Q.P. 70863
STRENGTH OF MATERIALS (DEC 2019 SEM3 MECHANICAL)
Q1)
a) Derive an expression for the strain energy due to suddenly applied load. (5)
Ans: Consider a member of length L, cross-sectional area A
And subjected to suddenly applied load P. let the instantaneous
Deformation of member be ๐ฟ๐ฟ๐ and the corresponding
Instantaneous stress developed be ๐๐ . The load verses deformation
Curve would be a straight line.
Now work done by the load
= area under load-deformation curve.
= ๐ x ๐ฟ๐ฟ๐ .
We know strain energy U=work done.
๐ = ๐ x ๐ฟ๐ฟ๐ .
๐ =๐2๐ด๐ฟ
2๐ธ.
๐ x ๐ฟ๐ฟ๐ = ๐๐
2๐ด๐ฟ
2๐ธ.
๐๐ = 2๐
๐ด. โฆโฆโฆโฆโฆ.(1)
But stress due to gradually applied load = f= ๐
๐ด .
๐๐ =2f. โฆโฆโฆโฆโฆ(2)
Also ๐ฟ๐ฟ๐ =2 ๐ฟ๐ฟ. Equations 1 and 2 indicate that sudden loads produce twice the stress thereby
resulting in twice the strains as compared to the same loads when gradually
applied.
P
P
Load
0 deformation
L
๐ฟ๐ฟ
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b) Derive the relation between load, shear force and bending
moment (5)
Ans:
Fig. (a) shows a beam carrying a general loading. Consider an element of the
beam formed by taking two cutting sections m-m and n-n distance dx apart. The
general loading over the element can be assumed to be a u.d.l. of intensity w per
unit run since dx is very small.
On the right face of the element the shear resistance force V and moment
resistance M is developed. On the left face the shear resistance force has an
increment of dV and becomes V + DV, while the moment resistance M increases
by dM and becomes M + dM. Since the beam is in equilibrium, an element of the
beam must also be in equilibrium.
Applying Conditions of Equilibrium to the element
Using Fy = 0 + ve
(v + dV) - V - w dx = 0
.: dV = w dx or dV
dx=W .......... 1
Equation 1 implies that the rate of change of shear force at any section represents
the rate of loading at that section.
Now using ฦฉM = 0 +ve
Taking moments about section m-m
(M + dM) - M - V dx โ (w dx) x โ ๐
2 =0.
m n dx/2 w dx
V+dV M
A B
M+dM V
m n
(a) (b)
dx dx
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dM - V dx - w (โ ๐)2
2 =0.
The square of a small quantity can be neglected, we therefore have
dM = Vdx or dM
dx=V ....... 2
Equation 2 implies that the rate of change of bending moment at any section
represents the shear force at that section.
c) Write the assumptions made in theory of pure torsion and derive torsional
formula. (5)
Ans: Assumptions made in pure torsion theory:
The material of the shaft is homogenous and isentropic throughout its length.
Circular sections remains circular and planar.
The twisting moment T acts in a plane perpendicular to the shaft axis.
At any section, the radial lines remain straight.
Hookeโs law is valid.
Derivation of Torsion Equation
Consider a solid cylindrical shaft of radius R and length L be subjected to an equal
and opposite twisting torques of magnitude T at the ends of the shaft such that the
axis of the torques coincide with the shaft axis o-o. The shaft is therefore in pure
torsion.
Consider a straight fibre AB on the surface of the shaft and parallel to the shaft
axis o-o. Now on application of the twisting torque T, the initially straight fibre AB
gets twisted into a helix AC. The corresponding angle BAC =ษธ represents the shear
strain of the fibre. Let ๐ be the corresponding shear stress developed in the fibre.
Shear strain of the fibre = ษต = ๐ต๐ถ
๐ฟ .
Now let ษต be the angular movement of the radial line OB to new position OC.
From geometry Arc BC = R x ษต.
.:Shear strain of the fibre = ษต = ๐ ๐ฅ ษต
๐ฟ .
We know , Modulus of Rigidity G=๐ โ๐๐๐ ๐ ๐ก๐๐๐ ๐
๐ โ๐๐๐ ๐ ๐ก๐๐๐๐ .
G= ๐
ษธ.
๐ = G. ษธ
A
C
B L
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๐ = G x ๐ ๐ฅ ษต
๐ฟ .
๐
๐ =
G.ษต
๐ฟ โฆโฆโฆโฆ.(1)
consider a cutting section m-m taken to cut the shaft into two parts. Consider the
equilibrium of the left part of the shaft.
