CE 1201 â MECHANICS OF SOLIDS
+ semester www.semesterplus.com 1
UNIT – I
STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS Rigid bodies and deformable solids – stability, strength, stiffness – tension, compression and shear stresses – strain, elasticity, Hooke’s law, limit of proportionately, modulus of elasticity, stress- strain curve, lateral strain – temperature stressres – deformation of simple and compound bars – shear modulus, bulk modulus, relationship between elastic constants – biaxial state of stress – stress at a point – stress on inclined plane – principal stresses and principal planes – Mohr’s circle of stresses.
S.NO
2 MARKS
PAGE NO
1
Define stress and its types
6
2
Define strain.
6
3
Define tensile stress and tensile strain.
6
4
Define the three Elastic moduli. (April/May 2005)
7
5
Define shear strain and Volumetric strain
7
6
A square steel rod 20 mm x 20 mm in section is to carry an axial load (compressive) of 100 KN. Calculate the shortening in a length of 50 mm. E =
7
7
2..14 10 8 KN / M 2 Define Poisson’s ratio. (Nov / Dec 04)
8
8
Find the force “P”acting on the bar given in fig.(Apr/May04)
8
9
Find the force P”acting on the bar given in fig.(Apr/ May 04)
9
10
Define principal plane and principal stress.
9
11
12 13
14
semester +
Write down the relation between modulus of elasticity and modulus of rigidity and that between modulus of elasticity and bulk modulus. (Nov/Dec 02) When a rod of diameter 20mm is subjected to a tensile force of 40 kN, thje extension is measured as 250 divisions in 200mm extension meter. Find the modulus of elasticity if each division is equal to 0.001mm. (Nov/Dec 02 What do you understand by the assumption, plane section remain plane even after the application of load?
A metal bar 50 mm x 50 mm section, is subjected to anaxial compressive load of 500 KN.The contraction of a 200 mm gauge length is found to be 0.5 mm and the increase in thickness 0.04 mm. find E and μ.
9
9 9
9
www.semesterplus.com
+ semester www.semesterplus.com 2
2
15
16
Brass bar, having C.S.A of 10 cm is subjected to axial forces as shown in fig. Find the total elongation of the bar Take E 0.8 10 3 t / cm 2 Steel bar is 900 mm long its two ends are 40 mm and 30 mm in diameter and the length of each rod is 200 mm.The middle portion of the bar is 15 mm in diameter and 500 mm long. If the bar is subjected to an axial tensile load of 15 KN, find the total extension. E 200 GN / m 2
17
18
19
20
1G 10 9
1 volume of the bar, if and E 2.0 10 6 kg / cm 2 4 A bronze specimen has a E 1.2 10 6 kg / cm 2 and C 0.47 10 6 kg / cm 2 . Determine
Two parallel walls 6 m apart, are stayed together by a steel rod 20 mm passing through metal plates and nuts at each end. The nuts are tightened home, when the rod is at a temp of 1000 C. Determine the stress in the rod, when the temperature falls down 200C if. A bar is subjected to a tensile stress of 1000 kg / cm2. Determine the normal and Tangential stresses on a plane making an angle of 600 with the axis of tensile stress.
16 MARKS
11
Bar of 2 m length, 2 cm breadth and 1.5 cm thickness is subjected to a tensile load of 3000 kg. Find the final
S.NO
semester +
G giga and
10
12
13
13
14
PAGENO
www.semesterplus.com
+ semester www.semesterplus.com
1
2
3
4
5
6
7
semester +
A hollow cast – iron cylinder 4m long, 300 mm outer diameter and thickness of metal 50 mm is subjected to a central load on the top when standing straight. The stress produced is 75000 KN / m2. Assume Young’s Modulus for cast iron as 1.5 10 8 KN / m 2 and find (i)Magnitude of the load(ii)Longitudinal strain produced and (iii)Total decrease in length. The following observations were made during a tensile test on a mild steel specimen 40 mm in diameter and 200 mm long.Elongation with 40 KN load (within limit of proportionality) l 0.0304 mm, yield load = 161 KN. Maximum load = 242 KN, Length of specimen at fracture = 249 mm. Determine: i. Young’s Modulus of Elasticity ii. Yield point stress iii. Ultimate stress iv. Percentage elongation. A steel 2m long and 3 mm in diameter is extended by 0.75 mm when a weight W is suspended from the wire. If the same weight is suspended from a brass wire, 2.5 m long and 2 mm in diameter, it is elongated by 4.64 mm. Determine the modulus of elasticity of brass if that of steel be 2.0 10 5 kN / m 2 A member formed by connecting a steel bar to an aluminium bar as shown in fig. Assuming that the bars are prevented form buckling sidewise; calculate the magnitude of force p1 that will cause the total length of the member to decrease 0.25 mm. The values of elastic modulus of steel and aluminum are 2101 KN / mm2 and 70 KN / mm2 (April/May 04) A steel tie rod 50 mm in and 2.5m long is subjected to a pull of 100 KN. To what length the rod should be bored centrally so that the total extension will increase by 15 % under the same pull, the bore being 25 mm ? A steel flat plate AB of 1 cm thickness tapers uniformly from 10 cm to 5 cm width in a length of 40 cm. From first principles, determine the elongation of the plate, if an axial tensile force of 5000 kg acts on it. Take E 2.0 10 6 kg / cm 2 A steel cube block of 50 mm side is subjected to a
3
18
19
20
22
23
24
26
www.semesterplus.com
+ semester www.semesterplus.com 4
force of 6 KN (Tension), 8 KN (compressive) and 4 KN (tension) along x, y and z directions. Determine the change in the volume of the block. e 200 KN / mm 2 and m
8
9
10
10 3
A bar of 30 mm is subjected to a pull of 60 KN. The measured extension on gauge length of 200 mm is 0.09 mm and the change in diameter is 0.0039 mm. Calculate and the values of the three module. (Nov/Dec 04) At a point within a body subjected to two mutually perpendicular directions, the stresses are 80 N/mm2 tensile and 40 N/mm2 tensile. Each of the above stresses is accompanied by a shear stress of 60 N/mm2. Determine the normal stress. Shear stress and resultant stress on an oblique plane inclined at an angle of 450 with the axis of minor tensile stress. A point in strained material is subjected to the stresses as shown in figure. Locate the principal planes and evaluate the principal tresses. (Apr/May 02)
TWO MARKS:
27
28
29
UNIT – I
1. Define stress and its types When a body is acted upon by some load (or) external force, it undergoes deformation (i.e. change in shape or dimensions)
semester +
www.semesterplus.com
+ semester www.semesterplus.com
5 Stress is defined as the internal resistance offered by the material to the extremely applied force, expressed per unit area.
P A
stress P applied load A Area of sec tion
Types of stresses: 1. Axial stress 2. Bearing stress 3. Bending stress 4. Shear stress Types of axial stress: 1. Tensile stress 2. Compressive stress. 2. Define strain. Strain is defined as the ratio of change in length to the original length of the member Strain =
Change in length (dl) Original length (l)
3. Define tensile stress and tensile strain. When the resistance offered by a section of a member is against an increase in length, the section is said to offer tensile stress. Internal resis tan ce p Tensile stress t C.S . A A Tensile strain: The strain corresponding to tensile stress is tensile strain.
Tensile strain e
l l
Increase length Original length
4. Define the three Elastic moduli. (April/May 2005) Young’s Modulus: It is the ratio between tensile stress and tensile strain (or) compressive stress and compressive strain.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 6
E
t e et
or
c ec
Modulus of Rigidity: It is defined as the ratio of shear stress ( ) to shear strain and is denoted by C, N or G It is also called shear stress modulus of elasticity. Bulk (or) Volume Modulus of Rigidity It is defined as the ratio of normal stress (on each face of a solid cube) to volumetric strain and is denoted by the letter K.
