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slega [8]
3 years ago
6

What is a shearing stress? Is there a force resulting from two solids in contact to which is it similar?

Engineering
1 answer:
Luba_88 [7]3 years ago
7 0

Answer:

Shearing stresses are the stresses generated in any material when a force acts in such a way that it tends to tear off the material.

Generally the above definition is valid at an armature level, in more technical terms shearing stresses are the component of the stresses that act parallel to any plane in a material that is under stress. Shearing stresses are present in a body even if normal forces act on it along the centroidal axis.

Mathematically in a plane AB the shearing stresses are given by

\tau =\frac{Fcos(\theta )}{A}

Yes the shearing force which generates the shearing stresses is similar to frictional force that acts between the 2 surfaces in contact with each other.  

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Because assembly language is so close in nature to machine language, it is referred to as a ____________.
My name is Ann [436]

Answer:

symbolic machine code.

Explanation:

The instructions in the language are closely linked to the machine's architecture.

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A 5­stage single pipeline computer uses the pipeline mentioned in the lecture (not the book). It resolves the direction and targ
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3 years ago
A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream
Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}

Substituting figures in the equation we have:

= \frac{1}{1+0.1(2.5-1)+0(2-1)}

= 0.75

Let's now calculate equivalent flow rate of the car using:

Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}

= \frac{7500}{0.9*4*0.75*1}

= 2777.7 pc/h/en

Calculating for traffic density, we have:

D = \frac{Vp}{FFS}

D = \frac{2777.7}{70}

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0
3 years ago
Two standard spur gears have a diametrical pitch of 10, a center distance 3.5 inches and a velocity ratio of 2.5. How many teeth
lubasha [3.4K]

Answer:50 , 20

Explanation:

Given

Diametrical Pitch\left ( P_D\right )=\frac{T}{D}

where T= no of teeths

D=diameter

module(m) of gears must be same

m=\frac{D}{T}=\frac{1}{P_D}=0.1

Let T_1 & T_2be the gears on two gears

Therefore Center distance is given by

m\frac{\left ( T_1+T_2\right )}{2}=3.5

thus

0.1\frac{\left ( T_1+T_2\right )}{2}=3.5

T_1+T_2=70----1

and Velocity ratio is given by

VR=\frac{No\ of\ teeths\ on\ Driver\ gear}{No.\ of\ teeths\ on\ Driven\ gear}

2.5=\frac{T_1}{T_2}----2

From 1 & 2 we get

T_1=50, T_2=20

6 0
3 years ago
The author uses the example of the flameless candle to illustrate that​
krek1111 [17]

Answer:

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Explanation:

4 0
3 years ago
Read 2 more answers
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