Engineering Fundamentals
Engineering Fundamentals
6th Edition
ISBN: 9780357112144
Author: Saeed Moaveni
Publisher: MISC PUBS
bartleby

Concept explainers

Question
Book Icon
Chapter 6, Problem 2P
To determine

Convert the given information in the accompanying table from U.S. Customary units to SI units.

Expert Solution & Answer
Check Mark

Answer to Problem 2P

The conversion of given information in the accompanying table from U.S. Customary units to SI units are tabulated as follows:

Convert from U.S. Customary UnitsTo SI Units
65mileshr104.6kmhr and 29ms
60000Btuhr1760W
120lbmft31922.25kgft3
30lbfin2206.8kPa
200lbm90.7kg
200lbf890N

Explanation of Solution

Given data:

Refer to Problem 6.2 in textbook for the accompanying table.

Formula used:

Convert hr to s,

1hr=3600s

Convert foot to meter,

1ft=0.3048m

Convert inches to foot,

1in=112ft

Convert mile to foot,

1mile=5280ft

Convert km to m,

1km=1000m

Convert W to ftlbfs,

1W=0.7375ftlbfs

Convert Btu to ftlbf,

1Btu=778.17ftlbf

Convert ftlbfs to hp,

1ftlbfs=1550hp

Convert lbm to kg,

1lbm=0.4536kg

Convert lbf to N,

1lbf=4.448N

Calculation:

Case 1:

Convert the given value of U.S. Customary units,

65mileshr=65(5280ft)hr             [1mile=5280ft]=343200fthr (1)

Substitute the unit 0.3048m for 1ft in equation (1).

65mileshr=343200(0.3048m)hr=343200×0.3048mhr=104607.36mhr (2)

Reduce the equation as follows,

65mileshr=104607.36(11000km)hr[1m=11000km]=104.6kmhr (3)

Substitute the unit 3600s for 1hr in equation (2).

65mileshr=104607.36m(3600s)=104607.363600ms=29ms (4)

From equation (3) and equation (4), the conversion of given value of U.S. Customary units to SI units are 104.6kmhr.and 29ms.

Case 2:

Convert the given value of U.S. Customary units,

60000Btuhr=60000(778.17ftlbf)hr[1Btu=778.17ftlbf]=60000×778.17ftlbfhr=46690200ftlbf(3600s)[1hr=3600s]=12969.5ftlbfs

Reduce the equation as follows,

60000Btuhr=12969.5×10.7375W[1ftlbfs=10.7375W]=17585.763W (5)

From equation (5), the conversion of given value of U.S. Customary units to SI units is 17585.763W.

Case 3:

Convert the given value of U.S. Customary units,

120lbmft3=120(0.4536kg)ft3[1lbm=0.4536kg]=120×0.4536kgft3=54.432kgft3 (6)

Substitute the unit 0.3048m for 1ft in equation (6).

120lbmft3=54.432kg(0.3048m)3=54.4320.30483kgft3=1922.25kgft3 (7)

From equation (7), the conversion of given value of U.S. Customary units to SI units is 1922.25kgft3.

Case 4:

Convert the given value of U.S. Customary units,

30lbfin2=30(4.448N)in2[1lbf=4.448N]=133.44N(112ft)2[1in=112ft]=133.44×122Nft2=19215.36Nft2 (8)

Substitute the unit 0.3048m for 1ft in equation (8).

30lbfin2=19215.36N(0.3048m)2=19215.360.30482Nm2=206832.414Nm2 (9)

Substitute the unit Pa for Nm2 in equation (9).

30lbfin2=206832.414Pa=206832.414×11000kPa[1Pa=11000kPa]=206.8kPa (10)

From equation (10), the conversion of given value of U.S. Customary units to SI units is 206.8kPa.

Case 5:

Convert the given value of U.S. Customary units,

200lbm=200×0.4536kg[1lbm=0.4536kg]=90.7kg (11)

From equation (11), the conversion of given value of U.S. Customary units to SI units is 90.7kg.

Case 6:

Convert the given value of U.S. Customary units,

200lbf=200×4.448N[1lbf=4.448N]=890N (12)

From equation (12), the conversion of given value of U.S. Customary units to SI units is 890N.

Conclusion:

Hence, the conversion of given information in the accompanying table from U.S. Customary units to SI units has been explained.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
I really need help on b
WU Example 6 For the exterior transverse frame of the flat slab floor shown in figure, and by using the Direct Design Method, find: a. Longitudinal distribution of the total static moment at factored loads. b. Lateral distribution of moment at exterior panel (column and middle strip moments at exterior support) D= 6.5 kN/m² L= 5.0 kN/m² تفكر وکھل flat slap ما لا يوجد bon حامل . 3000 1000 5000 160 + 2000+ +2000+ 5000 2608 300 2000 Drop Panal السعف
Example 4 For the transverse interior frame (Frame C) of the flat plate floor with edge beams shown in Figure, by using the Direct Design Method, find: 1) Longitudinal distribution of total static moment at factored loads. 2) Lateral distribution of moment at interior panel (column and middle strip moments atnegative and positive moments). 3) Lateral distribution of moment at exterior panel (column and middle strip moments atnegative and positive moments). Plat 5000-5000 5000 -Frame C لا بوجود deen 0009 0009 Slab thickness = 180 mm, d = 150 mm q₁ = 16.0 kN/m² All edge beams = 250x 500 mm All columns = 500x 500 mm 6000

Chapter 6 Solutions

Engineering Fundamentals

Knowledge Booster
Background pattern image
Civil Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Engineering Fundamentals: An Introduction to Engi...
Civil Engineering
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Cengage Learning
Text book image
Fundamentals Of Construction Estimating
Civil Engineering
ISBN:9781337399395
Author:Pratt, David J.
Publisher:Cengage,