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saw5 [17]
3 years ago
6

A bar of steel has the minimum properties Se = 40 kpsi, S = 60 kpsi, and S-80 kpsi. The bar is subjected to a steady torsional s

tress of 15 kpsi and an alternating bending stress of 25 kpsi Find the factor of safety guarding against a static failure, and either the factor of safety guard- ing against a fatigue failure or the expected life of the part. For the fatigue analysis use:
(a) Modified Goodman criterion.

(b) Gerber criterion.

(C) ASME-elliptic criterion.

Engineering
1 answer:
nlexa [21]3 years ago
5 0

Answer:

(a) Modified Goodman criterion:

Factor of safety against fatigue failure =  1.0529

(b) Gerber criterion:

Factor of safety against fatigue failure = 1.31

(c) ASME-elliptic criterion:

Factor of safety against fatigue failure = 1.315

Explanation:

See the attached file for the calculation.

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A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maxi
saw5 [17]

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, \tau_{max}, is given by the following formula;

\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )

t_w = 1 cm = 0.01

h = 29 cm = 0.29 m

h_w = 25 cm = 0.25 m

b = 15 cm = 0.15 m

I_c = The centroidal moment of inertia

I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )

I_c = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right )  = 1.2257083\bar 3 \times 10^{-4}

I_c = 1.2257083\bar 3 × 10⁻⁴ m⁴

From which we have;

4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )

Which gives;

W = 11,416.6879 N

\sigma _{b.max} = \dfrac{M_c}{I_c}

\sigma _{b.max} = 1500 N/cm² = 15,000,000 N/m²

M_c = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

M_{max} = \dfrac{W \cdot L}{4}

L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417

L ≈ 0.64417 m ≈ 64.417 cm.

4 0
2 years ago
What are the main causes of injuries when using forklifts?
Julli [10]

Answer:

1, 3, 4, 5

Explanation:

just did it

4 0
2 years ago
A piston–cylinder device containing carbon dioxide gas undergoes an isobaric process from 15 psia and 80°F to 170°F. Determine t
drek231 [11]

Answer:

See explanation

Explanation:

Given:

Initial pressure,

p

1

=

15

psia

Initial temperature,

T

1

=

80

∘

F

Final temperature,

T

2

=

200

∘

F

Find the gas constant and specific heat for carbon dioxide from the Properties Table of Ideal Gases.

R

=

0.04513

Btu/lbm.R

C

v

=

0.158

Btu/lbm.R

Find the work done during the isobaric process.

w

1

−

2

=

p

(

v

2

−

v

1

)

=

R

(

T

2

−

T

1

)

=

0.04513

(

200

−

80

)

w

1

−

2

=

5.4156

Btu/lbm

Find the change in internal energy during process.

Δ

u

1

−

2

=

C

v

(

T

2

−

T

1

)

=

0.158

(

200

−

80

)

=

18.96

Btu/lbm

Find the heat transfer during the process using the first law of thermodynamics.

q

1

−

2

=

w

1

−

2

+

Δ

u

1

−

2

=

5.4156

+

18.96

q

1

−

2

=

24.38

Btu/lbm

7 0
3 years ago
A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a
dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have volume of box = 25ft*10ft*12ft= 3000ft^{3}

Volume= 84.95m^{3}

Since 1ft^{3} =0.028m^{3}

Now the weight of water displaced = Weight =\rho \times Volumewhererho is density of water = 1000kg/m^{3}

Thus weight of liquid displaced = \frac{84.95X1000}{1000}tonnes=84.95 tonnes..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

3 0
3 years ago
Select a research proposal topic that relates to electrical and electronics engineering and write a proposal report taking into
fiasKO [112]

Expand your technical knowledge, form global networks and balance life & work commitments. Our advanced diplomas remain current with technological and industry developments.

8 0
3 years ago
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