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3 years ago

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Link AB is to be made of a steel for which the ultimate normal stress is 65 ksi. Determine the cross-sectional area of AB for wh

ich the factor of safety will be 3.20. Assume that the link will be adequately reinforced around the pins at A and B.

1 answer:

Jlenok [28]3 years ago

7 0

Explanation:

ddddjidirjejekejwjwkw

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Argon is compressed in a polytropic process with n = 1.2 from 100 kPa and 30°C to 1200 kPa in a piston–cylinder device. Determin

gulaghasi [49]

Answer:

<em>181 °C</em>

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Explanation:

Initial pressure $P_{1}$ = 100 kPa

Initial temperature $T_{1}$ = 30 °C = 30 + 273 K = 303 K

Final pressure $P_{2}$ = 1200 kPa

Final temperature $T_{2}$ = ?

n = 1.2

For a polytropic process, we use the relationship

( $T_{2}$ / $T_{1}$ ) = ( $P_{2}$ / $P_{1}$ )^γ

where γ = (n-1)/n

γ = (1.2-1)/1.2 = 0.1667

substituting into the equation, we have

( $T_{2}$ /303) = (1200/100)^0.1667

$T_{2}$ /303 = 12^0.1667

$T_{2}$ /303 = 1.513

$T_{2}$ = 300 x 1.513 = 453.9 K

==> 453.9 - 273 = 180.9 ≅ <em>181 °C</em>

5 0

3 years ago

Chapter 15 – Fasteners: Determine the tensile load capacity of a 5/16 – 18 UNC thread and a 5/16 – 24 UNF thread made of the sam

Roman55 [17]

Answer:

The 5/16 – 24 UNF is stronger because it has more tensile load capacity.

Tensile load capacity for M8 -1.25 = 5670 lb

Tensile load capacity for M8 -1 = 6067 lb

Explanation:

For 5/16 - 18 UNC thread:

D = 0.3125

n = 18

Therefore the tensile load capacity is = 100000 X (0.7854 X (0.3125 - 0.9743/ 18) ^2

= 5243 lb.

Similarly for 5/16 - 24 UNF , only the n value changes to 24

we get the tensile load capacity = 5806.6 lb

Hence the 5/16 – 24 UNF is stronger because it has more tensile load capacity.

For metric Bolts:

We have to consider all values in SI units

Strength = 689 MPa

We get for M8 -1.25:

Tensile load capacity as = 689 X 36.6 = 25223 N = 5670 lb

For M8 -1:

Tensile load capacity as = 689 X 39.167 = 26986 N = 6067lb

7 0

3 years ago

Read 2 more answers

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is

romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

$c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\$

$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\$

$\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998} =1.3266*10^{-6}T_{BC}$

$2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}$

$T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in$

b) Maximum shear stress in BC

$\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi$

Maximum shear stress in AB

$\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi$

8 0

3 years ago

3. What is the mechanical advantage of the pulley system shown below? HII TAI 190 O A1 O E.2 OC.3 OD 4

Contact [7]

Answer:

I don't know ☺️☺️☺️❌‼️

Explanation:

I don't understand this question

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2 years ago

Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series

Zina [86]

Answer:

0.245 m^3/s

Explanation:

Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s

5 0

3 years ago