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sergey [27]
3 years ago
10

HELP HELP HELP

Engineering
1 answer:
Fantom [35]3 years ago
8 0

Summary

Students learn about the variety of materials used by engineers in the design and construction of modern bridges. They also find out about the material properties important to bridge construction and consider the advantages and disadvantages of steel and concrete as common bridge-building materials to handle compressive and tensile forces.

This engineering curriculum aligns to Next Generation Science Standards (NGSS).

Engineering Connection

When designing structures such as bridges, engineers carefully choose the materials by anticipating the forces the materials (the structural components) are expected to experience during their lifetimes. Usually, ductile materials such as steel, aluminum and other metals are used for components that experience tensile loads. Brittle materials such as concrete, ceramics and glass are used for components that experience compressive loads.

Learning Objectives

After this lesson, students should be able to:

List several common materials used the design and construction of structures.

Describe several factors that engineers consider when selecting materials for the design of a bridge.

Explain the advantages and disadvantages of common materials used in engineering structures (steel and concrete).

Educational Standards

NGSS: Next Generation Science Standards - Science

Common Core State Standards - Math

International Technology and Engineering Educators Association - Technology

State Standards

Suggest an alignment not listed above

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Worksheets and Attachments

Strength of Materials Worksheet (doc)

Strength of Materials Worksheet (pdf)

Strength of Materials Worksheet Answers (doc)

Strength of Materials Worksheet Answers (pdf)

Strength of Materials Math Worksheet (doc)

Strength of Materials Math Worksheet (pdf)

Strength of Materials Math Worksheet Answers (doc)

Strength of Materials Math Worksheet Answers (pdf)

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Answer:

Newton per square meter (N/m2)

Explanation:

Required

Unit of ultimate tensile strength

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U = \frac{Ultimate\ Force}{Area}

The units of force is N (Newton) and the unit of Area is m^2

So, we have:

U = \frac{N}{m^2}

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Answer:

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Explanation:

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The complex power S is given as:

S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA

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The line current I is given as:

I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)}  =100\angle -36.87^o\ A

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V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV

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S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000

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The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

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