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# Solved Problems: Steady Stresses and Variable Stresses In Machine Members

Mechanical - Design of Machine Elements - Steady Stresses and Variable Stresses In Machine Members

The piston rod of a steam engine is 50 mm in diameter and 600 mm long. The diameter of the piston is 400 mm and the maximum steam pressure is 0.9 N/mm2. Find the compression of the piston rod if the Young's modulus for the material of the piston rod

is 210 kN/mm2.

Solution.

Given

d = 50 mm l = 600 mm D = 400 mm

p = 0.9 N/mm2

E = 210 kN/mm2 = 210 × 103 N/mm2

Let δl = Compression of the piston rod.

We know that cross-sectional area of piston, = (π/4) × D2

= (π/4) × (400)2 = 125680 mm2

Maximum load acting on the piston due to steam,

= Cross-sectional area of piston × Steam pressure

125 680 × 0.9

113 110 N

We also know that cross-sectional area of piston rod,

= (π/4) × d2

(π/4) × (50)2

1964 mm2

Young's modulus (E),

210 × 103       = P l / A×δl

113110 x 600 / 1964 δl

34555 / δl

δl  = 34555 / (210 × 103)

= 0.165 mm

A thin steel tyre is shrunk on to a locomotive wheel of 1.2 m diameter. Find the internal diameter of the tyre if after shrinking on, the hoop stress in the tyre is 100 MPa. Assume E=200kN/mm2. Find also the least temperature to which the tyre must be heated above that of the wheel before it could be slipped on. The coefficient of linear expansion for the

tyre is 6.5×10–6 per °C.

Given :

D = 1.2 m = 1200 mm

σ = 100 MPa = 100 N/mm2

E = 200 kN/mm2 = 200 × 103 N/mm2 α = 6.5 × 10–6 per °C

Internal diameter of the tyre

Let d = Internal diameter of the tyre.

We know that hoop stress (σ),

=E (D - d) / d =200x103 (D - d) / d

(D-d) = 1 / 2x103 = 1.0005

=1200/1.0005

= 1199.4 mm 1.1994 m

Least temperature to which the tyre must be heated

Let     t = Least temperature to which the tyre must be heated.

We know that

D  = π d + π d . α.t

π d (1 + α.t) α.t = (πD / πd ) - 1

= 1 / (α x 2 x 103)

77°C

A mild steel rod supports a tensile load of 50 kN. If the stress in the rod is limited to 100 MPa, find the size of the rod when the cross-section is 1. circular, 2. square, and 3. rectangular with width = 3 × thickness.

Solution.

Given :

P = 50 kN = 50 × 103 N

σt = 100 MPa = 100 N/mm2

1. Size of the rod when it is circular

Let d = Diameter of the rod in mm.

Area, A  = (π/4) × d2

= 0.7854 (d)2

We know that tensile load (P),

50 × 103 = σt × A

100 × 0.7854 d2

78.54 d2

d2 = 50 × 103 / 78.54 = 636.6

= 25.23 mm

Size of the rod when it i s square

Let x = Each side of the square rod in mm. Area, A = x × x = x2

We know that tensile load (P),

50 × 103 = σt × A = 100 × x2

x2  = 50 × 103/100

= 500

x = 22.4 mm

Size of the rod when it is rectangular

Let t = Thickness of the rod in mm, and b = Width of the rod in mm = 3 t

Area, A  = b × t

3t × t

3 t2

We know that tensile load (P),

50 × 103 = σt × A

= 100 × 3 t2

= 300 t2

t2      = 50 × 103 / 300

= 166.7

t        = 12.9 mm

= 3t

= 3 × 12.9

= 38.7 mm

Determine the thickness of a 120 mm wide uniform plate for safe continuous operation if the plate is to be subjected to a tensile load that has a maximum value of 250 kN and a minimum value of 100 kN. The properties of the plate material are as follows: Endurance limit stress = 225 MPa, and Yield point stress = 300 MPa. The factor of safety based on yield point may be taken as 1.5.

