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CHAPTER-1
Stress & Strain ………………..…….………………………………….……... 4-13
Answer Key ………………………………………………………………………………………………………………….. 9
Solution …………………………………………………………………………………………………………………... 9-13
CHAPTER-2
SFD & BMD …………………….………………….…………………………….. 15-25
Answer Key ………………………………………………………………………………………………………..……... 21
Solution ………………………………………………………………………………………………………………... 21-25
CHAPTER-3
Bending Stress ……………………..………………………………………. 27-41
Answer Key ……………………………………………………………………………………………………….………. 32
Solution …………………………………………………………………………………………………………..…….32-41
CHAPTER-4
Shear Stress …………………………….……………………………………. 43-55
Answer Key ………………………………………………………………………………………………………..………. 47
Solution ……………………………………………………………………………………………………………….. 47-55
CHAPTER-5
Transformation of Stress ……..………………..….………………. 59-66
Answer Key ………………………………………………………………………………………………………..………. 64
Solution ……………………………………………………………………………………………………………….. 64-66
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CHAPTER-6
Torsion …………….……..……….…….………………………………….……. 68-62
Answer Key ………………………………………………………………………………………………………………… 70
Solution …………………………………………………………………………………………………………………. 70-72
CHAPTER-7
Theories of Failure .…………….…………….………………………….. 74-77
Answer Key ………………………………………………………………………………………………………..……... 76
Solution ………………………………………………………………………………………………………………... 76-77
CHAPTER-8
Column ………..…………………………………………………………………. 79-88
Answer Key ……………………………………………………………………………………………………….………. 82
Solution …………………………………………………………………………………………………………..……. 82-88
CHAPTER-9
Spring, Shear Centre & Combined Stresses ..………… 91-94
Answer Key ………………………………………………………………………………………………………..………. 92
Solution ………………………………………………………………………………………………………………… 92-94
CHAPTER-10
Thick & Thin Shells …...….……….……………………………………… 96-98
Answer Key ………………………………………………………………………………………………………..………. 97
Solution ………………………………………………………………………………………………………………… 97-98
CHAPTER-11
Deflection …………………..….……….………………………………….… 100-110
Answer Key ……………………………………………………………………………………………….……..………. 104
Solution ………………………………………………………………………………………………………………. 104-110
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Chapter
1
1.
Stress and Strain
A rectangular base plate is fixed at each of
3.
During tension test on mild steel bar, two
its corners by a 20 mm diameter bolt and
points A and B are marked at equal distance
nut. The plate rests on washer of 22 mm
from centre. Calculate the minimum gauge
internal diameter, and 50 mm external
length if area of original cross-section is 16
diameter. Upper washer which are placed
mm2.
between nut and plate are of 22 mm
internal diameter and 44 mm external
diameter, the base plate carries a load of
4.
12t. What would be the stress in the lower
A. 80 mm
B. 40 mm
C. 22.6 mm
D. 45.12 mm
The bar shown in the figure subjected to a
washer when the nuts are tightened to
tensile load of 12000 kg. Find the length of
produce a tension of 0.5t on each bolt?
middle portion if there is stress is to be
A. 221 kg/cm2
B. 307.01 kg/cm2
limited
C. 175 kg/cm2
D. 189.51 kg/cm2
elongation of the bar is to be 0.016 cm.
2.
to
1000
kg/cm2.
If
the
total
Take E = 2 × 106 kg/cm2.
Figure shows a rigid bar ABC fixed at a and
5.
suspended at B and C by two man holding a
A steel tie Rod 10 cm in diameter and 8 m
the same pull. The bore being 5 cm in
CD = 20,000 kg/mm2 for BE.
D. 750 kg
D. 28.75 mm
total extension will increase by 10% under
elasticity of for BE = 7000 kg/mm2 and for
C. 1750 kg
C. 34.79 mm
length bar should be bored centrally so that
tension in rope BE and CD. Modulus of
B. 1498.5 kg
B. 20.16 mm
long is subjected to pull of 16t. To what
rope via pulley system. Determine the
A. 12220 kg
A. 15.22 mm
diameter. Give your answer in meter up to
one point of decimal use E = 2000 t/m2
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6.
A Rigid wheel is 4 metres in diameter, it is
11. Impact factor is ratio of deformation due to
desired to shrink on to the wheel a tin steel
impact load to deformation to static load
tyre. Find the internal diameter of the tyre
and its expression is given by
if after fitting the hoop stress in tyre is
A. I.F. = 1 + 1 +
L static
2H
B. I.F. = 1 + 1 −
.L static
2H
C. I.F. = 1 + 1 +
2H
.L static
D. I.F. = 1 + 1 −
2H
.L static
10000 kg/cm2. Take = E = 2 × 106 kg/cm2
7.
A. 3.98 m
B. 4.20 m
C. 3.99 m
D. 3.96 m
A steel bar 50 mm wide, 12 mm thick and
30 cm long is subjected to axial pull if the
change in volume is 0.04536 cm3 calculate
the axial pull in kg. Give your answer to
12. In UTM experiment a sample of length 200
nearest integer. Take E = 2 × 10 6 kg/cm2
8.
mm was loaded in tension until failure. The
poison’s ratio = 0.32
failure load was 80 kN. The displacement
Three vertical rods equal in length and each
measured using the cross-head at failure
12 mm in diameter are equally spaced in a
was 15 mm. The strain at failure in the
vertical plane and together support a load
sample is 2%. The compliance of the UTM is
of 800 kg. The rods being so adjusted as to
constant and is given by K × 10–8 m/N
share the load equally if now an additional
where K is ______.
load of 800 kg. be added determine the
13. A Metal rod of length L varies as L = L0 (T
stress in middle rod. Middle rod is made of
+ T2) calculate the deflection due to
copper and outer rods are of steel. Take Es
temperature change of T.
= 2 × 106 kg/cm2 and Ec = 1 × 106 kg/cm2.
A.
L0T2
(3 + 2T )
6
B.
L0T2
(2 + 3T )
6
C.
L0T2
6
D.
L0T2
2T + 3T2
6
Given you answer to nearest integer in
kg/cm2
9.
Relationship between true stress and true
strain is given as  T = knT where n is
A. strength coefficient
B. strain hardening exponent
(
)
14. Resistance to indentation is the property of
C. coefficient of elasticity
metals for ductile material this property of
D. modus of elastic curvature
metals can be said to be
10. The true strain of a mild steel sample is
0.99% the engineering strain of sample is
A. Toughness
A. 0.100%
B. 0.010%
B. hardness
C. 1.00%
D. 0.001%
C. Ability to absorb energy till elastic limit
D. Resistance to scratching
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15. Calculate the elongation of a metallic bar
A.
3
5
B.
2
5
C.
5
2
D.
5
3
subjected to a load of 5 kN. Who’s elastic
modulus varies as E(x) =
1
N
 105
where
x
mm2
20. Calculate the strain energy for the system
x is the length of bar. Area can be taken as
shown in figure E = 200 GPa
100 mm2 and bar of unit length
A. 0.050 mm
B. 0.025 mm
C. 0.50 mm
D. 0.25 mm
16. A steel bar of 50 mm × 50 mm square cross
section is subjected to axial compressive
load of 200 kN if the length of the bar is 3
m and E = 200 GPa the elongation of the
bar will be
A. 2.4 mm
B. 4.8 mm
C. 1.2 mm
D. 0.6 mm
A. 312.5 N-mm
B. 468.75 N-mm
C. 156.25 N-mm
D. 625 N-mm
21. The dimension for torsional rigidity is
17. A bar having cross-section area of 600
mm2 is subjected to axial loads at the
positions indicated. The value of stress in
A. ML2T–2
B. ML–1T–2
C. ML–2T–1
D. ML3T–2
22. When a specimen is loaded the slip band
the segment BC is
formation also takes places when unloaded
dislocation moment occur. In this process
yield stress point will go
A. 21.67 N/mm2
B. 25.11 N/mm2
A. Higher
C. 11.25 N/mm2
D. 31.625 N/mm2
C. At the same position D. None of these
23. In
18. A bar of diameter 40 mm is subjected to a
tensile
load
such
that
the
the
B. Lower
following
relationship
measured
is
figure
drawn
which
represents engineering curve.
extension on a gauge length is 0.12 mm and
the change in diameter is 0.0040 mm. The
poison ratio is 0.3. The gauge length of the
bar should be
A. 360 mm
B. 180 mm
C. 200 mm
D. 400 mm
19. The poison ratio of a material is 0.25
calculate the ratio of bulk modulus of
A. OP
B. OQ
elasticity to its modulus of rigidity.
C. RS
D. ST
6
stress-strain
portion
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24. A cubical box is subjected to tri-axial loading
A. 2.28 mm
B. 1.14 mm
as shown in figure what should be the
C. 3.72 mm
D. 1.86 mm
uniform lateral pressure σ, so that lateral
27. A solid metal bar tapers uniformly from 40
strain is prevented? Assume μ = 0.25
mm diameter to 25 mm over its length of
500 mm. The rod when held vertical is
subjected to an axial tensile load of 20 kN E
= 2 × 105 N/mm2. The extension of the rod
in mm would be
A. 75 N/mm2
B. 37.5 N/mm2
C. 150 N/mm2
D. 300 N/mm2
A.
4
5
B.
2
5
C.
3
5
D.
1
5
28. A steel rail is 15 m long and is laid at a
25.
temperature
of
20°C
.
The maximum
temperature in summer is expected is 45°C
. The minimum gap required between is two
rails is
α = 12 × 10–6°C
Calculate the reaction at support A? given
A. 9 mm
B. 4.5 mm
that area of BC = 1.25 area is AB, Area of
C. 5 mm
D. 3.5 mm
CD = 0.25 AAB, Area of DE = 0.75 area of
29. A copper bar 50 cm length is fixed by means
AB, Area of EF = Area of AB
of support at its ends supports can yield by
A. 76.58
B. 62.58
0.02 cm if temperature of bar is raised of
C. 85.58
D. 57.055
80°C then stress induced in the bar for αc =
26. A square steel bar of cross-section 50 × 50
2 × 10–6°C and Ec = 1 × 106 kg/cm2
mm is 2 m long. It is subjected to bi-axial
A. 660 kg/cm2
B. 240 kg/cm2
stress as shown in figure. What is the
C. –660 kg/cm2
D. –240 kg/cm2
2
elongation of the bar in y-direction. Given μ
= 0.3, E = 2  105 =
30. A square steel bar of 150 mm side and 4 m
Nq
long is subjected to a load where upon it
mm2
absorbs a strain energy of 250J. what is its
modulus of resilience?
7
A.
1 N − mm
125 mm3
B.
1 N − mm
360 mm3
C.
1 N − mm
120 mm3
D.
1 N − mm
720 mm3
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ANSWER
1. A
2. B
3. C
4. D
11. C
12.
13. A
21. D
22. A
23. D
5.
6. A
7.
8.
9. B
10. C
14. B
15. D
16. C
17. A
18. A
19. D
20. B
24. B
25. B
26. A
27. D
28. B
29. D
30. B
SOLUTION
1.
A
When
the
compressive
nuts
are
load
in
tightened
the

the
upper
 TBE =
washer=tension in the bolt = 0.5t = 500 kg.
TCD +

× (4.42 – 2.22) cm2 = 11.40 cm2
4

× (52 – 2.22) cm2 = 15.83 cm2
4
TCD = 4281.5 kg
Load transmitted to one lower washer
=
TBE =
12
= 3t = 3000 kg
4
3.
7
 4281.5 = 1498.5kg
20
C.
Total compressive load to on lower washer
Minimum gauge length during tension test
= 3000 + 500 kg
is given by = 5.65 A0
= 3500 kg
= 5.65 16
Stress intensity in the lower washer
= 22.6 mm
3500
=
= 221 kg/cm2
15.83
2.
7
T
= 5780
20 CD
 27 
TCD 
 = 5780 kg
 20 
Area of lower washer
=
7
T
20 CD
Also, TBE + TCD = 5780
Area of upper washer
=
TCD  L
TBE
2
=

6  7000 4  20, 000 3
4.
D.
Stress in end portion
B.
=
12000
kg
= 152.86
2

2
cm
 10
4
Extension in end portions + extension in
Middle = 0.016 cm
Let tension in BE and CD are TBE and TCD
152.86  (50 − x )
1

2
= 2  1 = 2
2
3
3
1 =
6
2  10
TCD  L
TBE  L
2 =
ABE  EBE
ACD  ECD
+
1000  x
2  10
6
= 0.016
9643 + 847.14x = 32000
x=
TCD  L
TBE  L
2

=

ABE  EBE
ACD  ECD 3
24357
847.14
x = 28.75 cm
9
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5.
7.
Longitudinal stress =
Longitudinal stress

A =  102 = 78.5cm2
4
f =
16
= 0.2038 t/m2
78.5
=
f
0.2038
L =
 800 = 0.08152 cm
E
2000
= el =
stess
P
=
E
5  1.2  2x106
Lateral
strain
=
poission’s
ratio
ed = 0.32 × e
volumetric strain = el – 2 × ed
V
= el − 2  ed
V

= 78.5 –  52 = 58.875
4
Extension after bore is made
0.04536
P
0.32  P  2
=
−
6
30  5  1.2 5  1.2  2  10
5  1.2  2  106
= 1.1 × 0.08152 = 0.089672
P = 8400 kg
8.
Let the length of bore be ‘l’

0.2038
16  l  100
(80 − l)  100 + 58.875
2000
 2000
Area of each bar =

 1.22 = 1.13 cm2
4
(800 / 3)
= 0.089672
Initial stress on each bar =
⇒ 0.01019 (8 – l) + 0.0033981 l
Stresses due to additional load of 800 kg
= 0.089672
Strain in copper = strain in steel
l = 2.398 m
Pc
P
= s  Ps = 2  Pc
Ec Es
l = 2.4 m
1.13
Ps As + Pc Ac = total load = P = 800
A.
2Pc (2 × 1.13) + Pc × 1.13 = 800
D − d
5.65 Pc = 800
Hoop stress P = 
 E
 d 
Pc =
For the given condition Hoop stress should
800
5.65
Pc = 141.59
be 10000 kg/cm2
Pc≅ 142 kg/cm2
D−d
P=
 E = 10000
d
9.
B.
Relationship between true stress and true
 400 − d 
6

  2  10 = 10000
d


strain is given as  T = knT where n is strain
hardening exponent.
400
10000
=1+
= 1.005
d
2  106
d=
×
longitudinal stain
A’ = Area of reduced section
6.
P P kg/cm2
=
A
5  1.2
10. C.
ET = ln(1 + )
400
= 398.0099 cm
1.005
0.99 / 100 = ln(1 + ε)
d = 3.98 m
ε = 0.009949 = 1%
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11. C.
16. C
12.
Strain =
Δδ=
L
= 0.02
L
P.L
200×1000×3×1000
=
AE 200×109×50×50×10-6
Δδ = 1.2 mm
Δ L = 0.02 × L
17. A
P
= 0.02 × L
AE
L
0.02  L
=
AE
P
L
0.02  200  10−3
=
AE
80  103
L
= 5  10−8
AE
R = 63 – 50 = 13 kN
L
= 5  10−8 m/N
AE
Compliance =
Stress in portion BC =
K=5
= 21.67 N/mm2
13. A.
18. A.
ΔL = α.L.ΔT.
dl = α.L.dT
T
l=
Poisson ratio =
 .L0 ( T + T
2
0
) .dT
0.3 =
 T2 T3 
l = L 0 
+

 2
3 

l=
L0T2
(3 + 2T )
6
0.12 0.004
=
L
40
⇒ l = 36 × 10 = 360 mm
19. D.
E = 2G (1 + μ)
15. D
=
P.L
AE
=
P.L
=
AE(x)
E = 3k (1 – 2μ)
L
2 G
(1+μ)
= ( )×
3 K
(1-2μ)
P.dx
1
A·  105
x
K 2 1.25 2
5
= ×
= ×2.5=
G 3 0.5 3
3
P
 A  105 x.dx
20. B.
0
=
lateral strain
longitudinal strain
−(−0.0040 / 40)
0.12 / L
 0.3 
14. B
=
13  1000
600
x2
5 2
A  10
P
1
Strain energy =
0
δ = δ1 + δ2 =
5  1000  1000
5  1000  1000
+
200  200  1000 100  200  1000
δ = 0.125 + 0.0625 = 0.1875 mm
strain energy = 1/2 × 5 × 1000 × 0.1875
= 0.25 mm
= 468.75 N-mm
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21. D.
R A×1 (R A -100)× (R A +175)×0.5
+
+
A×E
1.25×AE
0.25AE
(R A -100)×0.5 (R A +139)×1
+
+
=0
0.75×AE
AE
Torsional Rigidity = G.J
=
N
2
mm
 mm4
= mass 
length
Time2
R A − 100 (R A + 175)  2
+
1.25
1
(R A − 100)  2
+
+ R A + 139 = 0
3
 RA +
 length2
= ML3T −2
1
2
100
+ 2 + + 1) −
1.25
3
1.25
200
+ 350 −
+ 139 = 0
3
 R A (1 +
22. A.
As the initial deformation opposes the flow
of dislocation. The specimen is said to be
 RA = −
work hardened as a result, yield point gets
⇒ RA = – 62.58
higher.
26. A.
23. D.
ε y=

Δ y
Δ
24. B.
y
σ y μσ x
E
E
= 114  10−50
= 114  10−5  2  1000
= 2.28 mm
 m×50 m×100
=0
E
E
E
27. D.
σ – 0.25 (150) = 0
=
342.33
= –62.58
5.47
150 ×1
= 37.5 N/mm2
4
25. B.
Given that area of BC = 1.25 area is AB,
Area of CD = 0.25 AAB,
Δ=
4PL
4×20×1000×500
=
πd1d2E π×40×25×2×105
Δ=
10
25×2×π
Δ=
1
5π
28. B.
Area of DE = 0.75 area of AB,
Minimum Gap = lαΔT = 15000  12  10−6  25
Area of EF = Area of AB
= 4.5 mm (gap)
29. D.
Temperature strain = α × ΔT
= 2 × 10–6 × 80
= 160 × 10–6 (Tensile)
Strain due to yielding of support
=
δT =0=δAB+δBC+δCD+δDE+δEF=0
12
0.02
(compressive)
50
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Total strain = 160 × 10–6 –
Stress induced = (160  10−6 −
= 160 –
30. B.
0.02
50
Modulus of Resilience = strain energy
absorbed per unit volume upto elastic limit
0.02
)  1  106
50
2
 106
50  100
= –240 kg/cm2

