Page 175 - WCS - Electrical

π

Force

applied
π 2
d =
Stress =
4
4

π

area
Original

cross
sectional

Force
applied

π 2
d =
Stress =
4
4
cross
Original

sectional

500
500
Load
=
Stress =
=
π

N applied ce

forced For
Load

22
Area
16π
π 2

500
500
d
Load=
=
σ = Stress =
=
Stress = 4
= 4

Original
Area
area

cros
2 sectional s
7
Load

N
(or)
22
mm
Area
16π
=
16x
σ =
Area
7
2
mm

500x7 Load
500
500
=
=
Stress =
=
stress(or) Load

forced
N
22
500x716πArea
=
σ
16x
=
strein Area
2
7
16x22
mm

stress
load
Applied

strein

stiffness = 500x7

stress
shear
=

Deflection
Applied

load
G =
16x22

stress

stiffness =
shear
E =

shear

stress
Deflection
G =
10
strein
600
newtons
strain

shear

30 load ed
Appli
10
Sprin
30 g
stiffness =
mm

stress
volumetric

newtons
600

stress

shear
Deflection
K =
30
G =

strain
volumetric
mm

30

stress
volumetric
strain

shear
K =
Ultimate
stress

10

of
Safety =

Factor
strain

volumetric
600

newtons

Safe
stress
stress

Ultimate
30
Factor

of
Safety =
30
mm

volumetric
stress
strain

Lateral

Safe
stress
K =
s
Poisson'
ratio =

volumetric
Linear

strain

Ultimate

Lateral
stress

Safe
Ultimate
stress =

Poisson'
s
ratio =

Factor

of
Safe
stress

strain
Linear

Ultimate

stress
Safe

Safe
stress =

E = = (or) E = strain forced strain strain area Spring Spring 16x22 Safety = 16x stress stress
E

stress
Safe
strain
Poisson' s ratio = Lateral strain 2.5
μ =
1+
2G

=
E
Linear
μ =
1+
stress
2G

=
WORKSHOP CALCULATION & SCIENCE - CITS Safe 4 2.5 = Ultimate stress
Safe

stress
E
1
μ =
−
4
2G 1 E E Safe stress = Ultimate stress
μ = μ =+
−

2.5

of
safety
2G 2G 1 Factor Ultimate stress
=
Safe
4
Equanting this value in (i) 2.5 stress = Factor of safety
π
2

  E μ   = E −  1 E  = 4 × 4 (4.2)
 − 21  −1   2G  − + 2  Safe 2.5 × = π Ultima 2 te stress
= 3K
1
(4.2)


= stress
G
  2G

1
E= 3K  − 21  E   −1      = 3K   − E  + 2   4 4 Factor of safety


  2G    G  2.5 2.5 π 2
22
4.2

 
2.5 4
4
22
   − E  E  3G - E   − E  = 4 ×= 7× 4 × × 4.2× (4.2)
  2
1
3
= 3K

3K 
E= 3K  −      2G −1     = 3K 1 G + 2  = × × 4.2× 4.2


7×

    G    E     G  3G - E   4 Load 4 30× 1000
=
3K  − G    = 3K   G   Working stress = Area 22 600 30× 1000
3


Load
2.5
 


×
= 4.2×
EG= 3K (3G-E)=9KG-3KE  3G - E  Working = stress = 7× 4 × stress 4.2
Area
600

4
 9KG E

Ultimate

 

=
E
3
= 3K

 3K
 −
 

Therefore EG= 3KE = 9KG   G   Factor of safety = Working stress 100030×

 G
+ 3K G

Load

stress
 9KG  


Ultimate

=
E

Factor
of



safety =
Therefore E(G= 3K) = 9KG Working stress = Area = stress
600
 G
+ 3K
Working

 9KG  450 M.Pa Ultimate stress
Therefore E =   Factor = of safety =
450
 G + 3K  50 M.Pa M.Pa stress
Working
=
50 M.Pa
EXAMPLES
450 M.Pa
=
1 A load of 500 kg is hanging from an iron rod of 8 mm diameter and 3.6 metres length. The rod is extended by
M.Pa
50

7.5 mm in axial direction. Find the values of stress and strain caused by the load.
Solution: Original length = 3.5 metres = 3500 mm Change in length = 7.5 mm
Diameter d= 8 mm
Stress = Force applied applied Force applied π 2 π π 2 π π 2 π

d =
d =
Area of cross- section = x 8 x 8 sq. mm
Force
Stress =
4
4
d =
area

sectional
Stress =cross Original

4
Original
sectional
cross

area

4
Original cross sectional area Force applied = 16π sq. mm 4 4 π 2 π

π
Stress = Stress = Original cross sectional area Load 500 500 π 2 Load 4 d = 500
Force
applied

d =
500 4

Load (or) forced N Original cross sectional area Stress = Stress = 16πArea = Load = = Stress = 500500 4 4 = =
22
Load
forced

(or)
=

N
σ = Load (or) = forced N = Area 16x Area 16π 16x 22
22
σ =
16π
Area
σ = mm 2 = Area 2 7 16x Load 500 7 500
mm
Stress
Area mm Load (or) forced N Stress = Load = 500 = = 500 = 22
7
2
=
16π
Area
σ = Load σ = forced = N = 2 500x7 Area 16π 16x 22 16x
(or)
500x7
Area
=
E = stress Area stress mm 2 mm 16x22 500x7 = 16x22 7 7
=

16x22

500x7
strein stress E = strein = 9.943 kg/ mm Ans. 500x7 =
E =
2
strein stress Applied load = 16x22
stress E = Spring stiffness = Spring ed 16x22 Applied load

2 A helical spring is loaded with a force of 600 newton and is compressed by 30 mm. What would be the load
load
Appli
stiffness =
shear stress E = strein Spring stiffness =ion
Deflect
shear
Deflection
G = shear stress G = strein stress required to compress it to 10 mm.
Deflection
shea G = strain r shear strain Applied load
Spring

Applied
shear strain shear stress Solution: Spring stiffness = = stiffness = load 10
10
Spring

stiffness
G = shear G = 600 newtons newtons 30 600 10 Deflection Deflection
stress

newtons
600
shear

strain
K = volumetric stress shear strain stress 30 mm 30 30 30 10
mm

volumetric
10
K =
600
Stress = Force applied K = volumetric stress volumetric strain π 2 π 30 mm 600 newtons newtons 30

strain
volumetric
=

d =
4

Ultimate
Original cross sectional area volumetric strain volumetric stress 4 Factor of Safety = Ultimate stress 30 mm 30 stress
Ultimatmm 30
volumetric stress Factor of stress e Safety =
K =
K = volumetric strain Factor of Safety = stress Safe Safe stress Safe stress

Ultim
Lateral strain volumetric strain = Load = 500 = 500 = 20 newtons/ mm ate stress
Stres
Laterals

strain

Factor
Safety
s
Poisson'
ratio =
σ = Load (or) forced = N Poisson'L sin Poisson' Lateral s strain Area 16π 16x 22 Ultimate stress of Ultimate = stress stress
ratio =
of

Safety =
Factor

Safe
ratio = strain ear

Linear
strain

stress
Ultimate
Safe

stress

stress =
Area mm 2 Linear strain Lateral = Lateral strain Safe Load required to compress the spring by 10 mm

stress
Ultimate
7
Safe
stress =

stress = stress Safe
Safe

strain
Safe
stress
ratio
Poisson'
s
= spring stiffness x deflection

Linear
strain

1+ μ = E Poisson' s ratio = Linear strain 500x7 Safe stress stress Ultimate stress

E
Safe
stress
Ultimate =
=
= 20 x 10
Safe
Safe
stress 2G 1+ μ = E 1+ μ = 2G 16x22 = 2.5 2.5 stress = 2.5 stress stress
= Safe
E = 2G E 4 =

