EASWARI ENGINEERING COLLEGE, RAMAPURAM · 3. Define shear centre. The shear centre (for any...

MUTHUKUMAR.P/Asst.Prof./CIVIL-III

MAHALAKSHMI ENGINEERING COLLEGE

TIRUCHIRAPALLI - 621213.

QUESTION WITH ANSWERS

DEPARTMENT : CIVIL SEMESTER: V

SUB.CODE/ NAME: CE 2252 / Strength of Materials

UNIT–5 ADVANCED TOPICS IN BENDING OF BEAMS

PART - A (2 marks)

1. Define Unsymmetrical bending

The plane of loading (or) that of bending does not lie in (or) a plane that contains the

principle centroidal axis of the cross- section; the bending is called Unsymmetrical bending.

2. State the two reasons for unsymmetrical bending. (AUC May/June 2012)

(AUC Ap[r /May 2011)

(i) The section is symmetrical (viz. Rectangular, circular, I section) but the load line is

inclined to both the principal axes.

(ii) The section is unsymmetrical (viz. Angle section (or) channel section vertical web)

and the load line is along any centroidal axes.

3. Define shear centre.

The shear centre (for any transverse section of the beam) is the point of intersection

of the bending axis and the plane of the transverse section. Shear centre is also known as

“centre of twist”

4. Write the shear centre equation for channel section.

f

w

A

A

be

6

3

e = Distance of the shear centre (SC ) from the web along the symmetric axis XX

Aw = Area of the web

Af = Area of the flange

5. A channel Section has flanges 12 cm x 2 cm and web 16 cm x 1 cm. Determine the shear

centre of the channel.

Solution:

b= 12-0.5 = 11.5 cm

t1 = 2cm, t2 = 1cm, h= 18 cm


Af = bt1 = 11.5 x 2 = 23 cm2

Aw = ht2 = 18 x 1= 18 cm2

f

w

A

A

be

6

3

cme 086.5

23

186

)5.11(3

6. Write the shear centre equation for unsymmetrical I section.

xxI

bbhte

4

)( 212

21

e = Distance of the shear centre (SC) from the web along the symmetric axis XX

t1 = thickness of the flange

h = height of the web

b1 = width of the flange in right portion.

b2 = width of the flange in left portion.

Ixx = M.O.I of the section about XX axis.

7. State the assumptions made in Winkler’s Bach Theory.(AUC Nov / Dec 2012)

(AUC Nov/Dec 2013) (AUC May/June 2012)

(1) Plane sections (transverse) remain plane during bending.

(2) The material obeys Hooke’s law (limit state of proportionality is not exceeded)

(3) Radial strain is negligible.

(4) The fibres are free to expand (or) contract without any constraining effect from

the adjacent fibres.

8. State the parallel Axes and Principal Moment of inertia.

If the two axes about which the product of inertia is found, are such , that the product

of inertia becomes zero, the two axes are then called the principle axes. The moment of

inertia about a principal axes is called the principal moment of inertia.

9. Define stress concentration. . (AUC Nov / Dec 2011)

The term stress gradient is used to indicate the rate of increase of stress as a stress

raiser is approached. These localized stresses are called stress concentration.

10. Define stress – concentration factor.

It is defined as the ratio of the maximum stress to the nominal stress.

nom

tK max

max = maximum stress

nom = nominal stress


11. Define fatigue stress concentration factor.

The fatigue stress – concentration factor (Kf ) is defined as the ratio of flange limit of

unnotched specimen to the fatigue limit of notched specimen under axial (or) bending loads.

)1(1 tf KqK

Value of q ranges from zero to one.

12. Define shear flow.

Shear flow is defined as the ratio of horizontal shear force H over length of the beam

x. Shear flow is acting along the longitudinal surface located at discharge y1.Shear flow is

defined by q.

z

zy

I

QV

x

Hq

H = horizontal shear force

13. Explain the position of shear centre in various sections.

(i) In case of a beam having two axes of symmetry, the shear centre coincides with

the centroid.