13
=
250  1000
N − mm
150  150  4000 mm3
=
25
1
=
150  15  4 360
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Chapter
SFD and BMD
2
1.
Calculate the maximum bending moment
for the loading as shown in figure
A. 7.5 kN-m
B. 6.95 kN-m
C. 1.875 kN-m
A.
C.
2.
WL2
24
2
WL
12
B.
D.
D. 9.375 kN-m
5WL2
12
4.
2
WL
8
At what distance from left support of the
given beam shear force is zero
Calculate the bending moment at point C. W
= 1 kN/m
5.
A. 4 − 2
B. 2 − 2
C. 2 + 2
D. 4 − 3
Calculate the maximum bending moment
for beam with over hangs shown below
A. 3.361 kN-m
B. 5.0415 kN-m
C. 5.125 kN-m
D. 5.1025 kN-m
3.
For a beam given below if midpoint of
member BC is hinged calculate bending
A. 60 kN-m
B. 40 kN-m
moment at midpoint of BE.
C. 50 kN-m
D. 70 kN-m
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6.
The reaction at roller end A is if W’ = W.L
A. 150 kN-m
B. 75 kN-m
C. 300 kN-m
D. Zero
10. For a simply supported beam shown in
figure given below at what distance from
support A bending moment is maximum.
7.
A.
2
WL
3
B.
3
WL
2
C.
WL
4
D.
WL
2
A.
Calculate bending moment at mid span of
simply supported beam as shown.
C.
L
2
L
3
B.
L
2
D.
L
3
11. Calculate fixed end moment for beam
shown in figure.
8.
A.
13
WL2
96
B.
12
WL2
96
C.
5
WL2 c
48
D.
25
WL2
48
Calculate fixed end moment at support A.
A. 25 kN-m
B. 50 kN-m
C. 100 kN-m
D. 40 kN-m
12. Calculate bending moment at C.
9.
A. 90 kN-m
B. 300 kN-m
C. 180 kN-m
D. 120 kN-m
What is the bending moment at fixed
support?
16
A. 30 kN-m
B. 15 kN-m
C. Zero
D. 7.5 kN-m
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13. Determine
the
location
of
maximum
16. Construct the bending moment diagram for
bending moment
the beam shown in figure.
A. 3 m
B. 5.0625 m
C. 2.0625 m
D. 2.625 m
14. Find reaction at support A.
A.
B.
A. Zero
B. 30 kN-m
C. 35 kN-m
D. 15 kN-m
C.
15. Consider the following statements regarding
D. None of these
bending moment. Which of the following
17. Which of the following statement is true for
statement is ‘FALSE’?
A.
For a cantilever beam clamped at A
a beam where only couple is acting in the
subjected to concentrated load at free
span.
end, shape of bending moment is
A.
triangular.
B.
moment
is
constant
everywhere in the span
In continuous beam, it is not necessary
B.
that at every support BM is hogging (–
Loaded span can be said to be shear
span
ve).
C.
Bending
C.
If force polygon is not close it will not
Shear force varies linearly with the
length
be in equilibrium.
D. Bending moment will have sudden
D. Hogging moment at any point shows
that a couple is acting in anticlockwise
change in sign and it is equal to couple
direction.
at that section.
17
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18. The resulting bending moment at midspan
20. What is the reaction at the support B.
R of the beam is
A. 27.5 kN-m-sagging
A. 100 Kn
B. 27.5 Hogging (kN-m)
B. 160 kN
C. 22.5 kN-m-sagging
C. 320 kN
D. 27.5 kN-m-sagging
D. 80 kN
21. A vertical pile AB is hinged at the base B.
19. The correct bending moment diagram for
assume end b is fixed and subjected to
the beam is
variable
load
due
to
earth
pressure.
Calculate the fixed end moment at B if pile
is anchored by a tie pin at C.
A.
A. 12 t-m
B. 6 t-m
C. 10 t-m
D. 9 t-m
22. Calculate the shear force at A.
B.
C.
D.
18
A. 7 kN
B. 5 kN
C. 14 kN
D. 2.5 kN
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23. Calculate shear force at right support
A. 5 3,5(1 − 3)
B. 5(1 − 3),5 3
C. 5 3,5 3
D. 5 3,5(1 + 3)
27. Which one of the following represent BMD
for this beam?
A. 6 kN
B. Zero
C. 12 kN
D. 24 kN
24. Calculate B.M. at midspan of beam
A. 5 kN-m
B. 20 kN-m
C. 10 kN-m
D. 15 kN-m
A.
25. Calculate vertical reaction at support D
B.
C.
A. 25 kN
B. 35 kN
C. 12.5 kN
D. Zero
D.
28. Intensity of loading on a simple supported
26. Calculate horizontal reaction at A and
beam is given by W = A sin π x. Beam of
vertical reaction at A respectively?
span L and x is distance along the axis
flexural rigidity of beam is EI. Bending
moment at a distance x is
A.
C.
20
W

W
2
B.
W2

D. π2W
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29. Which of the following statements is NOT
C.
correct?
A.
zero when shear force is constant
A SSB subjected to couple at ends
D. Shear force at any distance will be zero
produces a rectangular SF diagram
B.
Bending moment at any distance will be
when bending moment is constant
In the train load across a bridge only a
non-linear BMD is possible
ANSWER
1. C
2. B
3. D
11. A
12. B
13. C
21. B
22. A
23. B
4. A
5. A
6. B
7. C
8. D
9. D
10. C
14. D
15. D
16. B
17. B
18. C
19. C
20. B
24.C
25. D
26. A
27. C
28. C
29. C
SOLUTION
1.
C.
3.
Reactions at A and B must be equal
R A = RD
1
2R A= ×W×L
2
RA + RB + 20 = 10 + 5 × 1 = 15
R A=
RB = R C
(symmetry)
RA + RB = – 5 ...... (1)
WL
4
Moment of E should be equal to zero
Maximum B.M. will occur at mid-point
ME = RA × 3 + RB × 1 – 10 × 2 – 5 × 1 ×
L 1
L
1 L
B.M. at C = R A  −  W   (  )
2 2
2 3 2
0.5 = 0
=
3RA + RB = 22.5 ...... (2)
WL2 WL2
8
24
B.M. max. =
2.
D.
Solving equation (1) and (2)
–2RA = –27.5
wL2
12
RA = 13.75 kN
B.
R A + RD
RB = 22.5 – 3 × 13.75 = –18.75 kN
1
= (  1  0.5) + 10
2
Bending moment at midpoint of BE
RA + RB = 10.25
B.M. = RA × 2.5 + RB × 0.5 – 10 × 1.5 –
Taking moment about A; MA = 0
(5 × 0.5) × 0.25
⇒ RD  2 −
= (13.75 × 2.5) + (–18.75 × 0.5) – 15 –
1
2
 1  0.5   0.5 − 10  1 = 0
2
3
(2.5 × 0.25)
⇒ RD = 5.0415 kN
= 34.375 – 9.375 – 15 – 0.625
⇒ MC = RD × 1 = 5.0415 kN-m
= 9.375 kN-m
21
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4.
A.
B.M. at x – x = 40 × (1 + x) – 30 × x
R A + RB =
⇒ 40 + 40x – 30 × 2 = 60
1
 30  4 = 60
2
1
ΣMB = 0  R A  8 = (  30  4)  6
2
RA = 45 kN
Shear force at distance x
B.M.maximum = 60 kN-m
1
from A=R A -( ×30×2) where x = 2m
2
6.
B.
Taking moment about C = 0
= 45 – 30 = 15 kN
Shear force at distance x from midpoint of
−
WL2
+ W  2L − R A  L = 0
2
−
WL2
+ 2WL2 = R A  L
2
AC:
1
1
=R A -30-({ ×30×2}-{ ×Wx×(2-x)})=0
2
2
where,
RA =
wx w
w(2-x)
= Þwx=
=15(2-x)
2-x 2
2
7.
1
= R A - 30 - 30 + 15(2 - x)(2 - x) = 0
2
 45 - 30 - 30 +
15
(2 - x)2 = 0
2
RA =
WL
1
L 3
+ 2   W  = WL
2
2
4 4
3
WL
8
B.M. at Midspan :
15
= 2  (2 - x) = 2 x = 2 - 2
7.5
⇒ Hence distance from
support A = 2 + 2 − 2 = 4 − 2
5.
Ans. C.
2R A =
 -15 + 7.5(2 - x)2 = 0
 (2 - x)2 =
3
WL
2
A.
8.

3
L 1
L
L 2 L
WL 1
WL  −  W   ( +  ) −

8
2 2
4 4 3 4
4
8

3WL2 5WL2 WL2
−
−
16
96
32

5WL2
48
D.
Taking moment about B from right side
Vc × 4 – 10 × 2 = 0
R1 + R2 = 100
Vc = 5 kN, Also VA + VC
Taking moment about R2
= (5 × 6) + 10 = 40
RL × 2 – 40 × 3 + 60 × 1 = 0
MA = 5 × 6 × 3 + 10 × 8 – 5 × 10
MA = 90 + 80 – 50
R1 × 2 = 60
MA = 120 kN-m. x[chances of mistake being
R1 = 30 kN R2 = 70 kN
done]. This approach is wrong because
B.M. at R1 = 40 × 1 kN-m
there is internal Hinge in between A and c.
B.M. at midspan = 40 × 2 – 30 × 1
Hence take moment about B from left side
= 50 kN-m
and equate to zero.
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VA = 40 – VC = 40 – 5 = 35 kN
12. B.
MA + V A × 6 – 5 × 6 × 3 = 0
MA = 90 – (35 × 6)
= 90 – 210 = –120 kN-m
9.
D.
Since applied load is passing through fixed
support, so this will generate zero bending
moment. Also reaction supports at left and
right side must be equal and they counter
R1 = 3 kN, R3 – R2 = 5
the bending moment at fixed end. Hence
(R3 – 5) × 15 – 3 × 5 = 0
bending moment at fixed end should be
R3 = 6 kN, R2 = 1 kN, R1 = 3 kN
zero.
MC = R2 × 15 = 1 × 15 = 15 kN-m
10. C.
13. C.
B.M. is maximum where shear force is zero.
VA + VB =
WL
2
Taking moment about B = 0
VA×L=
WL L
×
2 3
Taking moment about A,
WL
VA=
6
3
Vc×6=8× +1.5×8
2
Shear force at a distance x from A.
Vc = 4t
WL 1
−  W  x = 0
6
2
VA = 8 + 1.5 – 4 = 5.5t
WX
W=
L
zero shear force.
Maximum bending moment will occur at
Assuming x from end A
WL 1 Wx2
−
=0
6
2 L
x=
8
S×Fx=5.5- ×x=0
3
L
x = 2.0625 m from A
3
14. D.
11. A.
Mfix + 10 × 4 – 10 × 2 - (5 × 1) × (0.5+0.5)
MA + 10 × 1 – 5 × 2 + 5 × 6 = 15
=0
MA = 10 – 10 – 15 + 30 = 0
Mfix = –25 kN-m
MA = – 15 kN-m
15. D.
Hogging moment at any point shows that a
couple is acting in clockwise direction.
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16. B.
 MA +
Taking moment about end A
P

2
= 0 Taking moment about left
side of C
Vb × 6 + 12 = 6 × 4 + 6 × 7
Vb = 9 kN, VA = 12 – 9 = 3 kN
MA = −
Bending moment calculations:
P
P
P
;MD =  =
2
2 2
4
⇒ So B.M. Diagram
B.M. at A = 0
B.M. just on the left hand side of C
= 3 × 2 = + 6 kN-m
B.M. at D = 3 × 4 – 12 = 0
20. B.
B.M. at B = –6 × 1 = –6 kN-m
Taking moment about A
And Hence option B is correct option.
RB × 2 + 40 × 4 – 80 × 6 = 0
17. B.
RB × 2 + 160 – 480 = 0
RB = 160 kN
21. B.
⇒ Moment at B =
Since No vertical load is acting
= 9 – 3 = 6 t-m
VA = – VB
22. A.
Hence shear force is constant, and span is
Taking moment about E
said to be shear span.
VD × 4 = 8 × 1 VD = 2 kN
18. C.
Rp + R Q
1
6
 1.5  6  − 1  3
2
3
VA + VB = 4 × 5 + 2 = 22
1
=  3  10 = 15kN
2
Taking moment about B.
VA × 4 – 4 × 4 × 2 + 4 × 1 × 0.5 + 2 × 1
Taking moment about Q
=0
1
Rp  10 = 5 + (  3  10)  5
2
VA × 4 = 32 – 4 = 28
VA = 7 kN
RP = 8 kN
1
2
23. B.
5
3
B.M. at R=Rp×5- ×3×5× -5
12 + 12 – 12 × 2 + RB × 4 = 0
= 8 × 5 – 12.5 – 5
(Taking moment about left end support)
= 35 – 12.5
RB = 0
= 22.5 kN-m-(sagging)
24. C.
19. C.
RB 
R1 + R2 = 20
=P
Taking moment about right support
2
R1 × 1 – 5 × 2 + 15 × 1 = 0
P
RB =
2
RA =
R1 = –5 kN
B.M. at Midspan = – 5 × 0.5 – 5 × 1.5
P
2
= –10 kN-m
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25. D.
For AC B.Mx = 2.5x
Since D is roller support hence reaction in
For CB B.M = 2.5 × 2 + 2.5x – 5 × x × x/2
horizontal direction is zero
= 5 + 2.5x – 2.5x2
Taking moment about E
B.M at B = 0
– VD × 1 + HD × 2 = 0
28. C.
HD = 0
VD = 0
dM
d2M dV
d2M
=V
=

=W
2
dx
dx
dx
dx2
26. A.
Taking moment about A :
d2M
dx2
10  3 = VB  2
dM
A
= A sin X  dx = cos x
dx

VB = 5 3
VA + VB = 5 kN
M=
VA = 5(1 − 3)
HA = 10 cos 30 = 10 
=W
3
=5 3
2
M=
27. C.
M=
VA + VB = 10
VA × 4 = 5 × 2 × 1, VA = 2.5 kN, VB
29. C.
= 7.5 kN

25
A
cos x  dx

A
2
W
2
sin X
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Chapter
3
1.
Bending Stress
A rectangular steel plates of dimensions 20
A. 0.70 M
mm × 3 mm is required to be bent in a form
B. M
of circular arch of radius 2 meters by
C. 1.41 M
applying the end couples. Compute the
D. 2 M
magnitude of the required couples (in
2.
4.
Newton meter, up to one decimal place)
rectangular beam of 200 mm × 300 mm is
assuming the material remains safe for any
given by:
applied load. The Young’s modulus of
Mx = 50x – 20x2 + 100 kN-m
elasticity for the material of the plates is
Where,
200 GPa.
x is the distance of the point considered
A hollow circular beam of external diameter
from one of the ends of beam.
250 mm and thickness of 20 mm is to be
The maximum bending stress (in N/mm2, up
used a beam in a steel structure. If the
to two decimal places) produced anywhere
maximum permissible stress for the beam
across beam is _______.
material
maximum
is
150
MPa,
bending
determine
moment
5.
the
A 3 m circular timber beam carry a
uniformly
carrying
distributed
the
load
span.
of
5
Determine
kN/m
capacity of the section.
throughout
A. 62.26 kN-m
minimum diameter required for a section if
the
permissible stresses are 12 MPa and 8 MPa
B. 114.56 kN-m
is compression and tension, respectively.
C. 115.53 kN-m
6.
D. 230.09 kN-m
3.
The bending moment at any point for a
The ratio of bending strength of the square
section to that of circular section is ______.
The moment carrying capacity of square
For the cross-sectional area for the both
cross-sectional beam when placed as its one
beams are same and they are composed of
of the diagonals being parallel to the
identical materials.
horizontal plane is recorded as ‘M’. The
A. 0.707
moment carrying capacity of the same
B. 0.845
section when placed as its one of the faces
C. 1.181
being parallel to the horizontal plane is
D. 1.414
_______.
27
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7.
Framed structure composes of a beam of
9.
In
which
of
the
following
cases,
a
modernized cross section meant to suit
particular/segment of the span is subjected
some architectural requirements. The cross
pure bending?
section of the beam is depicted below:
(i)
The permissible stress for the sectional
material is 200 MPa and it is supposed to
carry uniformly distributed load over it.
(ii)
Compute the maximum load intensity that
be safely applied on the 8 m span beam.
8.
A. 25 kN/m
B. 35 kN/m
C. 65 kN/m
D. 90 kN/m
(iii)
An engineer in order to save money over the
construction plans to install a fletched
rather than a regular one. As he knew the
(iv)
bending stress is primarily meant to be
A. i, ii and iii only
resisted at the top and bottom section, he
B. ii, iii and iv only
recommended the following section:
C. i, iii and iv only
D. i, ii, iii and iv
10. Which of the following is not an assumption
of theory of simple bending?
A.
The value of young’s modulus is same
in tension and compression
Maximum permissible stress for steel = 150
B.
MPa
Each layer of the beam is free to expand
or contract independently of the layer
Maximum permissible stress for timber:
above or below it.
(a) tension = 50 MPa
C.
(b) compression = 45 MPa
The radius of curvature is small as
compared to the dimensions of the
Determine the moment carrying of the
section
above section (in kN-m, rounded to the
D. The plane section remain plane before
nearest integer) assuming the modular ratio
and after bending
as 20
28
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11. A
square
cross-sectional
beam
of
dimensions b × d units is cut in ‘n’ number
of pieces along the depth and then stocked
up on each other giving a new section of the
A.
B.
C.
D.
same dimension. The moment carrying ratio
of the solid square case to that of the
stacked one is approximately equal to
______.
A. n
C. n
B. 1/n
14. The beam of an overall depth 300 mm used
D. 1/n2
2
in a high-tech biotechnology is subjected to
12. A circular section as shown in the figure
two different thermal environments for
shown in the figure below is subjected to
experiment propose. The temperature at
bending moment ‘m’
the top and bottoms surfaces of the beam
are 20°C and 30°C respectively. The vertical
deflection of the beam (in mm) at its midspan
due
to
temperature
gradient
is
___________.
(Assume
B. bc = bf =
C. bC = bf =
D. bc = bf =
coefficient
of
thermal
expansion for the beam material be 1.50 ×
What is the maximum bending stress?
A. bc = bf =
the
10–5 per °C)
64M
d3
32M
d3
16M
d3
15. A solid shaft of 200 mm diameter is
8M
transmitting a torque of 20 kN-m at the
d3
same time it is subjected to a bending
13. A cantilever beam loaded with the point load
moment of 15 kN-m and axial thrust of 200
‘w’ at the free end is to be designed for
kN.
attaining uniform strength at any section
across the beam.
Which of the following shape of beam is
The bending stress at the point B is
recommended
_______.
cantilever
for
beam
if
the
above
the
width
loaded
is
kept
constant?
29
A. 15.04 N/mm2
B. 19.09 N/mm2
C. 25.46 N/mm2
D. 44.56 N/mm2
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16. Two cantilever beams identical to each
19. The
maximum
section
modulus
for
a
other in terms of shape and size are loaded
rectangular section cut that can be obtained
with same point load at their free ends the
from a circular section of diameter ‘d’ is
first beam is made up of steel and the other
_________.
one is made up of aluminium alloy. The
value of Torque’s modulus of elasticity for
steel 2 times that of aluminium alloy. If the
maximum bending stress in the steel beam
is ‘fs’ and that of aluminium alloy is ‘f a’.
calculate the ratio of fs/fa.
A. 1/4
B. 1/2
C. 2
D. 1
B.
3
C. d
3
D. d
3
17. An I-section beam as shown in the figure is
6 3
so designed that the extreme fibre stresses
2
3
A. d
3
d3
9 3
20. For a beam span, the tensile as well as
in the ratio of 4 : 3 in the beam.
compressive strains is recorded at every
section. The compressive strains recorded
at the top of the beam and 100 mm below
it across the depth are 4 × 10–6 and 1.5 ×
10–6 respectively. The total depth of the
beam is 250 mm.
If the area under the respectively, the ratio
The width ‘b’ of the upper flange (b < 20
of A1 and A2 will be _______.
cm) of the beam section should be
______.
A. 10 cm
B. 11.2 cm
C. 12.4 cm
D. 15 cm
A. 0.56
B. 0.75
C. 1.25
D. 1.77
21. For a rectangular beam of dimension 60 mm
× 120 mm shown in the figure below.
18. A beam consists of a symmetrically rolled
steel joist. The beam is simply supported
the centre of the span. If the maximum
stress due to bending is 150 N/mm2, Find
the ratio of the depth of the section to span
in order that the central deflection may not
exceed 1/450 of the span.
The percentage of bending moment resisted
(Take E = 2 × 10 N/mm )
by the shaded portion of the section is
A. 1/36
B. 2/43
_______.
C. 1/56
D. 2/71
(Assume up to decimal places)
5
2
30
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22. A simply supported beam of span ‘L’ carries
section about the neutral axis x-x is given
a uniformly distributed load ‘w’ along the
by _____.
whole span. If the width ‘b’ of the beam is
constant throughout the span then when
the
permissible
bending
stress
‘f’
the
beam’s mid-span depth will be _______.
A.
3WL
2bf
3WL
4bf
B.
1 3W
C.
2 fb
3W
D. L
2fb
23. A cantilever beam of constant depth carries
a concentrated load at the mid span of the
bd3 (m − 1)bd3
+
12
24
C.
bd3 mbd3
+
12
48
D.
bd3 (m − 1)bd3
+
12
48  12
at its ends so that its length ‘l’ equal to 30
section a distance x from the free end
cm, when bent, as a circular arc, subtends,
should be proportional to _______.
C. x
B.
0.05 cm × 2.5 cm is bent by couples applied
sections the same, the breadth of the
2
bd3 (m + 1)d3b
+
12
12
25. A thin steel ruler having its cross section of
beam. To make the maximum stress at all
A. x
A.
a central angel θ = 60°. The maximum
B.
x
stress
D. x
3
magnitude is ________. (Take E = 2 ×
24. The figure below shows the structure of
induced
is
the
ruler
and
the
105 N/mm2)
fletched beam composed of timber and
A. 4.36 N/mm2
B. 8.72 N/mm2
steel. The second moment of area of the
C. 13.09 N/mm2
D. 26.18 N/mm2
ANSWER
1.
2. C
3. C
11. A
12. B
13. C
21.
22. C
23. A
4.
5.
6. C
7. C
8.
9. D
10. C
14.
15. B
16. D
17. B
18. D
19. D
20. D
24. D
25. B
SOLUTION
1.
.
f = bending stress at a distance y from NA
From the bending/flexural equation
axis
f M E
=
=
y
I R
y = the distance of layer from NA with stress
‘f’
Where,
32
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m = bending or resisting moment at the
External diameter (d0) = 250 mm
Thickness (t) = 20 mm
section.
Internal diameter (di) = 250 – 2 × t
I = moment of inertia/second moment of
= 210 mm
area bout the NA
Second moment of area (I) =
E = young’s modulus of elasticity of the
material
∴ I=
Given: E = 200 GPa,
∴ I = 9.628 × 108 mm4
20  33
= 45mm4
12
Now,
M=
Now,
M=
2.
17

