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3 years ago

18. What is being shown in the above Figure?

Engineering

2 answers:

slavikrds [6]3 years ago

8 0

D. Camshaft gear backlash is being checked

hope this helps :)

umka2103 [35]3 years ago

3 0

Answer: O D

Explanation:

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Geological process lead to melting

viktelen [127]

Answer:

ok so what !! ha

Explanation:

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3 years ago

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

3 years ago

A 35 ft simply supported beam is loaded with concentrated loads 15 ft in from each support. On one end, the dead load is 8.0 kip

Dafna11 [192]

Answer:

From the load equation

F=stress*Area

Given stresses are 8 kips and 9 kips.

Hence the minimum weight supported=6.695 lbs.

4 0

3 years ago

Read 2 more answers

A successive approximation A/D converter has a full-scale output of 0 to 10 V and uses an 8-bit register. An input of 6.185 V is

Komok [63]

Answer:

158

Explanation:

The input is 6.185/10.000 = 0.6185 of the full-scale of the A/D converter. If a 10V input is represented by a register value of 255, then the expected register value for this input is about ...

0.6185·255 = 157.7175 ≈ 158

_____

In hexadecimal, that is 9E.

5 0

3 years ago

A cylindrical 4340 steel bar is subjected to reversed rotating–bending stress cycling, which yielded the test results presented

yawa3891 [41]

Based on the maximum applied load, the factor of safety, and the distance between the loadbearing points, the minimum allowable bar diameter is <u>18.6 mm. </u>

<h3>What is the minimum allowable bar diameter?</h3>

The diameter is included in the following formula:

Maximum stress / Factor of safety = (16 x Maximum applied load x distance between loadbearing points x 10⁻¹⁰) / (π x diameter³)

Solving gives:

(490 x 10⁶) / 2.25 = (16 x 5,000 x 55.0 x 10⁻¹⁰) / (π x diameter³)

217,777,777.78 = 0.00044 / (π x diameter³)

diameter = 18.6 mm

Find out more on diameter at brainly.com/question/16874040.

#SPJ1

3 0

2 years ago