K
n V
5. Define shear strain and Volumetric strain Shear strain is defined as the ratio of transverse displacement to the distance from the lower force. Shear strain
Transverse displacement
=
Distance from lower face
Volumetric strain: body
It is defined as the ratio between change in volume and original volume of the
v
change in volume v Original volume V
6. A square steel rod 20 mm x 20 mm in section is to carry an axial load (compressive) of 100 KN. Calculate the shortening in a length of 50 mm. E = 2..14 10 8 KN / M 2 Solution: Area A = Length
0.02 0.02 0.0004
m2
l 50 mm (or) 0.05 m P 100
KN
E 2.14 10 8 KN / m2
Shortening of the rod l :
semester +
www.semesterplus.com
+ semester www.semesterplus.com 7
Stress 1 E
100 P 250000 KN / m 2 A 0.0004
Stress Strain
Strain
Stress E E
250000 2.14 10 8 l 250000 l 2.14 10 8
l
250000 0.05 2.14 10 8
= 0.0000584 m (or) 0.0584 mm Hence the shortening of the rod = 0.0584 mm. 7. Define Poisson’s ratio. (Nov / Dec 04) The ratio of lateral strain to the longitudinal strain is a constant for a given material, when the material is stressed within the elastic limit. This ratio is called Poisson’s ratio and it is generally denoted by 1/m (or) .
Lateral strain Longitudinal strain
8. Find the force “P” acting on the bar given in fig. (April / May 04)
Sum of all left direction force = Sum of all right direction force 50 + P = 45 + 15 P = 60 – 50 = 10 kN 9. What type of stress will be induced in a bar when the ends are restrained and subjected to i) rise in temperature and ii) a fall in temperature? (April / May 04) Thermal stresses: These are the stresses induced in a body due to change in temperature. Thermal stresses are set up in a body, when the temperature of the body is not allowed to expand or contract freely. i) Rise in temperature – compressive stress ii) a fall in temperature -tensile stress.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 8
10. Define principal plane and principal stress. The plane, which have no shear stress, are known as principal planes. Hence principal planes are the planes of zero shear stress. These planes carry only normal stresses. The normal stresses, acting on a principal plane, are known as principal stresses.
11. Write down the relation between modulus of elasticity and modulus of rigidity and that between modulus of elasticity and bulk modulus. (Nov/Dec 02) The relation between modulus of elasticity and bulk modulus is given by E = 3K (1-2/m) The relation between modulus of elasticity and modulus of rigidity E = 2G (1+(1/m)) 12. When a rod of diameter 20mm is subjected to a tensile force of 40 kN, thje extension is measured as 250 divisions in 200mm extension meter. Find the modulus of elasticity if each division is equal to 0.001mm. (Nov/Dec 02) Solution: Diameter = 20mm ; P = 40 kN = 40 x 103 N ; Extension = 250 divisions 1 division = 0.001 mm; L = 200 mm δl = 250 x 0.001 = 0.25 mm P / A 40 x10 3 / 314.16 E = σ/e = 0.25 / 200 l / l Modulus of Elasticity = 101.85 x 103 N/mm2 13. What do you understand by the assumption, plane section remain plane even after the application of load? When some external load acts on the beam, the shear force and bending moments are set up at aii sections of the beam. Due to the shear force and bending moment, the beam undergoes certain deformation. After removal of the load the beam will come to its original position based on that assumption. 14. A metal bar 50 mm x 50 mm section, is subjected to an axial
compressive load of 500 KN. The contraction of a 200 mm gauge length is found to be 0.5 mm and the increase in thickness 0.04 mm. find E and μ. Solution: b = 50 mm,
t = 50 mm
Area = 50 50 2500 mm 2 P = 500 KN Length, l = 200 mm, l 0.5mm. Increase in thickness, t 0.04mm Young’s Modulus:
semester +
www.semesterplus.com
+ semester www.semesterplus.com 9
l
pl AE
500 10 3 200 0.5 E 80 KN / mm 2 2500 E Poission’s Ratio:
Lateral strain Linear strain
Linear strain = 0.0025
t Lateral strain thickness 0.04
1 0.0025 50 m
1 0.32 m
15. brass bar, having C.S.A of 10 cm2 is subjected to axial forces as shown in fig.
Find the total elongation of the bar Take E 0.8 10 3 t / cm 2 Given, Area A = 10 cm2 E 0.8 10 3 t / cm 2
l Total elongation of the bar. For the sake of simplification, the force of 8 tonnes acting at B, may be split up into three forces of 5 tones, 2 tonnes and 1 tone. Now it will be seen that the part AB of the bar is subjected to a tension of 5 tonnes part BC is subjected to a compression of 2 tonnes and the part BD is subjected to a compression of 1 tonne as shown in fig.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 10
Using the relation,
l
1 p1l1 p 2 l 2 p3 l 3 AE
1 5 60 2 100 1 220 10 0.8 10 3
12 0.015cm 0.15mm 800
16. Steel bar is 900 mm long its two ends are 40 mm and 30 mm in diameter and the length of each rod is 200 mm. the middle portion of the bar is 15 mm in diameter and 500 mm long. If the bar is subjected to an axial tensile load of 15 KN, find the total extension. E 200 GN / m 2
G giga and
1G 10 9
P = 15 kN A1 = 1256 mm2 A2 = 176.625 mm2 A3 = 706.50 mm2 l1 = 200 mm l 2 = 500 mm l 3 = 200 mm Total extension of the bar:
l1 , l 2
l1
and l 3
pl1 A1 E
l2
l l1 l 2 l 3
pl pl1 pl 2 3 A1 E A2 E A3 E
P E
pl2 A2 E
l3
pl3 A3 E
l1 l l2 3 A1 A2 A3
= 0.2454 mm.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 11
17. bar of 2 m length, 2 cm breadth and 1.5 cm thickness is subjected to a tensile load of 3000 kg. Find the final volume of the bar, if 1 4
and E 2.0 10 6 kg / cm 2 L = 2 m = 200 cm, b = 2 cm, t = 1.5 cm.
Vol
V 200 2 1.5 600cm 3 P = 3000 kg.
1 1 m4 3 4
E 2 10 6 kg / cm 2
v
2 stress e1 e ; v E m P 3000 1000 kg / cm 2 A bt
1000 1 6 2000 2 10
v v
v
18.
A
1 2 1 2000 4
1 4000
1 600 0.15cm 3 4000
bronze
specimen has 6 2 C 0.47 10 kg / cm . Determine
a
E 1.2 10 6 kg / cm 2
and
Solution:
C
m 1.2 10 6 0.47 10 2m 1 6
mE 2m 1
0.94m 1 1.2m 0.94m 0.94 1.2m 0.94 1.2m 0..94m 0.94 m = 0.26 m
semester +
www.semesterplus.com
+ semester www.semesterplus.com 12
1 0.26 1 0.277 m 0.94 m
19. Two parallel walls 6 m apart, are stayed together by a steel rod 20 mm passing through metal plates and nuts at each end. The nuts are tightened home, when the rod is at a temp of 1000 C. Determine the stress in the rod, when the temperature falls down 200C if. a. b.
The ends do not yield and The ends yield by 1mm.
