EBK ENGINEERING FUNDAMENTALS: AN INTROD
EBK ENGINEERING FUNDAMENTALS: AN INTROD
5th Edition
ISBN: 9780100543409
Author: MOAVENI
Publisher: YUZU
bartleby

Concept explainers

Question
Book Icon
Chapter 6, Problem 35P
To determine

Convert the strength of selected materials from MPa to ksi.

Expert Solution & Answer
Check Mark

Answer to Problem 35P

The conversion for the strength of selected materials given in accompanying table from MPa to ksi as follows:

MaterialUltimate strength (MPa)Ultimate strength (ksi)
Aluminum alloys10055014.579.7

Concrete

(compression)

1070

1.410.1

Stee1  
Machine55086079.7124.7
Spring7001,900101.5275.5
Stainless4001,00058.0145.0
Tool900130.5
Structural Steel34083049.3120.3
Titanium alloys9001,200130.5174.0
Wood (Bending)  
Douglas fir50807.211.6
Oak501007.214.5
Southern pine501007.214.5

Explanation of Solution

Given data:

Refer to Problem 6.35 in textbook for the accompanying table.

1ksi=1000lbfin.2 (1)

Formula used:

Convert N to lbf,

1N=224.809×103lbf

Convert meter to foot,

1m=3.2808ft

Convert foot to inches,

1ft=12in.

Calculation:

Rearrange the equation (1) for conversion of unit as follows,

1lbfin.2=11000ksi (2)

Case 1:

For Aluminum alloys:

Ultimate strength=100MPa=100×106Pa (3)

Substitute the unit Nm2 for Pa in equation (3).

Ultimate strength=100×106Nm2 (4)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (4).

Ultimate strength=100×106(224.809×103lbf)(3.2808ft)2=(100×106)(224.809×103)3.28082lbfft2=2088594.746lbfft2 (5)

Substitute the unit 12in. for 1ft in equation (5).

Ultimate strength=2088594.746lbf(12in.)2=2088594.746122lbfin.2=14504.13lbfin.2 (6)

Substitute the unit 11000ksi for 1lbfin.2 in equation (6).

Ultimate strength=14504.13×11000ksi=14.5ksi (7)

For Aluminum alloys:

Ultimate strength=550MPa=550×106Pa (8)

Substitute the unit Nm2 for Pa in equation (8).

Ultimate strength=550×106Nm2 (9)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (9).

Ultimate strength=550×106(224.809×103lbf)(3.2808ft)2=(550×106)(224.809×103)3.28082lbfft2=11487271.1lbfft2 (10)

Substitute the unit 12in. for 1ft in equation (10).

Ultimate strength=11487271.1lbf(12in.)2=11487271.1122lbfin.2=79772.716lbfin.2 (11)

Substitute the unit 11000ksi for 1lbfin.2 in equation (11).

Ultimate strength=79772.716×11000ksi=79.7ksi (12)

Case 2:

For Concrete (compression):

Ultimate strength=10MPa=10×106Pa (13)

Substitute the unit Nm2 for Pa in equation (13).

Ultimate strength=10×106Nm2 (14)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (14).

Ultimate strength=10×106(224.809×103lbf)(3.2808ft)2=(10×106)(224.809×103)3.28082lbfft2=208859.475lbfft2 (15)

Substitute the unit 12in. for 1ft in equation (15).

Ultimate strength=208859.475lbf(12in.)2=208859.475122lbfin.2=1450.413lbfin.2 (16)

Substitute the unit 11000ksi for 1lbfin.2 in equation (16).

Ultimate strength=1450.413×11000ksi=1.4ksi (17)

For Concrete (compression):

Ultimate strength=70MPa=70×106Pa (18)

Substitute the unit Nm2 for Pa in equation (18).

Ultimate strength=70×106Nm2 (19)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (19).

