mos unit -3.pptx

DNR College of Engineering and
Technology
MECHANICS OF SOLIDS
by
M.THAMBI BABU
Assistant professor
Department of Mechanical Engineering
Department of Mechanical Engineering DNR College of Engineering and Technology

UNIT-3
FLEXURAL STRESSES
Theory of simple bending – Assumptions – Derivation of
bending equation: M/ I = f/y = E/R Neutral axis –
Determination bending stresses – section modulus of
rectangular and circular sections (Solid and Hollow), I,T,
Angle and Channel sections – Design of simple beam
sections.
SHEAR STRESSES: Derivation of formula – Shear stress
distribution across various beams sections like
rectangular, circular, triangular, I, T angle sections.
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Theory of simple bending
(Assumptions)
1.The material of the beam is perfectly homogeneous (i.e., of the same kind
throughout) and isotropic (i.e., of equal elastic properties in all directions).
2.The beam material is stressed within its elastic limit and thus, obeys Hooke’s law.
3.The transverse sections, which were plane before bending, remains plane after
bending also.
4.Each layer of the beam is free to expand or contract, independently, of the layer above
or below it.
5.The value of E (Young’s modulus of elasticity) is the same in tension and
compression.
6.The beam is in equilibrium i.e., there is no resultant pull or push in the beam
section.
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Key Points:
1. Internal bending moment causes beam to deform.
2. For this case, top fibers in compression, bottom in tension.
Bending in beams
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Key Points:
1. Neutral surface – no change in length.
2. Neutral Axis – Line of intersection of neutral surface with the transverse
section.
3. All cross-sections remain plane and perpendicular to longitudinal axis.
Bending in beams
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Engineering and Technology

Key Points:
1.Bending moment
causes beam to
deform.
2.X = longitudinal axis
3.Y = axis of symmetry
4.Neutral surface –
does not undergo a
change in length
Bending in beams
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Theory of Simple Bending
Consider a small length of a simply supported beam subjected to a bending moment as shown in Fig. 14.1 (a). Now consider two
sections AB and CD, which are normal to the axis of the beam RS. Due to action of the bending moment, the beam as a whole will bend
as shown in Fig. 14.1 (b).
Since we are considering a small length of dx of the beam, therefore the curvature of the beam in this length, is taken to be circular. A
little consideration will show that all the layers of the beam, which were originally of the same length do not remain of the same length
any more. The top layer of the beam has suffered compression and reduced to AC. As we proceed towards the lower layers of the
beam, we find that the layers have no doubt suffered compression, but to lesser degree; until we come across the layer RS, which has
suffered no change in its length, though bent into RS. If we further proceed towards the lower layers, we find the layers have suffered
tension, as a result of which the layers are stretched. The amount of extension increases as we proceed lower, until we come across the
lowermost layer BD which has been stretched to B D.
Simple bending
Now we see that the layers above have been compressed and those below RS have been stretched. The amount, by which layer is
compressed or stretched, depends upon the position of the layer with reference to RS. This layer RS, which is neither compressed nor
stretched, is known as neutral plane or neutral layer. This theory of bending is called theory of simple bending.
Bending Stress
Consider a small length dx of a beam subjected to a bending moment as shown in Fig. 14.2 (a). As a result of this moment, let this
small length of beam bend into an arc of a circle with O as centre as shown in Fig. 14.2 (b).
LetM= Moment acting at the beam,
= Angle subtended at the centre by the arc and
R= Radius of curvature of the beam.
Department of Mechanical Engineering DNR College of Engineering
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Department of Mechanical Engineering DNR College of Engineering and
Technology

Position of Neutral Axis
The line of intersection of the neutral layer, with any normal cross-section of a beam, is known as neutral axis of that section. We have seen
that on one side of the neutral axis there are compressive stresses, whereas on the other there are tensile stresses. At the neutral axis,
there is no stress of any kind.
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Modulus of Section
We have already discussed in the previous article, the relation for finding out the bending stress on
the extreme fibre of a section, i.e.,
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Flexure
Formula
M E 
 
I R y
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Beam subjected to 2
BM
In this case beam is subjected to
moments in two directions y and
z. The total moment will be a
resultant of these 2moments.