 (2504 − 2104 ) =
 (254 − 214 )  104
64
64
R = radius of curvature
R = 2m, I =

 (d04 − di4 )
64
E
2  105  45
= 4500 N-mm
I =
R
2000
3.
f
150
I =
 9.6289  108 = 115.53 kNm
y
125
C.
The design criteria for a beam is bending
= 4.5 Nm
moment.
C.
equation:
From the bending/flexural equation
According
to
bending/flexural
f
M E
=
=
y
I R
f
I
M =  I  M = f   m = f  z  mz
y
y
f M E
=
=
y
I R
Where,
Where,
f = bending stress at a distance y from NA
Z is the section modulus
axis
∴ moment carrying capacity depends upon
y = the distance of layer from NA with stress
the section modulus of the section.
‘f’
Now,
m = bending or resisting moment at the
Case I :
section.
One of the diagonal parallel to horizontal
I = moment of inertia/second moment of
plane.
area bout the NA
E = young’s modulus of elasticity of the
material
R = radius of curvature
For the given section
Section modulus for case 1:
I
z1 =
=
y
33
2
2a
a
 ( )3
a3
12
2
=
a
6 2
2
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Case 2:
fmax =
One of face being parallel to horizontal
131.25  106  6
200  3002
∴ fmax = 43.75 N/mm2
plane
5.
.
From the bending/flexural equation
f M E
=
=
y
I R
Where,
Section modulus for case 2:
f = bending stress at a distance y from NA
a7
I
a3
z2 = = 12 =
a
y
6
2
axis
y = the distance of layer from NA with stress
‘f’
a3
M
Z
M
1
Now, 1 = 1  1 = 6 32 =
M2 Z2
M2
2
a
6
M = bending or resisting moment at the
If M1 = M
I = moment of inertia/second moment of
section.
area bout the NA
∴ M2 = 2  M1 = 1.41M
4.
E = young’s modulus of elasticity of the
.
The bending stress would be maximum
material
where the bending moment is maximum.
R = radius of curvature
∵M=f×z
For the given span:
For the given moment equation:
mx = 50x – 20 x2 + 100
The optimum value would take place at the
value of x obtained from
dm
=0
dx
w  L2
8
Now,
mmax =
dm
= 50x − 40x = 0
dx
 mmax =
d2m
dx3
= −40  0
5  32
8
= 5.625 kN/m
∴ At x = 1.25 m, maximum value of bending
Since the section is symmetrical, adopting
moment occurs at this point.
lower value of the permissible stress
∴ Mmax = 50 × 1.25 – 20 × 1.252 + 100
Now,
∴ Mmax = 131.25 kN-m
f M
m
=
z=
y
I
f
Now,
fmax =
mmax mmax  6
=
z
bd2
34
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For circular section,
m = bending or resisting moment at the
section.
d3 5.625  106
=
32
8
I = moment of inertia/second moment of
area bout the NA
⇒ d = 192.76 mm
6.
E = young’s modulus of elasticity of the
C.
material
Assume square of side ‘a’ units and circular
section of diameter ‘d’ units.
R = radius of curvature
For equal area,
From the flexural equations:
a2 =
f M
I
=
M=f
y
I
y

 d2  a = 0.886d
_
4
Now,
Moment of inertia for the above section is as
The bending strength of any section is
follows:
directly to proportional to section modulus
Isection = Irectangle section – Isemi circle × 2
of the section.
 Isec tion =
 MsquareZsquare & McircleZcircle
Msquare

3
Mcircle
a
= 63
d
32
∴ Isection = 8.2146 × 108 mm4
(0.886  d)3
6
=
d3
32
Bending moment (m) is given as:
M=
d
W =
∴ square section is 1 = 18 times stronger
than a solid circular section, made of
7.
8.2146  108  100
 10−6 = 547.64 kN-m
150
Now, for udl, M =
3
3
∴ Ratio (r) = 32  0.886  d3 = 1.181
6
400  3003
17  1004
−2
12
8
8.
8m
2
L
=
wL2
8
547.64  8
= 68.46 kN/m
88
.
identical material & having same cross-
For beam, the major design criteria are
section area.
bending moment.
C.
From the bending/flexural equation
For a beam, the bending moment is the
major design criteria.
f M E
=
=
y
I R
From bending equation:
Where,
From the bending/flexural equation
f = bending stress at a distance y from NA
f M E
=
=
y
I R
axis
y = the distance of layer from NA with stress
Where,
‘f’
f = bending stress at a distance y from NA
m = bending or resisting moment at the
axis
section.
y = the distance of layer from NA with stress
I = moment of inertia/second moment of
‘f’
area bout the NA
35
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E = young’s modulus of elasticity of the
due to bending moment only and that
length of the beam is said to be is pure
material
bending.
R = radius of curvature
The fletched beam comprises of more than
one material, to simplify the calculation, we
are required
to correct the composite
section to single section for calculating
moment of inertia.
I=
400  3003 20  1003
−
= 8.9833  108mm4
12
12
y = 150
Z=
8.9833  108
= 5.99  106 mm4
150
Now, for stress calculation:
∴ All of cases given are that of pure bending
As the section is symmetric, the timber
∴ Answer is i, ii, iii and iv
permissible is lower of the compressive &
10. C.
tensile stress.
The assumptions of pure/simple bending
∴ Permissible stress is 40 MPa
theory are as follows:
Now,
∴ Statement C is an incorrect statement.
40  150
Maximum stress (f) =
= 120 MPa <
50
THEORY
SIMPLE
BENDING
WITH
ASSUMPTIONS MADE
150 MPa
Before dismissing the theory of simple
∴ f = 120 MPa
bending, let us see the assumptions made
The maximum moment capacity (M) is
in the theory of simple bending. The
given as
following are the important assumptions:
M = f × z = 120 × 5.99 × 106 × 10–6
1.
= 720 kN-m
9.
OF
The
material
of
the
beam
is
homogeneous and isotropic.
D.
2.
Pure bending : If a length of a beam is
The
value
of
Young’s
modulus
of
subjected to a constant bending moment
elasticity is the same in tension and
and no shear force (i.e zero shear force)
compression.
then the stress set up in that length will be
36
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3.
The transverse sections which were
Moment of inertia for case 2:
plane before bending, remain plane
d
b  ( )3
n
I = n
12
after bending also.
4.
The beam is initially straight, and all
d
n  b  ( )3
nb  d2 bd2
n
z=
=
=
12
6n
6n2
longitudinal filaments bend into circular
arcs
with
a
common
Centre
of
bd2
M
1
Now, 1 = 62 = = n
1
M2
bd
n
???
curvature.
5.
The
radius
of
curvature
is
large
compared with the dimensions of the
12. B.
cross-section.
6.
From the flexural equation,
Each layer of the beam is free to expand
f M E
=
=
y
I R
or contract, independently of the layer,
above or below it.
For
11. A.
symmetrical
section
the
distanced
maximum stress point for tensile stress and
For the given equation:
compression stress is are equal from neutral
axis.
∴ yb = yt
fb
M
=

yb
 d4
64
The strength of a bending member suns up
 fb =
if the section is placed parallel to each other.
The moment carrying capacity of a B section
32m
d3
 fb = ff =
is the function of its section modulus
32m
d3
13. C.
MαZ
For bending member, the bending strength
For case 1:
is the major strength parameter.
bd3
I
bd
Z = = 12 =
d
y
6
2
For uniform strength, the bending stress at
any section across the span should remain
constant.
For case 2:
Bending stress (f) =
Moment of inertia for a particular portion
about the neutral axis of which section is

given as
In = Ix + Ah2
M
6M
= cons tan t  2 = constant
Z
bt
 t2 =
The term Ah is neglected for simplify
2
6m
6W
=
x
b(a = cons tan tvalue)
ab
calculation and its lesser value as compared
∴ t x
to Ix
∴ t2 α x
37
M
M
y  f =
I
Z
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∴ The recommended shape should be (c)
Computation:
=
1.5  10−5  10  50002
= 1.5625mm
8  300
15. B.
The shaft is subjected to axial, bending and
14. .
For a beam subjected to temperature
torsion load, the behavior is each case is
variation the beam wraps down
depicted below:
(i) Axial load
Nature of stress: axial
Average variation of temperature from
centroidal axis to extreme fibers =
(ii) Torsion
I
2
Maximum strain at extreme
I
2
fibers () = ( ) from bend equation
Nature of stress: shear stress
E f
=
k y
(iii) Bending moment
Now,
For a circular bent
Nature of stress: bending stress
∴ Bending
stress
is
only
produced
by
bending moment.
⇒ From flexural equations:
Ey y h / 2
h
R =
=
=
=
I
f
E
dT
( )
2
f M E
=
=
y
I R
From the geometry of circle
f =
L L
 = (2R − )  
2 2
16. D.
δ2 can be neglected as ‘δ’ is small
∴ control deflection () =
=
2
M 15  106  32
=
= 19.09N / mm2
Z
  2003
For the beam, the major design criteria are
L2
8R
bending moment
From flexural equation,
2
L
T
TL
( ) =
8
h
8h
From the bending/flexural equation
f M E
=
=
y
I R
TL2
 [ =
]
8h
38
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Where,
 3 − 11.833   0 + 1.833 
= 20  5  
+
  7.857  5
2
2

 

f = bending stress at a distance y from NA
 b = 11.219cm
axis
18. D.
y = the distance of layer from NA with stress
For the given condition:
‘f’
m = bending or resisting moment at the
section.
I = moment of inertia/second moment of
area bout the NA
Smax =
E = young’s modulus of elasticity of the
material
Now,
R = radius of curvature
According to flexural equation,
Now, The bending stress depends upon the
bending
moment
WL3
WL
,M
=
48EI max
4
not
on
the
M f
E
M f
= = 
=
....... (i)
I
y R
I
y
young’s
modulus of elasticity of material.
Smax =

 s =1
a
WL3
W  L2
ML2
....... (ii)
=
=
48EI 4  12EI 12EI
From equation (i) & (ii)
∴ the ratio is 1.
17. B.
12EI
=
max
For the given section,
The bending stress in compression to
tension is 4: 3


12EI  max
L2
I
=
12E  max
f
f

=
2
y
y
L
L
450 = f  y = f  450
y
L
12  E
L2
12E 
y d 2  150  450
1
1
2
=
=

=
LL
35.55 35.50 71
12  2  105
19. D.
2
The section modulus is Z = I = xy
y
f
3
x
 t = =
 12 − 4x = 3x
fb 3 30 − x
z=
⇒ x = 17.143 cm
Now, compressive force should be equals to
d2 – 3x2 = 0
Stress at (1) – (1) = 4 units
12.143
= 2.833
12.143
Stress at (3) – (3) = 3 
7.857
= 1.833
12.857
x(d2 − x2 )
6
For max value of section modulus (z),
that of tensile force
Stress at (2) – (2) = 4 
6
x=
d
y=
Stress at (4) – (4) = 3
 4 + 2.833   0 + 2.833 
 b5 
+
  12.143  5
2
2

 

Zmax
39
3
d2 −
d2
=
3
2
d
3
d 2d2

2d3
d3
3
= 3
=
=
6
36 3 9 3
dz
=0
dx
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20. D.
Mresisted 12  2
y3 60
=
 ( )20
3
M
3
120
For the given section & details:
Mresisted 12  2
y3 60
=
 ( )20
3
M
3
120
22. C.
For the given beam:
The value of can be x can be obtaining
similar triangle theorems
x
4  10
−6
=
x − 100
1.5  10−6
 2.5x = 400  x =
 1.5x = 4x − 400
Maximum bending moment at mid-span
400
= 160mm
2.5
Mx =
Now,
Stress () =
P
1
= EE  A
A
E
WL2
8
From flexural equation, M = f × z
E
A
E
A
 1 = 2  T = C
A2 E1
AC ET
 d2 =
3 WL2
4 bf
Now, Ec = 4 × 10–6
 d2 =
3 WL2
4 bf
d=
3 WL2
4 bf
d=
L
3W

2
bf
ET = 4  10−6 

90
= 2.25  10−6
160
AT
4  10−6
=
= 1.77
AC 2.25  10−6
21. .
Moment = Force × distance
23. A.
= stress × Area × distance
As per the given data:
depth ‘d’ remains constant and the width ‘b’
is variable.
For maximum bending stress same at all
section,
bending
stress
‘f’
Assuming a thin portion of the section of
constant
beam of thickness ‘dy’.
⇒ Now, from flexural equation:
60
Mresisted = 2 
W
x
f
M
f
=

= 2
d b  d3
y
I
n
2
12
   60  dy  y
20
60
Mresisted = 2 
M
 y  60  ydy
I
20

60
Mresisted = 2 

20
M
60  1203
12
 bn =
 60  y2dy
 bxx
40
3w
fd2
x
should
be
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24. D.
25. B.
From flexural equation,
M f
E
= =
........ (i)
I
y R
From circular property, l = Rθ ....... (ii)
From (i) & (ii)
E
Ixx
b  d3
=
+
12
Ixx =
E
= y =
y =
(x  60
0
60  0
d
(mb − b)  ( )3
3
3
4 = bd + (m − 1)bd
12
12
48  12
∴ f = 8.72 N/mm2
bd3
1
(1 + (m − 1) 
)
12
48