Ultimate strength=70×106(224.809×103lbf)(3.2808ft)2=(70×106)(224.809×103)3.28082lbfft2=1462016.322lbfft2 (20)

Substitute the unit 12in. for 1ft in equation (20).

Ultimate strength=1462016.322lbf(12in.)2=1462016.322122lbfin.2=10152.89lbfin.2 (21)

Substitute the unit 11000ksi for 1lbfin.2 in equation (21).

Ultimate strength=10152.89×11000ksi=10.1ksi (22)

Case 3:

For Steel-Machine:

Ultimate strength=550MPa=550×106Pa (23)

Substitute the unit Nm2 for Pa in equation (23).

Ultimate strength=550×106Nm2 (24)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (24).

Ultimate strength=550×106(224.809×103lbf)(3.2808ft)2=(550×106)(224.809×103)3.28082lbfft2=11487271.1lbfft2 (25)

Substitute the unit 12in. for 1ft in equation (25).

Ultimate strength=11487271.1lbf(12in.)2=11487271.1122lbfin.2=79772.716lbfin.2 (26)

Substitute the unit 11000ksi for 1lbfin.2 in equation (26).

Ultimate strength=79772.716×11000ksi=79.7ksi (27)

For Steel-Machine:

Ultimate strength=860MPa=860×106Pa (28)

Substitute the unit Nm2 for Pa in equation (28).

Ultimate strength=860×106Nm2 (29)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (29).

Ultimate strength=860×106(224.809×103lbf)(3.2808ft)2=(860×106)(224.809×103)3.28082lbfft2=17961914.82lbfft2 (30)

Substitute the unit 12in. for 1ft in equation (30).

Ultimate strength=17961914.82lbf(12in.)2=17961914.82122lbfin.2=124735.519lbfin.2 (31)

Substitute the unit 11000ksi for 1lbfin.2 in equation (31).

Ultimate strength=124735.519×11000ksi=124.7ksi (32)

For Steel-Spring:

Ultimate strength=700MPa=700×106Pa (33)

Substitute the unit Nm2 for Pa in equation (33).

Ultimate strength=700×106Nm2 (34)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (34).

Ultimate strength=700×106(224.809×103lbf)(3.2808ft)2=(700×106)(224.809×103)3.28082lbfft2=14620163.22lbfft2 (35)

Substitute the unit 12in. for 1ft in equation (35).

Ultimate strength=14620163.22lbf(12in.)2=14620163.22122lbfin.2=101528.9113lbfin.2 (36)

Substitute the unit 11000ksi for 1lbfin.2 in equation (36).

Ultimate strength=101528.9113×11000ksi=101.5ksi (37)

For Steel-Spring:

Ultimate strength=1900MPa=1900×106Pa (38)

Substitute the unit Nm2 for Pa in equation (38).

Ultimate strength=1900×106Nm2 (39)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (39).

Ultimate strength=1900×106(224.809×103lbf)(3.2808ft)2=(1900×106)(224.809×103)3.28082lbfft2=39683300.18lbfft2 (40)

Substitute the unit 12in. for 1ft in equation (40).

Ultimate strength=39683300.18lbf(12in.)2=39683300.18122lbfin.2=275578.474lbfin.2 (41)

Substitute the unit 11000ksi for 1lbfin.2 in equation (41).

Ultimate strength=275578.474×11000ksi=275.5ksi (42)

For Steel-Stainless:

Ultimate strength=400MPa=400×106Pa (43)

Substitute the unit Nm2 for Pa in equation (43).

Ultimate strength=400×106Nm2 (44)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (44).

Ultimate strength=400×106(224.809×103lbf)(3.2808ft)2=(400×106)(224.809×103)3.28082lbfft2=8354378.985lbfft2 (45)

Substitute the unit 12in. for 1ft in equation (45).

Ultimate strength=8354378.985lbf(12in.)2=8354378.985122lbfin.2=58016.521lbfin.2 (46)

Substitute the unit 11000ksi for 1lbfin.2 in equation (46).