You can apply principle of
superposition to calculate stresses.
(topic covered in unit 1).
Resultant moments and
stresses
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Section Modulus
Section modulus is defined as ratio of moment of inertia about the
neutral axis to the distance of the outermost layer from the neutral axis
Z 
I
ymax
I

M 
y
I
M max

ymax
M  max
I
y max
M max Z
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Section Modulus of symmetrical
sections
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SECTION MODULUS
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1. A cantilever beam of length 2m fails when a load of 2KN is applied at the free end. If the section is
40mmx60mm, find the stress at the failure.
Solution:
Length of beam = 2m or 2000mm load at failure = 2KN
Section dimensions = 40mm X 60mm Calculation of moment of inertia
I = bd3/12
= (40) (603)/12
= 7.2X105mm4
Calculation of bending moment about fixed end M = WL
= (2)(2)
= 4KN-m
Calculation of bending stress
M /I= σ / y
Substitute for above (where y = depth /2= 60/2 = 30mm)
There fore
σ = 166.67N/mm2
2.A rectangular beam 200mm deep and 300mm wide is simply supported over the span of 8m. What uniformly
distributed load per metre the beam may carry, if the bending stress is not exceed 120N/mm2.
Solution:
Length of beam = 8m or 8000mm
Section dimensions = 300mm X 200mm
maximum bending stress = σ = 120N/mm2.
Calculation of bending moment for the above
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condition M = wL2/8
= w (8)2/8
= 8wX106
Calculation of moment of inertia I = bd3/12
= (300) (2003) /12
= 2X108 mm4
Calculation of Udl
M /I= σ / y
Substitute for above (where y = depth /2= 200/2 =100mm) 8wX106 /2X108= 120 / 100
w =3X104 N/m or 30 N/mm
3.A beam is simply supported and carries a uniformly distributed load of 40KN/m run over the whole span. The section of the beam is
rectangular having depth as 500mm.If the maximum stress in the material of the beam is 120N/mm2and moment of inertia of the
section is 7x108mm4, find the span of the beam.
Solution:
Depth of beam = 500mm
maximum bending stress = σ = 120N/mm2. moment of inertia =7x108mm4
Calculation of bending moment for the above condition
M = wL2/8
= 40(L)2/8
= 5L2
Calculation of length of beam
M /I= σ / y
Substitute for above (where y = depth /2= 500/2 =250mm) 5L2 /7x108= 120 / 250
L=8197.56 mm
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4. Calculate the maximum stress induced in a cast iron pipe of external diameter 40mm,of internal diameter 20mm and length
4m when the pipe is supported at its ends and carries a point load of 80N at its centre.
Solution:
Length of beam = 4m or 4000mm Internal diameter = 20mm
External diameter = 40mm
Point load for simply supported beam Calculation of maximum bending moment
M= W L /4
M = 80 X 4000 /4 M = 80 KN-m
Calculation of moment of inertia
I = π (D 4 –d4)/64
I = π (40 4 –20 4)/64
I = 117809.7mm4
M /I= σ / y
Substitute for above (where y = depth /2= 40/2 =20mm) 80X1000/117809.7 = σ / 20
σ = 13.58 N/mm2
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5.A rectangular beam 300mm deep is simply supported over a span of 4m. Determine the uniformly distributed load per
meter which the beam may carry, if the bending stress should not exceed 120N/mm2.Take I=8x106mm4.
Solution:
Length of beam = 4m or 4000mm Depth of the beam = 300mm
maximum bending stress = σ =120N/mm2 condition: udl for simply supported
beam I=8x106mm4
Calculation of maximum bending moment M= W L2 /8
M= W (4000)2 /8 M= 2 X106 W
Calculation of udl
M /I= σ / y
2 X106 W /8x106= 120 / 150 W = 3.2N/mm2
6.A square beam 20mmx20mm in section and 2m long is supported at the ends. The beam fails when a point load of 400N is
applied at the centre of the beam. What uniformly distributed load per meter length will break a cantilever of the same
material 40mm wide,60mm deep and 3m long?
