Page 176 - WCS - Electrical

π

applied

Force
π 2
d =
π
Stress =

applied
Force
4
d = 4

sectional
area

Original
cross
Stress =
π
4

Force
cross

sec
Original

d =
Stress =
4

cross
Original

sectional

500
500
Load
=
=
Stress =
Load

forced
N
Load
(or)

22
Area
= 16π
=
π
16x
σ =

Load
applied
forced

Fo

(or)rce
500
22
π 2
Area
Load 16π

d =
Area
σ = =
=
Stress
=
=
16x
Stress =
mm
4
4
(or)
N
Load
cross
2 area
Original
22
sectional
Area
Area
mm
=
σ =
16x
500x7
Area forced N = applied area tional 2 2 area Stress = π 2 π 2 4 500 4 16π 500 7 500 7 7
mm
=
500x7
500
Load
500
16x22

stress
= =
σ = Load (or) forced = N Stress = Area 16x22 = π 16x 22 WORKSHOP CALCULATION & SCIENCE - CITS
500x7
E =

16π
stress
=
strein
E =
stress
7
Area
d =
mm
π
Stress = strein Force 2applied π 2 π 2 16x22 Applied load
E = Force

applied
4
4 =
d
Original
Stress = Original shear cross sectional area Spring 500x7 4 Applied load
Spring

stiffness =
strein
4

stress
cross

Deflection
area

sectional
stiffness =
G =
Applied
stress
shear

load

G = stress F shear orce applied = π 16x22 π Deflection 500 2
Spring
stiffness = 500Load
strain
2
3 If ultimate stress is 2.5 kg/mm and factor of safety is 4. Find safe stress for suspending a
d =
Stress = Orig Lo G = cross inal (or) ad shear stress N Stress = 4 500 Deflection 10
=
=
strain

shear
Load
500
E =
4

forced
22
Area
newtons
=
=

σ =
0.4 mm diameter wire. 16x
=
π
(or)

strain
F
shear
Load strein forcedapplie orce sectional area Stress = 600 = 600 16π 16π 22 10 30
Nd
π 2
Area
newtons

d
Area =
Stress Orig Areacross inal volumetric mm 2 Spring stiffness = 4 App mmload ied 16x 7 30 7 10 2
σ = =
30l

stress
4

50
2 nal
Load
area
newtons 5000
sectio
K =
Solution: Ultimate stress = 2.5 kg/mm
600 mm 0
3
stress
volumetric
=
stress
Deflection

volumetric
strain
G = shear forced mm Stress = Area 500x7 = 22 30
Lo
N
K =(or) ad
16π
mm

30
stress

volumetric
σ = shear strain strain 2 Factor of safety = 4 Ultimate stress
=
volumetric =
16x
500x7 500Load
500
K =

=
Ultimate 7
Safety =

Area stress

=
=
volumetric
N
E = forced )

Safe
Load (or stress mm strain Stress = Factor of16x22 10 stress
stress
Factor
of
16x22

22
Safety =

16

600 Area
newtonsπ
σ = E = strein = Factor0x Safety = Ultimate stress
16x
Safe
stress
30

of7
2
strein
mm
30

mm
Area s stress ratio = Lateral strain 50 Applie 7 stress
volumetric
=
Safe load d
Poisson'
K = volumetric stress ar Lateral strain Spring 16x22 Applied Ultimate stress
stiffness =
Linear
Poisson' stress
strain

s
ratio =
load
strain
she
Safe
stiffness
strain
Linear

Ultimate
stress
E = G = stress Lateral strain Spring 500x7 = stress = Deflection stress
=
ratio =
Poisson's shear
strein
Safes

Ultimate
stres
stress = Deflection
Safe

shear
G = stress strain strain Factor of Safety = Safe stress stress
16x22

Ultimate
Linear

Safe

E = shear strain E Spring stiffness = newtons 600 stress = stress Safe load Applied 10
1+
μ =
10
strein
E

shear

Deflection
π 2
30
Poisson' s ratio = stress 2G strain Force applied 600 newtons d 2.5 π Safe stress
μ = Lateral
= =
1+
G = Stress =
2.5
30
mm load plied
Ap
30
volumetric
E stress
4 4
2G
4
Origi
shear
cross

1+ Linear
K =

μ =

30
mm
volumetric

stress
2.5
stress

K = shear vo strainnal strain sectional area Spring stiffness = = Ultimate stress 2
4
10
E

stress = Deflection
Safe
2G strain ric
lumet

= 0.625 kg/mm Ans.
=
1
G = she μ = strain ar μ = strain 600 newtons Safe 4 Ultimate stress Ultimate 500500 30 stress
−
stress