Given :

b        = 120 mm ;

Wmax = 250 kN;

Wmin = 100 kN ;

σe   = 225 MPa = 225 N/mm2 ;

σy  = 300 MPa = 300 N/mm2; F.S. = 1.5

Let t = Thickness of the plate in mm.

Area, A = b × t

= 120 t mm2

We know that mean or average load,

Wm = (Wmax + Wmin) / 2 = (250 + 100) / 2

=175kN = 175 × 103 N

Mean stress, σm   = Wm / A

= 175x103 / 120 t N/mm2

Variable load, Wv          = (Wmax - Wmin) / 2

= (250 - 100) / 2

=75kN = 75 × 103 N

Variable stress, σv  = Wv / A

= 75x103 / 120 t N/mm2

According to Soderberg’s formula,

(1/FoS) = (σm / σv) + (σv / σe)

(1/1.5) = (175x103 / 120t x 300) + (75x103 / 120t x 225)

= 7.64 × 1.5

= 11.46 say 11.5 mm

A 50 mm diameter shaft is made from carbon steel having ultimate tensile strength of 630 MPa. It is subjected to a torque which fluctuates between 2000 N-m to – 800 N-m. Using Soderberg method, calculate the factor of safety. Assume suitable values for any other data needed.

Solution.

Given :

d = 50 mm

σu = 630 MPa = 630 N/mm2 Tmax = 2000 N-m

Tmin = – 800 N-m

We know that the mean or average torque,

Tm = (Tmax + Tmin) / 2

[2000 + (- 800)] /2

600 N-m = 600x103 N-mm

Mean or average shear stress,

τm  = 16Tm / πd3

(16 x 600 x 103) / π(50)3

24.4 N/mm2

Tv = (Tmax - Tmin) / 2

[2000 - (- 800)] /2

1400 N-m = 1400x103 N-mm

Variable shear stress,

τv  = 16Tv / πd3

(16 x 1400 x 103) / π(50)3

57 N/mm2

Since the endurance limit in reversed bending (σe) is taken as one-half the ultimate tensile strength (i.e. σe = 0.5 σu) and the endurance limit in shear (τe) is taken as 0.55 σe, therefore

τe    = 0.55 σe

0.55 × 0.5 σu

0.275 σu

0.275 × 630

173.25 N/mm2

Assume the yield stress (σy) for carbon steel in reversed bending as 510 N/mm2, surface finish factor (Ksur) as 0.87, size factor (Ksz) as 0.85 and fatigue stress concentration factor (Kfs) as 1.

Since the yield stress in shear y) for shear loading is taken as one-half the yield stress in reversed bending y), therefore

τy = 0.5 σ y = 0.5 × 510 = 255 N/mm2

Let      F.S. = Factor of safety.

We know that according to Soderberg'sformula,

(1 / FOS) = (τm / τv) + ( (τv x Kf) / (τe x Ksur x Ksz) ) = 0.096 + 0.445

0.541 F.S. = 1 / 0.541

1.85

A circular bar of 500 mm length is supported freely at its two ends. It is acted upon by a central concentrated cyclic load having a minimum value of 20 kN and a maximum value of 50 kN. Determine the diameter of bar by taking a factor of safety of 1.5, size effect of 0.85, surface finish factor of 0.9. The material properties of bar are given by, ultimate strength of 650 MPa, yield strength of 500 MPa and endurance strength of 350 MPa.

Solution.

Given :

l = 500 mm

Wmin = 20 kN = 20 × 103 N

Wmax = 50 kN = 50 × 103 N ;

F.S. = 1.5 ;

Ksz = 0.85 ;

Ksur = 0.9 ;

σ u   = 650 MPa = 650 N/mm2

σy = 500 MPa = 500 N/mm2

σe = 350 MPa = 350 N/mm2

Let    d                               = Diameter of the bar in mm.