E 2 10 6 kg / cm 2
and
12 10 / c 6
Solution:
Length of rod = l = 6m = 600 cm
of rod = d = 20 mm = 2 cm. Temperature t = 1000 - 200 = 800 c E 2 10 6 kg / cm 2 ; 12 10 6 / 0 C
when the ends do not yield.
1 stress in the rod
tE
1 12 10 6 80 2 10 6 1920 kg / cm 2 when the ends yield by 1mm
2 stress in the rod
2 12 10 6 8
0.1 6 2 2 10 158 kg / cm 600
20. A bar is subjected to a tensile stress of 1000 kg / cm2. Determine the normal and Tangential stresses on a plane making an angle of 600 with the axis of tensile stress. 30 0
n cos2 750kg / cm 2 t sin cos 433kg / cm 2 866 kg / cm 2
semester +
www.semesterplus.com
+ semester www.semesterplus.com 21.
13
A point in a stained material is subjected to two mutually perpendicular stresses of 2000 kg/cm2 and 1000 kg/cm2. Determine the intensities of normal and resultant stresses on a plane inclined at 300 to the axis of the minor stress. Major stress = 1 2000 kg / cm 2 Minor stress = 2 1000 kg / cm 2
Angle of plane, which it makes with the axis of minor, principle
stress 30 0 Normal stress
n
1 2 2
1 2 2
cos 2
= 1750 kg/ cm2 Tangential stress
t
1 2 2
Resultant stress
sin 2 433kg / cm 2
r n 2 t 2 1802 .8kg / cm 2
22. At a point in a strained material the principal stresses are 100 N/mm2 (tensile) and 60 N/mm2 (comp). Determine normal stress, shear stress, resultant stress on a plane inclined at an 500 to the axis of major principal stress. Also determine the maxi shear stress at the point. 1 100 N / mm 2 ,
2 60 N / mm 2 ,
90 0 50 0 40
t 78.8N / mm 2
n 33.9 N / mm 2 , R 85.8 N / mm 2 Maximum shear stress;
t max
semester +
1 2 2
80 N / mm 2
www.semesterplus.com
+ semester www.semesterplus.com 14
23. A point in a strained material is subjected to a compressive stress of 800 kg/cm2 and a shear stress of 560 kg/cm2. Determine graphically or otherwise, the maximum and minimum intensity of stress. 560 kg / cm 2
800 kg / cm 2 , Maximum intensity of stress: 2 tan 2
2 560 800
= 1.4 2 54 0 28'
27 014' 2
n1 2 2 2 = 288.2 kg/cm2 (Tensile)
n2
Minimum intensity of direct stress. 2
n 2 2 1088 .2kg / cm 2 2 2 24. A steel bolt 2.5 cm diameter is subjected to a direct tension of 1500 kg and a shearing forces of 1000 kg. Determine the intensities of normal and shear stress across a plane inclined at an angle of 60 0 to the longitudinal axis of the bolt. Also determine the resultant stress. Solution:
2.5cm.
Area = 4.91 cm2
Direct tension = 1500 kg.
Direct stress on the bolt =
1500 305.5kg / cm 2 4.91
Shearing force = 1000 kg.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 1000 203.7kg / cm 2 4.91
Shear stress =
15
Angle which the plane makes with the longitudinal axis of the bolt = 60 0 bolt
Angle, which the plane makes with the normal to the longitudinal axis of the
90 0 60 0 30 0
n
Normal stress:
2
1 cos 2 sin 2
= 405.5 kg/ cm2
1 2
t sin 2 e cos 2
Shear stress:
= 30.5 kg/cme+
r
Resultant stress:
n 2 t 2
405.52 30.52
= 406.6 kg/cm.2
25. A point is subjected to a tensile stress of 60 N/mm2 and a compressive stress of 40 N/mm2, acting on two mutually perpendicular planes and a shear stress of 10 N/mm2 on these planes. Determine the principal stresses as well as maxi shear stress. Also find out the value of maxi shear stress. 1 60 N / mm 2 and 2 40 N / mm 2
10 N / mm 2
Principal stresses:
n1
n2
semester +
2
1 2
2 2 2 1 61N / mm 2
1 2
2 2 2 1 41N / mm 2
2
2
2
www.semesterplus.com
+ semester www.semesterplus.com 16
Maximum shear stress
t
semester +
n1 n 2 2
61 41 51N / mm 2 2
www.semesterplus.com
+ semester www.semesterplus.com 17
SIXTEEN MARKS QUESTION AND ANSWERS: 1.A hollow cast – iron cylinder 4m long, 300 mm outer diameter and thickness of metal 50 mm is subjected to a central load on the top when standing straight. The stress produced is 75000 KN / m2. Assume Young’s Modulus for cast iron as 1.5 10 8 KN / m 2 and find i. ii. iii.
Magnitude of the load Longitudinal strain produced and Total decrease in length.
Solution: Outer diameter, D = 300 mm = 0.3 m Thickness ,
t = 50 mm = 0.05 m
Length,
l=4m
Stress produced 75,000 KN / m2
E 1.5 10 8
KN / m2
d D 2t
Inner diameter of the cylinder
0.3 2 0.05 = 0.2 m i.
Magnitude of the load P: Using the relation,
(or)
P A
P A 75000
75000 ii.
0.3 4
2
D 4
2
d2
0.2
2
P = 2945.2 KN. Longitudinal strain produced e: Using the relation,
semester +
www.semesterplus.com
+ semester www.semesterplus.com strain, e iii.
Stress 75,000 0.0005 E 1.5 10 8
18
Total decrease in length , l : using the relation,
strain
change in length Original length
0.0005
l 4
l 0.0005 4 0.002 m = 2 mm Hence decrease in length = 2 mm.
2. The following observations were made during a tensile test on a mild steel specimen 40 mm in diameter and 200 mm long. Elongation with 40 KN load (within limit of proportionality) l 0.0304 mm, yield load = 161 KN Maximum load = 242 KN Length of specimen at fracture = 249 mm. Determine: v. Young’s Modulus of Elasticity vi. Yield point stress vii. Ultimate stress viii. Percentage elongation. Solution: i.
Young’s Modulus of Elasticity (E) :
P 40 3.18 10 4 KN / m 2 A 0.042 4
Stress,
Strain, e
l l
0.0304 0.000152 200
Stress 3.18 10 4 Strain 0.000152 2.09 10 8 KN / m2 E
semester +
www.semesterplus.com
+ semester www.semesterplus.com 19
(ii)
Yield point stress: Yield point stress
Yield po int load area
4
(iii)
0.042
12.8 10 4 KN / m 2
Ultimate stress: Maximum load
Ultimate stress =
Area
4
(iv).