(ii) In case of sections having one axis of symmetry, the shear centre does not

coincide with the centroid but lies on the axis of symmetry.

14. State the principles involved in locating the shear centre.

The principle involved in locating the shear centre for a cross – section of a beam is

that the loads acting on the beam must lie in a plane which contains the resultant shear

force on each cross-section of the beam as computed from the shearing stresses.

15. Determine the position of shear centre of the section of the beam shown in fig.

Solution:

t1 = 4 cm, b1 = 6 cm, b2 = 8 cm

h1 = 30 – 4 = 26 cm

xxI

bbhte

4

)( 212

21

Ixx = 43

33

2085212

222)13(414

12

4142 cm

xx

x

cmx

e 9077.020852(4

)68(264 22

16. State the stresses due to unsymmetrical bending.

VVUU

bI

u

I

vM

sincos

σb = bending stress in the curved bar

M = moment due to the load applied

IUU = Principal moment of inertia in the principal axes UU

IVV = Principal moment of inertia in the principal axes VV


17. Define the term Fatigue.

Fatigue is defined as the failure of a material under varying loads, well below the

ultimate static load, after a finite number of cycles of loading and unloading.

18. State the types of fatigue stress.

(i) Direct stress

(ii) Plane bending

(iii) Rotating bending

(iv) Torsion

(v) Combined stresses

(a) Fluctuating or alternating stress

(b) Reversed stress.

19. State the reasons for stress- concentration.

When a large stress gradient occurs in a small, localized area of a structure, the high

stress is referred to as a stress concentration. The reasons for stress concentration are (i)

discontinuities in continuum (ii) contact forces.

20. Define creep.

Creep can be defined as the slow and progressive deformation of a material with

time under a constant stress.

21. Define principal moment of inertia. (AUC Nov/Dec 2013)

The perpendicular axis about which the product of inertia is zero are called“principal axes” and the

moments of inertia with respect to these axes are called as principal

moments of inertia.

The maximum moment of inertia is known as Major principal moment of inertia and

the minimum moment of inertia is known as Minor principal moment of inertia.

PART B (16 MARKS)

1. Explain the stresses induced due to unsymmetrical bending.

Fig. shows the cross-section of a beam under the action of a bending moment M

acting in plane YY.

Also G = centroid of the section,

XX, YY = Co-ordinate axes passing through G,

UU, VV = Principal axes inclined at an angle θ to XX and YY axes respectively

The moment M in the plane YY can be resolved into its components in the planes

UU and VV as follows:

Moment in the plane UU, M’ = M sinθ

Moment in the plane VV, M’ = M cosθ

The components M’ and M” have their axes along VV and UU respectively.

The resultant bending stress at the point (u,v) is given by,

UUVVUUVV

bI

M

I

M

I

vM

I

uM cossin"'


vvUUb

I

uSin

I

VCosM

At any point the nature of σb will depend upon the quadrant in which it lies. The equation of

the neutral axis (N.A) can be found by finding the locus of the points on which the resultant

stress is zero. Thus the points lying on neutral axis satisfy the condition that σb = 0

0vvUU I

uSin

I

VCosM

0vvUU I

uSin

I

VCos

uCos

Sin

I

Iv

vv

UU (or) uI

Iv

vv

UU tan

This is an equation of a straight line passing through the centroid G of the section and

inclined at an angle with UU where

tantanvv

UU

I

I

Following points are worth noting:

i. The maximum stress will occur at a point which is at the greatest distance form the

neutral

ii. All the points of the section on one side of neutral axis will carry stresses of the same

nature and on the other side of its axis, of opposite nature.

iii. In the case where there is direct stress in addition to the bending stress, the neutral

axis will still be a straight line but will not pass through G (centroid of section.)

2. Derive the equation of Shear centre for channel section. (AUC April/May 2005)

Fig shows a channel section (flanges: b x t1 ; Web h x t2) with XX as the horizontal symmetric axis.