Now consider an elemental area dA at a distance r from o subjected to shear stress
q. Let ๐ be the shear stress in a fibre on the shaft surface i.e.at distance R from o.
we know that the shear stress distribution along the radial line varies linearly with
the radial distance from the axis of the shaft.
Shear stress on the elemental area= q= ๐
๐ x ๐ .
Hence, shear resistance force developed by elemental area = qdA = ๐
๐ x ๐ dA.
Now. Moment of resistance developed by the elemental area = ๐
๐ x ๐ dA x r.
Total moment of resistance offered by the cross section of the shaft
=ฯ
Rโซ r2 =
ฯ
Rร J.
Here, โซ r2 โ A = J is the polar moment of inertia of the shaft section about the axis
of the shaft.
Now
Torque transmitted by shaft = Moment of resistance t developed in the Shaft
T =ฯ
Rร J
T
R=
I
J โฆโฆโฆ..(2)
Comparing eqn 1 and 2, we get
ฯ
Rโ
T
J=
Gฮธ
L
Above equation is referred to as the equation for pure torsion.
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d) Draw shear stress distribution diagram for symmetry I section, T section
and rectangular section. (5)
Ans: 1. Rectangular Section.
Consider a rectangular section of dimensionsโ b x d.
The N.A passes through the centroid at d/2 from the base.
Consider a layer m-m at distance y from the N.A.
area above layer m-m, A=b x (d
2โ y)
Distance of the centroid of shaded area from N.A,
yฬ = y +1
2(
d
2โ y)
= y
2+
d
4=
1
2(
d
2+ y)
Using shear stress equation
๐ =vร(Axyฬ )
Iโ b.
๐mโm =v[b(
โ
2xy)ร
1
2(
โ
2+y)]
b โ 3
12รb
.
๐mโm = 6V
bd3 (d2
4โ y2) โฆโฆ.(1) General relation for shear stress
The above equation indicates parabolic nature of shear stress variation
At top and bottom edge y =โ
2 , ๐=0.
Also ๐ would be maximum when y=0
๐ฬ max =3v
2bd ... (2) Relation for maximum shear stress
For a rectangular section (bxd) subjected to vertical shear V.
Average shearing stress = ๐๐๐ฃ๐๐๐๐๐ =๐ โ๐๐๐ ๐๐๐๐๐
๐ โ๐๐๐ ๐๐๐๐ . ๐๐๐ฃ๐๐๐๐๐ =
๐
๐ ๐ โ โฆ..(3)
From equations (2) and (3)
๐๐๐๐ฅ=1.5 ๐๐๐ฃ๐๐๐๐๐
This shows that for a rectangular section the maximum shear stress which occurs at
the NA is 1.5 times the mean or average shear stress resisted by the section.
d
yฬ y
b
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2. symmetry I section
Consider a symmetrical I section of flange width b,
Web width b, web depth d and overall depth D.
The N>A passes through the centroid at D/2
From the base.
Shear stress analysis in the flange.
Consider a layer m-m at a distance y from N.A
Width of layer m-m=B
area above layer m-m, A=B(๐ท
2โ ๐ฆ).
Distance of centroid of shaded area from N.A
C =D
2โ
(D
2โy)
2=
(D
2+y)
2 .
Using ฯ =๐ ๐ (๐ด ๐ ๐)
๐ผ.๐ .
ฯ ๐โ๐(๐๐๐๐๐๐)=vxB
IxB(
D
2โ y)
(D
2+y)
2.
ฯ ๐โ๐(๐๐๐๐๐๐)= v
2I(
D2
4โ y2). โฆ.general relation for shear stress in flange.
Shear stress analysis in the web.
Consider a layer nโn at a distance y from the N.A
Width of layer n-n= b.
Shaded area above layer n-n, A=๐ด1+ ๐ด2.
Where ๐ด๐ด1 = ๐ต (๐ท
2โ
โ
2) and ๐ด2 = ๐ (
โ
2โ ๐ฆ)
Distance of centroid of ๐ด1 from N.A
Y1ฬ =๐ท
2โ
1
2(
๐ท
2โ
โ
2) =
๐ท
4+
โ
4 =
๐ท
2+
๐
2
2 .
Distance of centroid ๐ด2 from N.A yฬ 2 = ๐ฆ +๐
2โ๐ฆ
2=
๐ฆ+๐
2
2 .
Using ฯ =V x (A x yฬ )
๐ผ.๐ .
Using ฯ๐โ๐ = ๐ ร๐ด1ร ๏ฟฝฬ ๏ฟฝ1+๐ด2ร๏ฟฝฬ ๏ฟฝ2
๐ผ.๐ .