161
242
0.042
19.2 10 4 KN / m 2
Percentage elongation: Percentage elongation =
Length of specimen at fracture original length Original length
249 200 0.245 200
= 24.5 %
3. A steel 2m long and 3 mm in diameter are extended by 0.75 mm when a weight W is suspended from the wire. If the same weight is suspended from a brass wire, 2.5 m long and 2 mm in diameter, it is elongated by 4.64 mm. Determine the modulus of elasticity of brass if that of steel be 2.0 10 5 N / n Solution: Given: l s 2m,
d s 3mm, l s 0.75mm
Es 2.0 10 5 N / mm 2 ,
l b 2.5m,
d b 2mm
l b 4.64m Modulus of Elasticity of brass, Eb :
semester +
www.semesterplus.com
+ semester www.semesterplus.com 20
From Hooke’s law,
l
Wl AE
where l Extension, l = length, A = Cross sectional area E = Modulus of Elasticity. Case 1: For steel wire
l3
Wl3 As E s
0.75 W 2 1000
2 3 10 5 4
32 W 0.75 4
1 2 10 5 2000
(1)
Case 2: For brass wire
lb 4.64
W lb Ab E b W 2.5 1000
4
2 E b 2
22 W 4.64 4
1 Eb 2500
(2)
Equating equation (i) and (ii), we get
0.75
semester +
32 4
2 1 1 2 10 4.64 Eb 4 2500 2000 5
2
www.semesterplus.com
+ semester www.semesterplus.com 21
Eb 0.909 10 5 N / mm 2 4. A member formed by connecting a steel bar to an aluminium bar as shown in fig. Assuming that the bars are prevented form buckling sidewise; calculate the magnitude of force p1 that will cause the total length of the member to decrease 0.25 mm. The values of elastic modulus of steel and aluminum are 2101 KN / mm2 and 70 KN / mm2 Solution:
Given:
As 50 50 2500 mm 2 AA 100 100 10,000 mm 2 Length of steel bar = l3 = 300 mm. = l 3 = 380 mm
l 0.25mm
E A 70 10 3 N / mm 2
Es 210 10 3 N / nn 2 :
P = Magnitude of the required force.
l1 l 2 A1 E1 A2 E 2
l P
P = 224.36 KN
semester +
www.semesterplus.com
+ semester www.semesterplus.com 22
Area at the reduced section
0.05 4
2
0.025 2 0.001472 m 2
1 Stress in the reduced section,
100 1000 0.001472
67.93 10 6 N / m 2
Elongation of the rod
2.5 x .x E
E
0.731 10 3 50.92 10 6 2.5 x 67.9 10 6 x 0.731 10 3 9 9 200 10 200 10 50.92 10 6 2.5 x 67.9 10 6 x 200 10 9 0.731 10 3
50.92 10 6
2.5 x 1.33x 2.87 x = 1.12 m.
5. A steel tie rod 50 mm in and 2.5m long is subjected to a pull of 100 KN. To what length the rod should be bored centrally so that the total extension will increase by 15 % under the same pull, the bore being 25 mm ? Solution: Dia of steel tie rod = 50 mm = 0.05 m Length of steel tie rod l = 2.5 m P = 100 KN. Dia of bore = 25 mm = 0.025 m
semester +
www.semesterplus.com
+ semester www.semesterplus.com 23
e 200 10 9 N / m 2 Length of the bore x: Stress in the solid rod,
P 100 1000 50.92 10 6 N / m 2 A 0.052 4
l
50.92 10 6 2.5 Elongation of the solid rod l E 200 10 9 = 0.000636 m (or) 0.636 mm. Elongation after the rod is bored 1.15 0.636 = 0.731 mm.
6. A steel flat plate AB of 1 cm thickness tapers uniformly from 10 cm to 5 cm width in a length of 40 cm. From first principles, determine the elongation of the plate, if an axial tensile force of 5000 kg acts on it. Take E 2.0 10 6 kg / cm 2 Given:
Width of the plate at a distance x from A
10 10 5
x x 10 40 8
C.S.A of the bar
x x Ax 110 10 cm 2 8 8 Stress,
semester +
x
P 5000 kg / cm 2 x Ax 10 8
www.semesterplus.com
+ semester www.semesterplus.com
24
5000 1 x x 6 40010 10 2 10 8 8
x Strain, x E
Elongation of the elementary length
ex.dx
dx x 40010 8
cm
The total extension of the bar may be found out by integrating the above equation between the limits 0 and 40.
1 l (ie.) 400
40
0
dx 10
x 8 40
1 x log 10 e 8 0 1 400 8
1 log e 5 log e 10 50
1 log e 10 log e 5 50
1 10 log e 50 5
1 log e 2cm 50
1 2.3 log10 2cm 50
2.3 0.3010 50
log e
2.3 log 10
= 0.014 cm = 0.14 mm.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 25
7. A steel cube block of 50 mm side is subjected to a force of 6 KN (Tension), 8 KN (compressive) and 4 KN (tension) along x, y and z directions. Determine the change in the volume of the block. e 200 KN / mm 2 and m
10 3
Side of steel cube = 50 mm
A 50 50 2500 mm 2
and v 50 50 50 125000 mm 3
PX 6KN; PY 8KN; Pz 4KN
v V
e X eY e Z
eX
X
x E
Y mE
Z mE
P 6 10 3 2.4 N / mm 2 A 2500
P 8 10 3 Y 3.2 N / mm2 A 2500
Z
PZ 4000 1.6 N / mm2 A 2500
eX
2.4 3.2 3 1.6 3 2.88 10 E 10 E E E
eY
3.2 2.4 3 1.6 3 4.4 E 10 E 10 E E
eZ
1.6 2.4 3 3.2 3 1.84 10 E 10 E E E
v v
semester +
e X eY e Z
www.semesterplus.com
+ semester www.semesterplus.com 26
v 125 10 3
2.88 4.4 1.84 E E E
v 125 10 3
0.32 3 0 . 2 mm 200 10 3
8. A bar of 30 mm is subjected to a pull of 60 KN. The measured extension on gauge length of 200 mm is 0.09 mm and the change in diameter is 0.0039 mm. Calculate and the values of the three module. Solution: i.
Young’s Modulus:
E
e
P 60 10 3 84.9 N / mm 2 A 302 4
e
l l
0.09 0.00045 200
E 188.67 KN / mm 2
ii.
Poisson’s ratio:
Lateral strain d / d Linear strain e
d / d iii.
semester +
0.00039 0.00013 3
0.00013 13 45 m 0.00045 45 13
Modulus of rigidity:
www.semesterplus.com
+ semester www.semesterplus.com 45 188.67 10 3 mE C 13 73.19 KN / mm 2 2m 1 45 2 1 13 iv.
27
Bulk Modulus:
45 188 .67 10 3 mE K 13 148 .95 KN / mm 2 3m 2 45 3 2 13 9. At a point within a body subjected to two mutually perpendicular directions, the stresses are 80 N/mm2 tensile and 40 N/mm2 tensile. Each of the above stresses is accompanied by a shear stress of 60 N/mm2. Determine the normal stress. Shear stress and resultant stress on an oblique plane inclined at an angle of 450 with the axis of minor tensile stress. Solution: Given,
Major tensile stress, 1 80 N / mm 2 Minor tensile stress, 2 40 N / mm 2 Shear stress, q = 60 N/mm2 Angle of oblique plane, with the axis of minor tensile stress. Normal stress ( n )
i.
n
1 2 2
1 2 2
cos 2 sin 2
80 40 80 40 cos 2 45 0 sin 2 60 sin 2 45 2 2
n 120 N / mm 2 ii.
semester +
Shear (or tangential) stress
t
www.semesterplus.com
+ semester www.semesterplus.com t
1 2 2
28
sin 2 cos 2
80 40 sin 2 45 60 cos2 45 2
t 20 N / mm2 iii..
Resultant stress
R
n 2 t 2
R
= 121.665 N / mm2
10. A point in strained material is subjected to the stresses as shown in figure. Locate the principal planes and evaluate the principal tresses. Solution: Given,
Stress on the face BC and AD = 600 kg / cm2 Inclination of the stress = 600
Stress normal on the face BC or BD
1 600 cos 30 0
600 sin 30 0 Location:
2 400
lan2
2R 0 39 0 21' 1 2
Principal stress
n1
semester +
1 2 2
2
2 1 2 2
www.semesterplus.com
+ semester www.semesterplus.com 29
= 766 kg/cm2
n2
1 2 2
2
2 2 1 2
= 154 kg/cm2
semester +
www.semesterplus.com
UNIT – I
STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS Rigid bodies and deformable solids – stability, strength, stiffness – tension, compression and shear stresses – strain, elasticity, Hooke’s law, limit of proportionately, modulus of elasticity, stress- strain curve, lateral strain – temperature stressres – deformation of simple and compound bars – shear modulus, bulk modulus, relationship between elastic constants – biaxial state of stress – stress at a point – stress on inclined plane – principal stresses and principal planes – Mohr’s circle of stresses.