Let S = Applied shear force. (Vertical downward X)

(Then S is the shear force in the web in the upward direction)

S1 = Shear force in the top flange (there will be equal and opposite shear force in

the bottom flange as shown.)

Now, shear stress ( ) in the flange at a distance of x from the right hand edge (of the top

flange)

tI

ySA

xa

2.1

hxtyA (where t = t1 , thickness of flange)

xx

xh

xx I

Sh

tI

xSt

22.

.

.

1

1


Shear force is elementary area

dztddxtd AA 11 ..

Total shear force in top flange

dxt

b

..

0

1 (where b = breadth of the flange)

b

xx

b

xx

xdxI

shtdxt

I

hSS

0

11

0

12

.;2

(or) 4

.2

11

b

I

ShtS

xx

Let e = Distance of the shear centre (sc) from taking moments of shear forces about the

centre O of the web,We get

hSeS .. 1

xxxx I

bhtSh

b

I

Sht

4

..

4.

221

21

xxI

thbe

4

122

(1)

Now, 122

.12

23

2

2

1

31 hth

tbtb

Ixx 122

.

6

32

21

31 hthtbbt

122

32

21 hthbt

(neglecting the term 3

3

1bt, being negligible in comparison to other

terms)(or) 12

2

12bbtht

hI xx

Substitute the value of Ixx in equation (1) we get,

12

12

122

122

6

3

6

12

4 htht

tb

bthth

thbe

Let bt1 = Af (area of the flange)

ht2 = A (area of the web)

Then


f

wfw

f

A

A

b

AA

bAe

6

3

6

3

i.e

3. Derive the equation of Shear center for unequal I-section

Solution:

Fig. shows an unequal I – section which is symmetrical about XX axis.

Shear stress in any layer,

It

ySA

where I = IXX = 1212

23

121

31

21

hxtbb

tbb

Shear force S1 :

2

... 11

hxtyAdxtdA

S1 =

1

0

11

1

2

..b

XX

dxxth

tI

txSdA

=

1

0

12

..b

XX

dxth

I

xS =

XX

b

XX I

bShtx

I

Sht

422

211

0

21

1

Similarly the shear force (S2) in the other part of the flange,

S2 =XXI

bSht

4

221

Taking moments of the shear forces about the centre of the web O, we get

S2. h = S1. h + S .e (S3 = S for equilibrium)

(where, e = distance of shear centre from the centre of the web)

or, (S2 – S1) h = S.e

eSI

bbtSh

XX

.4

)( 21

221

2

xxI

bbhte

4

21

22

21

f

w

A

A

be

6

3


4. Derive the stresses in curved bars using Winkler – Bach Theory.

The simple bending formula, however, is not applicable for deeply curved beams where the

neutral and centroidal axes do not coincide. To deal with such cases Winkler – Bach Theory is

used.

Fig shows a bar ABCD initially; in its unstrained state. Let AB’CD’ be the strained position of

the bar.

Let R = Radius of curvature of the centroidal axis HG.

Y = Distance of the fiber EF from the centroidal layer HG.

R’ = Radius of curvature of HG’

M = Uniform bending moment applied to the beam (assumed

positive when tending to increase the curvature)

= Original angle subtended by the centroidal axis HG at its

centre of curvature O and

’ = Angle subtended by HG’ (after bending) a t the center of curvature ’

For finding the strain and stress normal to the section, consider the fibre EF at a distance y

from the centroidal axis.

Let σ be the stress in the strained layer EF’ under the bending moment M and e is strain in the

same layer.

Strain, )(

)(')''('

yR

yRyR

EF

EFEFe or 1

'.

''

yR

yRe

e0 = strain in the centroidal layer i.e. when y = 0

1'

.'

R

R or

'.

''1

yR

yRe --------- (1)

and 1+e = '

.'