Solution:
Step 1: Data: case 1: point load application at centre of the beam Length of beam = 2m or
2000mm
Cross section of the beam = 20mmx20mm simply supported beam
Calculation of maximum bending moment M= W L /4
M= (400) (2000) /4
M= 200x103
Calculation of moment of inertia
I = bd3/12
= (20) (203)/12
= 13333.33mm4
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M /I= σ / y
2 X105 /13333.33= σ / 10
σ = 150N/mm2
Calculation of magnitude of udl when dimensions of the beam are changed Length of beam =3m or 3000mm
Width of beam = 40mm Depth of beam = 60mm
Condition: cantilever beam
Calculation of maximum bending moment M= W L2 /2
M= W (3000)2 /2
Calculation of moment of inertia I = bd3/12
= (40) (603 ) /12
= 72x104mm4
Calculation of load
M /I= σ / y
W (3000)2 /2 /72x104= 150 / 30
W = 800N/m
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7. A timber beam of rectangular section is to support a load of 20KN uniformly distributed over a span of 3.6m when beam is
simply supported. If the depth is to be twice the breadth, and the stress in timber is not exceed 7N/mm2, find the dimensions
of the cross section. How could you modify the dimensions with 20KN of concentrated load is present at centre with same
breadth and depth ratio.
when simply supported beam of length 3.6m carries udl of 20KN and depth is twice the width We know that W = w L
= 20 X 1000X3.6
= 5.56N
Moment = WL/8
M = 5.56 X 1000X 3.6 /8
M = 2499.75 N-mm
Calculation of cross sectional dimensions of the beam
σ = 7N/mm2
M /I= σ / y
2499.75/(bd3/12) = 7/(d/2)
b = 8.12mm
d =2b = 16.24mm
when simply supported beam of length 3.6m carries point load of 20KN and depth is twice the width Moment = WL/4
M = 20 X 106X 3.6 /4 M = 18X 106 N-mm
Calculation of cross sectional dimensions of the beam
σ = 7N/mm2
M /I= σ / y
18X 106 /(bd3/12) = 7/(d/2)
b = 156.82mm
d = 2b = 313.65mm
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1.Two wooden planks 150 mm × 50 mm each are connected to form a T-
section of a beam. If a moment of 6.4 kN-m is applied around the horizontal neutral axis, induc- ing tension
below the neutral axis, find the bending stresses at both the extreme fibres of the cross- section.
Given: Size of wooden planks = 150 mm × 50 mm and
moment (M) = 6.4 kN-m = 6.4× 106
N-mm.
Two planks forming the T-section are shown in Fig. . First of all, let us find out the centre of gravity of the beam section. We
know that distance between the centre of gravity of the section and its bottom face,
Distance between the centre of gravity of the section and the upper extreme fibre,
yt = 20 – 125 = 75 mm
and distance between the centre of gravity of the section and the lower extreme fibre,
yc = 125 mm
We also know that Moment of inertia of the T section about an axis passing through its c.g. and parallel to the botom face,
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A rectangular beam 60 mm wide and 150 mm deep is simply supported
over a span of 4 metres. If the bneam is subjected to a uniformly distributed load of 4.5 kN/m,
find the maximum bending stress induced in the beam.
Given : Width (b) = 60 mm ; Depth (d) = 150 mm ; Span (l) = 4 m = 4 × 103
mm and
uniformly distributed load (w) = 4.5 kN/m = 4.5 N/mm.
We know that section modulus of the rectangular section,
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3.shows a rolled steel beam of an unsymmetrical I-section If the maximum bending stress
in the beam section is not to exceed 40 MPa, find the moment, which the beam can
resist.
Given: Maximum bending stress (max) = 40 MPa = 40 N/mm .
We know that distance between the centre of gravity of the section and bottom face,
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Shearing Stresses in Beams
Shearing Stress at a Section in a Loaded Beam
Consider a small portion ABDC of length dx of a beam loaded with uniformly distributed load as shown in Fig.
Shearing stress
We know that when a beam is loaded with a uniformly distributed load, the shear force and bending moment vary at every
point along the length of the beam.
LetM= Bending moment at AB,
M + dM= Bending moment at CD, F= Shear force at AB,
F + dF= Shear force at CD, and
I= Moment of inertia of the section about its neutral axis.
