Chapter 6, Problem 35P

To determine

Convert the strength of selected materials from MPa to ksi.

Answer to Problem 35P

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

Material	Ultimate strength (MPa)	Ultimate strength (ksi)
Aluminum alloys	$100-550$	$14.5-79.7$
Concrete (compression)	$10-70$	$1.4-10.1$
Stee1
Machine	$550-860$	$79.7-124.7$
Spring	$700-1,900$	$101.5-275.5$
Stainless	$400-1,000$	$58.0-145.0$
Tool	$900$	$130.5$
Structural Steel	$340-830$	$49.3-120.3$
Titanium alloys	$900-1,200$	$130.5-174.0$
Wood (Bending)
Douglas fir	$50-80$	$7.2-11.6$
Oak	$50-100$	$7.2-14.5$
Southern pine	$50-100$	$7.2-14.5$

Explanation of Solution

Given data:

Refer to Problem 6.35 in textbook for the accompanying table.

$1\text{ksi}=1000\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ (1)

Formula used:

Convert N to lbf,

$1\text{N}=224.809\times {10}^{-3}\text{lbf}$

Convert meter to foot,

$1\text{m}=3.2808\text{ft}$

Convert foot to inches,

$1\text{ft}=12\text{in}\text{.}$

Calculation:

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

$1\frac{\text{lbf}}{\text{in}{\text{.}}^2}=\frac{1}{1000}\text{ksi}$ (2)

Case 1:

For Aluminum alloys:

$\begin{array}{c}\text{Ultimate strength}=100\text{MPa}\\ =\text{100}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (3)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (3).

$\text{Ultimate strength}=\text{100}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (4)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (4).

$\begin{array}{c}\text{Ultimate strength}=\text{100}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{100}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =2088594.746\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (5)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (5).

$\begin{array}{c}\text{Ultimate strength}=2088594.746\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{2088594.746}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =14504.13\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (6)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (6).

$\begin{array}{c}\text{Ultimate strength}=14504.13\times \frac{1}{1000}\text{ksi}\\ =\text{14}\text{.5}\text{ksi}\end{array}$ (7)

For Aluminum alloys:

$\begin{array}{c}\text{Ultimate strength}=550\text{MPa}\\ =\text{550}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (8)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (8).

$\text{Ultimate strength}=\text{550}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (9)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (9).

$\begin{array}{c}\text{Ultimate strength}=\text{550}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{550}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =11487271.1\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (10)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (10).

$\begin{array}{c}\text{Ultimate strength}=11487271.1\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{11487271.1}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =79772.716\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (11)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (11).

$\begin{array}{c}\text{Ultimate strength}=79772.716\times \frac{1}{1000}\text{ksi}\\ =\text{79}\text{.7}\text{ksi}\end{array}$ (12)

Case 2:

For Concrete (compression):

$\begin{array}{c}\text{Ultimate strength}=10\text{MPa}\\ =\text{10}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (13)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (13).

$\text{Ultimate strength}=\text{10}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (14)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (14).

$\begin{array}{c}\text{Ultimate strength}=\text{10}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{10}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =208859.475\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (15)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (15).

$\begin{array}{c}\text{Ultimate strength}=208859.475\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{208859.475}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =1450.413\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (16)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (16).

$\begin{array}{c}\text{Ultimate strength}=1450.413\times \frac{1}{1000}\text{ksi}\\ =\text{1}\text{.4}\text{ksi}\end{array}$ (17)

For Concrete (compression):

$\begin{array}{c}\text{Ultimate strength}=70\text{MPa}\\ =7\text{0}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (18)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (18).

$\text{Ultimate strength}=\text{70}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (19)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (19).

$\begin{array}{c}\text{Ultimate strength}=\text{70}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{70}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =1462016.322\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (20)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (20).

$\begin{array}{c}\text{Ultimate strength}=1462016.322\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{1462016.322}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =10152.89\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (21)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (21).

$\begin{array}{c}\text{Ultimate strength}=10152.89\times \frac{1}{1000}\text{ksi}\\ =10.1\text{ksi}\end{array}$ (22)

Case 3:

For Steel-Machine:

$\begin{array}{c}\text{Ultimate strength}=550\text{MPa}\\ =\text{550}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (23)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (23).

$\text{Ultimate strength}=\text{550}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (24)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (24).

$\begin{array}{c}\text{Ultimate strength}=\text{550}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{550}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =11487271.1\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (25)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (25).

$\begin{array}{c}\text{Ultimate strength}=11487271.1\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{11487271.1}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =79772.716\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (26)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (26).

$\begin{array}{c}\text{Ultimate strength}=79772.716\times \frac{1}{1000}\text{ksi}\\ =\text{79}\text{.7}\text{ksi}\end{array}$ (27)

For Steel-Machine:

$\begin{array}{c}\text{Ultimate strength}=860\text{MPa}\\ =\text{860}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (28)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (28).

$\text{Ultimate strength}=\text{860}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (29)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (29).

$\begin{array}{c}\text{Ultimate strength}=\text{860}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{860}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =17961914.82\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (30)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (30).

$\begin{array}{c}\text{Ultimate strength}=17961914.82\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{17961914.82}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =124735.519\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (31)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (31).

$\begin{array}{c}\text{Ultimate strength}=124735.519\times \frac{1}{1000}\text{ksi}\\ =124.\text{7}\text{ksi}\end{array}$ (32)

For Steel-Spring:

$\begin{array}{c}\text{Ultimate strength}=700\text{MPa}\\ =\text{700}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (33)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (33).

$\text{Ultimate strength}=\text{700}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (34)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (34).

$\begin{array}{c}\text{Ultimate strength}=\text{700}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{700}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =14620163.22\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (35)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (35).

$\begin{array}{c}\text{Ultimate strength}=14620163.22\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{14620163.22}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =101528.9113\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (36)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (36).

$\begin{array}{c}\text{Ultimate strength}=101528.9113\times \frac{1}{1000}\text{ksi}\\ =101.5\text{ksi}\end{array}$ (37)

For Steel-Spring:

$\begin{array}{c}\text{Ultimate strength}=1900\text{MPa}\\ =\text{1900}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (38)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (38).

$\text{Ultimate strength}=\text{1900}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (39)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (39).

$\begin{array}{c}\text{Ultimate strength}=\text{1900}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{1900}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =39683300.18\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (40)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (40).

$\begin{array}{c}\text{Ultimate strength}=39683300.18\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{39683300.18}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =275578.474\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (41)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (41).

$\begin{array}{c}\text{Ultimate strength}=275578.474\times \frac{1}{1000}\text{ksi}\\ =275.5\text{ksi}\end{array}$ (42)

For Steel-Stainless:

$\begin{array}{c}\text{Ultimate strength}=400\text{MPa}\\ =\text{400}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (43)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (43).

$\text{Ultimate strength}=\text{400}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (44)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (44).

$\begin{array}{c}\text{Ultimate strength}=\text{400}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{400}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =8354378.985\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (45)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (45).

$\begin{array}{c}\text{Ultimate strength}=8354378.985\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{8354378.985}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =58016.521\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (46)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (46).

$\begin{array}{c}\text{Ultimate strength}=58016.521\times \frac{1}{1000}\text{ksi}\\ =58.0\text{ksi}\end{array}$ (47)

For Steel-Stainless:

$\begin{array}{c}\text{Ultimate strength}=1000\text{MPa}\\ =\text{1000}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (48)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (48).

$\text{Ultimate strength}=\text{1000}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (49)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (49).

$\begin{array}{c}\text{Ultimate strength}=\text{1000}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{1000}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =20885947.46\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (50)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (50).

$\begin{array}{c}\text{Ultimate strength}=20885947.46\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{20885947.46}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =145041.302\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (51)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (51).

$\begin{array}{c}\text{Ultimate strength}=145041.302\times \frac{1}{1000}\text{ksi}\\ =145.0\text{ksi}\end{array}$ (52)

For Steel-Tool:

$\begin{array}{c}\text{Ultimate strength}=900\text{MPa}\\ =\text{900}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (53)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (53).

$\text{Ultimate strength}=\text{900}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (54)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (54).

$\begin{array}{c}\text{Ultimate strength}=\text{900}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{900}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =18797352.72\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (55)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (55).

$\begin{array}{c}\text{Ultimate strength}=18797352.72\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{18797352.72}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =130537.172\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (56)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (56).

$\begin{array}{c}\text{Ultimate strength}=130537.172\times \frac{1}{1000}\text{ksi}\\ =130.5\text{ksi}\end{array}$ (57)

For Steel-Structural Steel:

$\begin{array}{c}\text{Ultimate strength}=340\text{MPa}\\ =\text{340}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (58)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (58).

$\text{Ultimate strength}=\text{340}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (59)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (59).

$\begin{array}{c}\text{Ultimate strength}=\text{340}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{340}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =7101222.137\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (60)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (60).

$\begin{array}{c}\text{Ultimate strength}=7101222.137\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{7101222.137}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =49314.043\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (61)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (61).

$\begin{array}{c}\text{Ultimate strength}=49314.043\times \frac{1}{1000}\text{ksi}\\ =49.3\text{ksi}\end{array}$ (62)

For Steel-Structural Steel:

$\begin{array}{c}\text{Ultimate strength}=830\text{MPa}\\ =\text{830}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (63)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (63).

$\text{Ultimate strength}=\text{830}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (64)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (64).

$\begin{array}{c}\text{Ultimate strength}=\text{830}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{830}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =17335336.39\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (65)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (65).

$\begin{array}{c}\text{Ultimate strength}=17335336.39\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{17335336.39}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =120384.281\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (66)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (66).

$\begin{array}{c}\text{Ultimate strength}=120384.281\times \frac{1}{1000}\text{ksi}\\ =120.3\text{ksi}\end{array}$ (67)

For Steel-Titanium alloys:

$\begin{array}{c}\text{Ultimate strength}=900\text{MPa}\\ =\text{900}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (68)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (68).

$\text{Ultimate strength}=\text{900}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (69)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (69).

$\begin{array}{c}\text{Ultimate strength}=\text{900}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{900}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =18797352.72\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (70)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (70).

$\begin{array}{c}\text{Ultimate strength}=18797352.72\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{18797352.72}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =130537.172\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (71)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (71).

$\begin{array}{c}\text{Ultimate strength}=130537.172\times \frac{1}{1000}\text{ksi}\\ =130.5\text{ksi}\end{array}$ (72)

For Steel-Titanium alloys:

$\begin{array}{c}\text{Ultimate strength}=1200\text{MPa}\\ =\text{1200}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (73)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (73).

$\text{Ultimate strength}=\text{1200}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (74)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (74).

$\begin{array}{c}\text{Ultimate strength}=\text{1200}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{1200}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =25063136.96\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (75)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (75).

$\begin{array}{c}\text{Ultimate strength}=25063136.96\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{25063136.96}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =174049.562\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (76)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (76).

$\begin{array}{c}\text{Ultimate strength}=174049.562\times \frac{1}{1000}\text{ksi}\\ =174.0\text{ksi}\end{array}$ (77)

Case 4:

For Wood (Bending)-Douglas fir:

$\begin{array}{c}\text{Ultimate strength}=50\text{MPa}\\ =\text{50}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (78)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (78).

$\text{Ultimate strength}=\text{50}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (79)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (79).

$\begin{array}{c}\text{Ultimate strength}=\text{50}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{50}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =1044297.373\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (80)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (80).

$\begin{array}{c}\text{Ultimate strength}=1044297.373\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{1044297.373}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =7252.065\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (81)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (81).

$\begin{array}{c}\text{Ultimate strength}=7252.065\times \frac{1}{1000}\text{ksi}\\ =7.2\text{ksi}\end{array}$ (82)

For Wood (Bending)-Douglas fir:

$\begin{array}{c}\text{Ultimate strength}=80\text{MPa}\\ =\text{80}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (83)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (83).

$\text{Ultimate strength}=\text{80}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (84)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (84).

$\begin{array}{c}\text{Ultimate strength}=\text{80}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{80}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =1670875.797\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (85)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (85).

$\begin{array}{c}\text{Ultimate strength}=1670875.797\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{1670875.797}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =11603.304\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (86)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (86).

$\begin{array}{c}\text{Ultimate strength}=11603.304\times \frac{1}{1000}\text{ksi}\\ =11.6\text{ksi}\end{array}$ (87)

For Wood (Bending)-Oak:

$\begin{array}{c}\text{Ultimate strength}=50\text{MPa}\\ =\text{50}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (88)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (88).

$\text{Ultimate strength}=\text{50}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (89)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (89).

$\begin{array}{c}\text{Ultimate strength}=\text{50}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{50}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =1044297.373\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (90)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (90).

