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vodomira [7]
3 years ago
12

Unless a structural element is intended to be battered, sloped, or cambered (or some other non-conforming position), wood, steel

, and concrete should always be installed and maintained __________ , __________ , and __________. These conditions are critical in order to ensure maximum stability, structural performance and user satisfaction. Please insert the answer into the spaces provided below.
Engineering
1 answer:
SSSSS [86.1K]3 years ago
5 0

Answer:

stability; resistance; rigidity.

Explanation:

Okay, let us fill in the gap in the question above. Please note that the capitalized words are the missing words to fill in the gap.

"Unless a structural element is intended to be battered, sloped, or cambered (or some other non-conforming position), wood, steel, and concrete should always be installed and maintained STABILITY , RESISTANCE , and RIGIDITY. These conditions are critical in order to ensure maximum stability, structural performance and user satisfaction."

This question has to deal with buildings or say structural (civil) engineering. The definition to the missing words are given below:

STABILITY: Stability occurs when we have the center of gravity coinciding with the base of the structure.

RESISTANCE : Resistance simply means the 'tension' that is is how much the structure can resist an applied force.

RIGIDITY : RIGIDITY can also be associated with resistance and it is the property of a structure to resist bending.

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Answer:

B.

Explanation:

A safety-critical system (SCS) or life-critical system is a system whose failure or malfunction may result in one (or more) of the following outcomes: death or serious injury to people. loss or severe damage to equipment/property.

7 0
3 years ago
What is the primary water source for a water cooled recovery unit's condensing coll?
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A) chilled water from evaporator
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Rubber bushings are used on suspensions to
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D. All of the above
4 0
3 years ago
How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor
zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0
3 years ago
Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given
Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = \frac{1}{6}*334.8 = 55.8 ft^3

Volume of sand = \frac{2}{6}*334.8 = 111.6 ft^3

Volume of gravel = \frac{3}{6}*334.8 = 167.4 ft^3

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

4 0
4 years ago
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