S.NO
2 MARKS
PAGE NO
1
Define stress and its types
6
2
Define strain.
6
3
Define tensile stress and tensile strain.
6
4
Define the three Elastic moduli. (April/May 2005)
7
5
Define shear strain and Volumetric strain
7
6
A square steel rod 20 mm x 20 mm in section is to carry an axial load (compressive) of 100 KN. Calculate the shortening in a length of 50 mm. E =
7
7
2..14 10 8 KN / M 2 Define Poisson’s ratio. (Nov / Dec 04)
8
8
Find the force “P”acting on the bar given in fig.(Apr/May04)
8
9
Find the force P”acting on the bar given in fig.(Apr/ May 04)
9
10
Define principal plane and principal stress.
9
11
12 13
14
semester +
Write down the relation between modulus of elasticity and modulus of rigidity and that between modulus of elasticity and bulk modulus. (Nov/Dec 02) When a rod of diameter 20mm is subjected to a tensile force of 40 kN, thje extension is measured as 250 divisions in 200mm extension meter. Find the modulus of elasticity if each division is equal to 0.001mm. (Nov/Dec 02 What do you understand by the assumption, plane section remain plane even after the application of load?
A metal bar 50 mm x 50 mm section, is subjected to anaxial compressive load of 500 KN.The contraction of a 200 mm gauge length is found to be 0.5 mm and the increase in thickness 0.04 mm. find E and μ.
9
9 9
9
www.semesterplus.com
+ semester www.semesterplus.com 2
2
15
16
Brass bar, having C.S.A of 10 cm is subjected to axial forces as shown in fig. Find the total elongation of the bar Take E 0.8 10 3 t / cm 2 Steel bar is 900 mm long its two ends are 40 mm and 30 mm in diameter and the length of each rod is 200 mm.The middle portion of the bar is 15 mm in diameter and 500 mm long. If the bar is subjected to an axial tensile load of 15 KN, find the total extension. E 200 GN / m 2
17
18
19
20
1G 10 9
1 volume of the bar, if and E 2.0 10 6 kg / cm 2 4 A bronze specimen has a E 1.2 10 6 kg / cm 2 and C 0.47 10 6 kg / cm 2 . Determine
Two parallel walls 6 m apart, are stayed together by a steel rod 20 mm passing through metal plates and nuts at each end. The nuts are tightened home, when the rod is at a temp of 1000 C. Determine the stress in the rod, when the temperature falls down 200C if. A bar is subjected to a tensile stress of 1000 kg / cm2. Determine the normal and Tangential stresses on a plane making an angle of 600 with the axis of tensile stress.
16 MARKS
11
Bar of 2 m length, 2 cm breadth and 1.5 cm thickness is subjected to a tensile load of 3000 kg. Find the final
S.NO
semester +
G giga and
10
12
13
13
14
PAGENO
www.semesterplus.com
+ semester www.semesterplus.com
1
2
3
4
5
6
7
semester +
A hollow cast – iron cylinder 4m long, 300 mm outer diameter and thickness of metal 50 mm is subjected to a central load on the top when standing straight. The stress produced is 75000 KN / m2. Assume Young’s Modulus for cast iron as 1.5 10 8 KN / m 2 and find (i)Magnitude of the load(ii)Longitudinal strain produced and (iii)Total decrease in length. The following observations were made during a tensile test on a mild steel specimen 40 mm in diameter and 200 mm long.Elongation with 40 KN load (within limit of proportionality) l 0.0304 mm, yield load = 161 KN. Maximum load = 242 KN, Length of specimen at fracture = 249 mm. Determine: i. Young’s Modulus of Elasticity ii. Yield point stress iii. Ultimate stress iv. Percentage elongation. A steel 2m long and 3 mm in diameter is extended by 0.75 mm when a weight W is suspended from the wire. If the same weight is suspended from a brass wire, 2.5 m long and 2 mm in diameter, it is elongated by 4.64 mm. Determine the modulus of elasticity of brass if that of steel be 2.0 10 5 kN / m 2 A member formed by connecting a steel bar to an aluminium bar as shown in fig. Assuming that the bars are prevented form buckling sidewise; calculate the magnitude of force p1 that will cause the total length of the member to decrease 0.25 mm. The values of elastic modulus of steel and aluminum are 2101 KN / mm2 and 70 KN / mm2 (April/May 04) A steel tie rod 50 mm in and 2.5m long is subjected to a pull of 100 KN. To what length the rod should be bored centrally so that the total extension will increase by 15 % under the same pull, the bore being 25 mm ? A steel flat plate AB of 1 cm thickness tapers uniformly from 10 cm to 5 cm width in a length of 40 cm. From first principles, determine the elongation of the plate, if an axial tensile force of 5000 kg acts on it. Take E 2.0 10 6 kg / cm 2 A steel cube block of 50 mm side is subjected to a
3
18
19
20
22
23
24
26
www.semesterplus.com
+ semester www.semesterplus.com 4
force of 6 KN (Tension), 8 KN (compressive) and 4 KN (tension) along x, y and z directions. Determine the change in the volume of the block. e 200 KN / mm 2 and m
8
9
10
10 3
A bar of 30 mm is subjected to a pull of 60 KN. The measured extension on gauge length of 200 mm is 0.09 mm and the change in diameter is 0.0039 mm. Calculate and the values of the three module. (Nov/Dec 04) At a point within a body subjected to two mutually perpendicular directions, the stresses are 80 N/mm2 tensile and 40 N/mm2 tensile. Each of the above stresses is accompanied by a shear stress of 60 N/mm2. Determine the normal stress. Shear stress and resultant stress on an oblique plane inclined at an angle of 450 with the axis of minor tensile stress. A point in strained material is subjected to the stresses as shown in figure. Locate the principal planes and evaluate the principal tresses. (Apr/May 02)
TWO MARKS:
27
28
29
UNIT – I
1. Define stress and its types When a body is acted upon by some load (or) external force, it undergoes deformation (i.e. change in shape or dimensions)
semester +
www.semesterplus.com
+ semester www.semesterplus.com
5 Stress is defined as the internal resistance offered by the material to the extremely applied force, expressed per unit area.
P A
stress P applied load A Area of sec tion
Types of stresses: 1. Axial stress 2. Bearing stress 3. Bending stress 4. Shear stress Types of axial stress: 1. Tensile stress 2. Compressive stress. 2. Define strain. Strain is defined as the ratio of change in length to the original length of the member Strain =
Change in length (dl) Original length (l)
3. Define tensile stress and tensile strain. When the resistance offered by a section of a member is against an increase in length, the section is said to offer tensile stress. Internal resis tan ce p Tensile stress t C.S . A A Tensile strain: The strain corresponding to tensile stress is tensile strain.
Tensile strain e
l l
Increase length Original length
4. Define the three Elastic moduli. (April/May 2005) Young’s Modulus: It is the ratio between tensile stress and tensile strain (or) compressive stress and compressive strain.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 6
E
t e et
or
c ec
Modulus of Rigidity: It is defined as the ratio of shear stress ( ) to shear strain and is denoted by C, N or G It is also called shear stress modulus of elasticity. Bulk (or) Volume Modulus of Rigidity It is defined as the ratio of normal stress (on each face of a solid cube) to volumetric strain and is denoted by the letter K.