E tric
volume

Load
1 2G
−
Safe of or
Safety
π
stress =
Fact Stress

Ultimate
stress

E Force
mm
30

volume 2G stress tric (or) forced applied Facto 4 If ultimate stress is 2.5 kg/mm2 and factor of safety is 4. Find out the safe load for the suspension of 4.2 mm
E
stress = = == π 2
Ultimate 10

of
N
Safety d =
Load
Safe

Factor
safety
stress 22
stress = = Area
−
μ =
μ = =
newtons
K = 1+ Stress 2Gric 2G cross inal 1 = sectional area Safe of r 2.54 Factor 16π safety
600

σ =
16x
4
stress
diameter wire. Ultimates Safe
stres
volumet Orig

strain
of30
Lateral

stress =
Area

strain 2
mm
30

π

volumetric
stress
Poisson'
s
4 Ultimate
2

ratio =
of
safety

Factor
Poisson' s  E ratio =  E Linear strain E  Factor of Safe = mm 2.5 Safe (4.2) stress 7
strain

Lateral
=
×
Solution: Diameter of wire d = 4.2 mm
K =
π
2.5=
Safety

Ultimate
stress
2
 


500
Load
×4
4500
(4.2)
strain
=
volumetric

500x7
stress =
Safe

stress
E  −
1
= 3K
+ 2
 −
Line

−1 ar
 E  2 1−
Ultimate
stress
 μ =
=
 



2.5 4
4 =

× stress teat
stress =Ulti

(4.2) stress Safeess e
str
Area of original cross – section
−1 2G forced (or)
Safe
N

1
 − 

 

= 3K  
Area
16π
Poisson  − 21  2G Load  E stress       strain  G 1 1 + 2   G  + 2     Fa Safe ctor stress =y of Stress = = Ultimam π = 4 stress fe safety 2 16x 22
Safet = 16x22
 strain ral
=
 =
Late
 σ

E
4 Sa
 2G

 
 ratio s'
stress
Safe

= 3K
E
of
 − 21 = E = Area
Factor
−1
7
 −2



Linear
 2G strein
μ =
22 stress te
Ultima
1+

 
= = 2.5

Poisson' 1+ ratio = E Lat 2G strain eral strain mm G  Safe stress 2.5 22 × 2 7× stress e Applied load
4.2
×
4.2×
μ =
2.
2.5 π5
=

s
2.5 500x7

4Saf
 
Spring
(4.2) 4 =
=
3K  Linear
4 4 Ultima
E
  E  2G   E  strain  3G - E  sq. mm 4.24.2×××= ×= 4 = stiffness 22 stress te 4

2.5 47×

= 3K
E  −3
 3G 

4
 −1 E E
Deflection
1
 − 21  3K  μ = = shear = 3K     stress - + 2 E    Safe stress = = 16x22 Load 4.2× 4.2 1000
  = 3K

×
−
×
  stress

G
 −3  
30×


  G 1−
G
μ EE =

 3G 
 1   2G =+  μ =  3K  G 2G    strein   strain G   G - E   Working 2.5 4Saf 7× str e 4ess = 1000
stress =mate
30× stress
Ulti
E  r
  shea
Load
2G 1−


=
600
 
Area
Working

= stress
Ultimate
10
 2G E  −3 G  = 3K  G  Safe stress Safe stress stress = = Area Factor of safety 1000
4
30×
=
Load 600
Applied
1+ μ  = E 2G      2.5 600 newtons = load 30
22
Working

2.5
stress =

Facto
= Spring
× stiffness
4.2× safety of r=
4.2
×
 9KG
stress
Ultimate

Area
=
600
1 shear
4

stress
of × 2.
Deflection
7
=

2
 9KG
= 4
E
 μ = K G − E volumetric -   stress Factor Ult 4 π5 mm 30 Ultimate stress
  3G
 

E =
safety = stress imate (4.2)×
 2G =
 + 3K
stress
3
3K  E − E =    E  9KG strain    E  Safe 2.5 = × π 2 4 Working stress
of

 G 
Factor
safety = 2

 = 3K   strain shear
4 (4.2)
 volumetric
=
Ultimate
 −1
×ety
kg/mm 30Load of Factor

