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4 years ago

1. Round off 2553 N to three significant figures.

Engineering

1 answer:

Marizza181 [45]4 years ago

8 0

Answer:

(1) 2553 N = 2550 N

(2) 58342 m = 58300 m

(3) 68.534 s = 68.5 s

Explanation:

To round off a number to any significant number start from the last digit, round it off to 1 if the number is up to 5 and to 0 if the last digit is less than 5. Add this 1 or 0 to the preceding digit and continue the process until you are left with three non zero digits, if you are rounding off to three significant figures.

(1) Round off 2553 N to three significant figures.

= 2550 N

(2) Round off 58342 m to three significant figures.

= 58300 m

(3) Round off 68.534 s to three significant figures.

= 68.500 s (zero after decimal point is insignificant)

= 68.5 s

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

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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:

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

c) the safety factor of the resulting assembly is calculated as:

safety factor = $\frac{Yield \ strength }{walking \ stress}$

safety factor = $\frac{400}{62.8}$

safety factor = 6.4

Thus, the safety factor in the resulting assembly = 6.4

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