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= v
Ib[B (
D
2โ
d
2)
(D
2+
โ
2)
2+ b (
d
zโ y)
(y+โ
2)
2] .
= ๐
8๐ผ๐[ ๐ต(๐ท2 โ ๐2) + ๐(๐2 โ 4๐ฆ2)] . โฆโฆโฆ. general relation
for shear stress
The maximum shear stress is at y=0 i.e. at N.A.
ฯ๐๐๐ฅ =๐
8๐ผ๐[ ๐ต(๐ท2 โ ๐2) + ๐2] โฆโฆ relation for max shear stress.
Shear stress in flange at the flange and web junction i.e. at ๐ฆ =โ
2 .
ฯ๐ฝโ๐ฝ(๐๐๐๐๐๐) =๐
2๐ผ(
๐ท2
2โ
โ 2
2) =
๐
8๐ผ(๐ท2 โ ๐2) .
Shear stress in web at the flange and web junction i.e. at ๐ฆ =โ
2 .
ฯ๐ฝโ๐ฝ(๐ค๐๐) =๐
8๐ผ๐[๐ต(๐ท2 โ ๐2) + ๐ (๐2 โ
4โ 2
4)
= ฯ๐ฝโ๐ฝ(๐ค๐๐) =๐
8๐ผ๐[๐ต(๐ท2 โ ๐2)]. โฆโฆโฆ (2)
From equations (1) and (2) we understand that the stress in web drastically
increases by ๐ต
๐ times the corresponding stress in the flange in the junction
Layer J-J of flange and web.
e) write the assumption for simple bending and derive and derive the flexural
formula. (5)
Ans: assumptions made in theory of simple bending:
The material of beam is homogenous and isotropic.
The beam is initially straight and all the longitudinal fibres bend in circular
arcs with a common centre of curvature.
Members have symmetric cross sections and are subjected to bending in the
plane of symmetry.
The beam is subjected to pure bending and the effect of shear is neglected.
Plane(transverse) sections through a beam taken normal to the axis of the
beam remain plane after the beam is subjected to bending.
The radius of curvature is large as compared to the dimensions of the beam.
Flexural formula:
Consider a beam section under pure sagging bending. Here the layers above
The N.A are subjected to compressive forces. The magnitude of this
Compressive force is proportional to the location of the layer from the N.A
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similarly, the layers below the N.A are subjected to the tensile forces in all the
Layers about the N.A is known as the moment of resistance of the section.
The moment of resistance is a result of bending and is a resisting moment to
The applied resistance is result of bending and is a resisting moment to the
Applied bending moment.
We know that, force on layer=๐ธ
๐ . ๐ฆ. ๐๐ด .
Moment of this force about N.A =๐ธ
๐ . ๐ฆ. ๐๐ด. ๐ฆ.
= ๐ธ
๐ . ๐ฆ2. ๐๐ด.
Total moment of forces in different layers
= moment of resistance = โซ ๐ธ
๐ . ๐ฆ2. ๐๐ด.
= ๐ธ
๐ โซ ๐ฆ2. ๐๐ด.
If M is the external bending moment, then the moment of resistance should be
equal to the bending M.
๐ =๐ธ
๐ โซ ๐ฆ2. ๐๐ด.
But โซ ๐ฆ2. ๐๐ด. represents the second moment of area or the moment of inertia I
Of the section about the N.A
๐ =๐ธ
๐ ๐ ๐ผ or
๐
๐ผ=
๐ธ
๐ โฆโฆโฆโฆ.(1)
But ๐
๐ฆ=
๐ธ
๐
Comparing it with equation (1), we get.
๐
๐ฆ=
๐
๐ผ=
๐ธ
๐ .
Since the stress f is due to bending we denote it as ๐๐ and would refer to it as
Bending stress.
๐๐
๐ฆ=
๐
๐ผ=
๐ธ
๐ . โฆโฆ equation of bending or flexure equation.
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f) find the maximum power that can be transmitted through 50mm diameter
shaft at 150 rpm, if the maximum permissible shear stress is 80 N/๐๐๐.(5)
Ans: N= 150 rpm.
๐ =2๐๐
60 =
2๐ร150
60= 15.70 rad/s.
D= 50 mm.
๐ = 80 ๐/๐๐2.
Polar moment of inertia ๐ฝ =๐
32๐4.
โด ๐ฝ =๐
32ร (50)4.
๐ฝ = 245.43 ๐๐4.
Using torsions equations
๐
๐ =
๐
๐ฝ
80
25=
๐
245.43.
โด T= 785.376 N-mm.