K
n V
5. Define shear strain and Volumetric strain Shear strain is defined as the ratio of transverse displacement to the distance from the lower force. Shear strain
Transverse displacement
=
Distance from lower face
Volumetric strain: body
It is defined as the ratio between change in volume and original volume of the
v
change in volume v Original volume V
6. A square steel rod 20 mm x 20 mm in section is to carry an axial load (compressive) of 100 KN. Calculate the shortening in a length of 50 mm. E = 2..14 10 8 KN / M 2 Solution: Area A = Length
0.02 0.02 0.0004
m2
l 50 mm (or) 0.05 m P 100
KN
E 2.14 10 8 KN / m2
Shortening of the rod l :
semester +
www.semesterplus.com
+ semester www.semesterplus.com 7
Stress 1 E
100 P 250000 KN / m 2 A 0.0004
Stress Strain
Strain
Stress E E
250000 2.14 10 8 l 250000 l 2.14 10 8
l
250000 0.05 2.14 10 8
= 0.0000584 m (or) 0.0584 mm Hence the shortening of the rod = 0.0584 mm. 7. Define Poisson’s ratio. (Nov / Dec 04) The ratio of lateral strain to the longitudinal strain is a constant for a given material, when the material is stressed within the elastic limit. This ratio is called Poisson’s ratio and it is generally denoted by 1/m (or) .
Lateral strain Longitudinal strain
8. Find the force “P” acting on the bar given in fig. (April / May 04)
Sum of all left direction force = Sum of all right direction force 50 + P = 45 + 15 P = 60 – 50 = 10 kN 9. What type of stress will be induced in a bar when the ends are restrained and subjected to i) rise in temperature and ii) a fall in temperature? (April / May 04) Thermal stresses: These are the stresses induced in a body due to change in temperature. Thermal stresses are set up in a body, when the temperature of the body is not allowed to expand or contract freely. i) Rise in temperature – compressive stress ii) a fall in temperature -tensile stress.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 8
10. Define principal plane and principal stress. The plane, which have no shear stress, are known as principal planes. Hence principal planes are the planes of zero shear stress. These planes carry only normal stresses. The normal stresses, acting on a principal plane, are known as principal stresses.
11. Write down the relation between modulus of elasticity and modulus of rigidity and that between modulus of elasticity and bulk modulus. (Nov/Dec 02) The relation between modulus of elasticity and bulk modulus is given by E = 3K (1-2/m) The relation between modulus of elasticity and modulus of rigidity E = 2G (1+(1/m)) 12. When a rod of diameter 20mm is subjected to a tensile force of 40 kN, thje extension is measured as 250 divisions in 200mm extension meter. Find the modulus of elasticity if each division is equal to 0.001mm. (Nov/Dec 02) Solution: Diameter = 20mm ; P = 40 kN = 40 x 103 N ; Extension = 250 divisions 1 division = 0.001 mm; L = 200 mm δl = 250 x 0.001 = 0.25 mm P / A 40 x10 3 / 314.16 E = σ/e = 0.25 / 200 l / l Modulus of Elasticity = 101.85 x 103 N/mm2 13. What do you understand by the assumption, plane section remain plane even after the application of load? When some external load acts on the beam, the shear force and bending moments are set up at aii sections of the beam. Due to the shear force and bending moment, the beam undergoes certain deformation. After removal of the load the beam will come to its original position based on that assumption. 14. A metal bar 50 mm x 50 mm section, is subjected to an axial
compressive load of 500 KN. The contraction of a 200 mm gauge length is found to be 0.5 mm and the increase in thickness 0.04 mm. find E and μ. Solution: b = 50 mm,
t = 50 mm
Area = 50 50 2500 mm 2 P = 500 KN Length, l = 200 mm, l 0.5mm. Increase in thickness, t 0.04mm Young’s Modulus:
semester +
www.semesterplus.com
+ semester www.semesterplus.com 9
l
pl AE
500 10 3 200 0.5 E 80 KN / mm 2 2500 E Poission’s Ratio:
Lateral strain Linear strain
Linear strain = 0.0025
t Lateral strain thickness 0.04
1 0.0025 50 m
1 0.32 m
15. brass bar, having C.S.A of 10 cm2 is subjected to axial forces as shown in fig.
Find the total elongation of the bar Take E 0.8 10 3 t / cm 2 Given, Area A = 10 cm2 E 0.8 10 3 t / cm 2
l Total elongation of the bar. For the sake of simplification, the force of 8 tonnes acting at B, may be split up into three forces of 5 tones, 2 tonnes and 1 tone. Now it will be seen that the part AB of the bar is subjected to a tension of 5 tonnes part BC is subjected to a compression of 2 tonnes and the part BD is subjected to a compression of 1 tonne as shown in fig.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 10
Using the relation,
l
1 p1l1 p 2 l 2 p3 l 3 AE
1 5 60 2 100 1 220 10 0.8 10 3
12 0.015cm 0.15mm 800
16. Steel bar is 900 mm long its two ends are 40 mm and 30 mm in diameter and the length of each rod is 200 mm. the middle portion of the bar is 15 mm in diameter and 500 mm long. If the bar is subjected to an axial tensile load of 15 KN, find the total extension. E 200 GN / m 2
G giga and
1G 10 9
P = 15 kN A1 = 1256 mm2 A2 = 176.625 mm2 A3 = 706.50 mm2 l1 = 200 mm l 2 = 500 mm l 3 = 200 mm Total extension of the bar:
l1 , l 2
l1
and l 3
pl1 A1 E
l2
l l1 l 2 l 3
pl pl1 pl 2 3 A1 E A2 E A3 E
P E
pl2 A2 E
l3
pl3 A3 E
l1 l l2 3 A1 A2 A3
= 0.2454 mm.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 11
17. bar of 2 m length, 2 cm breadth and 1.5 cm thickness is subjected to a tensile load of 3000 kg. Find the final volume of the bar, if 1 4
and E 2.0 10 6 kg / cm 2 L = 2 m = 200 cm, b = 2 cm, t = 1.5 cm.
Vol
V 200 2 1.5 600cm 3 P = 3000 kg.
1 1 m4 3 4
E 2 10 6 kg / cm 2
v
2 stress e1 e ; v E m P 3000 1000 kg / cm 2 A bt
1000 1 6 2000 2 10
v v
v
18.
A
1 2 1 2000 4
1 4000
1 600 0.15cm 3 4000
bronze
specimen has 6 2 C 0.47 10 kg / cm . Determine
a
E 1.2 10 6 kg / cm 2
and
Solution:
C
m 1.2 10 6 0.47 10 2m 1 6
mE 2m 1
0.94m 1 1.2m 0.94m 0.94 1.2m 0.94 1.2m 0..94m 0.94 m = 0.26 m
semester +
www.semesterplus.com
+ semester www.semesterplus.com 12
1 0.26 1 0.277 m 0.94 m
19. Two parallel walls 6 m apart, are stayed together by a steel rod 20 mm passing through metal plates and nuts at each end. The nuts are tightened home, when the rod is at a temp of 1000 C. Determine the stress in the rod, when the temperature falls down 200C if. a. b.
The ends do not yield and The ends yield by 1mm.
E 2 10 6 kg / cm 2
and
12 10 / c 6
Solution:
Length of rod = l = 6m = 600 cm
of rod = d = 20 mm = 2 cm. Temperature t = 1000 - 200 = 800 c E 2 10 6 kg / cm 2 ; 12 10 6 / 0 C
when the ends do not yield.