  = 3
  −
Working
+ 3K  
stress


G E
E   2 1
  μ = G −  G −1 + 3K  1K  + 2 G   + 2   Working Factor 60 Area = stress saf 1000 stress
 E
= 
 
safety = Ultimate 10
4
4

of

Factor
stress
stress
= Ultimate

 −1
1
−1
 G    −= 3K
 2 2G
 2G  
newtons 0
of


Safety =
Working
stress


stress =
600

Safe
2.5
 2G
G
π
30
 

2

stress
Safe

450 safety of r
×Facto
M.Pa
=
(4.2) mm 30
   E volumetric stress   Safe load = Safe stress x Area of original cross- section
=22
E
4
4 2.5

 

K =
 − 21 E Poisson' volu  = 3K Lateral  strain Factor of 2.5 π2.5 = 450 2 M.Pa stress

+ 2
Ultimate
−1
1

M.Pa 4.24.2×
 9KG  
×
 −
×
50
= 22

=
 ratio = metric
strain


 s
 2G
4
G
7×
450 4.24.2××
 

50
M.Pa

 E
-
 G 

4

Linear

stress stress timate stress ltimate
Ul U
=

   E + 3K E       = 3K E  3G strain = = safety = 4 (4.2)× × 4 7× Working M.Pa

3K 
4
4
  3G
Safe of r
=
stress =
 −3   = 3K1
 − 21  3K     −3 − E   G = 3K  − -  E  2 G    Facto Safety Load 30× 1000
1
+
50
M.Pa

  

 
Safe
2.5
Safe
22
G
=

    2G G          G   Working × stress = 4.2× 3d 4.210000× stress stress
Loa
×

=
=
4
450 47×
  E   μ =o E Lateral  strain Working stress = M.Pa Area 600 600
 3G
-
E
Area
Poisson
=
22
2.5
3K  −3 1+ratis'    9KG   strain 2.5 = × 50 × = 4.2× 4.2
= 3K


2GLinear
=

M.Pa Ultimate stress Ultimate
G 
1000 stress
= 8.6625 kg. Ans.
Load

 
E
=
4


= =4

Ultimate
E
     9KG   3G G - E   Working Fa stress = safety of ctor 74Safe × stress = 30× stress
 
Safe

stress
+ 3K
600 stress rking
Wo
3K E 3 = −      G     Factor 5 A motor vehicle steering tie-rod has a cross sectional area of 600 mm2 and it is subjected to a tensile load of
Area

= 3K
of
safety =


E

+ 3K
Working ×
1000
30
stress
 
E

   G G μ =   1− G  Working Load =Ultimate stress
30 kN. If the tensile strength of steel is 45o MPa (mega pascal) , find the factor of safety provided for in this
stress =
 9KG μ1+ 2G Safe stress 600
 =
stress
Ultimate

E =  2G Factor of rod. Area = 2.5 M.Pa of safety
450 Factor
safety = =

 G
+ 3K
4

M.Pa
 9KG   450 = Working stress
M.Pa stress te
Ultima

= 2.5
50π
E =  μ =  E − 1 Factor Solution: Ultimate Stress = Tenlile strength
safety =

2
of
M.Pa (4.2)×
=

 G + 3K  E    E  50 Working stress stress ate

4 Ultim
4
 2G
1
 − 21  −1    = 3K  − + 2  Safe stress = 450 MPa
=
450
M.Pa

  2G    G  = Factor of safety

50
M.Pa
Load = 30 kN = 30 x 1000 newtons

450 2.5 22 2
M.Pa π

= 2.5
(4.2)
×
=
4.2
4.2×
=
×
×
  E    E  Area of cross- section = 600 mm 2

50
M.Pa 44
 − 21    −1     = 3K  3G - E + 2  4 7× 4
1
 −
E 


 −3
 3K  2G      = 3K   G  

  G    G  Working stress = Load = 30× 1000
Working stress = 222.5 × Area 4.2× 4.2

600
×
=
7×
4
4
  E    3G - E  = 50 newtons per mm 2


3K E  −3  9KG = 3K   Ultimate stress
 
=

Factor




G
Load
1000
30×
   G     G  = 50 MPa of safety = Working stress
+ 3K
Working

stress =
=
Area
600
[ 1 MPa = 1 newton per mm MPa means mega pascal]
2
 9KG  450Ultimate stress

M.Pa
E =   Factor of safety =
Factor
of

safe
=ty =
 G + 3K  50 Work stress
M.Paing
450 M.Pa
=
50 M.Pa
163
CITS : WCS - Electrical - Exercise 16