But
P=T x ๐
= 785.376 x 15.70
= 12330.4 W.
Power developed by the shaft = 1.23 kW
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Q.2)
a) A bar of brass 20mm is enclosed in a steel tube of 40mm external diameter
and 20mm internal diameter. The bar and the tubes are initially 1.2 m long
and are rigidly fastened at both ends. If the temperature is raised by 60ยฐC,
find the Stresses induced in the bar and the tube.
Given: ๐ฌ๐= ๐ ร ๐๐๐ N/๐๐๐.
๐ฌ๐ = ๐ ร ๐๐๐ ๐ต/๐๐๐.
๐ถ๐ = ๐๐. ๐ ร ๐๐โ๐/ยฐ๐ช.
๐ถ๐ = ๐๐. ๐ ร ๐๐โ๐/ยฐ๐ช. (10)
Ans:
Steel tube
Brass bar steel tube
d= 20mm. ๐0= 40mm.
L=1.2mm=1200mm. ๐๐= 20mm.
A=๐
4ร ๐2. A=
๐
4ร (๐0
2 โ ๐๐2).
=๐
4ร (20)2. = 942.47 N/๐๐2.
= 314.15 ๐๐2. ๐ธ๐ = 2 ร 105 ๐/๐๐2.
๐ธ๐ = 1 ร 105 ๐/๐๐2. ๐ผ๐ = 11.6 ร 10โ6/ยฐ๐ถ.
๐ผ๐ = 18.7 ร 10โ6/ยฐ๐ถ.
T=60ยฐ๐ถ.
Load sharing relation:
Force in steel tube= force in brass bar.
๐๐ = ๐๐.
๐๐ ๐ด๐ = ๐๐๐ด๐.
๐๐ ร 942.47 = ๐๐ ร 314.15.
๐๐ = 0.333๐๐. โฆโฆโฆ.. (1)
Brass bar
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Strain relation:
๐ฟ๐ฟ๐ ๐ก๐๐๐ = ฮด๐ฟ๐๐๐๐ ๐ .
[๐๐๐๐ ๐๐ฅ๐๐๐๐ ๐๐๐ + ๐๐๐๐ข๐๐๐ ๐๐ฅ๐๐๐๐ ๐๐๐]๐ ๐ก๐๐๐ =
[๐๐๐๐ ๐๐ฅ๐๐๐๐ ๐๐๐ โ ๐๐๐๐ข๐๐๐ ๐๐๐๐ก๐๐๐๐ก๐๐๐]๐๐๐๐ ๐ .
[๐ผ๐๐ฟ +๐๐ฟ
๐ด๐ธ]
๐ ๐ก๐๐๐= [๐ผ๐๐ฟ +
๐๐ฟ
๐ด๐ธ]
๐๐๐๐ ๐ .
[(11.6 ร 10โ6) ร 60 +๐๐
2ร105] = [(18.7 ร 10โ6) ร 60 โ
๐๐
105].
0.5๐๐ +๐๐=42.6.
Substituting value of ๐๐ from eqn 1 in eqn 2, we get.
0.5(0.333๐๐) + ๐๐ = 42.6.
๐๐ = 36.51 ๐/๐๐2.
๐๐ = 0.333 ร ๐๐
๐๐ = 0.333 ร 36.51.
๐๐ = 12.15 ๐/๐๐2.
b) the state of stress at a point in a strained material is as shown in fig.
Determine
(1) the direction of principal planes.
(2) the magnitude of maximum shear stress .
Indicate the direction of all the above by a sketch. (10)
Ans: given:
๐๐ฅ = 200 N/๐๐2.
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๐๐ฆ = 150 N/๐๐2.
๐๐ฅ๐ฆ = 100 N/๐๐2.
๐1,2 = (๐๐ฅ+๐๐ฆ
2) ยฑ โ(
๐๐ฅโ๐๐ฆ
2)
2+ ๐๐ฅ๐ฆ
2 .
= 175 ยฑ โ252 + 1002 .
= 175 ยฑ 103.077.
๐1 = 278.077 ๐/๐๐2.
๐2 = 71.923 ๐/๐๐2 .
๐๐๐๐ฅ = 103.077 ๐/๐๐2.
tan 2๐๐ =2 ๐๐ฅ๐ฆ
๐1โ ๐2.
tan 2๐๐ =2ร100
50 = 4.
โด ๐๐ 1 = 37.98ยฐ and ๐๐2 = 90 + 37.98= 127.98ยฐ .