1 stress in the rod
tE
1 12 10 6 80 2 10 6 1920 kg / cm 2 when the ends yield by 1mm
2 stress in the rod
2 12 10 6 8
0.1 6 2 2 10 158 kg / cm 600
20. A bar is subjected to a tensile stress of 1000 kg / cm2. Determine the normal and Tangential stresses on a plane making an angle of 600 with the axis of tensile stress. 30 0
n cos2 750kg / cm 2 t sin cos 433kg / cm 2 866 kg / cm 2
semester +
www.semesterplus.com
+ semester www.semesterplus.com 21.
13
A point in a stained material is subjected to two mutually perpendicular stresses of 2000 kg/cm2 and 1000 kg/cm2. Determine the intensities of normal and resultant stresses on a plane inclined at 300 to the axis of the minor stress. Major stress = 1 2000 kg / cm 2 Minor stress = 2 1000 kg / cm 2
Angle of plane, which it makes with the axis of minor, principle
stress 30 0 Normal stress
n
1 2 2
1 2 2
cos 2
= 1750 kg/ cm2 Tangential stress
t
1 2 2
Resultant stress
sin 2 433kg / cm 2
r n 2 t 2 1802 .8kg / cm 2
22. At a point in a strained material the principal stresses are 100 N/mm2 (tensile) and 60 N/mm2 (comp). Determine normal stress, shear stress, resultant stress on a plane inclined at an 500 to the axis of major principal stress. Also determine the maxi shear stress at the point. 1 100 N / mm 2 ,
2 60 N / mm 2 ,
90 0 50 0 40
t 78.8N / mm 2
n 33.9 N / mm 2 , R 85.8 N / mm 2 Maximum shear stress;
t max
semester +
1 2 2
80 N / mm 2
www.semesterplus.com
+ semester www.semesterplus.com 14
23. A point in a strained material is subjected to a compressive stress of 800 kg/cm2 and a shear stress of 560 kg/cm2. Determine graphically or otherwise, the maximum and minimum intensity of stress. 560 kg / cm 2
800 kg / cm 2 , Maximum intensity of stress: 2 tan 2
2 560 800
= 1.4 2 54 0 28'
27 014' 2
n1 2 2 2 = 288.2 kg/cm2 (Tensile)
n2
Minimum intensity of direct stress. 2
n 2 2 1088 .2kg / cm 2 2 2 24. A steel bolt 2.5 cm diameter is subjected to a direct tension of 1500 kg and a shearing forces of 1000 kg. Determine the intensities of normal and shear stress across a plane inclined at an angle of 60 0 to the longitudinal axis of the bolt. Also determine the resultant stress. Solution:
2.5cm.
Area = 4.91 cm2
Direct tension = 1500 kg.
Direct stress on the bolt =
1500 305.5kg / cm 2 4.91
Shearing force = 1000 kg.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 1000 203.7kg / cm 2 4.91
Shear stress =
15
Angle which the plane makes with the longitudinal axis of the bolt = 60 0 bolt
Angle, which the plane makes with the normal to the longitudinal axis of the
90 0 60 0 30 0
n
Normal stress:
2
1 cos 2 sin 2
= 405.5 kg/ cm2
1 2
t sin 2 e cos 2
Shear stress:
= 30.5 kg/cme+
r
Resultant stress:
n 2 t 2
405.52 30.52
= 406.6 kg/cm.2
25. A point is subjected to a tensile stress of 60 N/mm2 and a compressive stress of 40 N/mm2, acting on two mutually perpendicular planes and a shear stress of 10 N/mm2 on these planes. Determine the principal stresses as well as maxi shear stress. Also find out the value of maxi shear stress. 1 60 N / mm 2 and 2 40 N / mm 2
10 N / mm 2
Principal stresses:
n1
n2
semester +
2
1 2
2 2 2 1 61N / mm 2
1 2
2 2 2 1 41N / mm 2
2
2
2
www.semesterplus.com
+ semester www.semesterplus.com 16
Maximum shear stress
t
semester +
n1 n 2 2
61 41 51N / mm 2 2
www.semesterplus.com
+ semester www.semesterplus.com 17
SIXTEEN MARKS QUESTION AND ANSWERS: 1.A hollow cast – iron cylinder 4m long, 300 mm outer diameter and thickness of metal 50 mm is subjected to a central load on the top when standing straight. The stress produced is 75000 KN / m2. Assume Young’s Modulus for cast iron as 1.5 10 8 KN / m 2 and find i. ii. iii.
Magnitude of the load Longitudinal strain produced and Total decrease in length.
Solution: Outer diameter, D = 300 mm = 0.3 m Thickness ,
t = 50 mm = 0.05 m
Length,
l=4m
Stress produced 75,000 KN / m2
E 1.5 10 8
KN / m2
d D 2t
Inner diameter of the cylinder
0.3 2 0.05 = 0.2 m i.
Magnitude of the load P: Using the relation,
(or)
P A
P A 75000
75000 ii.
0.3 4
2
D 4
2
d2
0.2
2
P = 2945.2 KN. Longitudinal strain produced e: Using the relation,
semester +
www.semesterplus.com
+ semester www.semesterplus.com strain, e iii.
Stress 75,000 0.0005 E 1.5 10 8
18
Total decrease in length , l : using the relation,
strain
change in length Original length
0.0005
l 4
l 0.0005 4 0.002 m = 2 mm Hence decrease in length = 2 mm.
2. The following observations were made during a tensile test on a mild steel specimen 40 mm in diameter and 200 mm long. Elongation with 40 KN load (within limit of proportionality) l 0.0304 mm, yield load = 161 KN Maximum load = 242 KN Length of specimen at fracture = 249 mm. Determine: v. Young’s Modulus of Elasticity vi. Yield point stress vii. Ultimate stress viii. Percentage elongation. Solution: i.
Young’s Modulus of Elasticity (E) :
P 40 3.18 10 4 KN / m 2 A 0.042 4
Stress,
Strain, e
l l
0.0304 0.000152 200
Stress 3.18 10 4 Strain 0.000152 2.09 10 8 KN / m2 E
semester +
www.semesterplus.com
+ semester www.semesterplus.com 19
(ii)
Yield point stress: Yield point stress
Yield po int load area
4
(iii)
0.042
12.8 10 4 KN / m 2
Ultimate stress: Maximum load
Ultimate stress =
Area
4
(iv).