Now consider an elementary strip at a distance y from the neutral axis as shown in Fig. 16.1 (b).
Now let  = Intensity of bending stress across AB at distance y from the neutral axis and
a = Cross-sectional area of the strip.
We have already discussed that
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Distribution of Shearing Stress
we shall discuss the distribution of shear stress over the following sections:
1.Rectangular sections,
2.Triangular sections,
3.Circular sections,
4.I-sections,
5.T-sections
Distribution of Shearing Stress over a Rectangular Section
Consider a beam of rectangular section ABCD of width and depth as shown in Fig. . We know that the shear
stress on a layer JK of beam, at a distance y from the neutral axis,
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F= Shear force at the section,
A= Area of section above y (i.e., shaded area AJKD),
y = Distance of the shaded area from the neutral axis,
A y = Moment of the shaded area about the neutral axis,
I= Moment of inertia of the whole section about its neutral axis, and
b= Width of the section.
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1.A wooden beam 100 mm wide, 250 mm deep and 3 m long is carrying a uniformly distributed
load of 40 kN/m. Determine the maximum shear stress and sketch the variation of shear stress
along the depth of the beam.
SOLUTION. Given: Width (b) = 100 mm ; Depth
(d) = 250 mm ; Span (l) = 3 m = 103
mm and
uniformly distributed load (w) = 40 kN/m = 40 N/mm.
We know that shear force at one end of thebeam,
and area of beam section,
A=b · d = 100 × 250 = 25 000 mm
Average shear stress across the section,
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Distribution of Shearing Stress over an I-Section
I-section.
Consider a beam of an I-section as shown in Fig. 16.8 (a) LetB= Overall width of the
section,
D= Overall depth of the section,
d= Depth of the web, and
b= Thickness of the web.
We know that the shear stress on a layer JK at a distance y from the neutral axis,
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1.An I-sections, with rectangular ends, has the following dimensions:Flanges =150 mm × 20 mm,
Web = 300 mm 10 mm.Find the maximum shearing stress developed in the beam for a shear
force of 50 kN.
SOLUTION. Given: Flangewidth (B) = 150 mm ; Flangethickness= 20 mm ;
Depth of web (d) = 300 mm; Width of web = 10 mm;
Overall depth of the section (D) = 340 mm and
shearing force (F) =50 kN = 50  103
N.
We know that moment of inertia of the I-section about its centre of gravity
and parallel to x-x axis,
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2.An I-section beam 350 mm  200 mm has a web thickness of 12.5 mm and
a flange thickness of 25 mm. It carries a shearing force of 200 kN at a section. Sketch the shear stress
distribution across the section.
Given:
Overall depth (D) = 350 mm ; Flange width (B) = 200 mm ;
Width of Web =12.5 mm ; Flange thickness = 25 mm and
the shearing force (F) = 200 kN = 200  103
N.
We know that moment of inertia of the I-section about it centre of gravity and parallel to x-x axis,
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1. Moment of inertia acting on a semi-circle about symmetrical axes is given as
_______
a. 1.57 r4
b. 0.055 r4
c. 0.392 r4
d. 0.11 r4
Answer: c
2. What is the moment of inertia acting on a rectangle of width 15 mm and depth
40 mm about base by using theorem of parallel axes?
a. 320 x 103 mm4
b. 300 x 103 mm4
c. 240 x 103 mm4
d. 80 x 103 mm4
Answer: a
3. What is the S.I. unit of sectional modulus?
a. mm4
b. mm3
c. mm2
d. m
Answer: b
4. What is the moment of inertia acting on a circle of diameter 50 mm?
a. 122.71 x 103 mm4
b. 306.79 x 103 mm4
c. 567.23 x 103 mm4
d. 800 x 103 mm4
Answer: b
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5. Which of the following relations is used to represent theorem of perpendicular
axes? (H = Vertical axis, I = Moment of inertia and K = Radius of gyration)
a. IPQ = Ixx + AH2
b. IPQ = Ixx + Ak2
c. Izz = Ixx + Iyy
d. Izz + Ixx + Iyy = 0
Answer: c
6. What is the moment of inertia acting on a semicircle of radius 20 mm about the
asymmetrical axes?