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Q3)
a) find slope at point A and B, deflections at points C and D for a beam as
shown in fig. also find the maximum deflection. Take E= 200GPa and
I=๐๐๐ ๐๐๐ . (10)
Ans: 12 kN/m 5kN
๐ป๐ต 3m 3m 3m 2m
๐ ๐ด ๐ ๐ต
Finding support reactions:
Taking moment about point A: +ve.
12 ร 3 + 5 ร 9 โ ๐ ๐ฉ ร 11 = 0.
โด ๐ ๐ฉ = 7.36 ๐๐.
โ๐น๐ = 0. +ve.
๐ ๐จ + ๐ ๐ฉ โ 12 โ 5 = 0.
๐ ๐จ + 7.36 โ 12 โ 5 = 0
โด ๐ ๐จ = 9.64 ๐๐.
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x
2(๐ฅโ8)2
2 kN/m 5kN
๐ป๐ต (x-8)m 3m 3m 2m
np
๐ ๐ด x ๐ ๐ต
Taking section x-x near support A.
Distance of section x-x from point B=X
Taking moment about section x-x.
โด โ ๐ ๐ต ร ๐ + 5 ร (๐ โ 2) +2(๐ โ 8)
2ร (๐ โ 8) = ๐๐ฅ๐ฅ
Now we have,
๐ธ๐ผ๐2๐ฆ
๐๐ฅ2= ๐
๐ธ๐ผ๐2๐ฆ
๐๐ฅ2= โ7.36 ร ๐ + 5 ร (๐ โ 2) + (๐ โ 8)2
Integrating both the sides w.r.t X, we get.
๐ธ๐ผโ ๐ฆ
โ ๐ฅ= โ
7.36๐2
2+
5(๐โ2)2
2+
(๐โ8)3
3+ ๐1. โฆ..(eqn for slope).
Integrating both sides w.r.t X, we get.
๐ธ๐ผ. ๐ฆ = โ7.36๐3
2ร3+
5(๐โ2)3
6+
(๐โ8)4
4+ ๐ถ1๐ + ๐ถ2. โฆ..(eqn for deflection).
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applying boundary conditions:
At X=0 y=0.
๐ธ๐ผ. ๐ฆ = โ7.36๐3
2 ร 3+
5(๐ โ 2)3
6+
(๐ โ 8)4
4+ ๐ถ1๐ + ๐ถ2.
โด ๐ธ๐ผ(0) = 0 + 0 + ๐ถ2
โด ๐ถ2 = 0. Now
At X=11 y=0.
๐ธ๐ผ. ๐ฆ = โ7.36๐3
2 ร 3+
5(๐ โ 2)3
6+
(๐ โ 8)4
4+ ๐ถ1๐ + ๐ถ2.
0 = โ7.36(11)3
6+
5(6)3
6+ 0 + 8๐ถ1 + 0
๐ถ1 = 181.58
Now slope equation :
๐ธ๐ผ๐๐ฆ
๐๐ฅ= โ
7.36๐2
2+
5(๐ โ 2)2
2+
(๐ โ 8)3
3+ 181.58.
Slope at points A and B.
At point A X=0
โด ๐ธ๐ผ๐๐ฆ
๐๐ฅ= 0 + 181.58
โ y
โ x=
181.58
๐ธ๐ผ
= 181.58ร109
200ร108= 9.079๐๐๐
At point B X=11
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๐ธ๐ผ๐๐ฆ
๐๐ฅ= โ
7.36(11)2
2+
5(11 โ 2)2
2+
(11 โ 8)3
3+ 181.58.
dy
dx= โ2.91๐ad
Deflection:
๐ธ๐ผ. ๐ฆ = โ7.36๐3
2 ร 3+
5(๐ โ 2)3
6+
(๐ โ 8)4
4+ ๐ถ1๐ + ๐ถ2.
Deflection at point C
X=6
๐ธ๐ผ. ๐ฆ = โ7.36(6)3
2 ร 3+
5(6 โ 2)3
6+ 181.58 ร 6
y=43.89mm.
Deflection at point D.
X=9
๐ธ๐ผ. ๐ฆ = โ7.36(9)3
2 ร 3+
5(9 โ 2)3
6+
(1)4
4+ 181.58 ร 6
y= 47.74mm.
maximum deflection .
X= 11 2
๐ธ๐ผ. ๐ฆ = โ7.36(5.5)3
2 ร 3+
5(5.5 โ 2)3
6+ 181.58 ร 6
๐ฆ๐๐๐ฅ = 46.056 ๐๐.
