Can someone draw a electric circuit with electrical symbols for this

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0

2 years ago

Millions of years ago, the Sierra Nevada region began to be uplifted along a crack in Earth's crust. The region on the other sid

Evgen [1.6K]

Answer: The correct answer is : Fault block mountain with rough edges and steep cliffs

Explanation: Snowy saws are an example of a mountain chain blocked by faults. The snowy mountains were formed because the tectonic movement forced some segments of the earth's crust up into irregular pieces and others down.

8 0

3 years ago

A double-pane insulated window consists of two 1 cm thick pieces of glass separated by a 1.8 cm layer of air. The window measure

Elanso [62]

Answer:

(b). T = 22.55 ⁰C

(c). q = 557.8 W

Explanation:

we take follow a step by step process to solving this problem.

from the question, we have that

The two glass pieces is separated by a 1.8 cm distance layer of air.

the thickness of glass piece is 1 cm

width = 4 m

the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b). the thermal resistance due to the conduction in the first glass plane is given thus;

R₁ = Lg / Kg A ................(1)

given that Kg rep. the thermal conductivity of the glass plane

A = conduction surface area

Lg = Thickness of glass plane4

taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

inputting values into equation (1) we have,

R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]

R₁ = 1.068 ˣ 10 ⁻³ k/w

Being that we have same thermal resistance in the first and second plane,

therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w

⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus

R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

A = surface area

La = thickness of air

substituting values into the equation we have

R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]

R₂ = 5.73 ˣ 10⁻² k/w

Given the thermal resistance on the outer surface due to convection, we have

R₄ = 1/hA

inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w

R₄ = 6.94 ˣ 10⁻³k/w

Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄

R-total = 0.0663 kw

From this we can calculate the rate of heat loss

using q = Ti - To / R-total ..............(3)

given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C

from equation (3),

q = 27- (-10) / 0.0063 = 557.8 W

q = 557.8 W

⇒ Applying the heat transfer formula for inside surface glass temperature gives;

q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

cheers i hope this helps

8 0

3 years ago

List all the qualities of an engineer?

Dennis_Churaev [7]

Answer:

Math and Computer Skills. A qualified engineer should be good at math, at least through the level of calculus and trigonometry, and understand the importance of following the data when making design decisions.

Organization and Attention to Detail.

Curiosity.

Creativity.

Critical Thinking.

Intuition.

Explanation:

6 0

2 years ago

cThe Mars Rover Spirit got stuck in the Martian sand. The wheels kept slipping. Attempts to free it were futile. Discuss the typ

IgorC [24]

Answer:

Improved/ advanced types of Actuators include servo systems, create a large range of actuator motion in response to the changing needs of the operational environment or process.

Actuators are local or automated suppliers of working motion.

Hydraulic and air cylinders can be classified as single-acting cylinders, meaning that the energy source result in movement in one direction and a spring is used for the other direction.

Explanation:

An actuator control system is referred to as any electronic, electrical, or electromechanical system often used to activate an actuator, control the direction as well as extent and duration of its output. Actuator control systems could take the form of extremely simple, manually-operated, start-and-stop stations, either sophisticated or programmable computer systems. The more improved/ advanced types include servo systems that produce a large range of actuator motion in response to the changing needs of the operational environment or process. This type of actuator control system uses an interface arrangement that assimilates feedback from the process or mechanism and adjusts the actuator in the right way. Most actuator systems will include at least a set of travel limits that prevent the actuator destroying itself or the secondary mechanism.

Actuators are local or automated suppliers of working motion. They are used to changes, adjust, or move a secondary mechanism, where a physical operator cannot intervene directly. They are denoted by a large range of varying types using electrical and electromagnetic, hydraulic, or pneumatic power sources to create linear or rotary outputs. One element they all have in common is the actuator control system used to start, stop, and adjust the range, speed, and duration of the working motion.

Actuators can produce a linear motion, rotary motion or oscillatory motion which means they can create motion in one direction, in a circular motion or in opposite directions at regular intervals. Hydraulic and air cylinders can be classified as single-acting cylinders, meaning that the energy source result in movement in one direction and a spring is used for the other direction.

7 0

3 years ago