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Korvikt [17]
3 years ago
14

List these materials from least effective to most effective in terms of bridge construction. Explain why the worst is the worst

and the best is the best.
Brick
Aluminum
Wood
Plastic
Concrete
Reinforced concrete
Steel
Iron
Engineering
1 answer:
spayn [35]3 years ago
7 0

Explanation:

The four primary materials used for bridges have been wood, stone, iron, and concrete. Of these, iron has had the greatest effect on modern bridges. Steel is used to make reinforced and prestressed concrete. Modern bridges are almost exclusively built with steel, reinforced concrete, and prestressed concrete.

Wood and Stone: Wood is relatively weak in both compression and tension, but it has almost always been widely available and inexpensive. Civil Engineers now incorporate laminated wooden beams and arches into some modern bridges.

Stone is strong in compression but weak in tension. Its primary application has been in arches, piers, and abutments.

Iron and Steel: Cast iron is strong in compression but weak in tension. Wrought iron has much greater tensile strength. Steel is superior to any iron in both tension and compression. Steel can be made to varying strengths, some alloys being five times stronger than others. The civil engineer refers to these as high-strength steels.

Concrete: Concrete is an artificial stone made from a mixture of water, sand, gravel and cement. It is strong in compression and weak in tension. Concrete with steel bars embedded in it is called reinforced concrete. Reinforcement allows for less concrete to be used because the steel carries all the tension; also, the concrete protects the steel from corrosion and fire.

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Under normal conditions, gauge pressure is ______ absolute pressure.
levacccp [35]

Answer:

Option B  Lower than

Explanation:

Gauge pressure is a relative measurement based on atmospheric pressure. Gauge pressure can be positive if it is above atmospheric pressure or it can also be negative it is below.  On another hand, absolute pressure is an actual pressure in a space and its value has always to be zero or above. Basically absolute pressure is zero if it is in a perfect vacuum. So the measurement of absolute pressure is gauge pressure + atmospheric pressure.  This is the reason in normal condition the gauge pressure = absolute pressure - atmospheric pressure and therefore is lower than absolute pressure

3 0
4 years ago
Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a ba
Tpy6a [65]

Answer:

Among all three tools, the ceramic tool is taking the least time for the production of a batch, however, machining from the HSS tool is taking the highest time.

Explanation:

The optimum cutting speed for the minimum cost

V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)

Where,

C,n = Taylor equation parameters

T_h =Tool changing time in minutes

C_e=Cost per grinding per edge

C_m= Machine and operator cost per minute

On comparing with the Taylor equation VT^n=C,

Tool life,

T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)

Given that,  

Cost of operator and machine time=\$40/hr=\$0.667/min

Batch setting time = 2 hr

Part handling time: T_h=2.5 min

Part diameter: D=73 mm =73\times 10^{-3} m

Part length: l=250 mm=250\times 10^{-3} m

Feed: f=0.30 mm/rev= 0.3\times 10^{-3} m/rev

Depth of cut: d=3.5 mm

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, C_e= \$20/15+2=\$3.33/edge

Tool changing time, T_t=3 min.

C= 80 m/min

n=0.130

(a) From equation (i), cutting speed for the minimum cost:

V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}

\Rightarrow 47.7 m/min

(b) From equation (ii), the tool life,

T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}

\Rightarrow T=53.4 min

(c) Cycle time: T_c=T_h+T_m+\frac{T_t}{n_p}

where,

T_m= Machining time for one part

n_p= Number of pieces cut in one tool life

T_m= \frac{l}{fN} min, where N=\frac{V_{opt}}{\pi D} is the rpm of the spindle.

\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc

So, the number of parts produced in one tool life

n_p=\frac {T}{T_m}

\Rightarrow n_p=\frac {53.4}{4.01}=13.3

Round it to the lower integer

\Rightarrow n_p=13

So, the cycle time

T_c=2.5+4.01+\frac{3}{13}=6.74 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc

(e) Total time to complete the batch= Sum of setup time and production time for one batch

=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times4.01}{457}=0.4387=43.87\%

Now, for the cemented carbide tool:

Cost per edge,

C_e= \$8/6=\$1.33/edge

Tool changing time, T_t=1min

C= 650 m/min

n=0.30

(a) Cutting speed for the minimum cost:

V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min [from(i)]

(b) Tool life,

T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min [from(ii)]

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc

n_p=\frac {7}{0.53}=13.2

\Rightarrow n_p=13 [ nearest lower integer]

So, the cycle time

T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc

(e) Total time to complete the batch=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.53}{275.5}=0.0962=9.62\%

Similarly, for the ceramic tool:

C_e= \$10/6=\$1.67/edge

T_t-1min

C= 3500 m/min

n=0.6

(a) Cutting speed:

V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}

\Rightarrow V_{opt}=2105 m/min

(b) Tool life,

T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc

n_p=\frac {2.33}{0.091}=25.6

\Rightarrow n_p=25 pc/tool\; life

So,

T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc

(e) Total time to complete the batch

=2\times60+ {50\times 2.63}=251.5 min=4.19 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.091}{251.5}=0.0181=1.81\%

3 0
4 years ago
At what depth in water is the increased pressure five times greater than atmospheric pressure (101 kPa)?​
umka21 [38]

Explanation:

40.4m

Explanation:

Pressure at depth is given as

P = P, + pgh

Final pressure at depth h= 5 Po

5Po= Po + pgh

pgh = 4Po = 4 x 1.01 x 10^5

h = (4.04×10^5)/ (1000x10)

h=40.4m

8 0
3 years ago
Can someone put each letter by the correct word for my automotive class !
Oksana_A [137]

Answer:

L = spindle

M = lower ball joint

part without the letter showing = steering knuckle

Explanation:

6 0
3 years ago
A carpenter uses a hammer to strike a nail. Approximate the hammer's weight of 1.8lbs, as being concentrated at the head, and as
bija089 [108]

Answer:

The average force F exherted by the nail over the hammer is 178.4 lbf.

Explanation:

The force F exherted by the nail over the hammer is defined as:

F = |I|/Δt

Where I and Δt are the magnitude of the impact and the period of time respectively. We know that the impact can be calculated as the difference in momentum:

I = ΔP = Pf - Pi

Where Pf and Pi are the momentum after and before the impact. Recalling for the definition for momentum:

P = m.v  

Where m and v are the mass and the velocity of the body respectively. Notice that final hummer's momentum is zero due to the hammers de-acelerate to zero velocity. Then the momentum variation will be expressed as:

ΔP =  - Pi = -m.vi

The initial velocity is given as 50 mph and we will expressed in ft/s:

vi = 50 mph * 1.47 ft/s/mph = 73.3 ft/s

By multiplyng by the mass of 1.8 lbs, we obtain the impulse I:

|I|= |ΔP|= |-m.vi| = 1.8 lb  * 73.3 ft/s = 132 lb.ft/s

Dividing the impulse by a duration of 0.023 seconds, we finally find the force F:

F =  132 lb.ft/s / 0.023 s = 5740 lb.ft/s^2  

Expressing in lbf:

F = 5740 lb.ft/s^2 * 0.031 lbf/lb.ft/s^2  =  178.4 lbf

7 0
4 years ago
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