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3 years ago

Whats the best used for cantilever bridge

Engineering

1 answer:

Luden [163]3 years ago

6 0

Answer:

A cantilever bridge is a bridge built using cantilevers ,structures that project horizontally into space ,supported on only one end.

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On what kind of sectional drawing would you find the maximum length of a building?

Genrish500 [490]

Answer:

A longitudinal section

Explanation:

A longitudinal section is a section drawn along the length of an object, as opposed to a cross section, which is drawn across the width or diameter of an object.

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2 years ago

John’s engineering team has just finished testing prototypes. What should be their next step in the design process?

VladimirAG [237]

Answer:

A. implementation

Explanation:

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3 years ago

A gas cylinder is connected to a manometer that contains water. The other end of the manometer is open to the atmosphere, which

horrorfan [7]

Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa

Explanation:

Given that;

patm = 79 kPa, h = 13 in of H₂O,

A sketch of the problem is uploaded along this answer.

Now

pA = patm + 13 in of H₂O ( h × density × g )

pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)

pA = 82.23596 kPa

the absolute static pressure in the gas cylinder is 82.23596 kPa

4 0

4 years ago

A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal

nirvana33 [79]

Answer:

a) the temperature to which the pin must be cooled for assembly is $T_2 = -101.89^ \ ^0}C$

b) the radial pressure at room temperature after assembly is $P_f = 62.8 \ MPa$

c) the safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion $\alpha = 12.3*10^{-6} \ ^0 C$

Yield strength $\sigma_y$ = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature $T_1$ = 20°C

outer diameter of the collar $D_o = 95 \ mm$

inner diameter of the collar $D_i = 60 \ mm$

pin diameter $D_p$ = $60.03 \ mm$

Clearance c = 0.06 mm

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

$(D_i - c )-D_p = \alpha * D_p(T_2-T_1)$

$(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}$ $(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}T_2$ $\ \ - \ \ 0.01476738$

$-0.09 + 0.01476738$ = $7.38369*10^{-4}T_2$

−0.07523262 = $7.38369*10^{-4}T_2$

$T_2 = \frac{-0.07523262}{7.38369*10^{-4}}$

$T_2 = -101.89^ \ ^0}C$

To determine the radial pressure at room temperature after assembly ;we have:

$P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\ P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2} \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa$