I should over-tighten bolts so that my creations do not fall apart.

What are the rigging devices used to move loads such as steel plates and sheet piles without the use of slings, but grip the loa

scoundrel [369]

The rigging device which are used to move loads without the use of slings, but grip the load by biting down and using jaw tension to secure the load, is lifting clamps.

<h3>What are the rigging devices?</h3>

The rigging devices are used to lift the objects and items when the safety is required. This device is used in the industries.

Types of rigging devices

Rigging hooks-These rigging device is used when the heavy load need to be lift.
Lifting clamps-Lifting clamp are used to lift the device with jaw tension to secure the load. In this, there is no use of slings.
Pulley and blocks-In the load is lifts with the help of block and pulley arrangement. This is a widely used rigging device.

Thus, the rigging device which are used to move loads without the use of slings, but grip the load by biting down and using jaw tension to secure the load, is lifting clamps.

Learn more about the rigging devices here;

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4 0

2 years ago

When calculating a resolved shear stress within a single unit cell, the angle between the applied stress direction and the slip

givi [52]

Answer: 35.3 °

Explanation:

Body-centered cubic lattice (bcc or cubic-I), just like all lattices, has lattice points at the eight corners of the unit cell with an additional points at the center of the cell. It has unit cell vectors a = b = c and interaxial angles α=β=γ=90°.

The simplest crystal structures are those that have present only a single atom at each lattice point.

body-centered cubic unit cell has atoms at each of the eight corners of a cube (like the cubic unit cell) plus one atom in the center of the cube. Each of the corner atoms is the corner of another cube so the corner atoms are shared between eight unit cells. It is said to have a coordination number of 8. The bcc unit cell consists of a net total of two atoms; one in the center and eight eighths from corners atoms

With the use of BCC unit cell, if a applied stress is in [110] direction, but slip applies in [111] direction, the angle between applied direction and slip direction is given as:

[1 1 0] [1 1 1]

λ = Cos^-1 ( 1×1 + 1×1 + 0×1 ÷ (1^2 + 1^2 +0^2) (1^2 + 1^2+ 1^2))

Cos^-1 2/ sqrt 6

= 35.386°

6 0

3 years ago

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo

SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out / W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0

3 years ago

Which of the following statements are true about staying safe while

Tatiana [17]

Answer:

What you should do changes based on the hazards you're exposed to.

You should be aware of any hazards that are present.

3 0

3 years ago

A mitochondrial membrane complex consisting of ATP synthase, adenine nucleotide translocase (ATP–ADP translocase), and phosphate

Fed [463]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

8 0

3 years ago

I should over-tighten bolts so that my creations do not fall apart.​

I should over-tighten bolts so that my creations do not fall apart.