R

R --------- (2)

Dividing equation (1) and (2) , we get

01

1

e

e

'.

''

R

R

yR

yR or

R

yR

ye

R

y

R

ye

e

1

'

'

'

'. 00

According to assumption (3) , radial strain is zero i.e. y = y’

Strain,

R

yR

ye

R

y

R

ye

e

1

''. 00

Adding and subtracting the term e0. y/R, we get

R

yR

ye

R

ye

R

ye

R

y

R

ye

e

1

.''

. 0000


R

y

yRR

e

ee

1

)1

'

1)(1( 0

0 ------------- (3)

From the fig. the layers above the centroidal layer is in tension and the layers below the centroidal

layer is in compression.

Stress , σ = Ee = )

1

)1

'

1)(1(

(0

0

R

y

yRR

e

eE ___________ (4)

Total force on the section, F = dA.

Considering a small strip of elementary area dA, at a distance of y from the centroidal layer HG, we

have

dA

R

y

yRR

e

EdAeEF

1

)1

'

1)(1(

.0

0 dA

R

y

y

RReEdAeEF

1

)1

,

1(1. 00

dA

R

y

y

RReEAeEF

1

)1

,

1(1. 00 ____________ (5)

where A = cross section of the bar

The total resisting moment is given given by

dA

R

y

yRR

e

EydAeEdAyM

1

)1

'

1)(1(

...

20

0

dA

R

y

y

RReEeEM

1

)1

,

1(10.

2

00 (since )0ydA

M = E (1+e0) dA

R

y

y

RR1

1

'

12

Let 22

1

AhdA

R

y

y

Where h2 = a constant for the cross section of the bar

M = E (1+e0)21

'

1Ah

RR ----------- (6)

Now, dAyR

yydA

yR

RydA

R

y

y 2

..

1

= dAyR

yydA .

2

dA

R

y

y

1

dA

R

y

y

R.

1

10

2

= 21Ah

R ---------- (7)

Hence equation (5) becomes


F = Ee0 .A – E (1+e0 )R

Ah

RR

21

'

1

Since transverse plane sections remain plane during bending

F = 0

0 = Ee0 .A – E (1+e0 )R

Ah

RR

21

'

1

E e0 .A = E (1+e0 )R

Ah

RR

21

'

1

e0 = (1+e0 )R

Ah

RR

21

'

1 (or)

2

0

h

Re(1+e0 )

RR

1

'

1

Substituting the value of 2

0

h

Re(1+e0 )

RR

1

'

1 in the equation (6)

M = E 2

2

0 Ahh

Re = e0 EAR

Or EAR

Me0 substituting the value of e0 in equation (4)

2

0*

1

*h

Re

R

y

yE

AR

M (or)

EAR

M

h

R

R

y

yE

AR

M**

1

*2

2

1*

1

*h

R

y

Ry

AR

M

AR

M

yR

y

h

R

AR

M2

2

1 (Tensile)

yR

y

h

R

AR

M2

2

1 (Compressive)

5. The curved member shown in fig. has a solid circular cross –section 0.01 m in

diameter. If the maximum tensile and compressive stresses in the member are not to

exceed 150 MPa and 200 MPa. Determine the value of load P that can safely be

carried by the member.

Solution:

Given,

d = 0.10 m; R = 0.10 m; G = 150 MPa = 150 MN / m2 (tensile )

2 = 200 MPa = 200 MN / m2 (Compressive)


Load P:

Refer to the fig . Area of cross section,

232

2

10854.710.044

md

A

Bending moment, m = P (0.15 + 0.10) =0.25 P

2

422

10.0

10.0.