Ultimate strength=58016.521×11000ksi=58.0ksi (47)

For Steel-Stainless:

Ultimate strength=1000MPa=1000×106Pa (48)

Substitute the unit Nm2 for Pa in equation (48).

Ultimate strength=1000×106Nm2 (49)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (49).

Ultimate strength=1000×106(224.809×103lbf)(3.2808ft)2=(1000×106)(224.809×103)3.28082lbfft2=20885947.46lbfft2 (50)

Substitute the unit 12in. for 1ft in equation (50).

Ultimate strength=20885947.46lbf(12in.)2=20885947.46122lbfin.2=145041.302lbfin.2 (51)

Substitute the unit 11000ksi for 1lbfin.2 in equation (51).

Ultimate strength=145041.302×11000ksi=145.0ksi (52)

For Steel-Tool:

Ultimate strength=900MPa=900×106Pa (53)

Substitute the unit Nm2 for Pa in equation (53).

Ultimate strength=900×106Nm2 (54)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (54).

Ultimate strength=900×106(224.809×103lbf)(3.2808ft)2=(900×106)(224.809×103)3.28082lbfft2=18797352.72lbfft2 (55)

Substitute the unit 12in. for 1ft in equation (55).

Ultimate strength=18797352.72lbf(12in.)2=18797352.72122lbfin.2=130537.172lbfin.2 (56)

Substitute the unit 11000ksi for 1lbfin.2 in equation (56).

Ultimate strength=130537.172×11000ksi=130.5ksi (57)

For Steel-Structural Steel:

Ultimate strength=340MPa=340×106Pa (58)

Substitute the unit Nm2 for Pa in equation (58).

Ultimate strength=340×106Nm2 (59)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (59).

Ultimate strength=340×106(224.809×103lbf)(3.2808ft)2=(340×106)(224.809×103)3.28082lbfft2=7101222.137lbfft2 (60)

Substitute the unit 12in. for 1ft in equation (60).

Ultimate strength=7101222.137lbf(12in.)2=7101222.137122lbfin.2=49314.043lbfin.2 (61)

Substitute the unit 11000ksi for 1lbfin.2 in equation (61).

Ultimate strength=49314.043×11000ksi=49.3ksi (62)

For Steel-Structural Steel:

Ultimate strength=830MPa=830×106Pa (63)

Substitute the unit Nm2 for Pa in equation (63).

Ultimate strength=830×106Nm2 (64)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (64).

Ultimate strength=830×106(224.809×103lbf)(3.2808ft)2=(830×106)(224.809×103)3.28082lbfft2=17335336.39lbfft2 (65)

Substitute the unit 12in. for 1ft in equation (65).

Ultimate strength=17335336.39lbf(12in.)2=17335336.39122lbfin.2=120384.281lbfin.2 (66)

Substitute the unit 11000ksi for 1lbfin.2 in equation (66).

Ultimate strength=120384.281×11000ksi=120.3ksi (67)

For Steel-Titanium alloys:

Ultimate strength=900MPa=900×106Pa (68)

Substitute the unit Nm2 for Pa in equation (68).

Ultimate strength=900×106Nm2 (69)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (69).

Ultimate strength=900×106(224.809×103lbf)(3.2808ft)2=(900×106)(224.809×103)3.28082lbfft2=18797352.72lbfft2 (70)

Substitute the unit 12in. for 1ft in equation (70).

Ultimate strength=18797352.72lbf(12in.)2=18797352.72122lbfin.2=130537.172lbfin.2 (71)

Substitute the unit 11000ksi for 1lbfin.2 in equation (71).

Ultimate strength=130537.172×11000ksi=130.5ksi (72)

For Steel-Titanium alloys:

Ultimate strength=1200MPa=1200×106Pa (73)

Substitute the unit Nm2 for Pa in equation (73).

Ultimate strength=1200×106Nm2 (74)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (74).