41
0.5
2
300  360
60  d
2  105 
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42
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Chapter
4
1.
Shear Stress
A beam of rectangular section is subjected
4.
to traverse loads. The shear stress at a
supported over the span of 5 meters is
particular section at 1/3
of the depth of
loaded with a central load of 100 kN. The
beam is 120 MPa, the maximum shear
maximum principal stress at a cross-section
stress is _______.
distant 2 meters from the support, at a
A. 135 MPa
point 100 mm from neutral axis is _______.
B. 160 MPa
A. 0 N/mm2
B. 2.811 N/mm2
C. 200 MPa
C. 9.375 N/mm2
D. 10.15 N/mm2
rd
5.
D. 240 MPa
2.
A rod of circular section is subjected to a
A 300 mm × 150 mm I-girder is installed to
shearing force on a plane perpendicular to
support the heavy rolling loads over the
its axis. If the rod is used as a simply
span. The girder is subjected to a shear
supported beam and load with a central load
force of 200 kN at a particular section,
‘ω’. The expression of the free length of rod
calculate the maximum shear stress in the
in
flange (in N/mm2) if the flange thickness
maximum shearing stress, due to shear
A laminate wooden beam 120 mm wide and
diameter
for
A.
1
D
3
B.
C.
2
D
3
D. D
240 deep is made up of three 120 mm × 80
planks
of
which
the
stress.
respectively.
mm
terms
force is one-third the maximum direct
and web thickness are 10 mm and 8 mm
3.
A 200 mm × 400 mm beam is simply
glued
together
to
resist
longitudinal shear. The beam is support
6.
1
D
2
A hollow circular beam of thickness 12 .5
over a span of 3 meters. The safe U.D. L the
mm and outer diameter 100 mm is simply
wooden beam can carry, if the allowable
supported at the ends. The beam is loaded
shear stress in the glued joints is 0.5
with a uniformly distributed load of 20 kN/m
N/mm is ________.
throughout the span of 10 meters. Calculate
A. 2.4 kN/m
the maximum shear stress at any point in
B. 3.6 kN/m
the beam. (in N/mm2, up to one decimal
C. 7.2 kN/m
places)
2
D. 14.4 kN/m
43
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7.
A beam has an equilateral triangular crosssection, having the side length ‘a’. If a
section of the beam is subjected to a shear
force ‘F’, the maximum stress at the level of
neutral axis in the cross-section is given by
A.
______.
A.
C.
8.
4F
B.
2
3a
4F
D.
3 3a2
8F
3a2
8F
3 3a2
A square cross section of side ‘a’ unit is
B.
placed such that one of its diagonals is
parallel to the horizontal. The location of the
maximum shear stress from the neutral axis
will be at distance of ______.
A. Zero
C.
9.
B.
a
D.
4 2
C.
a
2
a
8
A rectangular beam of width 100 mm is
subjected to a maximum shear force of 100
D.
kN. The corresponding maximum shear
11. What is the shear stress at the neutral axis
stress in the cross section is 5 N/mm2. The
depth of beam should be ___.
in a beam of isosceles triangular section
A. 100 mm
B. 150 mm
with a base of 50 mm and height 30 mm
C. 200 mm
D. 250 mm
subjected to shear of 5 kN?
10. A horizontal beam shown in the figure given
is below is subjected to traverse load.
A. 6.67 MPa
B. 8.87 MPa
C. 10 MPa
D. 12 MPa
12. If a beam of rectangular cross-section is
subjected to vertical shear force V, the
shear force carried by the upper one-fourth
of the cross-section is _______.
Which
one
of
the
following
diagrams
represents the shear force along the crosssection?
44
A.
3v
16
B.
5v
16
C.
7v
21
D.
5v
32
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13. Match the List-I (Different forms of cross-
A. a-3, b-2, c-1, d-4
section) with the List-II (Suitable shear
B. a-3, b-4, c-1, d-2
stress distribution diagram across section)
C. a-1, b-4, c-3, d-2
and select the most appropriate option.
D. a-1, b-3, c-2, d-3
List-I :
14. A hallow circular section is one of the most
efficient section to resist the loading due to
its outspread from the neutral axis thus
a)
increasing the moment of inertia. At a
diameter 200 mm and thickness 25 mm is
resisting the shear load of 50 kN. Calculate
b)
the shear stress (in N/mm2) at the inner
edge of the hollow section. (Assume y =
87.5 mm)
c)
15. For a thin circular tube, the maximum shear
stress is ______ the average shear stress
over the cross section.
d)
A. 1.33
List-II:
B. 1.5
C. 1.67
D. 2
16. Which of the following expression is an
1)
appropriate one to calculate the shear
stress across a section?
2)
F is the shear force at the section
3)
4)
45
A.
F 2
(D − y2 )
I
B.
F 2
(D − y2 )
2I
C.
F D2
(( ) − y2 )
I 2
D.
F D 2
(( ) − y2 )
2I 2
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46
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17. A circular cross-sectional beam of radius ‘R’
19. A simply supported timber beam is 10 cm
is subjected to the loading perpendicular to
wide and 20 cm deep carries a point load w
its span axis. The shear stress at a depth ‘y’
at the middle point of the span. The
above the neutral axis at any section is
permissible stress in flexure and shear are
given by _________.
100 kg/cm2 and 15 kg/cm2 respectively.
Ignoring the self-weight of the beam,
‘F’ is the shear force considered at the given
calculate the maximum length of the span
section
A.
F
(R2 − y2 )
2EI
B.
above which the bending stress will govern
F
(R2 − y2 )
EI
the safe load.
3F
(R2 − y2 )
D.
4EI
F
(R2 − y2 )
C.
4EI
stress
being
B. 7.47 cm
C. 12.5 cm
D. 25 cm
D. 1 meter
The
30
ratio
of
maximum
shear
stress
produced in the square section to that of
kg/cm2. Find the depth of beam.
A. 8.33 cm
C. 0.75 meters
resisting equal shear load ‘w’ in each case.
to a maximum shear force of 5000 kg, the
shear
B. 0.67 meters
20. The two-beam section of the same area are
18. A rectangular beam 10 cm wide is subjected
corresponding
A. 0.5 meters
circular section is _________.
A. 1.5
B. 1.33
C. 1.13
D. 0.89
ANSWER
1. A
2.
3. C
11. C
12. D
13. B
4. D
14.
5. D
6.
7. C
8. D
9. B
10. D
15. D
16. D
17. B
18. D
19. B
20. C
SOLUTION
1.
A.
b = width of section at the considered depth
The shear stress at any section is given as:
=
VAy
Ib
Where,
V = Traverse (shear) load at the section
A = Area above the depth at which shear
The axis at 1/3rd depth 00’ and the neutral
stress is calculated
y = distance of C.G of the area above
axis is represented by NA. The stress at the
considered depth from the Neutral axis.
Neutral axis is the maximum shear stress
I = moment of Inertia of the section
depth
47
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The shear stress is maximum is the flange
VA1y1
max
Ib1
A y

=
= 1
VA2y2
 
A2y
00
Ib2

max
 
00
web.
d d
b 
2
4 =9
=
d d 8
b 
3 3
 max =
2.
when calculated the function of flange and
=
VA 200  103  300  10  70
=
Ib
3.09  107  8
τ = 169.822 N/mm2
3.
9
 120 = 135 N/mm2
8
C.
For the given section:
.
The shear stress at any section is given as:
=
VAy
Ib
Where,
The critical design model would be the
V = Traverse (shear) load at the section
glued joint between the two wooden plates
A = Area above the depth at which shear
120  2403
= 1.382 × 108 mm4
12
Ixx =
stress is calculated
y = distance of C.G of the area above
For simply supported beam loaded with
considered depth from the Neutral axis.
u.d.l.
I = moment of Inertia of the section
b = width of section at the considered depth
Max shear (V) =
L
2
= 1.5ω kN
Now,
Shear stress at glued joint is as follows:
INA
Z
300  1503 292  1303
=
=
12
12
AA
=
1.5  120  80  80
1.382  108  120
ZAA’ = 0.069 ω N/mm2
INA = 3.09 × 107 mm4
Now permissible strength = 0.5 N/mm2
V = 200 kN
⇒ 0.5 = 0.069 ω
For I-section the shear stress is in the
⇒ ω = 7.2 N/mm
following manner
= 7.2 kN/m
4.
D.
For a beam subjected to bending & shear
load the principal stress are given as
follows:
48
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9.375
9.375 2
+ (
) + 2.8112
2
2
f
f
max/min = ( )  ( )2 + ()2
2
2
max =
Where,
= 10.15 N/mm2
f = bending stress produced at a point of
5.
considered section
D.
The direct stress is forming the bending
τ = shear stress produced at a point of
stress due to loading given by flexural
considered section
equation.
Computation:
For, simply supported beam
Maximum shear force (Vf ) =
Ixx =
200  4003
= 1.067 × 109 mm4
12

2
Maximum bending moment (m) =
A = 200 × 400 = 80000 mm2
L
4
Bending stress
Now,
(fb ) =
L
M
d
8L
 y(m) = 44  = (m) =
I
2
d
d3
64
…… (i)
Shearing stress
(Z) =
100
V=
= 50 kN
2
Moment at section AA'
z=
M = 50 × 2 = 100 kNm
Bending
stress
considered
point
at
section
is
given
AA'
and
a
8
3d2
..... (ii)
by
flexural
From (i) & (ii)
8
N
100  106
y =
 100 = 9.375 N/mm2
I
1.067  109
3d2
=
1 8L

3 d3
⇒L=d
Shear stress is given as follows:
Length should be equal to the diameter
50  103  80000  150
1.067  109  200
= 2.811 N/mm
1
f
3 b
Z=
equation.
z=
Ib
 
4d
  d2 4
2
8
3
2

d4
d
64
Now, as per given condition,
Now,
f =
VA y
6.
2
.
The shear stress at any section is given as:
=
49
VAy
Ib
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Where,
Now, z =
V = Traverse (shear) load at the section
A = Area above the depth at which shear
stress is calculated
∴ Zmax =
= distance of C.G of the area above
considered depth from the Neutral axis.
F
1
b 4
2
4
F

1
3
b 4
2
For equilateral triangle, b = a and
I = moment of Inertia of the section
h=
b = width of section at the considered depth
3
a
2
For hollow circle
 Zmax =
8.
4
F
4F

=
3 1
3
3 3a2
 a
a
2
2
D.
For a square section of side ‘a’ unit averaged
d0 = 100 mm and di = 75 mm
such that a diagonal is horizontal. The shear
moment of inertia (I)
stress distribution is given as follows
I=
4
4
  100
  75
−
64
64
∴ I = 33.56 × 104 mm4
Area above Neutral axis =

 (1002 − 752 )
8
∴ A = 1718.06 mm2
Now, the distance of C.G. of the area from
Here,
NA is given as follows:
y=
A1y1 + A2y2
A1 + A2

4  100 
4  75
 1002 
−  752 
6
8
6
y= 8


 1002 −  752
8
8
¯
D
D
 ( )2 + ( )2 = a
2
2
∴ y = 28.05 mm
Now, V =
∴Z=
20  10
 103 = 100000 N
2
100000  1718.06  28.05
33.56  104  100
= 14.52 N/mm2
7.

C.
For a actual section:
The maximum shear stress (Zmax) is 4/3rd of
the average shear stress (Z)

9.
D
D
=aa=
12
2
D 1
a
=  2a =
8 8
4 2
B.
The maximum shear stress (τmax)
(i) for rectangular section
max =
3
 avg
2
(ii) For triangular section
@SolutionsAndTricks
https://t.me/SolutionsAndTricks
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max =
From the two distribution shown above, for
4
 avg
3
as unsymmetric the max. shear stress does
Computation:
Hence, 5 =
not occurs along neutral axis.
6
Thus, option A is eliminated and thus ‘d’
3 100  10

2 1000  b
option is correct.
3 1
300
 b = 103  
=
= 150 mm
2 10
2
11. C.
The shear stress at any section is given as:
10. D.
=
The shear stress at any section is given as:
=
VAy
Ib
VAy
Ib
Where,
V = Traverse (shear) load at the section
Where,
A = Area above the depth at which shear
V = Traverse (shear) load at the section
stress is calculated
A = Area above the depth at which shear
y = distance of C.G of the area above
stress is calculated
considered depth from the Neutral axis.
y = distance of C.G of the area above
I = moment of Inertia of the section
considered depth from the Neutral axis.
b = width of section at the considered depth
I = moment of Inertia of the section
Then,
b = width of section at the considered depth
The maximum shear stress (τmax)
The shear stress is the function of the width
(i) for rectangular section
of the section at the considered point:
3
 avg
2
Thus, as the width increases, thus the shear
max =
stress value decreases and vice versa.
(ii) For triangular section
Thus, the option is directly eliminated as the
max =
stress is increasing with increase in width
and hence choice.
4
 avg
3
∴ more shear stress (τmore) =
For a rectangular section, the shear stress
distribution is a follow:
avg =
3

2 avg
V
5  103
=
= 6.67 N/mm2
A 1
 50  30
2
τavg = 10 N/mm2
12. D.
Thus, the option ‘b’ is eliminated as there
The shear stress at any section is given as:
only increase recorded for lower bar portion.
=
The shear stress distribution for some
sections are given below:
VAy
Ib
Where,
V = Traverse (shear) load at the section
A = Area above the depth at which shear
stress is calculated
y = distance of C.G of the area above
considered depth from the Neutral axis.
51
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I = moment of Inertia of the section
b = width of section at the considered depth
Computation:
For the above problem, we are required a
small of the area from the upper portion.
The shear stress thus obtained would be
integrated along the considered depth to
obtain
total
shear
stress
resisted
by
considered depth.
Thus from the above, the correct match of
List-I and List-II is a-3, b-4, c-1 and d-4.
The shear stress for x-x’ is given by

x − x
14. .
d y
V  b  dy  ( − )
2
2
=
bd3
b
12
The shear stress at any section is given as:
=
VAy
Ib
The overall shear force resisted to desired
Where,
depth is as follows:
V = Traverse (shear) load at the section
A = Area above the depth at which shear
Vxx = xx  b  y
D/4

 Vxx =
0
Vxx =
Vxx =
Vxx =
12
d3
D/4

0
stress is calculated
d y
Vb(dy )  ( − )
2 2 b y
bd3
b
12
y = distance of C.G of the area above
considered depth from the Neutral axis.
I = moment of Inertia of the section
d y
 ( − )dy
2 2
b = width of section at the considered depth
d
y2
y3
 [  ( )D/4
− ( )D/4
]
0
2
2
6 0
d
12
3
d3
d3
12V 5d3
5V

[
−
]
=

=
3
3
64
384
384
32
d
d
12
13. B.
The shear stress distribution for various
standard sections are as follows:
=
VA y
Ixx =
Ib

 (2004 − 1504 ) = 53.69 × 106 mm4
64
The width can be worked as follows:
52
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b
( )2 = 2  (1002 − 752 ) ⇒ b = 132.29 mm
2
Area is given as below:
Ay =

[(R + t)3 − R3]
6
Ay =
 3
[R + t3 + 3R2t + 3R2t − R3]
6
Neglecting higher terms
Ay − = 2R2t
Maximum shear stress
=
Area = Area of sector – Area of triangle
Average shear stress =
82.8 
1
  2002 −  2  75  66.14
360 4
2
∴ A = 265.163 mm
V
V  2R2t
V
Ay =
=
3
Ib

Rt
R t  2t
V
V
=
area 2Rt
Thus, the maximum shear stress is twice
2
the average shear stress is the section.
16. D.
3
 875
∴  = 50  10  2265.163
= 1.39 N/mm2
6
83.69  10  132.29
The shear stress at this level is given by
15. D.
=
Consider the thin tubular section is the
figure below :
F  Ay
Ib
For the given section:
For shaded area,
d
2
Area = π[(R + t)2 – R2]
A = ( − y)  b
= π[R2 + t2 + 2Rt – R2]
A = Area of the section above y
Neglecting terms of higher orders in t, as t
y = Distance of the C.G of area A from
is very small.
neutral axis
Area = 2πrt
Now,
y=y+
Moment of Inertia (Ixx) is given as follows:
Ixx =
=
y=

[(R + t)4 − R 4 ]
4
1
d
(y + )
2
2
b = actual width of the section at the level

[R4 + t4 + 4R3t + 6R2t2 + 4 Rt3 – R4]
4
I = MOI of the while section about N.A.
Neglecting higher the terms order is ‘t’

∴ Ixx = πR3t
= F(

Moment of the shaded area about N.A
Ay =
1 d
d y y d
( − y) = y + − = +
2 2
4 2 2 4

4(R + t) 
4R
(R + t)2
−  R2 
4
3
2
3
 [ =
53
d
1
d
− y)  b   (y + )
2
2
2
=
F d2
(
− y2 )
2I 4
F d2 2
(( ) − y2 )
2I 2
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17. B.
Substituting the value of Ay is equation (i),
FAy
... (i)
Ib
=
we get
Where A y = Moment of the shaded area
t=
F
about the neutral axis (N.A.)
2 2
(R − y2 )3/2
3
Ib
But b = EF = 2 × EB = 2 ×
I = Moment of inertia of the whole circular
Substituting this value of b in the above
section
b = Width of the beam at the level EF.
equation, we get
Consider a strip of thickness dy at a distance
2
F(R2 − y2 )3/2
F
= 3
=
(R2 − y2 )
2
2
EI
I2 R − y
y from H.A. Let dA is the area of strip.
Then dA = b × dy = EF × dy (∵ b = EF)
18. D.
= 2 × EB × dy (∵ EF = 2 × EB)
2
The shear stress at any section is given as:
2
= 2  R − y  dy
=
(∵ In rt. Angled triangle OEB, side EB
VAy
Ib
Where,
R2 − y2 )
=
V = Traverse (shear) load at the section
Moment of this area dA about N.A.
A = Area above the depth at which shear
= y × dA
stress is calculated
= y  2 R2 − y2  dy( dA = 2 R2 − y2 dy)
y = distance of C.G of the area above
considered depth from the Neutral axis.
= 2y R2 − y2 dy
I = moment of Inertia of the section
Moment of the whole shaded area about the
b = width of section at the considered depth
N.A. is obtained by integrating the above
for rectangle section,
equation between the limits y and R
Maximum shear stress = 3/2 × average
R
∴ Ay = 2y R2 − y2 dy
stress
y
∴ 30 =
R
= − (−2y) R2 − y2 dy
19. B.
Now (–2y) is the differential of (R2 – y2).
the
integration
3 5000

2 10  d
∴ d = 25 cm
y
Thence,
R2 − y2
of
the
If ‘w’ is the safe load at the mid span.
above
Let the span be ‘l’ meters
equation becomes as
Maximum shear force (S) = w/2kg
R
 (R 2 − y2 )3/2 
Ay = 

3/2

 y
=−
2
[(R2 − R2 )3/2 − (R3 − y2 )3/2 ]
3
=−
2
2
[0 − (R2 − y2 )3/2 = (R2 − y2 )3/2 ]
3
3
Maximum shear stress = 3/2 × average
shear stress = 15 kg/cm
3


= 15
2 10  20
∴ ρ = 2000 kg
∴ w = 23 = 2 × 2000 = 4000 kg
54
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Maximum bending moment : (m)
=
y = distance of C.G of the area above
4000 
 100
4
∴ M = 100000
considered depth from the Neutral axis.
I = moment of Inertia of the section
kg cm
b = width of section at the considered depth
Equating the maximum bending moment to
the moment of resistance of
Now,
1
fbd2 , we
6
For square/rectangle section:
max =
have,
100000 =
1
 100  10
6
For circular section:
max =
2
l = meters
3
∴ If
the
span
exceeds 2/3 meters
3
 avg
2
4
 avg
3
Now,
the
For square section:
bending stress will govern the safe load.
20. C.
For circular section:
The shear stress at any section is given as:
max,squares =
VAy
τ =
Ib
r=
Where,
V = Traverse (shear) load at the section
A = Area above the depth at which shear
stress is calculated