a. 125.663 x 103 mm4
b. 17600 mm4
c. 1500 mm4
d. 8800 mm4
Answer: b
7. What is the product of sectional modulus and allowable bending stress called as?
a. Moment of inertia
b. Moment of rigidity
c. Moment of resistance
d. Radius of gyration
Answer: c
8. A uniformly distributed load of 20 kN/m acts on a simply supported beam of
rectangular cross section of width 20 mm and depth 60 mm. What is the maximum
bending stress acting on the beam of 5m?
a. 5030 Mpa
b. 5208 Mpa
c. 6600 Mpa
d. Insufficient data
Answer: b
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9. The bending formula is given as _____
a. (M/E) = (σ/y) = (R/I)
b. (M/y) = (σ/I) = (E/R)
c. (M/I) = (σ/y) = (E/R)
d. none of the above
Answer: c
10. Which of the following laminas have same moment of inertia (Ixx = Iyy), when passed
through the centroid along x-x and y-y axes?
a. Circle
b. Semi-circle
c. Right angle triangle
d. Isosceles triangle
Answer: a
11. What is the average shear stress acting on a rectangular beam, if 50 N/mm2 is the
maximum shear stress acting on it?
a. 31.5 N/mm2
b. 33.33 N/mm2
c. 37.5 N/mm2
d. 42.5 N/mm2
Answer: b
12. The ratio of maximum shear stress to average shear stress is 4/3 in _______
a. circular cross-section
b. rectangular cross-section
c. square cross-section
d. all of the above
Answer: a
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13. What is the shear stress acting along the neutral axis of triangular beam section, with base 60 mm and
height 150 mm, when shear force of 30 kN acts?
a. 15.36 N/mm2
b. 10.6 N/mm2
c. 8.88 N/mm2
d. Insufficient data
Answer: c
14. A circular pipe is subjected to maximum shear force of 60 kN. What is the diameter of the pipe if
maximum allowable shear stress is 5 Mpa?
a. 27.311 mm
b. 75.56 mm
c. 142.72 mm
d. 692.10 mm
Answer: c
15. A square object of 4 mm is subjected to a force of 3000 N. What is the maximum allowable shear
stress acting on it?
a. 250.14 mm2
b. 281.25 mm2
c. 400.32 mm2
d. 500 mm2
Answer: b
16. The average shear stress in a beam of circular section is _______ times the maximum shear stress.
a. 0.75
b. 1.5
c. 4/3
d. Equal
Answer: a
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17. What is the shear stress acting along the neutral axis, over a triangular section?
a. 2.66 (S/bh)
b. 1.5 (S/bh)
c. 0.375 (S/bh)
d. None of the above
Answer: a
18. Maximum shear stress in a triangular section ABC of height H and base B occurs at _________
a. H
b. H/2
c. H/3
d. neutral axis
Answer: b
19. The shear stress acting on the neutral axis of a beam is _____
a. maximum
b. minimum
c. zero
d. none of the above
Answer: a
20. Which of the machine component is designed under bending stress?
a. Shaft
b. Arm of a lever
c. Key
d. Belts and ropes
Answer: b
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1.Derive the formula for section modulus for a beam circular cross section ?
Oct/Nov(2019)
Ans: CIRCULAR SECTION :
For a circular x-section, the polar moment of inertia may be computed in the following
manner
Consider any circular strip of thickness r located at a radius 'r'.
Than the area of the circular strip would be dA =
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2.Derive the formula for section modulus for a beam of rectangular cross section? Oct/Nov(2019)
Rectangular Section:
For a rectangular x-section of the beam, the second moment of area may be computed as below :
Consider the rectangular beam cross-section as shown above and an element of area dA , thickness dy
breadth B located at a distance y from the neutral axis, which by symmetry passes through the centre of
section. The second moment of area I as defined earlier would be
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Thus, for the rectangular section the second moment of area about the neutral axis i.e., an axis through
the centre is given by
Similarly, the second moment of area of the rectangular section about an axis through the lower edge of
the section would be found using the same procedure but with integral limits of 0 to D .
Therefore
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