Ultimate strength=1200×106(224.809×103lbf)(3.2808ft)2=(1200×106)(224.809×103)3.28082lbfft2=25063136.96lbfft2 (75)

Substitute the unit 12in. for 1ft in equation (75).

Ultimate strength=25063136.96lbf(12in.)2=25063136.96122lbfin.2=174049.562lbfin.2 (76)

Substitute the unit 11000ksi for 1lbfin.2 in equation (76).

Ultimate strength=174049.562×11000ksi=174.0ksi (77)

Case 4:

For Wood (Bending)-Douglas fir:

Ultimate strength=50MPa=50×106Pa (78)

Substitute the unit Nm2 for Pa in equation (78).

Ultimate strength=50×106Nm2 (79)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (79).

Ultimate strength=50×106(224.809×103lbf)(3.2808ft)2=(50×106)(224.809×103)3.28082lbfft2=1044297.373lbfft2 (80)

Substitute the unit 12in. for 1ft in equation (80).

Ultimate strength=1044297.373lbf(12in.)2=1044297.373122lbfin.2=7252.065lbfin.2 (81)

Substitute the unit 11000ksi for 1lbfin.2 in equation (81).

Ultimate strength=7252.065×11000ksi=7.2ksi (82)

For Wood (Bending)-Douglas fir:

Ultimate strength=80MPa=80×106Pa (83)

Substitute the unit Nm2 for Pa in equation (83).

Ultimate strength=80×106Nm2 (84)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (84).

Ultimate strength=80×106(224.809×103lbf)(3.2808ft)2=(80×106)(224.809×103)3.28082lbfft2=1670875.797lbfft2 (85)

Substitute the unit 12in. for 1ft in equation (85).

Ultimate strength=1670875.797lbf(12in.)2=1670875.797122lbfin.2=11603.304lbfin.2 (86)

Substitute the unit 11000ksi for 1lbfin.2 in equation (86).

Ultimate strength=11603.304×11000ksi=11.6ksi (87)

For Wood (Bending)-Oak:

Ultimate strength=50MPa=50×106Pa (88)

Substitute the unit Nm2 for Pa in equation (88).

Ultimate strength=50×106Nm2 (89)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (89).

Ultimate strength=50×106(224.809×103lbf)(3.2808ft)2=(50×106)(224.809×103)3.28082lbfft2=1044297.373lbfft2 (90)

Substitute the unit 12in. for 1ft in equation (90).

Ultimate strength=1044297.373lbf(12in.)2=1044297.373122lbfin.2=7252.065lbfin.2 (91)

Substitute the unit 11000ksi for 1lbfin.2 in equation (91).

Ultimate strength=7252.065×11000ksi=7.2ksi (92)

For Wood (Bending)-Oak:

Ultimate strength=100MPa=100×106Pa (93)

Substitute the unit Nm2 for Pa in equation (93).

Ultimate strength=100×106Nm2 (94)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (94).

Ultimate strength=100×106(224.809×103lbf)(3.2808ft)2=(100×106)(224.809×103)3.28082lbfft2=2088594.746lbfft2 (95)

Substitute the unit 12in. for 1ft in equation (95).

Ultimate strength=2088594.746lbf(12in.)2=2088594.746122lbfin.2=14504.13lbfin.2 (96)

Substitute the unit 11000ksi for 1lbfin.2 in equation (96).

Ultimate strength=14504.13×11000ksi=14.5ksi (97)

For Wood (Bending)-Southern pine:

Ultimate strength=50MPa=50×106Pa (98)

Substitute the unit Nm2 for Pa in equation (98).

Ultimate strength=50×106Nm2 (99)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (99).

Ultimate strength=50×106(224.809×103lbf)(3.2808ft)2=(50×106)(224.809×103)3.28082lbfft2=1044297.373lbfft2 (100)

Substitute the unit 12in. for 1ft in equation (100).

