What learning is required to become a mechanical engineer?

Rigid bar ACB is supported by an elastic cir-cular strut DC having an outer diameter of 15 in. and inner diameter of 14.4 in. Th

Ne4ueva [31]

Answer:

The change in length of the circular strut DC = 0.0028 in.

The vertical displacement of the rigid bar at point B = 0.00378 in.

Explanation:

We have the following parameters or information in the question given above:

=> The outer diameter = 15 in., the inner diameter = 14.4 in., the modulus elasticity of E = 29,000 ksi, and the Point load P = 5kips.

The diagram showing the rigid bar ACB is supported by an elastic cir-cular strut DC is given in the attached picture below.

According to Newton's law of motion, it can be seen that the force on CD, that is FCD is equal and opposite to ACD. Hence, FCD = ACD.

Where FCD = p × [4 + 5] ÷ [sin Ф × 4].

kindly note that from the diagram sin Ф = 3/5, cos Ф = 4/5 and tan Ф = 3/4. Also p =5.

Hence, FCD =[ 5 × 9] ÷ [3/5 × 4] = 18.75 kip. So FCD = ACD.

The next thing here is to determine the area and length of CD, say the area of CD is G, thus, G = π/4 × [ 15² - 14.4²] = 13.854 in².

The lenght of CD is = √[4² + 3²] = √[16 + 9] = 5ft. Thus, 5 × 12 = 60in.

Hence, the change in length of the circular strut DC = [18.75 × 60] ÷ 13.854 × 29000 = 0.0028 in.

The vertical deflection of CD = 0.0028 × 3/5 = 0.00168 in.

We have that; 4 /CV = 9BV. Hence, BV = 9/4× CV.

(CV = vertical deflection of CD).

The vertical displacement of the rigid bar at point B = 9/ 4 × 0.00168 in = 0.00378 in.

7 0

3 years ago

A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp

Volgvan

Answer:

a) 22.5number

b) 22.22 m length

Explanation:

Given data:

Bridge length = 500 m

width of bridge = 12 m

Maximum temperature = 40 degree C

minimum temperature = - 35 degree C

Maximum expansion can be determined as

$\Delta L = L \alpha (T_{max} - T_{min})$

where , \alpha is expansion coefficient $= 12\times 10^{-6}$ degree C

SO, $\Delta L = 500\times 12\times 10^{-6}\times ( 40 - (-35))$

$\Delta L = 0.45 m = 450 mm$

number of minimum expansion joints is calculated as

$n = \frac{450}{20} = 22.5$

b) length of each bridge

$Length = \frac{500}{22.5} = 22.22 m$

8 0

3 years ago

Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You clo

agasfer [191]

Answer:

a.) I = 7.8 × 10^-4 A

b.) V(20) = 9.3 × 10^-43 V

Explanation:

Given that the

R1 = 20 kΩ,

R2 = 12 kΩ,

C = 10 µ F, and

ε = 25 V.

R1 and R2 are in series with each other.

Let us first find the equivalent resistance R

R = R1 + R2

R = 20 + 12 = 32 kΩ

At t = 0, V = 25v

From ohms law, V = IR

Make current I the subject of formula

I = V/R

I = 25/32 × 10^3

I = 7.8 × 10^-4 A

b.) The voltage across R1 after a long time can be achieved by using the formula

V(t) = Voe^- (t/RC)

V(t) = 25e^- t/20000 × 10×10^-6

V(t) = 25e^- t/0.2

After a very long time. Let assume t = 20s. Then

V(20) = 25e^- 20/0.2

V(20) = 25e^-100

V(20) = 25 × 3.72 × 10^-44

V(20) = 9.3 × 10^-43 V

8 0

3 years ago

Explain the advantages of continuous insulation in envelope assemblies and why it is better to have the continuous insulation on

Luden [163]

Answer:

Continuous insulation helps in eliminating the necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.

Continuous insulation helps building last much longer over a period of time,without the need to upgrade and repair.

Inside the placement of continuous insulation in the envelope needed includes the following; the foundation wall or slab insulation, balcony interface, bond joist insulation, insulation against sub floor, on the interior of masonry wall.

Explanation:

Solution:

The analytical path to success contains a well planned and designed building exterior, which avoids the loss of energy, control cost and the maximization of technology advancement in materials.

Property installed continuous insulation on the exterior can also execute as an air barrier. the flashing of wall penetrations can form into a drainage plane. this plane can stop potentially damaging moisture from entering into the wall assembly.

By making use of continuous insulation it helps in removing necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.

Continuous insulation helps building to withstand the test of time without the need to upgrade and repair

Now inside the placement of continuous insulation in the envelope is needed as follows:

Foundation wall or slab insulation
Insulation against sub floor
Balcony interfaces
On the interior of masonry wall
Bond joist insulation

6 0

4 years ago

Consider film condensation on a vertical plate. Will the heat flux be higher at the top or at the bottom of the plate? Why?

STALIN [3.7K]

Answer:

During film condensation on a vertical plate, heat flux at the top will be higher since the thickness of the film at the top, and thus its thermal resistance, is lower.

Explanation:

https://www.docsity.com/pt/cengel-solution-heat-and-mass-transfer-2th-ed-heat-chap10-034/4868218/

https://arc.aiaa.org/doi/pdf/10.2514/1.43136

https://arxiv.org/ftp/arxiv/papers/1402/1402.5018.pdf

8 0

3 years ago