$\begin{array}{c}\text{Ultimate strength}=1044297.373\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{1044297.373}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =7252.065\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (91)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (91).

$\begin{array}{c}\text{Ultimate strength}=7252.065\times \frac{1}{1000}\text{ksi}\\ =7.2\text{ksi}\end{array}$ (92)

For Wood (Bending)-Oak:

$\begin{array}{c}\text{Ultimate strength}=100\text{MPa}\\ =\text{100}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (93)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (93).

$\text{Ultimate strength}=\text{100}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (94)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (94).

$\begin{array}{c}\text{Ultimate strength}=\text{100}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{100}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =2088594.746\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (95)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (95).

$\begin{array}{c}\text{Ultimate strength}=2088594.746\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{2088594.746}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =14504.13\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (96)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (96).

$\begin{array}{c}\text{Ultimate strength}=14504.13\times \frac{1}{1000}\text{ksi}\\ =\text{14}\text{.5}\text{ksi}\end{array}$ (97)

For Wood (Bending)-Southern pine:

$\begin{array}{c}\text{Ultimate strength}=50\text{MPa}\\ =\text{50}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (98)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (98).

$\text{Ultimate strength}=\text{50}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (99)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (99).

$\begin{array}{c}\text{Ultimate strength}=\text{50}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{50}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =1044297.373\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (100)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (100).

$\begin{array}{c}\text{Ultimate strength}=1044297.373\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{1044297.373}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =7252.065\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (101)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (101).

$\begin{array}{c}\text{Ultimate strength}=7252.065\times \frac{1}{1000}\text{ksi}\\ =7.2\text{ksi}\end{array}$ (102)

For Wood (Bending)-Southern pine;

$\begin{array}{c}\text{Ultimate strength}=100\text{MPa}\\ =\text{100}\times {\text{10}}^{\text{6}}\text{Pa}\end{array}$ (103)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (103).

$\text{Ultimate strength}=\text{100}\times {\text{10}}^{\text{6}}\frac{\text{N}}{{\text{m}}^{\text{2}}}$ (104)

Substitute the units $224.809\times {10}^{-3}\text{lbf}$ for $1\text{N}$ and $3.2808\text{ft}$ for $1\text{m}$ in equation (104).

$\begin{array}{c}\text{Ultimate strength}=\text{100}\times {\text{10}}^{\text{6}}\frac{\left(224.809\times {10}^{-3}\text{lbf}\right)}{{\left(3.2808\text{ft}\right)}^{\text{2}}}\\ =\frac{\left(\text{100}\times {\text{10}}^{\text{6}}\right)\left(224.809\times {10}^{-3}\right)}{{3.2808}^2}\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\\ =2088594.746\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}\end{array}$ (105)

Substitute the unit $12\text{in}\text{.}$ for $1\text{ft}$ in equation (105).

$\begin{array}{c}\text{Ultimate strength}=2088594.746\frac{\text{lbf}}{{\left(12\text{in}\text{.}\right)}^{\text{2}}}\\ =\frac{2088594.746}{{12}^2}\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\\ =14504.13\frac{\text{lbf}}{\text{in}{\text{.}}^{\text{2}}}\end{array}$ (106)

Substitute the unit $\frac{1}{1000}\text{ksi}$ for $1\frac{\text{lbf}}{\text{in}{\text{.}}^2}$ in equation (106).

$\begin{array}{c}\text{Ultimate strength}=14504.13\times \frac{1}{1000}\text{ksi}\\ =\text{14}\text{.5}\text{ksi}\end{array}$ (107)

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

Table 1

Material	Ultimate strength (MPa)	Ultimate strength (ksi)
Aluminum alloys	$100-550$	$14.5-79.7$
Concrete (compression)	$10-70$	$1.4-10.1$
Stee1
Machine	$550-860$	$79.7-124.7$
Spring	$700-1,900$	$101.5-275.5$
Stainless	$400-1,000$	$58.0-145.0$
Tool	$900$	$130.5$
Structural Steel	$340-830$	$49.3-120.3$
Titanium alloys	$900-1,200$	$130.5-174.0$
Wood (Bending)
Douglas fir	$50-80$	$7.2-11.6$
Oak	$50-100$	$7.2-14.5$
Southern pine	$50-100$	$7.2-14.5$

Conclusion:

Hence, the conversion for the strength of selected materials from MPa to ksi has been explained.