128

1

16

dh = 7.031 x 10-4 m2

Direct stress, compA

pd

Bending stress at point 1 due to M:

yR

y

h

R

AR

Mb 2

2

1 1 (tensile)

Total stress at point 1,

11 bd

yR

y

h

R

AR

M

A

P2

2

1150 (tensile)

05.010.0

05.0

10031.7

10.01

10.010854.7

25.0

10854.7150

4

2

33

PP

= -127.32 P + 318.31 P x 5. 74

= 1699.78 P

KNP 25.8878.1699

10150 3

(i)

Bending stress at point 2 due to M:

12

2

2yR

y

h

R

AR

Mb (comp)

Total stress at point 2,

22 bd

12002

2

yR

y

h

R

AR

M

A

P

105.010.0

05.0

10031.7

10.0

10.010854.7

25.0

10854.7 4

2

33

PP

=127.32 P + 318. 31 P x 13.22

= 4335.38 P


MNP38.4335

200

KNP 13.4638.4335

10200 3

(ii)

By comparing (i) & (ii) the safe load P will be lesser of two values

Safe load = 46.13 KN.

6. Fig. shows a frame subjected to a load of 2.4 kN. Find (i) The resultant stresses at a

point 1 and 2;(ii) Position of neutral axis. (April/May 2003)

Solution:

Area of section 1-2,

A = 48 * 18*10-6 = 8.64 * 10-4m2

Bending moment,

M = -2.4*103*(120+48) * = -403.2 Nm

M is taken as –ve because it tends to decrease the curvature.

(i) Direct stress:

Direct stress σd = 26

4

3

/77.210*10*64.8

10*4.2mMN

A

P

23

2

2

2log R

DR

DR

D

Rh e

Here R = 48 mm = 0.048 m, D = 48 mm = 0.048 m

23

2 )048.0(048.0)048.0(2

048.0)048.0(2log

048.0

048.0eh

= 0.0482 (loge3 – 1) = 2.27 * 10-4 m2


(ii) Bending stress due to M at point 2:

yR

y

h

R

AR

Mb 2

2

2 1 ;

26

4

2

4/10*

024.0048.0

024.0

10*27.2

048.01

048.0*10*64.8

2.403mMN

= -9.722 (1-10.149) = 88.95 MN/m2 (tensile)

(iii) Bending stress due to M at point 1:

yR

y

h

R

AR

Mb 2

2

1 1

26

4

2

4/10*

024.0_048.0

024.0

10*27.2

048.01

048.0*10*64.8

2.403mMN

= -42.61 MN/m2 = 42.61 MN/m2 (comp)

(iv) Resultant stress:

Resultant stress at point 2,

σ2 = σd + σb2 = 2.77 + 88.95 = 91.72 MN/m2 (tensile)

Resultant stress at point 1,

σ1 = σd + σb1 = 2.77 -42.61 = 39.84 MN/m2 (comp)

(v) Position of the neutral axis:

42

4

22

2

10*27.2048.0

10*27.2*048.0y

hR

Rhy

= -0.00435 m = - 4.35 mm

Hence, neutral axis is at a radius of 4.35 mm


7. Fig. shows a ring carrying a load of 30 kN. Calculate the stresses at 1 and 2.

Solution:

Area of cross-section = 2222 01131.01.113124

mcmcmx

Bending moment M = 30*103 * (13.5*10-2)Nm = 4050 Nm

h2 = ......*128

1

16 2

42

R

dd

Here d = 12 cm, R = 7.5 +6 = 13.5 cm

h2 =2

42

5.13

12*

128

1

16

12 = 9.89 cm2 = 9.89*10-4 m2

Direct Stress σd = 263

/65.210*01131.0

10*30mMN

A

P

Bending stress at point 1 due to M,

yR

y

h

R

AR

Mb 2

2

1 1

6

4

2

1 10*06.0135.0

06.0

10*89.9

135.01

135.0*01131.0

4050b 2.65*6.67

= 17.675 MN/m2 (tensile)

Bending stress at point 2 due to M,

yR

y

h

R

AR

Mb 2

2

2 1

6

4

2

1 10*06.0135.0

06.0

10*89.9

135.01

135.0*01131.0

4050b 2.65*13.74

= 36.41 MN/m2 (comp)

Hence σ1 = σd + σb1 = -2.65 + 17.675

= 15.05 MN /m2 (tensile)

and σ2 = σd + σb2 = -2.65 – 36.41

= 39.06 MN/m2 (comp)


8. A curved bar is formed of a tube of 120 mm outside diameter and 7.5 mm thickness. The

centre line of this is a circular arc of radius 225 mm. The bending moment of 3 kNm tending

to increase curvature of the bar is applied. Calculate the maximum tensile and compressive

stresses set up in the bar.