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b) draw SF and BM diagram for the beam shown in fig. (10)
20kN/m 60kN 20kN
30kN
A B
2m 4m 3m 2m 2m
Ans:
40kN 60kN 20kN
30kN
1m 1m A ๐ป๐ด B
2m 4m 3m 2m 2m
๐ ๐ด ๐ ๐ต
Support reaction calculation:
โ๐๐ด = 0 +ve.
โ20 ร 11 + ๐ ๐ต ร 9 โ 30 โ 60 ร 4 + 40 ร 1 = 0.
๐ ๐ต = 50๐๐.
โ๐น๐ฆ = 0. +ve.
โ40 + ๐ ๐ด โ 60 + ๐ ๐ฉ โ 20 = 0.
๐ ๐ด = 70๐๐.
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Shear force calculation: -ve
๐๐น๐น(๐ฝ๐ ) = 0.
๐๐น๐น(๐ฝ๐ฟ) = 20๐๐.
๐๐น๐ต(๐ฝ๐ ) = 20๐๐.
๐๐น๐ฉ(๐ฝ๐ฟ) = โ๐ ๐ต + 20 = โ50 + 20 = โ30๐๐.
๐๐น๐ท(๐ฝ๐ ) = โ30๐๐.
๐๐น๐ท(๐ฝ๐ฟ) = 60 โ 30 = 30๐๐.
๐๐น๐ด(๐ฝ๐ ) = 30๐๐.
๐๐น๐ด(๐ฝ๐ฟ) = โ70 + 30 = โ40๐๐.
Bending moment calculation +ve
๐ต๐๐น = 0. ๐ต๐๐ต = โ20 ร 2 = โ40. ๐ต๐๐ธ(๐ฝ๐ ) = 50 ร 2 โ 20 ร 4 = 20.
๐ต๐๐ธ(๐ฝ๐ฟ) = 20 โ 30 = โ10.
๐ต๐๐ท = โ20 ร 7 + 50 ร 5 โ 30
= 80kN.
๐ต๐๐ด = โ20 ร 11 + 50 ร 9 โ 30 โ 60 ร 4. = โ40 .
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40kN 60kN 20kN
30kN
1m 1m A ๐ป๐ด B
2m 4m 3m 2m 2m
๐ ๐ด ๐ ๐ต
30 30 20 20
SFD
-30 -30
80
20
BMD
0 -10 0
P s
-40 -40
-40
0
0
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Q4)
a) A vertical column of rectangular section is subjected to a compressive
load of P=800kN as shown in fig. find the stress intensities at the four
corners of the column. (10)
Ans: ๐๐ฅ๐ฅ = 0.2๐ = 200๐๐. ๐๐ฆ๐ฆ = 0.6๐ = 600๐๐.
Direct stress= ๐
๐ด.
=800ร103
2000ร1000.
= 0.4 ๐/๐๐2.
Bending stress ๐๐๐ฅ๐ฅ due to eccentricity w.r.t x axis:
๐๐๐ฅ๐ฅ = ยฑ๐ ๐๐ฅ๐ฅ ๐ฆ๐ฅ๐ฅ
๐ผ๐ฅ๐ฅ.
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๐ผ๐ฅ๐ฅ =(๐โ )3
12=
2000ร(1000)3
12= 16.6 ร 1010๐๐4.
๐๐๐ฅ๐ฅ = ยฑ800ร103ร200ร500
16.66ร1010.
๐๐๐ฅ๐ฅ = ยฑ0.481 ๐/๐๐2 . (side AB=+ve, side CD=-ve).
Bending stress ๐๐๐ฆ๐ฆ due to eccentricity w.r.t y axis:
๐๐๐ฆ๐ฆ = ยฑ๐ ๐๐ฅ๐ฅ ๐๐ฆ๐ฆ
๐ผ๐ฆ๐ฆ.
๐ผ๐ฆ๐ฆ =๐โ 3
12=
1000ร(2000)3
12= 66.67 ร 1010 ๐๐4.
๐๐๐ฆ๐ฆ = ยฑ800ร103ร600ร1000
66.67ร1010
๐๐๐ฆ๐ฆ = ยฑ0.71 ๐/๐๐^2. (side CB=+ve, side AD =-ve).
The resultant stresses at the corners would be addition of direct stress and bending
stress along x any y axis.
๐๐ = ๐โ + ๐๐๐ฅ๐ฅ + ๐๐๐ฆ๐ฆ .
At corner A ๐๐ ๐ด= 0.4 + 0.481 โ 0.71 = 0.171 ๐/๐๐2 (compressive).
At corner B ๐๐ ๐ต=0.4 + 0.481 + 0.71 = 1.591 ๐/๐๐2 (compressive).