Ultimate strength=1044297.373lbf(12in.)2=1044297.373122lbfin.2=7252.065lbfin.2 (101)

Substitute the unit 11000ksi for 1lbfin.2 in equation (101).

Ultimate strength=7252.065×11000ksi=7.2ksi (102)

For Wood (Bending)-Southern pine;

Ultimate strength=100MPa=100×106Pa (103)

Substitute the unit Nm2 for Pa in equation (103).

Ultimate strength=100×106Nm2 (104)

Substitute the units 224.809×103lbf for 1N and 3.2808ft for 1m in equation (104).

Ultimate strength=100×106(224.809×103lbf)(3.2808ft)2=(100×106)(224.809×103)3.28082lbfft2=2088594.746lbfft2 (105)

Substitute the unit 12in. for 1ft in equation (105).

Ultimate strength=2088594.746lbf(12in.)2=2088594.746122lbfin.2=14504.13lbfin.2 (106)

Substitute the unit 11000ksi for 1lbfin.2 in equation (106).

Ultimate strength=14504.13×11000ksi=14.5ksi (107)

Thus, the conversion for the strength of selected materials from MPa to ksi is tabulated in Table 1.

Table 1

MaterialUltimate strength (MPa)Ultimate strength (ksi)
Aluminum alloys10055014.579.7

Concrete

(compression)

1070

1.410.1

Stee1  
Machine55086079.7124.7
Spring7001,900101.5275.5
Stainless4001,00058.0145.0
Tool900130.5
Structural Steel34083049.3120.3
Titanium alloys9001,200130.5174.0
Wood (Bending)  
Douglas fir50807.211.6
Oak501007.214.5
Southern pine501007.214.5

Conclusion:

Hence, the conversion for the strength of selected materials from MPa to ksi 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
Consider, M people (aka pax) who want to travel by car from O to D. They all start working at D at Q (e.g., Q-8am). If a person departs at time t, assume the time needed to go from O to D is given by c(t)=A+Bx(t), where x(t) is the flow of people departing at time t [car/unit of time]. In addition, a is the penalty for being early at work (E(t) is how early the person arrived when departing at time t), and ẞ is the penalty for being late at work (L(t) is how late the person arrived when departing at time t). Assume 0 < a < 1 < ß. Further assume the departure time choice problem under the equilibrium conditions. Prove that the arrival time of people who depart when most of the M people start their trips is equal to Q.
a. A b. A 3. Sketch normal depth, critical depth and the water surface profile. Assume at A and B the water is flowing at normal depth. Label and Identify all curves (i.e., M1, S2, etc.) Yn > Ye Уп Ye Уп> Ус y
2. Design a trapezoidal ditch to carry Q = 1000 cfs. The ditch will be a lined channel, gravel bottom with sides shown below on a slope of S = 0.009. The side slopes of the 20-ft wide ditch will be 1 vertical to 3 horizontal. a) Determine the normal depth of flow (yn) using the Normal value for Manning's n. b) If freeboard requirements are 25% of the normal depth, how deep should the ditch be constructed? c) Classify the slope. T b Уп Z 1 Yn + FB

Chapter 6 Solutions

EBK ENGINEERING FUNDAMENTALS: AN INTROD

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
Solid Waste Engineering
Civil Engineering
ISBN:9781305635203
Author:Worrell, William A.
Publisher:Cengage Learning,
Text book image
Construction Materials, Methods and Techniques (M...
Civil Engineering
ISBN:9781305086272
Author:William P. Spence, Eva Kultermann
Publisher:Cengage Learning
Text book image
Materials Science And Engineering Properties
Civil Engineering
ISBN:9781111988609
Author:Charles Gilmore
Publisher:Cengage Learning
Text book image
Fundamentals Of Construction Estimating
Civil Engineering
ISBN:9781337399395
Author:Pratt, David J.
Publisher:Cengage,
Text book image
Fundamentals of Geotechnical Engineering (MindTap...
Civil Engineering
ISBN:9781305635180
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Cengage Learning