161
242
0.042
19.2 10 4 KN / m 2
Percentage elongation: Percentage elongation =
Length of specimen at fracture original length Original length
249 200 0.245 200
= 24.5 %
3. A steel 2m long and 3 mm in diameter are extended by 0.75 mm when a weight W is suspended from the wire. If the same weight is suspended from a brass wire, 2.5 m long and 2 mm in diameter, it is elongated by 4.64 mm. Determine the modulus of elasticity of brass if that of steel be 2.0 10 5 N / n Solution: Given: l s 2m,
d s 3mm, l s 0.75mm
Es 2.0 10 5 N / mm 2 ,
l b 2.5m,
d b 2mm
l b 4.64m Modulus of Elasticity of brass, Eb :
semester +
www.semesterplus.com
+ semester www.semesterplus.com 20
From Hooke’s law,
l
Wl AE
where l Extension, l = length, A = Cross sectional area E = Modulus of Elasticity. Case 1: For steel wire
l3
Wl3 As E s
0.75 W 2 1000
2 3 10 5 4
32 W 0.75 4
1 2 10 5 2000
(1)
Case 2: For brass wire
lb 4.64
W lb Ab E b W 2.5 1000
4
2 E b 2
22 W 4.64 4
1 Eb 2500
(2)
Equating equation (i) and (ii), we get
0.75
semester +
32 4
2 1 1 2 10 4.64 Eb 4 2500 2000 5
2
www.semesterplus.com
+ semester www.semesterplus.com 21
Eb 0.909 10 5 N / mm 2 4. A member formed by connecting a steel bar to an aluminium bar as shown in fig. Assuming that the bars are prevented form buckling sidewise; calculate the magnitude of force p1 that will cause the total length of the member to decrease 0.25 mm. The values of elastic modulus of steel and aluminum are 2101 KN / mm2 and 70 KN / mm2 Solution:
Given:
As 50 50 2500 mm 2 AA 100 100 10,000 mm 2 Length of steel bar = l3 = 300 mm. = l 3 = 380 mm
l 0.25mm
E A 70 10 3 N / mm 2
Es 210 10 3 N / nn 2 :
P = Magnitude of the required force.
l1 l 2 A1 E1 A2 E 2
l P
P = 224.36 KN
semester +
www.semesterplus.com
+ semester www.semesterplus.com 22
Area at the reduced section
0.05 4
2
0.025 2 0.001472 m 2
1 Stress in the reduced section,
100 1000 0.001472
67.93 10 6 N / m 2
Elongation of the rod
2.5 x .x E
E
0.731 10 3 50.92 10 6 2.5 x 67.9 10 6 x 0.731 10 3 9 9 200 10 200 10 50.92 10 6 2.5 x 67.9 10 6 x 200 10 9 0.731 10 3
50.92 10 6
2.5 x 1.33x 2.87 x = 1.12 m.
5. A steel tie rod 50 mm in and 2.5m long is subjected to a pull of 100 KN. To what length the rod should be bored centrally so that the total extension will increase by 15 % under the same pull, the bore being 25 mm ? Solution: Dia of steel tie rod = 50 mm = 0.05 m Length of steel tie rod l = 2.5 m P = 100 KN. Dia of bore = 25 mm = 0.025 m
semester +
www.semesterplus.com
+ semester www.semesterplus.com 23
e 200 10 9 N / m 2 Length of the bore x: Stress in the solid rod,
P 100 1000 50.92 10 6 N / m 2 A 0.052 4
l
50.92 10 6 2.5 Elongation of the solid rod l E 200 10 9 = 0.000636 m (or) 0.636 mm. Elongation after the rod is bored 1.15 0.636 = 0.731 mm.
6. A steel flat plate AB of 1 cm thickness tapers uniformly from 10 cm to 5 cm width in a length of 40 cm. From first principles, determine the elongation of the plate, if an axial tensile force of 5000 kg acts on it. Take E 2.0 10 6 kg / cm 2 Given:
Width of the plate at a distance x from A
10 10 5
x x 10 40 8
C.S.A of the bar
x x Ax 110 10 cm 2 8 8 Stress,
semester +
x
P 5000 kg / cm 2 x Ax 10 8
www.semesterplus.com
+ semester www.semesterplus.com
24
5000 1 x x 6 40010 10 2 10 8 8
x Strain, x E
Elongation of the elementary length
ex.dx
dx x 40010 8
cm
The total extension of the bar may be found out by integrating the above equation between the limits 0 and 40.
1 l (ie.) 400
40
0
dx 10
x 8 40
1 x log 10 e 8 0 1 400 8
1 log e 5 log e 10 50
1 log e 10 log e 5 50
1 10 log e 50 5
1 log e 2cm 50
1 2.3 log10 2cm 50
2.3 0.3010 50
log e
2.3 log 10
= 0.014 cm = 0.14 mm.
semester +
www.semesterplus.com
+ semester www.semesterplus.com 25
7. A steel cube block of 50 mm side is subjected to a force of 6 KN (Tension), 8 KN (compressive) and 4 KN (tension) along x, y and z directions. Determine the change in the volume of the block. e 200 KN / mm 2 and m
10 3
Side of steel cube = 50 mm
A 50 50 2500 mm 2
and v 50 50 50 125000 mm 3
PX 6KN; PY 8KN; Pz 4KN
v V
e X eY e Z
eX
X
x E
Y mE
Z mE
P 6 10 3 2.4 N / mm 2 A 2500
P 8 10 3 Y 3.2 N / mm2 A 2500
Z
PZ 4000 1.6 N / mm2 A 2500
eX
2.4 3.2 3 1.6 3 2.88 10 E 10 E E E
eY
3.2 2.4 3 1.6 3 4.4 E 10 E 10 E E
eZ
1.6 2.4 3 3.2 3 1.84 10 E 10 E E E
v v
semester +
e X eY e Z
www.semesterplus.com
+ semester www.semesterplus.com 26
v 125 10 3
2.88 4.4 1.84 E E E
v 125 10 3
0.32 3 0 . 2 mm 200 10 3
8. A bar of 30 mm is subjected to a pull of 60 KN. The measured extension on gauge length of 200 mm is 0.09 mm and the change in diameter is 0.0039 mm. Calculate and the values of the three module. Solution: i.
Young’s Modulus:
E
e
P 60 10 3 84.9 N / mm 2 A 302 4
e
l l
0.09 0.00045 200
E 188.67 KN / mm 2
ii.
Poisson’s ratio:
Lateral strain d / d Linear strain e
d / d iii.
semester +
0.00039 0.00013 3
0.00013 13 45 m 0.00045 45 13
Modulus of rigidity:
www.semesterplus.com
+ semester www.semesterplus.com 45 188.67 10 3 mE C 13 73.19 KN / mm 2 2m 1 45 2 1 13 iv.
27
Bulk Modulus:
45 188 .67 10 3 mE K 13 148 .95 KN / mm 2 3m 2 45 3 2 13 9. At a point within a body subjected to two mutually perpendicular directions, the stresses are 80 N/mm2 tensile and 40 N/mm2 tensile. Each of the above stresses is accompanied by a shear stress of 60 N/mm2. Determine the normal stress. Shear stress and resultant stress on an oblique plane inclined at an angle of 450 with the axis of minor tensile stress. Solution: Given,
Major tensile stress, 1 80 N / mm 2 Minor tensile stress, 2 40 N / mm 2 Shear stress, q = 60 N/mm2 Angle of oblique plane, with the axis of minor tensile stress. Normal stress ( n )
i.
n
1 2 2
1 2 2
cos 2 sin 2
80 40 80 40 cos 2 45 0 sin 2 60 sin 2 45 2 2
n 120 N / mm 2 ii.
semester +
Shear (or tangential) stress
t
www.semesterplus.com
+ semester www.semesterplus.com t
1 2 2
28
sin 2 cos 2
80 40 sin 2 45 60 cos2 45 2
t 20 N / mm2 iii..
Resultant stress
R
n 2 t 2
R
= 121.665 N / mm2
10. A point in strained material is subjected to the stresses as shown in figure. Locate the principal planes and evaluate the principal tresses. Solution: Given,
Stress on the face BC and AD = 600 kg / cm2 Inclination of the stress = 600
Stress normal on the face BC or BD
1 600 cos 30 0
600 sin 30 0 Location:
2 400
lan2
2R 0 39 0 21' 1 2
Principal stress
n1
semester +
1 2 2
2
2 1 2 2
www.semesterplus.com
+ semester www.semesterplus.com 29
= 766 kg/cm2
n2
1 2 2
2
2 2 1 2
= 154 kg/cm2
semester +
www.semesterplus.com
