All categories

3 years ago

What is the correct statement regarding the stress over the section of a shaft in torsion?

Engineering

1 answer:

4vir4ik [10]3 years ago

7 0

Answer:

d) A shear stress that varies linearly over the section.

Explanation:

Given that shaft is under pure torsion

As we know that relationship between shear stress in the shaft and radius given as

$\dfrac{T}{J}=\dfrac{\tau}{r}$

For solid shaft

$J=\dfrac{\pi}{64}d^4$

Therefore shear stress given as

$\tau=\dfrac{T}{J}\times r$

T=Applied torque on the shaft

τ=Shear stress at any radius r

From the above equation we can say that shear stress is vaying linearly with the radius of the shaft

Therefore the answer will be d.

You might be interested in

ceramics must be heated in order to harden the clay and make it durable. the equipment used to heat the clay

SIZIF [17.4K]

Ceramics must be heated in order to harden the clay and make it durable. The tool used to heat the clay is called a Kiln.

<h3>What happens when ceramic is heated?</h3>

Cristobalite is a silica polymorph that is used in ceramics. Quartz particles in porcelain can change into cristobalite during burning. This has effects on the fired matrix's thermal expansion. When there is a high level of vitrification or a shape is unstable during the firing of ceramic ware, warping occurs.

Ceramic items are porous, brittle, and rigid. They are thus employed in the production of glass, ceramics, cement, and bricks. Additionally, ceramics are employed extensively in gas turbine engines. Artificial bones and dental implants are both made of bio-ceramics.

Any of the several tough, fragile, heat- and corrosion-resistant materials created by sculpting and then heating an inorganic, nonmetallic material like clay to a high temperature are known as ceramics.

In order to solidify the clay and make ceramics durable, heat must be applied. A kiln is the name of the device used to heat clay.

To learn more about ceramics refer to:

brainly.com/question/26247382

#SPJ4

5 0

2 years ago

Gshyevdnmdcicm gngjc

Len [333]

This is not a valid question. Please try again.

4 0

3 years ago

I am trying to create a line of code to calculate distance between two points. (distance=[tex]\sqrt{ (x2-x1)^2+(y2-y1)^2}) My li

k0ka [10]

Answer:

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Explanation:

The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted. For the general case, it appears you simply need to change how you have written the code.

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.

You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.

Cheers.

6 0

3 years ago

At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

$\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2} }{2})^{2}+\tau _{xy}^{2} }$

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

$32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2} }$

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

$\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}$

Where

σp1 = 20 ksi

σp2 = -30 ksi

$\sigma _{p1} +\sigma _{p2}=-10 ksi$ (equation 1)

$\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi$ equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0

4 years ago

compressors, the gas is often cooled while being compressed to reduce the power consumed by the compressor. explain how cooling

ASHA 777 [7]

The amount of work done by steady flow devices varies with the particular gas volume. The kinetic energy of gas particles decreases during cooling.

When the gas is subjected to intermediate cooling during compression, the gas specific volume is reduced, which lowers the compressor's power consumption. Compression is less adiabatic and more isothermal because the compressed gas must be cooled between stages since compression produces heat. The system's thermodynamic cycle's cold sink temperature is lowered by cooling the compressor coils. By increasing the temperature difference between the heat source and the cold sink, this improves efficiency.

Learn more about thermodynamics here-

brainly.com/question/1368306

#SPJ4

8 0

1 year ago