55
4 w

3 A
max,square
sec tion
max,square
sec tion
=
1.5
= 1.13
1.33
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57
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58
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Chapter
5
1.
Transformation of Stress
Plane stress at a point in a body is defined
6.
In hydrostatic condition of stress
A.
by principal stresses 3σ and σ. On the plane
Centre of Mohr’s circle coincides with
the origin
of max shear stress, the ratio of the normal
stress to the maximum shear stress will be
2.
A. 1
B. 2
C. 3
D. 4
C.
Principal stresses are of the opposite
D. Mohr’s circle is a point circle on x axis
In a state of pure shear what is the ratio of
7.
What is the value of shear stress acting on
a plane inclined at 42.30 to the cross section
A. 1
B. -1
of circular bar which is subjected to axial
C. 2
D. -2
tensile load of 100 kN? Diameter of bar =
The radius of Mohr circle for strain gives the
40mm.
value of
A. 58.73MPa
A. Maximum shear strain
B. 79.2MPa
B. Maximum normal strain
C. 39.6 MPa
D. Insufficient data
C. Half of maximum normal strain
8.
D. Half of maximum shear strain
4.
Mohr’s circle touches y axis
nature.
major and minor principal stress
3.
B.
The stress tensor at a point is given
 70
A body is subjected to normal strains of
as 
 −20
800×10-6 and
The
200×10-6 in
x
and
y
directions respectively and the shear strain
diameter
of
Mohr’s
circle
is
____N/mm2.
on this plane is 800×10-6. The maximum
9.
Pure state of shear is obtained by
A.
shear strain associated is A×10-6 units. Find
Equal tension in two directions at right
angles
the value of A
5.
−20
2
 N/mm .
40 
B.
Mohr circle for strain cannot be drawn for
Equal compression in two directions at
right angles
A. plane stress conditions
C.
B. plane strain conditions
Equal and opposite stresses at right
angles
C. Both A and B
D. None of these
D. None
59
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10. In case of biaxial strain, the maximum value
A.
fails only because of criterion 1
of shear strain is given by
B.
fails only because of criterion 2
A.
Difference of the normal strains
C.
fails because of both criteria 1 and 2
B.
Half the difference of normal strains
D. does not fail
C.
Sum of the normal strains
13. Assertion (A): If the state of a point is in
D. Half the sum of the normal strains
pure
11. For a rectangular rosette as shown in figure
shear,
then
the
principal
planes
through that point makes an angle of
what is the value of shear strain?
45Оwith plane of shearing stress carries
principal stresses whose magnitude is equal
to that of shearing stress.
Reason
stresses
(R): Complementary
are
equal
in
shear
magnitude
but
opposite in direction.
A.
ϵ0 = 10x10-4ϵ45 = 25x10-4ϵ90 = 15x10-4
A. ϒxy = 15x10-4
B. ϒxy = 50x10-4
C. ϒxy = 25x10-4
D. ϒxy = 35x10-4
Both A and R are individually correct,
and R is the correct explanation of A
B.
Both A and R are individually correct,
but R is not the correct explanation of A
12. A carpenter glues a pair of cylindrical
wooden logs by bonding their end faces at
C.
an angle of θ = 30degrees as shown in the
D. A is false but R is true.
figure
A is true, but R is false.
14. The readings obtained from strain rosette
are as follows
ϵ1 = 420μ ϵ2 = -45μ ϵ4 = 165μ
The glue used at the interface fails if
Criterion 1; the maximum normal stress
exceeds 3.5MPa.
What is the value of maximum in plane
Criterion 2; the maximum shear stress
shearing strain in microns?
exceeds 1.5 MPa.
15. If (80,20) and (60,40) N/mm2 are any two
Assume that the interface falls before the
logs fail.
point on Mohr circle, then calculate the
When a uniform tensile stress of 4MPa is
maximum shearing stress in N/mm 2.
applied, the interface.
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16. Select the correct statements:
1.
The
sum
of
perpendicular
stresses
20. Assertion (A): A plane state of stress does
in
direction
not represent plane state of strain as well.
mutually
is
Reason (R): Normal stress acting in x and
always
y directions will also result into normal
invariant.
2.
strain along the z-direction.
On plane of maximum shear normal
A.
stresses does not act.
3.
Both A and R are individually correct,
and R is the correct explanation of A
Radius of Mohr circle of stress is half the
B.
maximum shear stress.
Both A and R are individually correct, but
R is not the correct explanation of A
Select the correct code.
C. A is true, but R is false.
A. 1 only
B. 2 only
C. 1&2 only
D. 1&3 only
D. A is false but R is true.
21. An element in the plane stress is having one
of the principal stress (tension) as 30 MPa
17. A simply supported beam of 8m span is
and it is subjected to stress as σx = -50 MPa
loaded with uniform distributed load of
and τxy = 40 MPa then what will be the value
20KN/m over the whole span. The beam
of σy in MPa.
cross section is 250x500 mm find the
22. For a particular stress system, the Mohr’s
maximum shear stress at mid span in
circle is as shown in the figure. Calculate the
N/mm2.
values of the major and minor principal
18. A Mohr circle of stress cuts x-axis at 40 and
stresses in MPa respectively?
60 MPa on positive side. The absolute
maximum tangential stress is……. MPa
19. Select the correct statements:
1.
If principal stresses are of opposite
nature then the maximum shear stress
is
equal
to
the
half
the
sum
of
magnitude of principal stresses.
2.
3.
On the plane of maximum shear stress
A. 150.50, 30.67
B. 120.71, -20.71
C. 150.50, 40.67
D. 120.71, -96.71
23. The state of stress at a point when
normal stresses are zero.
completely
On the plane of maximum shear stress,
determine the
normal stress is equal to the mean of
principal stresses.
A. 1 only
B. 1 and 2 only
C. 1 and 3 only
D. None
specified
enables
one
to
1.
Maximum shearing stress at the point
2.
Stress components on any arbitrary
plane containing that point
Which of the above statements are correct?
61
A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2
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24. Direct stresses of 120MPa(C) and 90 MPa(T)
B.
exist on two perpendicular planes at certain
R is not the correct explanation of A
point in a body. They are also accompanied
C.
by shear stress on the planes. The greater
28. Angle of obliquity is defined as
150 MPa. The shear stress on these planes
A.
is equal to ………. MPa
given planes.
there are normal stresses of 60 MPa and 40
B.
MPa (both tensile) at right angles to each
shear
stress
and
C.
direction of its plane will be respectively
A. 22.36 MPa and 76.7
o
B. 22.36 MPa and 31.7
o
D. angle between resultant stress and
normal stress
29. In a steel flat plate a state of plane stress is
D. 27.64 MPa and 31.7o
available calculate major principal stress in
26. The magnitude of principal stresses at a
are
250
angle between resultant stress and
shear stress
C. 27.64 MPa and 76.7o
point
angle between resultant stress and the
plane of given normal stress
other,with positive shearing stress of 20
maximum
angle between the plane on which
stresses are evaluated and one of the
25. At a point in an elastic material under strain
The
A is true but R is false.
D. A is false but R is true.
principal stress at the point due to these is
MPa.
Both A and R are individually correct, but
MPa(tensile)
and
MPa if σx = 140 MPa, μ = 0.25, τxy = 40
150
MPa, ϵz = -3x10-6 and E = 2x105 MPa.
MPa(compressive).The magnitude of the
shearing stress on a plane on which the
30. As shown in the figure below the two planes
normal stress is 200 MPa(tensile) and the
PQ and QR are having shear stress of 40
normal stress on a plane at right angle to
MPa. The plane PQ carries the tensile stress
this plane are respectively are
of 60 MPa. Find the value of normal and
A.50 7 MPa and 100 MPa(tensile)
shear stress on the plane PR if the angle
B. 100 MPa and 100 MPa(compressive)
between the plane PR and QR is 60О.
C. 50 7 MPa and 100 MPa(compressive)
D. 100 MPa and 100 MPa(tensile)
27. Assertion (A): Mohr’s circle of stress can
be related to Mohr’s circle of strain by some
constant of proportionality.
Reasoning
(R): The
relationship
is
A. 79.64 MPa, 5.98 MPa
a
function of yield stress of the material.
B. 60.2 MPa, 7.45 MPa
A.
Both A and R are individually correct,
C. 40.5 MPa, 16.56 MPa
and R is the correct explanation of A
D. 30.4 MPa, 12.67 MPa
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ANSWER
1. B
2. B
3. D
4.
5. A
6. D
7. C
8.
9. C
10. A
11. C
12. B
13. B
14.
15.
16. A
17.
18.
19. C
20. A
21.
22. B
23. C
24.
25. A
26. C
27. C
28. D
29.
30. A
SOLUTION
1.
B.
7.
τmax = (σmax-σmin)/2 = 1
Shear stress at an angle ϴ with the direction
Normal stress on this plane
of the applied load is given as
= (σmax+σmin)/2 = 2
τ = P sin(2ϴ)/2A = 39.6 MPa
Required ratio = 2
2.
8.
B.
In
pure
shear
coincides
with
are
case
centre
origin,
equal
of
hence
in
D.
4.
.
2
principal
magnitude
and
= 25 N/mm2
= 50N/mm2
ϒmax/2 =
9.
2
 x − y 
  xy 

 + 

 2 
 2 
ϒmax = 1000×10
C.
10. A.
11. C.
-6
ϵx’ =
A = 1000
A.
x + y
2
+
x − y
2
cos 2 +
rxy
2
sin2 ϵx = ϵ0 =
10x10-4,
plane
stress
conditions
the
strain
ϵy = ϵ90 = 15x10-4,
components are ϵx, ϵy, ϵz and ϒxy hence
ϵx’ = ϵ45 = 25x10-4and θ = 45О
Mohr’s circle cannot be plotted in 2-D
Solving, we get ϒxy = 25x10-4
In
plane
components
6.
2
 x − y 
 xy 

 + 

2


 2 
Radius=R=
Hence diameter of Mohr’s circle
2
In
y = 40 τxy = -20
Mohr’s
opposite in nature.
3.
.
x = 70
stresses
5.
C.
strain
condition
the
are ϵx,ϵy and ϒxy hence
strain
12. B.
Mohr
σ = 4 MPa, θ = 30О
strain circle can be made.
Normal stress at the interface
D.
= σcos2θ = 3MPa <3.5 MPa
In
hydrostatic
condition
the
principal
Shear stress at the interface = σsinθcosθ
stresses are of alike nature hence Mohr’s
= 1.732 MPa>1.5 MPa
circle is a point circle on x axis.
Hence, interface fails by criterion 2
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16. A.
13. B.
In pure shear case the centre of Mohr’s
On plane of principal stresses shear stress
circle coincides with origin hence magnitude
acting is zero.
of principal stress is equal to shearing
Radius of Mohr circle is equal to maximum
stress.
shear stress.
Hence reason is not the correct
Hence, only 1 is correct.
17.
explanation.
BM at mid span = 20x82/8
14. .
Sum of strains in mutually perpendicular
= 160 kNm = 160x106Nmm
direction is invariant
Bending stress at this section, σ = My/I
ϵ1+ϵ3 = ϵ2+ϵ4
Y = 250mm, I = 250X5003/12
ϵ3 = -300µ = ϵy
σ = 15.36 N/mm2
ϵx’ =
x + y
2
− 45 =
+
x − y
2
cos 2 +
rxy
2
Shear stress at this section is zero
sin2
Maximum shearing stress = σ/2
= 7.68 N/mm2 Ans
420 + (−300) 420 − (−300)
+
2
2
 xy
cos(2  45) +
sin(2  45)
2
18. .
σ1 = 40 MPa , σ2 = 60 MPa and σ3 = 0

ϒxy = -210µ
R=
(
x − y
2
)2 + (
 xy
2
abs,max
= max [| 1 − 2 | , | 2 | , | 1 |] = 30 MPa
2
2
2
19. C.
)2 = 375µ
On the plane of maximum shear stress
Maximum in plane shearing strain
normal stresses can exist
= 2R = 750µ Ans
20. A.
15. .
21. .
The line given points is the chord of the
Radius of Mohr circle
circle and its perpendicular bisector will pass
2
 x − y 
[ − 2 ]
2
 + xy = 1
2
2


= R = 
through the centre of the circle.
Mid-point of the chord will be (70,30)
Where,
Slope of the chord = (40-20)/(60-80)
σ1,2 are
the
major
and
minor
principal stresses
=–1
2
2
 x − y 
  − 2 
2
i.e 
 + xy =  1

2
 2 


Hence, slope of its perpendicular bisector
=1
Let the centre be (σ,0)
2
2
 −50 − y 
 30 − 2 
2

 + 40 = 
 …….(1)
2
2




Slope of perpendicular bisector
= (σ-70)/(0-30) = 1
Solving, we get σ = 40 i.e. Centre is (40,0)
Also sum of the normal stresses remains
Its distance from any point on the circle will
invariant
give its radius.
i.e -50+σy = 30+σ2
R=
or σ2 = σy-80
(40 − 80)2 + (0 − 20)2 = 44.72 units
Putting the value of σ2 in (1) we get σy = 10
Maximum shearing stress = 44.72 N/mm2
MPa Ans
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22. B.
1 + 2 = x + y
250+ (-150) = 200+  y
 y = - 100 MPa
[(250-(-150)/2]2 = [(200-(-100))/2]2+ 2xy
τxy = 50 7 MPa
Triangles OAB and BCD are similar.
27. C.
Hence, OB = BC = 50
Mohr’s circle of stress can be related to
So, Radius of Mohr circle = 50 2
Mohr’s circle of strain by poisson’s ratio.
Principal stresses are 50  50 2
Hence A is correct, but R is wrong.
28. D.
i.e 120.71 Mpa and -20.71 Mpa
23. C.
29. .
Using
Standard
formula
of
stress
z − (x + y )
ϵz =
transformation maximum shearing stress at
E
that point, stress components on any
Since, it is a plane stress condition so
arbitrary plane containing that point can be
σz = 0
calculated.
Substituting the values,
24. .
we get σy = -137.6 MPa
Taking tensile as positive and compressive
σ1.2 =
as negative
150 = {(-120) +90}/2+
2
2
 x − y 
2
 
 + xy

2


2
 (−120) − 90 
2

 + xy
2


σ1.2 = 144.45 ± 1.2
Major principal stress = 145.65 MPa
xy = 127.3 MPa
30. A.
25. A.

2
 x − y 
2
Maximum shearing stress = 
 + xy

2



= 22.36 MPa
x
xy
=
x + y
=−
Plane
Direction of this plane is given by tan(2θs)
=-
x + y
2
+
x − y
x − y
2
PR
is
2
cos 2 + xy sin2
sin2 + xy cos 2
at
30О to
plane
PQ
in
anticlockwise sense
(x − y )
2xy
Hence, put σX = 60 MPa, σy = 0,  xy = 40
MPa and θ = 30О
θs = -13.3О or 76.7О
σx’ = 79.64 MPa and  x’y’ = 5.98 Mpa
26. C.
Normal stress on the plane is 200 MPa and
-100 Mpa as the sum of normal stress is
invariant.

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Chapter
6
1.
Torsion
Which of the following expression is correct
4.
for a soild circular shaft of minimum
surface of a hollow shaft subjected to a
diameter d for a given maximum shear
torque of 100 kNm? The inner and outer
stress when it has to transmit P horsepower
diameter of the shaft is 15 mm and 30 mm,
at N rpm? [where k is a constant]
respectively.
A. 50.26N/mm2
P
N
A. d = k 3 ( )
B. 40.24 N/mm2
C. 20.12 N/mm2
P
N
B. d = k 2.5 ( )
D. 8.74 N/mm2
C. d = k (P2/N2)1/3
5.
A solid steel shaft of 120 mm diameter and
D. d = k (N /P )
1.5 m long is used to transmit power from
A cylindrical shaft of diameter D and length
one pulley to another. What will be the max
L is subjected to a uniformly distributed
strain energy that can be stored in Nm? The
torque of T kN-m/m length of the shaft as
maximum allowable shear stress is 50MPa
shown below. Angle of twist at the free end
and shear modulus is 80 GPa.
2
2.
What is the shear stress acting on the outer
2 1/3
6.
is
Torque producing one radian twist in a
shaft of unit length represents
A. Torsional stress
B. Torsional rigidity
C. Flexural rigidity
A.
C.
3.
16TL2
GD4
TL2
GD4
B.
D.
D. Moment of resistance
32TL2
7.
GD4
A hollow shaft having an internal diameter
32TL
40% of its external diameter transmits
GD4
562.5 kw power at 100 rpm. Determine the
What is the maximum shear stress in MPa
external diameter of the shaft if the shear
induced in a solid shaft of 50 mm diameter
stress is not to exceed 60N/mm2 and the
which is subjected to both bending moment
twist in a length of 2.5m should not exceed
and torque of 300 kN-mm and 200 kN-mm,
1.3 degrees. Assume maximum torque =
respectively.
1.25 mean torque and modulus of rigidity =
9×104N/mm2.
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8.
A hollow steel shaft of 300 mm external
12. Consider the following statements:
1.
diameter and 200 mm internal diameter
section of a circular shaft subjected to
must be replaced by a solid alloy shaft.
twisting varies linearly.
Assuming same values of polar modulus for
2.
both. What will be ratio of their torsional
The shear stress at the centre of a
circular shaft under twisting moment is
rigidities (steel to alloy)? [Take G for steel
maximum.
as 2.4 times of G for alloy, where G is
9.
The shear stress distribution across the
3.
The shear stress at the extreme fibres
modulus of rigidity]
of
A hollow shaft is subjected to max shear
moment is maximum.
stress of 
max.
circular
shaft
under
twisting
Select the correct code.
If the maximum strain
energy stored per unit volume is 0.41 
max
a
/G where G is the modulus of rigidity
2
A. 1 only
B. 2 only
C. 1 and 2 only
D. 1 and 3 only
13. A tapered solid shaft of length L, has cross
find the ratio of internal diameter to
section with its diameter varying from D at
external diameter of the shaft.
one end to 2D at the other. What will be the
10. A hollow shaft is having steel shaft having
angle of rotation when the shaft is subjected
external and internal diameter as 80 mm
to pair of equal and opposite torques T
and 60 mm, respectively. Volume of shaft is
applied at its ends?
106 mm3. What will be maximum strain
energy
stored
in
the
shaft
in
Nm
A.
14 TL
3 πGD4
B.
C.
28 TL
3 πGD4
D.
if
maximum allowable shear stress is 80 MPa?
Take shear modulus as 100 GPa.
14TL
πGD 4
28TL
πGD 4
14. Select the correct statements:
11. The tapered shaft is confined by the fixed
1.
supports at A and B. If a torque of 378 kN-
Shear stress at corners in non- circular
section is zero.
2.
m is applied at its midpoint, determine the
Shear stress at corners in noncircular
section is maximum.
reactions at supports.
3.
Shear stress is maximum along centre
line of larger side.
4.
Shear stress is zero along centre line of
larger side.
Select the correct code.
A. TA = 300 kNm , TB = 78 kNm
A. 2 and 3 only
B. TA = 304 kNm, TB = 74 kNm
B. 2 and 4 only
C. TA = 74 kNm, TB = 304 kNm
C. 1 and 4 only
D. TA = 189 kNm,TB = 189 kNm
D. 1 and 3 only
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15. Which of the following assumptions of
3.
torsion formula are correct
Sections are necessarily circular to
prevent warping.
1.
Stress do not exceed elastic limit.
Select the correct code.
2.
Shaft is loaded by twisting couples in
A. 1,2 and 3 only
B. 2 and 3 only
planes that are perpendicular to the
C. 1 and 2 only
D. None
axis of shaft.
ANSWER
1. A
2. A
3.
4. C
11. C
12. D
13. C
14. D
5.
6. B
7.
8.
9.
10.
15. B
SOLUTION
1.
A.
3.
 T G
=
=
r
J
L
 max =
Polar modulus ∝ d3
M = 300 KN-mm and T = 200 KN-mm
J=

 d4
32
16
d3
X M2 + T 2
We get, 14.7 MPa
4.
d
r=
2
C.
From torsion formula we have
 T G
=
=
r
J
L
So
(T/  )∝ d3
r = 30/2 = 15mm, T = 100x1000 N-mm, J
Power P = T (2N / 60 )
=
Since shear stress is constant so T∝ d3
2.
.