Solution:

Outside diameter of the tube, d2 = 120 mm = 0.12 m

Thickness of the tube = 7.5 mm

Inside diameter of the tube, d1 = 120-2*7.5 = 105 mm = 0.105m

Area of cross-section,

222 00265.015.012.04

mA

Bending moment M = 3 kNm

Area of inner circle,

221 00866.0105.0

4mA

Area of outer circle,

222 01131.012.0

4mA

For circular section,

h2 = ......*128

1

16 2

42

R

dd

For inner circle,

h2 = ......*128

1

16 2

41

21

R

dd

h2 = 4

2

42

10*08.7225.0

105.0*

128

1

16

105.0

For outer circle,

h2 = ......*128

1

16 2

42

22

R

dd; h2= 4

2

42

10*32.9225.0

12.0*

128

1

16

12.0

211

222

2 hAhAAh

0.00265 h2 = 0.01131*9.32*10-4 – 0.00866*7.078*10-4

h2 = 0.00166 m2, and R2/h2 = 0.2252/0.00166 = 30.49

Maximum stress at A,

yR

y

h

R

AR

MA 2

2

1 (where, y = 60 mm = 0.06 m)

263

/10*06.0225.0

06.049.301

225.0*00265.0

10*3mMNA

σA = 37.32 MN/m2 (tensile)

Maximum stress at B,


yR

y

h

R

AR

MB 2

2

1

263

/10*06.0225.0

06.049.301

225.0*00265.0

10*3mMNB

σB = 50.75 MN/m2 (comp)

8. A curved beam has a T-section (shown in fig.). The inner radius is 300 mm. what is

the eccentricity of the section?

Solution:

Area of T-section, = b1t1 + b2t2

= 60*20 + 80*20 = 2800 mm2

To find c.g of T- section, taking moments about the edge LL, we get

21

2211

AA

xAxAx

)20*80()20*60(

)10*20*80)(20*80()202

60)(20*60(

x =27.14 mm

Now R1 = 300 mm; R2 = 320 mm; R= 327.14 mm; R3 = 380 mm

Using the Relation:

2

2

31

1

22

32 log.log. R

R

Rt

R

Rb

A

Rh ee

23

2 )14.327()320

380(log*20)

300

320(log*80

2800

)14.327(eeh


= 12503.8(5.16+3.44) – 107020.6 = 512.08

y = )(56.108.512)14.327(

08.512*14.327222

2

mmhR

Rh

where y = e (eccentricity) = distance of the neutral axis from the centroidal axis.

Negative sign indicates that neutral axis is locates below the centroidal axis.

10. Fig. shows a C- frame subjected to a load of 120 kN. Determine the stresses at A and B.

(AUC Nov/Dec 2011)

Solution:

Load (P) = 120 kN

Area of cross – section = b1t1 +b2t2+ b3t3

= 120*30 + 150*30 +180*30 = 0.0135 mm2

To find c.g of the section about the edge LL,

21

2211

AA

xAxAx

)30*180()30*150()30*120(

)120*30*180()15*30*150()225*20*120(1y =113 mm=0.113 m

y2 = 240 – 113 = 127 mm = 0.127 m

R1 = 225 mm = 0.225 m

R2 = 225 + 30 = 255 mm = 0.255 m

R = 225 + 113 = 338 mm = 0.338 m

R3 = 225 +210 = 435 mm = 0.435 m

R4= 225 + 240 = 465 mm = 0.465 m

2

3

41

2

33

1

22

32 logloglog R

R

Rb

R

Rt

R

Rb

A

Rh eee

23

2 338.0435.0

465.0log12.0

255.0

435.0log03.0

225.0

255.0log15.0

0135.0

)338.0(eeeh

= 2.86 (0.01877 +0.016 +0.008) – 0.1142 = 0.008122 m2

Direct stress, σd = )(/89.810*0135.0

10*120 263

compmMNA

P

Bending moment, M = P*R

Bending stress at A due to the bending moment,

2

2

2

2

1)(yR

y

h

R

AR

MAb

127.0338.0

127.0

008122.0

338.01

*)(

2

2

AR

RPAb

= 8.89 (1+3.842) = 43.04 MN/m2 (tensile)

Bending stress at B due to the bending moment:


1

1

2

2

1)(yR

y

h

R

AR

MAb

113.0338.0

113.0

008122.0

338.01

*)(

2

AR

RPAb

= 8.89 ( 1- 7.064)

= -53.9 MN /m2 = 53.9 MN/m2 (comp)

Stress at A, σA = σd + (σb)A

= -8.89 + 43.04 = 34.15 MN/m2 (tensile)

Stress at B, σB = σd + (σb)B

= -8.89 – 53.9 = 62.79 MN/m2 (comp)

11. Derive the formula for the deflection of beams due to unsymmetrical bending.

Solution:

Fig. shows the transverse section of the beam with centroid G. XX and YY are two

rectangular co-ordinate axes and UU and VV are the principal axes inclined at an angle θ to the XY

set of co-ordinates axes. W is the load acting along the line YY on the section of the beam. The

load W can be resolved into the following two components:

(i) W sin θ …… along UG

(ii) W cos θ …… along VG

Let, δu = Deflection caused by the component W sin θ along the line GU for its bending about VV

axis, and

Δv = Deflection caused by the component W cos θ along the line GV due to bending abodt

UU axis.

Then depending upon the end conditions of the beam, the values of δu and δv are given by

VV

uEI

lWK 3sin

UU

vEI

lWK 3cos

where, K = A constant depending on the end conditions of the beam

and position of the load along the beam, and

l = length of the beam

The total or resultant deflection δ can then be found as follows:

22vu

223 cossin

UUVV I

W

I

W

E

Kl

UUVV IIE

Kl2

2

2

23 cossin


The inclination β of the deflection δ, with the line GV is given by:

tan

tanVV

UU

I

I

v

u

12. A 80 mm x 80 mm x 10 mm angle section shown in fig. is used as a simply supported

beam over a span of 2.4 m. It carries a load of 400 kN along the line YG, where G is the

centroid of the section. Calculate (i) Stresses at the points A, B and C of the mid – section of

the beam (ii) Deflection of the beam at the mid-section and its direction with the load line (iii)

Position of the neutral axis. Take E = 200 GN/m2 (AUC Apr/May 2011)

Solution:

Let (X,Y) be the co-ordinate of centroid G, with respect to the rectangular axes BX1 and BY1.

Now X = Y = mm66.23700800

350032000

10*7010*80

5*10*7040*10*80

Moment of inertia about XX axis:

23

23

)66.2345(*10*7012

70*10)566.23(*10*80

12

10*80XXI

= (6666.66 + 278556) + (285833.33 + 318777) = 889833 mm4

= 8.898 * 105 mm4 = IYY (since it is an equal angle section)

Co-ordinates of G1 = + (40-23.66), - (23.66-5) = (16.34,- 18.66)

Co-ordinates of G2 = -(23.66-5). + (45 – 23.66) = (-18.66, + 21.34)

(Product of inertia about the centroid axes is zero because portions 1 and 2 are rectangular strips)

If θ is the inclination of principal axes with GX, passing through G then,


90tan2

2tanXXXY

XY

II

I (since Ixx =Iyy)

2θ = 90º

i.e. θ1 = 45º and θ2 = 90º + 45º = 135º are the inclinations of the principal axes GU and GV

respectively.