At corner C ๐๐ ๐ถ = 0.4 โ 0.481 + 0.71 = 0.629 ๐/๐๐2 (compressive).
At corner D ๐๐ ๐ท = 0.4 โ 0.481 โ 0.71 = โ0.791 ๐/๐๐2 (tensile).
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b) a propeller shaft is required to transmit 50kW power at 500rpm. It is a
hollow shaft, having an inside diameter 0.6 times of outside diameter and
permissible shear stress for shaft material is 90๐ต/๐๐๐. Calculate the
inside and outside diameters of the shaft. (10)
Ans: P=50Kw = 50 ร 103 ๐.
N=500 rpm.
๐ =2๐๐
60= 52.35 ๐๐๐/๐ .
๐๐ = 0.6 ร ๐๐. [๐] = 90 ๐/๐๐2 . ๐๐ =?
๐๐ =?
Polar moment of inertia =๐ฝ =๐
32(๐๐
4 โ ๐๐4).
๐ฝ =๐
32(๐๐
4 โ (0.6๐๐)4).
๐ฝ = 0.085 ๐๐4 .
Using torsion equation.
๐
๐ =
๐
๐ฝ
90
๐๐4
2=
955.10
0.085โ ๐44.
๐๐ = 3.966 ๐๐. ๐๐ = 0.6 ร 3.966 = 2.37 ๐๐.
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Q5)
a) A cylindrical shell is 3m long and 1.2m in diameter and 12mm thick is
Subjected to internal pressure of 1.8 N/๐๐๐ calculate change in
dimensions and Volume of shell. Take ๐ฌ = ๐๐๐ ๐๐ต/๐๐๐, 1/m=0.3. (10)
Ans: given:
L= 3m= 3000mm.
d= 1.2m=1200mm.
t= 12mm.
p= 1.8 ๐/๐๐2.
E= 210 ร 103 ๐/๐๐2.
1/m=0.3.
๐ฟโ =? (change in diameter).
๐ฟ๐ฟ=? (change in length).
ฮดv=? (change in volume).
Circumferential stress
๐๐ =๐โ
2๐ก=
1.8ร1200
2ร12= 90 ๐/๐๐2.
Longitudinal stress
๐๐ =๐โ
4๐ก=
2160
48= 45 ๐/๐๐2 .
Circumferential strain
๐๐ =๐๐
๐ธโ
1
๐
๐๐
๐ธ= 3.642 ร 10โ4 .
Longitudinal strain
๐๐ =๐๐
๐ธโ
1
๐ ๐๐
๐ธ= 8.57 ร 10โ5
also,
๐๐ =๐ฟโ
โ
โด ๐ฟ๐ = ๐๐ ร ๐
๐ฟ๐ = 0.437๐๐
Now,
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๐๐ =๐ฟ๐ฟ
๐ฟ
โด ๐ฟ๐ฟ = ๐๐ฟ ร ๐ฟ. ๐ฟ๐ฟ = 0.2571 ๐๐.
Volumetric strain ๐๐ฃ = 2๐๐ + ๐๐ฟ
๐๐ฃ = 2(3.642 ร 10โ4) + 8.57 ร 10โ5
๐๐ฃ = 8.141 ร 10โ4
Since,
๐๐ฃ =๐ฟ๐
๐
โด ๐ฟ๐ = ๐๐ฃ ร ๐.
๐ฟ๐ = ๐๐ฃ ร๐
4(๐)2๐ฟ
โด ๐ฟ๐ = 2.762 ร 1010
b) A simply supported beam of length 3m and a cross section of 100mm X
200mm carrying a UDL of 4kN/m, find
1) Maximum bending stress in the beam.
2) maximum shear stress in the beam.
3) the shear stress at point 1 m to the right of the left support and 25mm
below the top surface of the beam. (10)
Ans: 12 kN
3000mm
6kN 6kN
The maximum bending moment =๐ค๐2
8.
Using bending equation:
๐๐
๐ฆ=
๐
๐ผ
y = 100mm.
M= ๐ค๐2
8=
4(3000)
8= 1500 ๐ โ ๐๐.
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I=๐โ 3
12=
100(200)3
120= 6666666.67๐๐4.
Substituting these values in bending to get max bending stress ๐๐
๐๐
100=
1500
3000
โด ๐๐ = 50 ๐/๐๐2. Max shear stress
Average shear stress =๐ โ๐๐๐ ๐๐๐๐๐
๐ โ๐๐๐ ๐๐๐๐ =
๐ค๐ฟ
100ร200=
12000
20000
๐๐๐ฃ๐ = 0.6 ๐/๐๐2.