*(304-154)
32
Or P/N∝d3
Substituting the values, we get 
P
Or d = k 3 ( )
N
= 20.12 N/mm2
5.
A.
Maximum strain energy stored in the solid
Torque acting at a distance x from A
shaft
T x = Tx
Umax = (τmax2/4G)xVolume
D4
T dx
θB =  x
where J =
32
GJ
0
L
Solving, we get θB =
.
=
16TL2
502


x1202x1.5 x 1000
4  80  1000 4
= 132.535 Nm
GD4
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6.
B.
θ=
8.
Polar modulus = J/r
TL
For θ = 1 radian and L = 1unit
GJ
From the question , Jsteel/rsteel = Jalloy/ralloy
T = GJ = Torsional rigidity
7.
.
.

(3004 − 2004 )

32
Or
Dalloy3
=
(300 / 2)
16
Power transmitted is given by
Dalloy = 278.78 mm
P = 2ΠNT/60
Ratio of torsional rigidities
562.5×103 = 2π×100×Tmean/60
=
T
mean
= 53714.8 N-mm
= 2.4x150/139.39 = 2.58 Ans
TMax = 1.25×Tmean = 67143.5N-mm
i)
9.
Diameter of the shaft when maximum
 T G
=
=
r
J
L
 =  max, r = D0/2 and J = π /32 (D04-Di4)
Torque in case of hollow shaft
Total Strain energy, U = ½ Tθ
T = (π/16) ×  × [(D04-Di4)/D0]
Putting the values
T = Tmax = 67143.5 and Di = 0.4D0,
we get Umax = (  max2/4G) (1+Di2/Do2) (π/4)
 = 60N/mm2
ii)
.
We know that ,
shear stress is 60N/mm2.
67143.5 =
GsteelJsteel
G
r
= steel steel
Galloy Jalloy Galloy ralloy
(D02-Di2)L
π/16×60[D04-(0.4D0)4/D0]
= (τmax2/4G) (1+Di2/Do2) x Volume
D0 = 18 mm
Maximum strain energy per unit volume
Diameter of the shaft when the twist in
= (τmax2/4G)(1+Di2/Do2)
the hollow shaft is not to exceed
or 0.41τmax2/G = (τmax2/4G)(1+Di2/Do2)
1.3degrees.
or Di/DO = 0.8
10. .
J = (π/32) × [D04-DI4] ….…
In the above question
(Di = 0.4D0)
= 0.09566D0
Umax = (τmax2/4G)(1+Di2/Do2)xVolume
4
802
(1+9/16)x1000000/1000
4  100  100o
We know that,
=
T/J = GΘ/L
= 25 Nm
11. C.
Θ = 1.3degrees = 0.02269 rad.
TA+TB = T = 378 kNm
L = 2500mm G = 9×104Nmm2
Rx = R(1+x/L)
T = Tmax = 67143.5

R x4
2
Substituting these values,
Polar moment of inertia,J =
we get D0 = 17.12mm
Since, the shaft is fixed at both the ends
The external diameter of the shaft should be
so θBA = 0
18 mm (greater of the two values obtained)
i.eθBC+θCA = 0
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13. C.
Diameter of shaft at a distance x,
Dx = D(1+x/L)
Polar moment of inertia J =
𝐿/2
∫
0
𝐿
𝑇𝐵 𝑑𝑥
𝑇𝐴 𝑑𝑥
−∫
=0
𝐺𝐽
𝐿/2 𝐺𝐽
Angle of rotation  =
L

Dx 4
32
T
 GJ dx
0
[θBC and θCA are of opposite nature]
Substituting the values and solving the
L /2
integration we get, θ =

0
L
(T − TA )dx
T dx
=  A
GJ
GJ
L /2
28 TL
3 GD4
14. D.
Putting the value of J and solving the
In noncircular section warping occurs and
integration we get
shear stress at corners will be zero whereas
TA = 152T/189 = 304 kNm and
it will be maximum along the centre line of
TB = 37T/189 = 74 kNm
larger side.
12. D.
15. B.
The shear stress variation is as shown
Stress should not exceed proportional limit
below. From the variation it is clear that it
so as to obey Hooke’s law
varies linearly and maximum at the extreme
Rest assumptions are correct.
fibres.

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Chapter
7
1.
Theories of Failure
Which is the most conservative theory of
6.
failure.
strain energy theory and maximum shear
A. Rankine Theory
stress theory are
B. Saint Venant Theory
A. square and hexagon respectively
C. Guest Theory
2.
D. Beltrami-Haigh Theory
B. square and ellipse respectively
Which is the most suitable theory of failure
C. hexagon and ellipse respectively
for ductile material.
D. ellipse and hexagon respectively
A. Maximum distortion energy theory
7.
B. Maximum strain energy theory
3.
to pure bending moment of 3.5 kNm. Find
D. Rankine Theory
the maximum twisting moment (in kNm)
Under what condition all the theories of
that can be applied on this shaft such that
the material of the shaft does not yield. Use
A. Principal stresses are of opposite nature
Tresca’s theory of failure. The yield stress of
B. Loading is hydrostatic
C. Loading is in the form of pure shear
the
D. Loading is uniaxial
400N/mm2.
Principal stresses at a point in an elastic
material
are
80
N/mm2(T)
and
8.
40
material
in
uniaxial
tension
is
A solid steel shaft is required to carry a
torque of 40 kNm and a bending moment of
N/mm2(C). If the material yields at 230
20 kNm. What will be the radius of shaft
N/mm2 determine its factor of safety using
5.
A solid shaft of diameter 50mm is subjected
C. Coulomb Theory
failure will give similar result when
4.
The yield locus according to maximum
maximum shear stress theory.
according to maximum principal stress
At a point in a structure, there are three
theory? Take FOS = 2.0, E = 200 GPa and
mutually perpendicular stresses of 800
kg/cm2 tensile,
400
yield stress = 250 N/mm2.
kg/cm2 compressive
A. 148.15 mm
and 600 kg/cm2 tensile. If poisson’s ratio is
0.3, what would be the equivalent stress
B. 142.15 mm
(kg/cm2) in simple tension according to
C. 138.15 mm
maximum principal strain theory.
D. 128.15 mm
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9.
Match the following:
stress theory. Consider section m-n as
List-I
A.
Maximum principal
List-II
1.
St Vanant
2.
Betrami and
critical section.
stress theory
B.
Maximum shear
stress theory
C.
Maximum principal
Haigh
3.
Tresa
4.
Von-Mises
5.
Rankine
strain theory
D.
Maximum distortion
energy theory
13. The state of plane stress is shown by the
A. a-5 b-3 c-1 d-4
B. a-5 b-1 c-2 d-4
C. a-3 b-5 c-1 d-2
D. a-3 b-1 c-2 d-5
stress tensor below.
80 25 

 MPa. It has been found that the
25 −40
10. If maximum principal stresses of 90N/mm2,
tensile yield strength is 250 MPa. Determine
60 N/mm2 and 30 N/mm2of the same
the factor of safety with respect to yield
nature acts on a body. The maximum
using maximum distortion energy theory.
14. A body is under the action of two principal
distortion energy stored per unit volume is
stresses of 40 N/mm2 and -70 N/mm2, the
A/G where G is the modulus of rigidity.
third principal stress being zero. The elastic
Report the value of A.
limit in tension as well as in compression is
11. Select the correct statements:
1.
200 N/mm2. The factor of safetybased on
Shear strain energy does not leads to
the elastic limit according to the maximum
change in volume.
2.
strain energy theory will be …………Take
Volume changes are associated with
poisson’s ratio = 0.3
normal stresses.
15. A cube of 5mm side is subjected to load
A. 1 only
B. 2 only
C. Both 1 and 2
D. None
system
as
shown.
What
will
be
FOS
according to maximum shear stress theory
12. A load P of 5000kg on the crank pin of the
if yield strength of the material is 80 MPa?
crank shaft as shown in the figure is
required to turn shaft at constant speed.
The crank shaft is made of ductile steel
having yield strength of 2800kg/cm 2 as
determined in simple tensile test. Calculate
the diameter of the shaft in cm based on a
factor of safety of 2.Use maximum shear
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ANSWER
1. C
2. A
3. D
4.
11. C
12.
13.
14.
5.
6. D
7.
8. A
9. A
10.
15.
SOLUTION
1.
C.
Putting the values
τmax in reality should be less than fy/ 3 but
we get T ≤ 3.44 x 106Nmm
according to this theory τmax should be less
Maximum twisting moment
= 3.44 kNm.
than 0.5fy.
2.
A.
3.
D.
4.
.
8.
Maximum bending stress under combined
bending and torsion in a circular shaft is
given by
According to max shear stress theory,
5.
A.
16
[M + M2 + T 2 ]
τmax = (fy/2)/FOS
σmax =
(80-(-40))/2 = 230/(2*FOS)
According to maximum principal stress
FOS = 1.92
theory
.
σmax≤ fy/FOS
According to maximum principal strain
Given, M = 20x106Nmm, T = 40x106Nmm,
theory
fy = 250N/mm2, FOS = 2
σmax/E =
d3
Putting the values,
1 − (2 + 3 )
,
E
we get d = 296.348 mm
σ1 = 800, σ2 = 400, σ3 = 600
Hence, radius of the shaft = 148.174 mm
Substituting the values,
9.
A.
we get σmax = 740 kg/cm2
10. .
6.
D.
Maximum distortion energy per unit volume
7.
.
is given by the expression = 1/12G [(σ1-
Maximum shear stress under combined
σ2)2+(σ2-σ3)2+(σ3-σ1)2]
bending and twisting is given as
= 1/12G[(90-60)2+(60-30)2+(30-90)2]
= 450/G
τmax = τmax = [16/(πd3)]X(M2+T2)1/2
So, A = 450
As per Tresca theory,
11. C.
[16/(πd3)]X(M2+T2)1/2 ≤ fy/2
Distortion occurs due to shear but volume
d = 50mm, M = 3.5x106 Nmm and
does not change.
fy = 400 N/mm2
Normal stress leads to volumetric strain.
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12. .
14. .
M = P×20 = 5000×20 = 100000kg-cm
σ1 = 40 MPa, σ2 = -70 MPa,
T = P×15 = 5000×15 = 75000kg-cm
σ3 = 0, fy = 200 MPa, µ = 0.3
According to maximum strain energy theory
Bending stress at section mnσ=
32M
d3
=
32  105
d3
1
[σ12+σ22+σ32-2µ(σ1σ2+σ2σ3+σ3σ1)]
2E
kg/cm2
2
 fy 


FOS 


Putting the values we get
2E
Shearing stress at section m-n
τ = 16T/πd3 =
d3
kg/cm2
FOS = 2.211
15. .

( )2 + 2
2
τmax =
=
12  105
20  105
d3
Principal stresses are given by
kg/cm2
σ1,2 =
According to max shear stress theory,
x + y
2
Given, σx =
τmax = (fy/2)/FOS
σy =
Putting the values, we get d = 10.44 cm
13. .
σ2 = 27.83MPa
Principal stresses are given by
x + y
2
1  1000
= 40 MPa
55
Solving, we get σ1 = 122.17 MPa and
 xy = -40 MPa fy = 250 MPa
σ1,2 =
2.5  1000
= 100MPa ,
55
1.25  1000
= 50MPa and
55
 xy =
σx = 80MPa, σy = –40 MPa and
2
 x − y 
2
 
 + xy

2


σ3 =
2
 x − y 
2
 
 + xy

2



σ1 = 85 MPa, σ2 = –45MPa, σ3 = 0
0.625  1000
= 25 MPa
55
abs,max
= max[
| 1 − 2 | | 2 − 3 || 3 − 1 |
]
2
2
2
= 48.585 MPa
From maximum distortion energy theory
According to maximum shear stress theory
f
( y )2
1
2
2
2
FOS
[(σ1-σ2) +(σ2-σ3) +(σ3-σ1) ] 
6G
12G
τabs,max≤(fy/2)/FOS
FOS = 0.823
FOS = 2.186

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Chapter
8
1.
Columns
The deflected shape of a column fixed at one
correct regarding assumption in Euler’s
approximately by
column theory.
B. y = asin
A.
x
2L
B.
C.
x
)
2L
The failure of column occurs due to
crushing as well as buckling.
D. The length of column is small as
x
)
2L
If the length
compared
of column
subjected
to
original
length,
the
to
its
cross-section
dimensions.
compressive load is increased by three
the
Failure of column occurs due to buckling
alone.
D. y = a(1 − cos
times
Initially the column is perfectly straight,
and load applied is truly axial.
x
2L
C. y = a(1 − sin
5.
A 200 x 100 mm cross-section of a
stanchion is of length 5 m with one end fixed
critical
and other end free, then the Euler's buckling
buckling load becomes
load
A. one-fourth of the original value
(in
[Take
B. four times the original value
kN)
Ixx =
for
1696.6
the
cm4,
stanchion
Iyy =
is
115.4
cm and E = 210 kN/mm ]
4
C. 1/9 of the original value
2
A. 351.64
D. 1/27 of the original value
3.
Which of the following statements are
end and free at the other can be given
A. y = a cos
2.
4.
B. 95.68
A column clamped at both the ends has
C. 88.16
buckling load of 4000N.During service, one
D. 23.92
end gets detached from the clamp and
6.
A circular rod 2.5 m long, tapers uniformly
of
from 25 mm diameter to 12 mm diameter.
percentage change in buckling load due to
Determine the extension of the rod (in mm)
change in the end condition is
under a pull of 30 kN. [Take E = 200 GPa]
becomes
free
end.
The
magnitude
A. 2.40
A. 50
B. 1.60
B. 75
C. 1.20
C. 83.25
D. None of these
D. 93.75
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7.
A rigid bar AB of length L is supported by
10. A hollow cast iron column whose outside
hinge at one end A and two spring of
diameter is 200 mm has a thickness of 20
stiffness K and 2K at another end as shown
mm. It is 4.5 m long and is fixed at both
in figure, the critical load for bar will be
ends. Calculate the ratio of Euler’s to
Rankine’s critical load.
[For
cast
iron
c = 550MPa, a =
1
1600
= 94000 N/mm2]
A. 1.42
B. 1.87
C. 2.42
D. 2.98
11. The resultant cuts the base of a circular
column of diameter ‘d’ with an eccentricity
8.
A. 0.5 KL
B. KL
C. 1.5 KL
D. 3 KL
 d 
equal to 
 the ratio of maximum stress
 16 
to minimum stress.
Two identical rigid bars AB and BC are
pinned at B and C supported at A by a pin
in a frictionless roller that can only displace
A. 4
B. 3
C. 2
D. Infinity
12. The rectangular column shown in the figure
vertically, then determine the critical load of
below carries 50 kN load having eccentricity
the system.
50 mm & 25 mm along x and y axis
respectively the stress at point ‘a’ is
9.
A. 6.25 N/m2
B. 6.25 N/mm2
C. 12.5 N/m2
D. 12.5 N/mm2
13. If a circle is drawn by taking the maximum
ka
4
A.
ka
2
B.
C.
ka
8
D. None of these
eccentricity of a solid circular section of
diameter ‘d’ as radius so that there will be
only compressive stress, then find the area
A circular column of length 2 m has Euler’s
of the resultant circle.
crippling load of 1.5 kN. If the diameter of
the column is reduced by 20%, then the
reduction
in
the
crippling
load
A.
 d2
4
B.
 d2
8
C.
 d2
16
D.
 d2
64
will
be_________%
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14. An I section column is pin jointed at both
16. A uniform column of cross sectional area ‘A’
ends as shown in figure. If 20 kN is acting
and flexure rigidity EI is heated from initial
at a distance
temp to. Find the temp at which column
4
from top in x direction and
start to buckle.
30 kN load is acting at a distance
2
from
α = coefficient Of linear expansion.
top is y direction, Find max slenderness
K = spring const.
ratio.
L = Length of column
Area of cross section = 3000 mm ,
2
rxx = 50 mm ryy = 15 mm ℓ = 3 m
L  2 EI
1
A. t = to +  −

 K AE   L3
L  2 EI
1
B. t = to +  −

 K AE   L2
L  2 EI
1
C. t = to +  +

 K AE   L3
L  2 EI
1
D. t = to +  +

 K AE   L2
A. 30
B. 40
C. 50
D. 60
17. If the crushing stress in the material of a
mild steel column is 3000 kg/cm2, Euler’s
15. According to Rankine’s formula which of the
option is correct.
formula for crippling load is applicable for
(All Symbols have usual meaning)
slenderness ratio.
A. Psafe =
B. Psafe =
C. Psafe =
D. Psafe =
fC A
A. λ ≤ 82
(1 +  ) FOS
B. λ ≥ 82
fC A
C. λ ≤ 328
(1 +   ) FOS
2 2
(
D. λ ≥ 328
fC A
18. A stanchion fixed at both ends of length 6
)
1 + 2 FOS
m consists of a builtup twin box section
fC A
using ISMB 250 with two plate of 260 mm
(1 +   ) FOS
2
× 10 mm size.
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Rankine’s formula with values of fd = 560
N/mm2  =
1
and FOS = 3. If one end
1600
of the column is fixed and other end is free,
the safe load (in KN) will be
The geometrical properties of ISMB 250
20. Find the value of x + y + z?
are
Boundary conditions - Euler buckling load
h = 250 mm
(column)
b = 125 mm
A = 47.55 cm2
A. Pin – Pin – x
2 EI
Ixx = 5131.6 cm4
Iyy = 334.5 cm4
L2
B. Fixed – free – y
2 EI
Find λmax. use leff = 0.65ℓ
19. A 2.5 m long column has a circular cross-
C. Fixed – fixed – z
section of 120 mm diameter. Consider
L2
2 EI
L2
ANSWER
1. D
2. C
3. D
11. B
12. B
13. D
4. A,B
14.
5. D
6. B
7. D
8. B
9.
10. C
15. C
16. C
17. B
18.
19.
20.
SOLUTION
1.
D.
−d2y
2
dx
Usual bending equation gives the deformed
shape as shown in figure


