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MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 1
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 2
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 3
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 4
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"'
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 5
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 6
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 7
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 8
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 9
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 10
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)
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 11
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 12
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 13
(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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 14
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)
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 15
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,
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 16
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 17
= 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:
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 18
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
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 19
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,
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 20
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
u = x cos θ + y sin θ
= -23.66 x cos 45º + (-23.66 x sin 45º ) = - 33.45 mm
v = y cosθ + x sin θ
= -23.66 cos 45º - (-23.66 x sin 45º) = 0
Point C ; x = 80 – 23.66 = 56.34, y = -23.66
u = x cos θ + y sin θ
= 56.34 cos 45º -23.66 x sin 45º = 23.1 mm
v = y cosθ + x sin θ
= -23.66 cos 45º - 56.34 sin 45º) =- 56.56 mm
UUVV
AI
vM
I
uM "'
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 21
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.
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 22
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 )
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 23
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)
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 24
MUTHUKUMAR.P/Asst.Prof./CIVIL-III Page 25