 40 mm M ax bending str ess = 483.13 M pa

34 )0625.05(1

)0625.05(1875.0

x x

x


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= ay / a

= 20025 (250+12.5) + 250 25 125

(200 25) +(250 25)

= 186.11mm

I NA= (200 253)/12 + 200 25 (88.89-12.5) 2

+(25x250 3 )/12 + 25 250 (186.11-125) 2

=85.32x10 6mm 4

Y Y

Y

200mm

25mm

25m

186.1mm

88.89mm


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t

Let us allow the permissible value of stress in tension t=100 N/mm 2

From similar trianglesc / t = 88.89/186.11

c / 100= 88.89/186.11

c =47.762N/mm 2 < 50 Hence safe.

The actual extreme fibre stress values are c = 47.762N/mm 2 &t = 100 N/mm 2

c

88.89 mm

186.11 mm


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Mmax=wl 2/8 = w 5000 2/8y=186.11 for t=100

y=88.89 for c =47.762

M/I NA = b/y

(wl 2)/(8 85.32 10 6) = 100/186.11= 47.72/ 88.89

w =14.67 N/mm=14.67 KN/m

PRACTI CEPROBL EM S


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1) Find the width x of the flange of a cast iron beam having

the section shown in fig. such that the maximum compressivestress is three times the maximum tensile stress, the member being in pure bending subjected to sagging moment.

( Ans: x= 225 mm)

25mm

N

25mm

A100mm

X

WEB

BS 1 PRACTI CE PROBL EM S


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2)A cast iron beam has a section as shown in fig. Find

the position of the neutral axis and the moment of inertiaabout the neutral axis. When subjected to bendingmoment the tensile stress at the bottom fibre is 25N/mm. Find, a) the value of the bending moment b) the

stress at the top fibre.( Ans: M= 25070 Nm, c =33.39 N/mm)

40

20

150

1202020

300mm


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3)A cast iron beam has a section as

shown in fig .The beam is a simplysupported on a span of 1.25 metersand is used to carry a downward pointload at midspan. Find the magnitudeof the load if the maximum tensilestress on the beam section is 30N/mm. Determine also the maximumcompressive stress.

(Ans. W= 174.22 N, c =40.73 N/mm)

120mm

80mm

30MM

BS 3

BS 4


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4)A groove 40mm40mm is cut symmetrically throughout

the length of the circular brass section as shown in fig. If the tensile stress shall not exceed 25 N/mm, find the safeuniformly distributed load which the brass can carry on asimply supported span of 4 meters.

( Ans: 5150 N/m)

100mm

40

40

BS 4


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BS 7

7) In an overhanging beam of wood shown in Fig., the

allowable stresses in bending and shear are 8MPa &0.80MPA respectively. Determine the minimum size of asquare section required for the beam.

A B

60KN 30 KN

3m 3m 2m

( Ans: 274mm 274mm )

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Bending Stress Ppt