Principal moment of inertia:

IUU = 22 )()2

()(2

1XY

XXYYYYXX I

IIII

= 25255

55 )10*226.5()2

10*898.810*895.8()10*898.810*895.8(

2

1

= (8.898 + 5.2266) *105 = 14.1245*105 mm4

IUU + IVV = IXX + IYY

IVV = IXX IYY – IUU

= 2*8.898 x 105 – 14.1246 x 105 = 3.67 x 105 mm4

(i) Stresses at the points A, B and C:

Bending moment at the mid-section,

Nmm

WlM 5

3

10*4.24

10*4.2*400

4

The components of the bending moments are;

M’ = M sin θ = 2.4 x 105 sin 45º = 1.697 x 105 Nmm

M’’ = M cos θ = 2.4 x 105 cos 45º = 1.697 x 105 Nmm

u,v co-ordinates:

Point A: x = -23.66, y = 80-23.66 = 56.34 mm

u = x cos θ + y sin θ

= -23.66 x cos 45º + 56.34 x sin 45º = 23.1 mm

v = y cosθ + x sin θ

= 56.34 cos 45º - (-23.66 x sin 45º) = 56.56 mm

Point B:

x = -23.66, y = -23.66


= -23.66 x cos 45º + (-23.66 x sin 45º ) = - 33.45 mm


= -23.66 cos 45º - (-23.66 x sin 45º) = 0

Point C ; x = 80 – 23.66 = 56.34, y = -23.66


= 56.34 cos 45º -23.66 x sin 45º = 23.1 mm


= -23.66 cos 45º - 56.34 sin 45º) =- 56.56 mm

UUVV

AI

vM

I

uM "'


2

5

5

5

5

/47.17101246.14

)56.56(10*697.1

1067.3

)1.23(10*697.1mmN

xxA

2

55

5

/47.15101246.14

0

1067.3

)45.33(10*697.1mmN

xxB

2

55

5

/788.3101246.14

56.56

1067.3

)1.23(10*697.1mmN

xxB

(ii) Deflection of the beam, δ:

The deflection δ is given by:

UUVV IIE

KWl2

2

2

23 cossin

where K = 1/48 for a beam with simply supported ends and carrying a point load at

the centre.

Load , W = 400 N

Length l = 2.4 m

E = 200 x 103 N/mm2

IUU = 14.1246 x 105 mm4

IVV = 3.67 x 105 mm4

Substituting the values, we get

25

2

25

233

)101246.14(

45cos

)1067.3(

45sin)104.2(400

48

1

xxE

xx

δ = 1.1466 mm

The deflection δ will be inclined at an angle β clockwise with the kine GV, given by

848.345tan1067.3

101246.14tantan

5

5

x

x

I

I

VV

UU

β = 75.43º - 45º = 30.43º clockwise with the load line GY’.

(iii) Position of the neutral axis:

The neutral axis will be at 90º - 30.43º = 59.57º anti-clockwise with the load line,

because the neutral axis is perpendicular to the line of deflection.


11. A cured bar of rectangular section , initially unstressed is subjected to bending moment

of 2000N –m tends to straighten the bar . the section is 5cm wide and 6cm deep in the plane

of bending and mean radius of curvature is 10cm . Find the position of neutral axis and the

stress at the inner and outer face. (AUC Apr /May 2010 )


12. A central horizontal section of book is a symmetrical trapezoid 60mm deep , the inner

width being 60mm the outer being 30mm . estimate the extreme intensities of stress when

the honk carries a load of 30kn . the load line passing 40mm from the inside edge of the

section and the centre of curvature being in the load line . also plot the stress distribution

across the section . (AUC Apr/May 2011)

EASWARI ENGINEERING COLLEGE, RAMAPURAM · 3. Define shear centre. The shear centre (for any...

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