4
strein E E 1+ μ = 4 = 200 newtons Ans.
μ = − 1 E 1+ μ = μ = E − 1 2G Applied load Ultimate stress 2.5 2.5
=
Another procedure

2G

Ultimate
stress
=
4

Safe
stress =
stress
Ultimate

G = shear stress 2G μ = 2G − 1 2G Spring stiffness = Deflection For 30 mm deflection ……. 600 newtons required For 10 mm deflection ……. ? required
Safe
stress =
4
E

SafeF

stress = safety of actor
Factor
of

safety

safety

shear strain μ = E − 1 μ = 2G − 1 2.5 π 2 Factor of Safe stress Ultimate stress

10
stress
2.5 Ultimate =
(4.2)
×
2
= =
π
stress
2.5 Safe
  E    E  2G    600 E newtons = x 600 = 200 newtons Ans. × π (4.2) Factor of safety
2

=
4
(4.2)
4
×
30
 E
safety

4 Fact
4 of or

1
=
 − 21
 −1
+ 2
 
E
1
−1


mm
K = volumetric stress   2G  E 3K −1   −     G  − 21   1   2G + 2      = 3K  − 30 + 2   4 4 2.5 π 2
 −
 = 3K 
 − 21 

 
G




volumetric strain   2G      G  E      E  2.5 × π= (4.2) 4 2 × 4 (4.2)
=
1
  E −1    2 −1  E = 3K   − + 2  4 4
 
stress

= 3K
 − 21    −1     2G Factor    of + 2 fety = Ultimate 2.5 × 22 22 4.2× 2.5 × 22 × 4.2× 4.2
1

 −
G

Sa
2.5 4.2×
=


=
G
 
4

  E    3G - E    2G  E     3G - E   Safe stress 7 = 4× 4 × 7× 4 × 4.2× 4.2 7× 4
4
= 3K
3
3K 3G

2.5
 strain
Poisson' s ratio = Lateral 3K  −        −3 E       G = 3K    −3 E- G    = 3K  G  Load 30× 2.5 22 = Load 22 30× 4.2×× 1000

1000
G
  
×


4.2
3K 






Working
stress =

Load =
×
G
stress = 4
= 4
Linear strain   G          E     3G - EUltimate stress Area = Working 30× 10004.2×× 7× 4.2

600
Working
stress =
= 4
  stress =
E e
  E 3K    −3  3G= 3K  Area 7× 4 Area 600 162
-   Saf
600


= 3K

 
1000
30×
Load
 


 9KG  3K  −3 G  9KG   G   G    G  Safe stress Ultimate stress Load CITS : WCS - Electrical - Exercise 16
1000 =

stress =

Working
30×

 



=
1+ μ = E E =   G + E 3K =   9KG   E =    Factor of safety = Working stress = Area Ultimate stress 600
Area
Ultimate
stress

of
Factor
safety =
600
safety = stress orking
 
2G  G + 3K   G + 3K   2.5 Factor of W Working stress Working stress
=

 9KG
Ultimate
 9KG =   4 Factor of Ultimate = stress stress
E

safety
Factor
μ = E − 1 E =   G + 3K  G + 3K  450 M.Pa of safety = Wor Working stress


450
2G Safe stress = Ultimate stress = 50 M.Pa M.Pa 450 M.Pa stress king
=
=

M.Pa
Factor of safety 50 M.Pa 50 450 M.Pa
=
2.5 π 2 450 M.Pa 50 M.Pa
=
  E    E  = 4 × 4 (4.2) 50 M.Pa
1
 − 21  −1    = 3K  − + 2 

  2G    G 
2.5 × 22 × 4.2× 4.2
=
  E    3G - E  4 7× 4
3K  −3    = 3K  

  G    G  Working stress = Load = 30× 1000
Area 600
 9KG  Ultimate stress
E =   Factor of safety =
 G + 3K  Working stress
450 M.Pa
=
50 M.Pa