๐๐๐๐ฅ = 1.5 ร ๐๐๐ฃ๐
โด ๐๐๐๐ฅ = 1.5 ร 0.6
๐๐๐๐ฅ = 0.9 ๐/๐๐2
Shear force at x=1m =1000mm.
๐๐น๐ฅ=1000๐๐ = 6000 โ 4 ร 1000 = 2000 ๐.=V
yฬ = 100๐๐.
Now ๐ =๐ร๐ดรyฬ
๐ผ.๐=
2000ร100ร200ร100
6666666.67ร100=5.97 ๐/๐๐2 .
๐ ๐๐ก 25๐๐ ๐๐๐๐๐ค ๐ก๐๐ ๐ ๐ข๐๐๐๐๐ ๐. ๐. 75๐๐ ๐๐๐๐ฃ๐ ๐. ๐ด
โด ๏ฟฝฬ ๏ฟฝ = 100 โ 100 + 75
2= 100 โ 87.5 = 12.5
โด ๐ =2000ร175ร100ร12.5
6666666.67ร100= 0.65 ๐/๐๐2.
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Q6)
a) A 400 mm long bar has rectangular cross section 10 X 30mm.this bar is
subjected to
1) 15 kN tensile force on 10 X 30 mm faces.
2) 80 kN compressive force on 10 X 400 mm faces.
3) 180kN tensile force on 30 X 400mm faces.
Find the change in volume if ๐ฌ = ๐ ร ๐๐๐ ๐ต/๐๐๐ and 1/m=0.3. (10)
E F
y
D C
z x G
A B
Ans: ๐ธ = 2 ร 105 ๐/๐๐2.
1/m=0.3
AB=400mm.
BC=30mm.
BG=10mm.
Stress in x-direction
๐๐ฅ =15ร103
10ร30. = 50 ๐/๐๐2 (tensile)
Stress in y-direction
๐๐ฆ =80ร103
10ร400= 20 ๐/๐๐2. (compressive).
Stress in z- direction
๐๐ฅ =180ร103
400ร30= 15 ๐/๐๐2.(compressive).
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Strain in x direction.
๐๐ฅ =๐๐ฅ
๐ธโ
1
๐
๐๐ฆ
๐ธโ
1
๐
๐๐ง
๐ธ
โด ๐๐ฅ = 3.025 ร 10โ4 .
Similarly
๐๐ฆ = โ1.525 ร 10โ4
๐๐ง = โ1.2 ร 10โ4
Now
Change in AB= L(AB) X ๐๐ฅ
= 400 ร 3.025 ร 10โ4
= 0.121mm (increase)
Change in BC= ๐ฟ(๐ต๐ถ) ร ๐๐ฆ
= 30 ร โ1.52 ร 10โ4
= -4.575ร 10โ3 ๐๐
Change in BG = ๐ฟ(๐ต๐บ) ร ๐๐ง
= 10 X -1.525ร 10โ4
= 1.2 ร 10โ3๐๐
New lengths:
AB= 4000+0.121=400.121mm.
BC= 30-4.575ร 10โ3=29.99mm
BG= 10 โ 1.2 ร 10โ3=9.99mm
๐โ๐๐๐๐ ๐๐ ๐ฃ๐๐๐ข๐๐ = ๐๐๐ค ๐ฃ๐๐๐ข๐๐ โ ๐๐๐๐๐๐๐๐ ๐ฃ๐๐๐ข๐๐. โด ๐ฟ๐ = (400.121 ร 29.99 ร 9.99) โ (400 ร 30 ร 10). ๐ฟ๐ = โ123.71๐๐3 (๐๐๐๐๐๐๐ ๐).
b) A hollow cylinder CI column is 4m long with both end fixed. Determine the
minimum diameter of the column, if it has to carry a safe load of 250kN
with a FOS of 5. Take internal diameter as 0.8 times the external diameter
๐ฌ = ๐๐๐ ๐ฎ๐ต/๐๐. (10)
Ans: ๐ฟ๐ = 0.5 ร 4 = 2๐ = 2000๐๐. P = 250๐๐ .
FOS=5.
๐๐ = 0.8 ร ๐๐ .
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Using Eulerโs equation:
๐๐ =๐2๐ธ๐ผ๐๐๐
๐ฟ๐ธ2 .
โด 250
5=
๐2ร200ร103ร๐
64(โ 0
4โโ ๐4)
(2000)2.
โด 250
5=
๐2ร200ร103ร๐
64(โ 0
4โ(0.8โ 04))
(2000)2.
By solving we get.
๐๐ = 43.24 ๐๐. Also
๐๐ = 34.59๐๐.