Let
=
EI
M
P(a − y)
=−
EI
EI
d2y
dx2
d2y
2
dx
+
= P(a − y)
Py Pa
=
EI EI
P
= n2
EI
Hence in operator form, the differential
equation reduces to
(D2 + n2)y = n2a
The solution of the above equation would
consists of complementary solution and
particular solution, therefore
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y = A cos (nx) + B sin (nx) + P.I
Critical buckling load, Pcr =
where, P.I is a particular value of y which
square of the effective length keeping other
∴ yP.I = a
factors invariant
∴ The complete solution becomes
Hence, critical load will become 1/16 times
y = A cos (nx) + B sin (nx) + a
the original critical load if one of the
Boundary conditions,
clamped ends becomes free.
(i) At x = 0, y = 0
So, percentage change in buckling load
⇒ A = -a
= (1-1/16) x100
dy
=0
dx
= 93.75%
⇒B=0

At
y = a(1 − cos
4.
x = L, y = a = a(1 − cos
column theory:
P
x)
EI
P
L =0
EI
⇒ cos
P
 3 5

L = ,
,
.(2n − 1)
EI
2 2 2
2
(i)
(ii)
column
is
initially
Material
is
uniform,
isotropic,
obeys Hooke law.
(iii)
Length of column is very large as
compared to x-section dimensions.
πx
y = a(1 − cos
)
2L
C.
(iv)
Self weight of column is neglected.
(v)
The column will fail by buckling only.
(vi)
The direct stress is very small as
compared with the bending stress
2EI
corresponding to buckling condition.
L2
5.
Since the length of the column is increased
D.
by three times the new length will become
End conditions are: one end fixed and other
9 times the original length. Hence critical
end free
buckling load becomes 1/9 of the original
 Lij = 2L
volume.
3.
perfectly
homogeneous, perfectly elastic and
P

=
EI 2L
Critical buckling load, Pcr =
The
straight and axially loaded.
The deflected shape of column is,
2.
A. B.
Following are the assumptions in Euler’s
P
x)
EI
⇒ cos
⇒
L2
Buckling load is inversely proportional to
satisfies the differential equation.
(ii) Atx = 0,
2EI
Euler's buckling load, P =
D.
Effective length of the column when both
Iyy  Ixx
ends are clamped = 0.5l
Effective length of the column when one end
Pe =
2EI
(2L)2
2EIy
4L2
(where, L = 5000 mm and
is free = 2l
84
=
2EI
4L2
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E = 210 x 103 N/mm2)
P =
Taking moment about A
2  210  1000  115.4  104
 (K)L + (2K)  L = (P )  
 P = 3KL
2
4  (5000)
8.
= 23917.99N
6.
B.
= 23.92kN
A free body diagram of the entire system of
B.
two rigid bars is shown below
The elongation of a tapering rod with
circular cross-section is given by,
L =
4PL
d1d2E
p = 30 kN
L = 2.5 m = 2.5 x 103 mm
d1 = 12 mm; d2 = 25 mm
E = 200 GPa
 L =
4  30  2.5  103  103
  12  25  200  103
= 1.59mm
7.
1.6mm
Take, ΣMA = 0
D.
The deflected shape of column can be
represented as

Hc  4a − Fs  3a = 0

Hc  4a − ka()  3a = 0

Hc =
3ka()
4
Now, for the calculation of critical load,
consider the free body diagram of lower bar
BC, shown below
FBD can be shown as
Take, ΣMB = 0
85

Hc  2a − Pcr  2a() − ka()  a = 0

Pcr =
ka
4
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9.
.
11. B
Euler’s crippling load,
When the column is subjected to eccentric
2EI
P=
loading,
2
L

2E
 d4
64
P=
L2
20% reduction in diameter,
P (Pe ) y
−
A
I
e = eccentricity =
So reduction in crippling load will be
d4
min =
P = Load on the column
 P1  0.4096d4
d4 − 0.4096d4
P (Pe ) y
+
A
I
Where,
 P1  (0.8d)4
=
max =
 100
I of (circular cross section) =
= 59.04 %
end = d/2

Area, A = (202 − 162 ) = 113.097cm2
4
Thus,
Moment of Inertia,
max =

(204 − 164 ) = 4637cm2
64
Radius of Gyration,
=
I
=
A
K=
41cm
Effective length, L eff
c A
1+
=
L2
( eff
)
2
K
550  113.097  10
2
 2.25  1000 
1


1600 
41  10 
L eff 2
=
( 4 + 2)
4P
2
d
−
Pd d 64
 
16 2  d4
( 4 − 2)
Load = 50 kN
ex = 50 mm, ey = 25 mm
2  94  109  4637  10−8
B = 200 mm, d = 100 mm
2.252
The stress at any point is equal to
⇒ Pe = 8497666 N

d2
+
Given,
Euler's critical load,
Pa =
P
d2
4
d
16  d
2
d4
64
P
12. B
⇒ P = 3510347.1 N
2EI
d2
P
max
=3
min
2
1+
P
min =
L 4.5
= =
= 2.25m
2
2
Rankine's Critical load P =
=
d4
64
y is distance of neutral axis from extreme
10. C.
I=
d
16
Pa
8497666
=
=2.42
P
3510347.1
a =
86
P (Pex ) x' (Pey ) y'
+
+
A
Iy
Ix
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At point A,
15. C
As per Rankine Theory,
x’ = 50 mm, y’ = 25 mm
(
)
1
1
1
=
+
P Pe Pc
50  103  50  50  12
50  103
+
200  100
100  2003
( a) =
+
50  103  25  12  25
P=
3
100  200
Pc Pe
Pc + Pe
Where,
= 6.25 N/mm2
Pe = Euler crippling load
13. D
Pc = Crushing load
P=
Pc
fc . A
=
Pc
f .A
1+
1 + 2c
Pe
 EAr2
l 2eff
=
There will be no any tensile stress.
0=
4P
d
2
(Px ) 64  d
−
d
4
 f 
1 +  2c  2
  E
P=
2
fc A
1 +  2
Psafe =
d
x=
8
Area of resultant circle = x 2 =
fC A
(1 +  ) FOS
2
16. C
d 2
64
Free expansion of column = α (t – to)L
14. .
Let Pbc the force exerted by spring on
column. Reduction is length of column due
Slenderness ratio is given by,
=
fc A
leff
rmin.
to spring =
So net expansion of bar =  ( t − to ) L −
Effective length does not depends upon
position of loading.
x =
y =
ryy
leff y
rxx
=
PL
AE
Net expansion of bar = reduction of spring
For both ends pinned, leff = l = 3 m
leff x
PL
AE
 ( t − to ) L −
3000
= 200
15
P=
3000
=
= 60
50
PL
P
=
AE K
 ( t − to ) L
L 
1
 K + AE 


=
2 EI
L2
L  2 EI
1
t = to +  +

 K AE   L3
Thus, slenderness ratio will be least of two
i.e. 60.
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17. B
As Iyy is min.
σc = 3000 kg/cm2
=
For mild steel, E = 2×105 N/mm2
Buckling stress should be less than crushing
2
 EA r
2
l 2eff
2 E
2

leff
6000  .65
=
= 79.88
rmin
48.82
Length of column = 2.5m,
Area =
 Pc
=

4
 (120) = 10178760.2 mm4
64
Radius of gyration
2 E
c
2 E
= 81.89
c

2
 (120 ) = 11309.74 mm2
4
Moment of Inertia I
 c
2 
A
23736167.83
= 48.82 mm
9955
Given,
 Pc
l 2eff
=
19. 114.98
stress
2 EI
Iyy
rmin =
9.81 N
= 3000 
= 294.3 N mm2
100 mm2
I
=
A
r=
82
10178760.2
= 30 mm
11309.74
For one end fixed and other free leff = 2l
= 2 × 2.5 = 5 m
18. 79.88
Crushing load (fd) = 560 N/mm2
Given,
For ISMB 250 section:
P=
h = 250 mm
b = 125 mm
=
A = 47.55 cm2
Ixx = 5131.6 cm4
fc A
1 +  2
560  11309.74
1
5000  5000 

3 1 +


1600
302


= 114979.44 N
Iyy = 334.5 cm4
= 114.98 kN
Size of plate = 260 × 10 mm
20. 5.25
Length = 6m
Euler’s crippling load for the following
Area of the combined section
= 47.55 × 100 + 2×260×10 = 9955 mm2
boundary conditions are as follows:
For the combined section:
for pin – pin −
Ixx = 5131.6  104 +
3
2  260  10
12
2 EI
for fixed – free −
= 80609333.33 mm4
2 EI
2
 260  103
125  

Iyy = 334.5 cm4 + 2 
+ 260  10  5 +
 
12
2  



for fixed – fixed −
= 23736167.83 mm4
a + b + c = 5.25

88
⇒a=1
L2
4L2
42 EI
L2
⇒ b = .25
⇒c=4
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90
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Chapter
Spring, Shear Centre
& Combined Stresses
9
1.
In a closely coiled helical spring having 100
4.
mm near diameter and made of 25 turns of
diameter as 100 mm and wire diameter as
20 mm diameter steel wire. The spring
12 mm consist of 16 coils. If it is subjected
carries an axial load of 200 N, modulus of
rigidly
100
GPa.
The
shearing
to an axial tension of 400 N, find maximum
stress
stress induced in the coil.
developed in the spring in MPa is ______
2.
A.
8

B.
16

C.
20

D.
32

5.
A. 24.9 N/mm2
B. 58.97 N/mm2
C. 31.5 N/mm2
D. 64.2 N/mm2
An elliptical type of leaf spring is 800 mm
long. Static deflection of spring under a load
A helical spring, with small slope of helix, is
supposed to transmit a maximum pull of 2
of 30 kN is 100 mm. Determine the
kN and to extend 10 mm for 400N load. If
maximum allowable stress if the leaves are
the mean diameter of the coil is 100 mm, G
75
is 80 GPa and allowable shear stress is 100
E=205 GPa
MPa than the required number of coils are
3.
A close-coiled helical spring, with the coil
A plane frame shown in the figure (not to
scale) has linear elastic springs at node H.
6.
The spring constants are kx = ky = 5 ×
mm
wide
and
8
mm
thick.
A. 781.5 N/mm2
B. 1500 N/mm2
C. 1025 N/mm2
D. 937.5 N/mm2
A rigid bar AB of length L is supported by
hinge at one end A and two spring of
105 kN/m3 and kθ = 3 × 105 kNm/rad.
stiffness K and 2K at other end as shown in
figure, the critical load for bar will be
For the externally applied moment of 30
kNm at node F, the rotation (in degrees,
round off to 3 decimals) observed in the
rotational spring at node H is _____
91
A. 0.5 KL
B. KL
C. 1.5 KL
D. 3 KL
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7.
Two closed coil springs of stuffiness s and
A. 12
B. 13
2s are arranged in series in one case and in
C. 14
D. 15
9.
parallel in the other case. The ratio of
8.
A close-cooled helical spring is to carry a
stiffness of springs connected in series to
load of 150 N. The mean coil diameter has
parallel is
to 10 times that of the wire diameter. If the
A. 1/3
B. 1/9
C. 2/3
D. 2/9
maximum shear stress is not to exceed 100
N/mm2, calculate the diameter of the coil.
For a helical spring the slope of helix may
be assumed small, is required to transmit a
A. 6 mm
B. 8 mm
C. 10 mm
D. 12 mm
10. A spring having diameter 8 mm and mean
maximum pull of 1 kN and to extend 10 mm
diameter of coil 100 mm and 12 turns. Find
for 100 N load. The mean diameter of the
the stiffness of spring if G = 80 GPa
coil is to be 100 mm. The diameter of wire
A. 3.41 N/mm2
B. 3.41 N/mm
is 100 mm. Calculate the no. of coils
C. 2.14 N/mm
D. 2.14 N/mm2
required? (G = 100 GPa and fs = 100 MPa)
ANSWER
1. C
2.
3.
4. B
5. C
6. D
7. D
8. B
9. A
10. B
SOLUTION
1.
C.
d = 20,
For springs:
R = 50 mm
(i) Maximum shear stress is given by:
max =
16T
d3
=
16  F 
d3
max =
D
2 = 16  W  R
d3
2.
2 3
4W D n
20

.
8WD
d3
d = diameter of spring wire
d3 =
8WD3n
Gd4
8  2000  100
  100
d = 17.21 mm
(iv) Stiffness is given by:
K=
=
Here, D = diameter of coil,
Gd4
(iii) Deflection is given by
=
  203
Shear stress = fs = 100 =
(ii) Strain energy stored is given by:
U=
16  20  50
Spring constant =
Gd4
8D3n
Given: W = 200 N,
92
W 400
=
= 40 N / mm

10
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5.
8WD3n
As,  =
Gd4
L = 800 mm, B= 75 mm
δ = 100 mm, W=30 kN, t=8 mm and E=
Gd4
So, n =
W
8   D3
  
n=
n
3.
.
20 x 103 N/mm2
=
80  103  (17.21)
4
8  40  1003
= 21.93
3WL3
8Enbt3
100 =
22
3  30  103  8003
8  205  103  n  75  83
n = 7.317
Now,
.
f =
f =
6.
3WL
2nbt2
3  30  103  800
2
2  7.317  75  8
= 1025 N/mm2
The deflected shape of column can be
represented as
R × 3 = 30
⇒ R = 10 kN
From right side of FBD
R × 3 = Kθ . θ
Therefore, rotation in the spring is,
=
3R
3  10 kNm
=
Ke 3  105 kNm/rad
= 10
–4
FBD can be shown as
rad = 0.0057 degree
= 0.006 degree
4.
Coil diameter D = 100 mm so radius,
R =50 mm
Wire diameter d=12 mm
Number of coils, n=16
Axial tension, W=400 N
=
=
16WR
Taking moment about A
d3
16  400  50
  123
 (K)L + (2K)  L = Pc  
2
= 58.97 N/mm
 Pc = 3KL
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7.
.
9.
Let stiffness of springs in series is K, Hence
stress due to torsional load and considering
1
1 1
= +
Ks s 2s
direct shear stress
 Ks =
2s2 2
= s
3s
3
And, Kp = 3s

8.
The maximum shear stress considering
Ks 2s / 3 2
=
=
KP
3s
9
Spring constant, K =
W

W
8WD3n
n=
16W (5d) 
d

1 + 4  10d 
3


d
80W
d2
(1 + 0.025)
d2 =
1.025  80W
fs
d2 =
1.025  80  150
= 39.152
  100
10. Deflection (δ) = =
δ = 100 mm for 100 N
n=
fs =
d = 6.25 mm ≈ 6 mm
Gd4
K =
16WR 
d 
1+
3 
4R 
d 
fs =
.
K =
fs =
100
= 10 N/m
10
U 64PR 3n
=
P
Gd4
Force(P) = stiffness(K) x deflection(δ)
Gd
W 3
8 D
  
Stiffness(K) = =
100  103  104
K=
4
P
Gd4
=
 64R3n
Radius = 100/2 = 50 mm
3
8  10  100
= 12.5  13

94
Gd4
64R3n
=
80  103  84
64  503  12
= 3.41 N/mm
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Chapter
10
1.
Thick and Thin Shells
A thin cylindrical shell is 1 m in diameter and
5.
2 m in length. A uniform fluid pressure of
pressure of 60 MPa. If the hoop stress on
5N/mm2 is
the outer surface is 150 Mpa, then the hoop
applied
on
the
cylinder.
stress on the internal surface is
Calculate the change in volume in cm3. Take
μ
=
0.3
and
E
=
2*105 N/mm2 and
thickness of material is 20 mm.
2.
6.
An iron pipe of 1 m diameter is required to
B. 180 MPa
C. 210 MPa
D. 135 MPa
A thin cylindrical tube closed at ends is
also applied to the tube. The principal
tensile stress of the pipe material is 22 MPa,
stresses are 100 and 25 units respectively.
then the minimum thickness of the pipe will
IF the yield stress is 300 units, then what is
be
the factor of safety according to maximum
A. 30 mm
B. 35 mm
C. 40 mm
D. 45 mm
shear stress theory?
7.
A thin cylindrical tube with closed ends is
A cylindrical tank subjected to internal
subjected to longitudinal stress σL = 14
pressure P is simultaneously compressed by
N/mm2, hoop stress σh = 20 N/mm2 and
an axial force F = 60 KN. The diameter is
shearing stress τ = 8 N/mm2. Then the
maximum shearing stress is
150 mm and thickness is 5 mm. Calculate
the maximum allowable internal pressure P
in N/mm2 based upon the allowable shear
8.
stress in the wall of the tank of 55 MPa.
4.
A. 105 MPa
subjected to internal pressure. A torque is
withstand 150 m head of water. If the
3.
A thick cylinder is subjected to an internal
A. 3 N/mm2
B. 6.50 N/mm2
C. 8.525 N/mm2
D. 8 N/mm2
A steel cylinder is 2000 mm long and 1000
mm in dia. It has 10 mm uniform thickness.
A thin walled spherical shell is subjected to
After being filled with water at atmospheric
an internal pressure. If the radius of the
pressure, some more water is pumped in
shell is increased by 2% and thickness is
until pressure is reached to 2 N/mm2, on
reduced by 1%, with the same internal
releasing the pressure the water is escaped
pressure, the percentage change in the
out which is measured as 2900 cc. If E s =
circumferential stress is
2×105 N/mm2 and μ = 0.Then calculate the
A. 1.01%
B. 2.02%
value of bulk modulus in GPa of water
C. 3.03%
D. 4.04%
assuming ends are flat and closed and there
is no bending.
96
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9.
A spherical shell of diameter 2 m is made of
10. A thin closed cylindrical shell of steel made
20 mm thick plate of steel whose yield
of 7.5 mm thick plate is filled with water
strength
the
under a fluid pressure 5 N/mm2. The
maximum depth in km of submergence up
internal diameter of cylinder is 250 mm and
to which vehicle can collect marine data.
its length is 2000 mm. Calculate the amount
is
350
MPa.
Determine
of water spilled when shell is opened in cc.
Take specific weight of water 10 KN/m3.
Take E = 2×105 N/mm2, μ = 0.3 and Kw =
2100 N/mm2.
ANSWER
1.
2. B
3.
4. C
5. C
6.
7. C
8.
9.
10.
SOLUTION
1.
.
= 1000 × 9.81 × 150 = 1.4715 MPa
Volumetric strain =
PD
(5-4µ)
4tE
So, 22 =
t = 33.44 mm say 35 mm
P = 5N/mm2
3.
D = 1000 mm
V=
.
F = 60 KN
  D2  L
= 1570.8×106 mm3
4
D = 150 mm; t = 5 mm
D/t = 150/5 = 30>10 so thin cylinder
Volumetric strain
=
τmax = (σh –σL)/2
V
5  1000
=
(5 − 4  0.3)
V
4  20  2  105
For maximum shear stress σL should be 0
= 1.1875×10-3
h =
ΔV = 1.1875×10-3×V
PD P  150
=
2t
25
So, 55 =
= 1.1875×10-3×1570.8×106
= 1865325 mm3
4.
= 1865.325 cm3
2.
1.4715  1000
2t
C.
C =
B.
Pr
t
C =
Since hoop stress is more than longitudinal
P  150
= P = 3.67 N/mm2
25
1
P  1.02  r
P r
= 1.0303
0.99  t
t
stress, so we check for hoop stress
h =
Percentage change =
PD
2t
P = 150 m of water
= 3.03%
97
(1.0303 − 1)
P r
t
P r
t  100
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5.
C.
max =
P = 60 MPa
c =
= 8.545 N/mm2
r 2 
  o2 +1
2
r

− ri


P  ri2
ro2
8.
=
ro2 − ri2  2
= 150 / 120
ro2 − ri2
ri2
V=
ri2
=
ri2
ro2
=(2900×1000)/(1570.79×106)
60  ri2
ro2
− ri2
= 1.846×10-3
 r2 
  o2 +1
r

 i

V =
5
 (9/5 +1)
4
P D
P
(5 − 4  ) +
4 t E
K
1.846  10−3 =
2  1000
5
4  10  2  10
= 210 MPa
K = 2232.14 N/mm2
.
= 2.23 GPa
max =
=
9.
P1 − P2
2
100 − 0
2
for maximum shear stress P2
L = H =
PD

4t 350
P  2000
 350
4  20
τy = fy/2
P = 14 N/mm2
= 300/2 = 150 unit
P = 10×1000×H
FOS = τY/τmax
H = 1400 m = 1.4 km
=150/50 = 3
10. .
C.
V =
0.5

h + l   h − l 
 
 + 2 
 2 

2


2
=
0.5

14 + 20   20 − 14  2
=
 
+ 64 



2
2



σ1 = 25.84 N/mm
2
K
.
= 50 units
1/2 =
(5 − 4  0.3) +
As yield strength is 350 N/mm2
should be 0
7.

