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goldfiish [28.3K]
2 years ago
15

determine the position d of the 6- kn load so that the average normal stress in each rod is the same.

Engineering
1 answer:
Zinaida [17]2 years ago
7 0

The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

<h3>What is meant by torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

$N_1 * x=N_2 *(L-x)$

also, we know that stress at both ends are same

$\frac{N_1}{12}=\frac{N_2}{8}$

$2 * N_1=3 * N_2$

Now from two equations we have

$\frac{3}{2} N_2 * x=N_2 *(L-x)

solving the above equation we have

$x=\frac{2}{5} L

so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$12mm^2 and $$8mm^2, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.

To learn more about torque refer to:

brainly.com/question/20691242

#SPJ4

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If the rotational speed of a pump motor is reduced by 35%, what is the effect on the pump performance in terms of capacity, head
FinnZ [79.3K]

Answer:

- the capacity of the pump reduces by 35%.

- the head gets reduced by 57%.

the power consumption by the pump is reduced by 72%

Explanation:

the pump capacity is related to the speed as speed is reduces by 35%

so new speed is (100 - 35) = 65% of orginal speed

speed Q ∝ N ⇒ Q1/Q2 = N1/N2

Q2 = (N2/N1)Q1    

Q2 = (65/100)Q1

which means that the capacity of the pump is also reduces by 35%.

the head in a pump is related by

H ∝ N² ⇒ H1/H2 = N1²/N2²

H2 = (N2N1)²H1

H2 = (65/100)²H1 = 0.4225H1

so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.

Now The power requirement of a pump is related as

P ∝ N³ ⇒ P1/P2 = N1³/N2³

P2 = (N2/N1)³P1

H2 = (65/100)²P1 = 0.274P1

So the reduction in power is 1 - 0.274 = 0.725 which is 72%

Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.  

8 0
3 years ago
A thick steel slab ( 7800 kg/m3 , c 480 J/kg K, k 50 W/m K) is initially at 300 C and is cooled by water jets impinging on one o
Nutka1998 [239]

Answer:

1791 secs  ≈ 29.85 minutes

Explanation:

( Initial temperature of slab )  T1 = 300° C

temperature of water ( Ts ) = 25°C

T2 ( final temp of slab ) = 50°C

distance between slab and water jet = 25 mm

<u>Determine how long it will take to reach T2</u>

First calculate the thermal diffusivity

∝  = 50 / ( 7800 * 480 ) = 1.34 * 10^-5 m^2/s

<u>next express Temp as a function of time </u>

T( 25 mm , t ) = 50°C

next calculate the time required for the slab to reach 50°C at a distance of 25mm

attached below is the remaining part of the detailed solution

5 0
3 years ago
4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The amb
Minchanka [31]

Answer:

a) 84.034°C

b) 92.56°C

c) ≈ 88 watts

Explanation:

Thickness of aluminum alloy fin = 12 mm

width = 10 mm

length = 50 mm

Ambient air temperature = 22°C

Temperature of aluminum alloy is maintained at 120°C

<u>a) Determine temperature at end of fin</u>

m = √ hp/Ka

   = √( 140*2 ) / ( 12 * 10^-3 * 55 )

   = √ 280 / 0.66 = 20.60

Attached below is the remaining answers

7 0
3 years ago
There are two identical oil tanks. The level of oil in Tank A is 12 ft and is drained at the rate of 0.5 ft/min. Tank B contains
Luba_88 [7]

Answer:

  16 minutes

Explanation:

This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.

The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."

Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...

  (4 ft)/(0.25 ft/min) = 16 min

The level in the two tanks will be the same after 16 minutes.

__

<em>Additional comment</em>

The oil levels at that time will be 4 ft.

You can write two equations for height:

  y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)

  y = 8 -0.25x . . . . . .  height in feet after x minutes (tank B)

These will be equal when ...

  y = y

  12 -0.5x = 8 -0.25x

  4 = 0.25x . . . . . . . . . . add 0.5x -8

  16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height

The graph shows when the tanks will have equal heights and when they will be drained.

4 0
2 years ago
Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1
babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A     ⇒    P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E)    ⇒  ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒  ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0
3 years ago
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