We know that the maximum bending moment,

Mmax = (W × l) / 4

max

= (50 × 103 × 500) / 4

= 6250 × 103 N- mm

And minimum bending moment,

Mmin  = ( Wmin × l) / 4

(20 × 103 × 500) / 4

2550 × 103 N – mm

Mean or average bending moment,

Mm  = (Mmax + Mmin) / 2

(6250 × 103 + 2500 × 103) / 2

4375 x 103   N- mm

and variable bending moment,

Mv = (Mmax Mmin ) / 2

(6250 × 103 2500 × 103) / 2

1875 x 103   N- mm

Section modulus of the bar,

= (π/32) x d 3

= 0.0982 d3  mm3

Mean or average bending stress,

σm = Mm / Z

4375 × 103 / 0.0982 d3

44.5 x 106 / d 3

and variable bending stress,

σv = Mv / Z

1875 x 103 / 0.0982 d3

19.1 x 106 / d3

We know that according to Goodman's formula,

(1 / FOS) = ( σm / σu) + ( (σv x Kf) / (σe x Ksur x Ksz) )

1 / 1.5 = (44.5x 106 / d 3 × 650) + ((19.1 × 106 × 1) / (d 3 × 350 × 0.9 × 0.85)) d3 = 139 × 103 × 1.5

= 209 × 103

= 59.3 mm

According to Soderberg's formula,

(1 / FOS)     = ( σm / σy) + ( (σv x Kf) / (σe x Ksur x Ksz) )

1 / 1.5         = (44.5x 106 / d 3 × 500) + ((19.1 × 106 × 1) / (d 3         × 350 × 0.9 × 0.85))

d3      = 160 × 103 × 1.5

= 240 × 103

= 62.1 mm

Taking larger of the two values, we have

d = 62.1 m

A simply supported beam has a concentrated load at the centre which fluctuates from a value of P to 4 P. The span of the beam is 500 mm and its cross-section is circular with a diameter of 60 mm. Taking for the beam material an ultimate stress of 700 MPa, a yield stress of 500 MPa, endurance limit of 330 MPa for reversed bending, and a factor of safety of 1.3, calculate the maximum value of P. Take a size factor of 0.85 and a surface finish factor of

0.9. Solution.

Given :

Wmin = P ;

Wmax = 4P

;

L = 500 mm;

d = 60 mm

;

σu = 700 MPa = 700 N/mm2 ;

σy = 500 MPa = 500 N/mm2 ;

σe = 330 MPa = 330 N/mm2 ; F.S. = 1.3 ;

Ksz = 0.85

; Ksur = 0.9

We know that the maximum bending moment,

Mmax = (Wmax × l) / 4

(4P × 500) / 4

500P N- mm

And minimum bending moment,

Mmin  = ( Wmin × l) / 4

(P × 500) / 4

125P N – mm

Mean or average bending moment,

Mm  = (Mmax + Mmin) / 2

(500P + 125P) / 2

312.5 P  N- mm

Variable bending moment,

Mv = (Mmax / Mmin ) / 2

(500P – 125P) / 2

187.5 P  N- mm

Section modulus of the bar,

= (π/32) x d 3

= 21.21 x 103  mm3

Mean or average bending stress, σm = Mm / Z

312.5P / 21.21 x 103

0.0147 P N/mm2

and variable bending stress, σv = Mv / Z

187.5P / 21.21 x 103

0.0088 P N/mm2

We know that according to Goodman's formula,

(1 / FOS) = ( σm / σu) + ( (σv x Kf) / (σe x Ksur x Ksz)

1 / 1.3 = (0.0147P / 700) + ((0.0088P× 1) / (330 × 0.9 × 0.85))

= ( 1/ 1. 3) × (106 × 55.8)

= 13785 N

According to Soderberg's formula,

(1 / FOS) = ( σm / σy) + ( (σv x Kf) / (σe x Ksur x Ksz) )

1 / 1.3 = (0.0147P / 500) + ((0.0088P× 1) / (330 × 0.9 × 0.85))

= ( 1/ 1. 3) × (106 × 64.2)

= 11982 N

From the above, we find that maximum value of P = 13.785 kN

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