×10002×2000
4
Volumetric strain = ΔV/V
9
5
= 60×
6.

×D2×L
4
= 1570.79×106 mm3
So at r = ri
c =
.
ΔV = 2900 cm3 = 2900*1000 mm3
150 when r = ro
150 = 60 
25.5 − 8.45
2
P D
P
(5 − 4  ) +
4 t E
K
5  250
4  7.5  2  105
(5 − 4  0.3) +
5
2100
= 3.1727*10-3
ΔV = 3.1727*10-3 ×
2
= 311479.09 mm3
σ2 = 8.45 N/mm2
= 311.48 cm3

98

×250×250×2000
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@SolutionsAndTricks
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Chapter
11
1.
Deflection
4.
What will be the vertical deflection and slope
at C;
The free and of a cantilever beam is
supported
by the free-end
of another
cantilever beam using a roller as shown in
the
figure
given
below.
What
is
the
deflection at the roller support B?
A. Deflection =
3ML2
EI
B. Deflection =
3ML2
2EI
ML
C. Slope =
EI
D. Slope =
2.
2ML
EI
A simply supported beam which carries a
5.
A.
8Pa3
3EI
B.
9Pa3
3EI
C.
64Pa3
35EI
D.
216Pa3
105EI
Two prismatic beams having the same
UDL over the whole-span is propped at the
flexural
centre of the span so that the beam is held
subjected to loading as shown below.
to the level of the end supports the reaction
rigidity
of
1000
KN-m2 are
of the prop will be equal to …………. Times
the
distributed
load.
(round
of
to
3
decimals).
3.
The vertical deflection of the middle hinge
of the given beam is:
If the slopes at the left support of these
beams
A. 0
576
m
B. EI
72
m
C. EI
36
m
D. EI
are
denoted
by
θ 1 and
θ2 (as
indicated in the figures) the correct option
is
100
A. θ1 = θ2
B. θ1 < θ2
C. θ1 > θ2
D. θ1 > > θ2
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6.
Vertical reaction developed at B in the frame
9.
A spring is connected to the free end of a
below due to the applied load of 200 kN
cantilever to reduce the deflection. When a
(with 200,000 mm2 cross-sectional area
moment of 2 kNm acts on the free end find
and 3.75 x 109 mm4 MOI for both member)
out the decrease in the deflection due to the
spring. Take EI = 20,000 kNm2.
is __ KN. E = constant for both members
A. 0.005 mm
B. 0.05 mm
C. 0.0025 mm
D. 0.025 mm
10. A cantilever beam ABC is subjected to a
7.
A cantilever beam PQ as shown in the figure
moment of 500 Nm at the free end as shown
has an extension QRS attached to its free
in figure.
The ratio of deflection at point C to point B
end. The ratio L/a so that the vertical
deflection
at
point
Q
will
is ________.
be zero is
[Take E = 2 x 105 N/mm2 and moment area
__________
method is preferable]
11. A square beam is subjected to a failure
8.
The deflection at D in beam shown below
loading condition. In First case the beam is
is Z/EI mm, The value of Z is. (up to 1
placed having square section horizontally
decimal place)
and in second case it is placed having
diagonal of square as horizontal. Under both
condition of section. Which case is more
stable and safe for design.
A. case 1
B. case 2
C. Both cases are same
D. none of these
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12. The rotation at C due to loading as shown in
C.
The deflection of beam at free end will
the figure below is
be
14wL3
81EI
D. The slope at the point of load applied
will be
A.
61PL3
432EI
B.
47PL3
216EI
C.
61L3
216EI
D.
47PL3
432EI
4wL2
9EI
16. A cantilever beam of length 6 m is subjected
to two point loads of magnitude 20 kN at a
distance of 2m and 4m from the free end.
What will be the peak ordinate in the
13. The deflection at C due to the loading as
conjugate beam if the flexural rigidity is
shown in the figure will be
constant throughout the beam?
7PL3
A.
48EI
5PL3
B.
48EI
7PL3
C.
96EI
5PL3
D.
96EI
A.
60
EI
B.
90
EI
C.
120
EI
D.
180
EI
17. The deflection at the free end of cantilever
of span L due to point load at free end is
14. A simply supported beam of span 4 m is
observed to be
subjected to a point load at its centre. If the
rotation at support due to the loading is
stored in the bar will be
0.0135 radian then the deflection at the
centre of beam will be
A. 18 mm
B. 25 mm
C. 38 mm
D. 46 mm
3L3
. The strain energy
2EI
15. In a cantilever beam of span L, subjected to
A.
27L3
8EI
B.
27L3
4EI
C.
21L3
8EI
D.
21L3
4EI
a concentrated load of ‘W’ acting at a
18. A cantilever beam of span L is carrying a
distance of L/3 from the free end. Then
uniformly distributed load in the one fourth
which of the following statements about the
portion of span from its free end. The
given cantilever beam is/are correct?
A.
be
B.
deflection at the free end due to the loading
The deflection of beam at free end will
will be
8wL3
81EI
The slope at the point of load applied
will be
2wL2
9EI
102
A.
144wL4
1024EI
B.
139wL4
1024EI
C.
155wL4
2048EI
D.
139wL4
2048EI
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19. A simply supported beam of span L is
subjected to an anticlockwise moment M at
the right support. The location of maximum
deflection from the left support will be
A.
PL2
144EI
B.
5PL2
144EI
C.
3PL2
144EI
D.
7PL2
144EI
23. The strain energy stored in a material per
unit volume subjected to pure shear q is
(All symbols have their usual meaning)
A.
C.
L
3
2L
3
B.
D.
A.
L
2 3
C.
4L
3
q2 (1 −  )
E
q2 (1 +  )
E
B.
D.
q2 (1 −  )
2E
q2 (1 +  )
2E
24. For the loading shown in the figure below,
20. A cantilever beam of square cross section of
the reaction at the support B in kN will be
400 mm size and span of 4 m carries UDL
(Take flexural rigidity of the beam equal to
of 3 KN/m over its full length. Strain energy
50000 kN-m2)
stored in the beam (in KNm) is_____. Take
E = 2.1 x 105 N/mm2
21. The value of ‘x’ if the deflection at C is
x
m. (up to 2 decimal places)
EI
25. A cantilever beam as shown in the figure
below is subjected to a concentrated load of
magnitude 20 kN at the free end. The
Moment
generated
be______kN-m
22. The slope at point C due to the loading as
shown in figure will be
103
at
fixed
end
will
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ANSWER
1. B, D
2.
3. D
11. A
12. A
13. D
21.
22. B
23. C
4. D
5. B
6.
7.
8.
9. A
10.
14. A
15. B,C
16. C
17. A
18. D
19. A
20.
24.
25.
SOLUTION
1.
B, D
The vertical deflection at point C will be
equal to zero.
Thus,
Downward deflection due to UDL = Upward
Slope at B due to moment at C:
θB =
deflection due to reaction Rc
ML
EI
R L3
5 wL4
5
= c  R c = wL = 0.625wL
384 EI
48EI
8
Deflection at C;
3.
δv(c) = Deflection at C due to moment M +
Due to symmetry, the 8 kN load will be
θB × L
V =
c
equally distributed on both beams. So the
ML2
+ (B  L)
2EI
=
ML2  ML

+
L
2EI  EI

=
3ML2
2EI
above beam can also be shown as given
below:
Downward
Slope at C,
deflection
at
free
end
of
cantilever due to point load is given by
θc = Slope at C due to moment M + θB
ML
2ML
C =
+ B =
EI
EI
2.
D
=
4.
0.625
PL3 4  33 36
=
=
m
3EI
3EI
EI
D
Let us assume the reaction on the support
Let the length of span AB is L and intensity
B due to the loading is RB. Thus, the beam
of UPL acting on the beam is w per meter
can be illustrated as
length.
104
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6.
2.78
Given,
A = 200,000 mm2
Deflection in AB:
AB =
I = 3.75 × 109 mm4
(P − RB )(3a)3
3EI
The frame can be illustrated as shown
Deflection in BC:
CB =
below:
3
RB(2a)
3EI
Due to continuity,
δAB = δBC
(P-RB) 33 = RB 23
⇒ 27 P = (27 + 8)RB = 35 RB
 RB =
27P
35
Now, CB =
5.
27P 8a3 72 Pa3 3 216Pa3

=
 =
35 3EI 35 EI 3
105EI
B
For Beam 1:
w = 5 kN/m
L=4m
EI = 1000 kN – m2
For the simply supported beam subjected to
Using Continuity
uniformly distributed load of 5 KN/m.
Deflection at A in beam AB = Compression
1 =
in column AC
wL3
5  43
=
24EI 24  1000

1
1 =
= 0.0133radian
75

For Beam 2:
W = 100 kN
L=2m
EI = 1000 kN – m2
For the simply supported beam subjected to
(200 − R)L3 RL
=
3EI
AE
(200 − R)  (2000)2
3  3.75  109
=

(200 − R) = 0.0141R

R =
R
200000
200
= 197.22kN
1.0141
Using equilibrium condition
point load at centre.
R + VB = 200
PL2
100  22
1
2 =
=
=
= 0.025radian
16EI 16  1000 40

Hence θ1 < θ2
VB = 200 − 197.22
= 2.78 kN
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7.
3
9.
A
The beam can also be represented as shown
Given:
below
EI = 20,000 kNm2
L = 5m
M = 2 kNm
Ks = 2 kN/m
Case 1:
When Spring is attached
Let the reaction developed in the spring is R
and the final deflection is δ1.
Deflection at B due to moment – Deflection
due to reaction = Resultant compression of
spring
Vertical Deflection at point Q:
ML2 RL3
−
= 1
2EI 3EI
ΔB = Upward deflection due to moment Downward deflection due to point load
B =
ML2 Ks1L3
−
=
2EI
3EI
(2Wa)L2 WL3
−
2EI
3EI
2  1  53
2  52
−
= 1
2  20000 3  20000
For ΔB = 0
(2Wa)L2 WL3
−
=0
2EI
3EI
25 − 83.3331 = 200001
1 = 1.2448mm
WaL2 WL3
=
EI
3EI
8.
Case 2:
L
=3
a
When spring is not present
1533.34
Deflection at B =
The deflection at the centre of beam due to
ML2
= 1.25mm
2EI
Reduction in deflection = 1.25 – 1.2448 =
loading at a point at a distance b from the
0.0052 mm.
support is
10. 3.62
Pb
=
(3L2 − 4b2 )
48EI
Given,
b = 4m
E = 2 × 105 N/mm2
L=12m
M = 500 Nm
=
IBC = 4×106 mm4, ICB = 2 IBC = 8×106 mm4
50  4
(3  122 − 4  42 )
48E1
LBC = 3m, LAB = 4m
4600 1533.34
=
=
3EI
EI
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Drawing M/EI of the beam:
Second case:
When beam is placed with diagonal
horizontal
3
 a 
2a 2 

a4
a
2

I=
=
’ ymax =
12
12
2
Z2(NA) =
Now
Deflection at C:
ΔC = tC/A = Moment of M/E1 diagram
c =
7250
6
2  10  4  10  10
−6
As per moment area theorem rotation at C
will be area of M/EI diagram between AC
The M/EI diagram for the above beam is
ΔB = tB/A = Moment of M/EI diagram
shown in the figure
between A and B about B
250
42
EIBC
2000
5
2  10  4  106  10−6
m
B = 2.5  10−3 m = 2.5mm

= 2
12. A.
m
Deflection at B:
=
Z2(NA)
Area of M/EI diagram under AC
c
9.06
=
= 3.62
B
2.5
1  PL 2PL  L 1  PL
PL  L 61PL3

+
+
 + 
 =
2  3EI 9EI  3 2  4EI 3EI  6 432EI
11. A.
Thus,
Let us assume that the side of square is
slope at C =
‘a’.
First case:
As per moment area theorem deflection at
a4
a
I=
, ymax =
2
12
I
ymax
=
61PL3
432EI
13. D.
When beam is placed horizontally
Z1(NA) =
6 2
Therefore case 1 is more safe
c = 9.0625  10−3m = 9.06mm
now, tB/A = B =
z1(NA)
a3
structure would be.
500
250
7250
 3  1.5 +
 45 =
EIBC
EIBC
EIBC
5
ymax
=
Higher the section modulus, safer the
between A and C about C
c =
I
C will be moment of area of M/EI diagram
between AC about C.
a3
6
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The M/EI diagram for the above beam is
3
 2L 
w 
3
2wL L
c =   +

3EI
9EI 3
shown in the figure
c =
14wL3
81EI
16. C.
The conjugate beam of the given beam is
shown in the figure below:
PL 
 PL
2
+
1  PL
PL  L  2EI 4EI  L 5PL3
C = 
+
 =
 
PL  6 96EI
2  2EI 4EI  2  PL
+


 2EI 4EI 
14. A.
Rotation at the support due to point load at
the centre
PL2
= 0.0135
16EI
The maximum ordinate of conjugate beam
=
PL2

= 16  0.0135 = 0.216
EI
17. A.
Now,
Deflection at the free end of a cantilever
Deflection at the centre of beam
=
120
.
EI
beam due to point load of intensity P,
3
PL
0.216  4
=
= 0.018m = 18mm
48EI
48
=
PL3
3EI
15. B. C.
Given deflection =
3L3
2EI
On comparing,
PL3 3L3
=
3EI 2EI
Rotation at B due to loading,
2
 2L 
w 
3
2wL2
B =   =
2EI
9EI
Which gives
P = 4.5 kN
Now, strain energy stored in the bar,
Deflection at free end i.e. C
L
L
c = B + B 
3
U=
108
M2dx
 2EI =
0
L
(4.5x)2 dx 27L3
 2EI = 8EI
0
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18. D.
20. 0.000515 kNm
The loading can also be placed as shown in
Given,
the figure below:
Span (L) = 4 m
Size of beam (b) = 400 mm
w = 3 kN/m, E = 2.1 × 105 N/mm2
I=
b4 4004
=
= 2.13  109 mm4
12
12
Moment at a distance x from free end
Deflection at free end:
3
 3L 
 3L 
w 
w 
4
4
wL
L
=
−    −  
8EI
6EI
4
8EI
4
4
=
Strain energy stored in the beam,
139wL4
2048EI
L
U=
19. A.
M2x dx
 2EI
0
Reaction at A
RA
L
U=
M
=
L
Writing deflection equation
EI
d2 y
dx2
=−
U=
M
x ......(i)
L
w2x4dx w2L5
=
8EI
40EI
0

32  45
40  2.1  105  2.13  109  10−9
= 0.000515 KNm
dy
M 2
 EI
=−
x + C1......(ii)
dx
2L
 EIy = −
wx2
2
Mx =
21. – 23.33
Since the loading is symmetrical, it can also
M 3
x + C1x + C2.....(iii)
6L
be represented as
At both supports, deflection will be zero
which gives
C1 =
ML
and C2 = 0
6
Slope at C will be zero i.e.
Putting the values in equation (ii)
−
dy
M 2 ML
 EI
=−
x +
dx
2L
6
M = 5 Nm
Deflection will be maximum if slope is zero.
Thus, equating
−
Deflection at C,
dy
equal to zero.
dx
M 2 ML
x +
=0
2L
6
x=
5  22 M  2
+
=0
2EI
EI
L
3
109
C = −
5  23 5  22
+
3EI
2EI
C = −
70
23.33
m=−
m
3EI
EI
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22. B.
Using Area moment theorem,
=
The M/EI diagram of the beam is shown in
the figure below,
=
PL3 ML2
+
3EI 2EI
35  2.53 5  2.52
+
3EI
2EI
= 3.95 × 10-3 m = 3.95 mm
Since deflection is more than 2 mm, there
will be some reaction at the support
RL3
= (3.95 − 2)  10−3
3EI
C
D
= C − D = Area between C and D in
R =
M/EI diagram
Since the slope at D is zero
1.95  10−3  3  50000
(2.5)3
= 18.72 kN
25. .
1 L  PL
PL 
5PL2
C − 0 =   
+
=
2 6  4EI 6EI  144EI
Deflection at B due to 20 kN loading at C =
Area Moment of M/EI diagram between A
23. C.
and B
For a material subjected to pure shear
1 = q, 2 = −q
Thus,
Strain in x direction,
1 =
1

− 2
E
E
1 =
q
q
+
E
E
100 40 

2
+

EI
EI 
1  100 40 

1 =  
+

3

100 40
2  EI
EI 
+
EI
EI
Strain energy stored in x direction,
q2 (1 +  )
1
U1 =  1  1 =
2
2E
1 =
Similarly strain in y direction,
Upward deflection at B due to reaction at
q (1 +  )
1
 2  2 =
2
2E
2
U2 =
support
2 =
Thus, total strain energy in the material
U = U1 + U2 =
360
EI
q2 (1 +  )
R  33 9R
=
3EI
EI
Since total deflection at B will be zero.
E
So, δ1 = δ2
24. 18.72
9R 360
=
EI
EI
P = 35 kN, M = 5 kNm,
L = 2.5 m
R = 40 kN
Flexural rigidity (EI) = 50000 kN-m
2
Now, Bending moment at free end
Total downward deflection due to loading
M = 40 × 3 – 20 × 5 = 20 kN-m

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