Chapter 3, Problem 3.54P

(a)

To determine

The minimum weight of a $1\text{m}$ long tension member of stainless steel.

Answer to Problem 3.54P

The minimum weight of a $1\text{m}$ long tension member stainless steel is $0.1328kg$ .

Explanation of Solution

Given:

The length of the wire is $l=1\text{m}$ .

The load supported by the member is $P=4\text{kN}$ .

Formula used:

The expression for area of the specimen is given as,

  $A=\frac{P}{S_y}$

Here, $S_y$ is the yield strength and $P$ is the load applied.

The expression for mass of the specimen is given as,

  $m=\rho \times A\times l$

Here, $m$ is the mass of the specimen, and $\rho$ is the density of the specimen.

Calculation:

Refer to table 3.4 “Room temperature mechanical properties and typical applications of annealed stainless steels” value obtained from this table is,

  $S_y=240\text{MPa}$

The expression for area of the specimen can be calculated as,

  $\begin{array}{c}A=\frac{P}{S_y}\\ A=\frac{4000\text{N}}{\left(240000\times {10}^3\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)}\\ =1.66\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The expression for mass of the specimen can be calculated as,

  $\begin{array}{c}m=\rho \times A\times l\\ m=8000\text{kg}/{\text{m}}^3\times 1.66\times {10}^{-5}{\text{m}}^2\times 1\text{m}\\ =0.1328\text{kg}\end{array}$

Conclusions:

Therefore, the minimum weight of a $1\text{m}$ long tension member stainless steel is $0.1328kg$ .

(b)

To determine

The minimum weight of a $1\text{m}$ long tension member of normalized 8620 steel.

Answer to Problem 3.54P

The minimum weight of a $1\text{m}$ long tension member of normalized 8620 steel is $0.0848kg$ .

Explanation of Solution

Given:

The length of the wire is $l=1\text{m}$ .

The load supported by the member is $P=4\text{kN}$ .

Formula used:

The expression for area of the specimen is given as,

  $A=\frac{P}{S_y}$

Here, $S_y$ is the yield strength and $P$ is the load applied.

The expression for mass of the specimen is given as,

  $m=\rho \times A\times l$

Here, $m$ is the mass of the specimen, and $\rho$ is the density of the specimen.

Calculation:

Refer to strength and density data book the value obtained as

  $S_y=360\text{MPa}$

The expression for area of the specimen can be calculated as,

  $\begin{array}{c}A=\frac{P}{S_y}\\ A=\frac{4000\text{N}}{\left(360000\times {10}^3\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)}\\ =1.111\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The expression for mass of the specimen can be calculated as

  $\begin{array}{c}m=\rho \times A\times l\\ m=7850\text{kg}/{\text{m}}^3\times 1.081\times {10}^{-5}{\text{m}}^2\times 1\text{m}\\ =0.0848\text{kg}\end{array}$

Conclusions:

Therefore, the minimum weight of a $1\text{m}$ long tension member normalized 8620 steel is $0.0848kg$ .

(c)

To determine

The minimum weight of a $1\text{m}$ long tension member of rolled 1080 steel.

Answer to Problem 3.54P

The minimum weight of a $1\text{m}$ long tension member of rolled 1080 steel is $0.0526kg$ .

Explanation of Solution

Given:

The length of the wire is $l=1\text{m}$ .

The load supported by the member is $P=4\text{kN}$ .

Formula used:

The expression for area of the specimen is given as,

  $A=\frac{P}{S_y}$

Here, $S_y$ is the yield strength and $P$ is the load applied.

The expression for mass of the specimen is given as,

  $m=\rho \times A\times l$

Here, $m$ is the mass of the specimen, and $\rho$ is the density of the specimen.

Calculation:

Refer to strength and density data book the value obtained as

  $S_y=585\text{MPa}$

The expression for the area of the specimen can be calculated as,

  $\begin{array}{c}A=\frac{P}{S_y}\\ A=\frac{4000\text{N}}{\left(585000\times {10}^3\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)}\\ =0.6837\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The expression for mass of the specimen can be calculated as

  $\begin{array}{c}m=\rho \times A\times l\\ m=7700\text{kg}/{\text{m}}^3\times 0.6837\times {10}^{-5}{\text{m}}^2\times 1\text{m}\\ =0.0526\text{kg}\end{array}$

Conclusions:

Therefore, the minimum weight of a $1\text{m}$ long tension member rolled 1080 steel is $0.0526kg$ .

(d)

To determine

The minimum weight of a $1\text{m}$ long tension member of 5052-O aluminum alloy.

Answer to Problem 3.54P

The minimum weight of a $1\text{m}$ long tension member of 5052-O aluminum alloy is $0.0833kg$ .

Explanation of Solution

Given:

The length of the wire is $l=1\text{m}$ .

The load supported by the member is $P=4\text{kN}$ .

Formula used:

The expression for area of the specimen is given as,

  $A=\frac{P}{S_y}$

Here, $S_y$ is the yield strength and $P$ is the load applied.

The expression for mass of the specimen is given as,

  $m=\rho \times A\times l$

Here, $m$ is the mass of the specimen, and $\rho$ is the density of the specimen.

Calculation:

Refer to table 3.5 “Properties of various aluminum alloy” value obtained from this table is,

  $S_y=215\text{MPa}$

The expression for the area of the specimen can be calculated as,

  $\begin{array}{c}A=\frac{P}{S_y}\\ A=\frac{4000\text{N}}{\left(215000\times {10}^3\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)}\\ =1.860\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The expression for mass of the specimen can be calculated as,

  $\begin{array}{c}m=\rho \times A\times l\\ m=2710\text{kg}/{\text{m}}^3\times 3.0769\times {10}^{-5}{\text{m}}^2\times 1\text{m}\\ =0.0833\text{kg}\end{array}$

Conclusions:

Therefore, the minimum weight of a $1\text{m}$ long tension member 5052-O aluminum alloy is $0.0833kg$ .

(e)

To determine

The minimum weight of a $1\text{m}$ long tension member of AZ31B-F magnesium.

Answer to Problem 3.54P

The minimum weight of a $1\text{m}$ long tension member of AZ31B-F magnesium is $0.7555kg$ .

Explanation of Solution

Given:

The length of the wire is $l=1\text{m}$ .

The load supported by the member is $P=4\text{kN}$ .

Formula used:

The expression for area of the specimen is given as,

  $A=\frac{P}{S_y}$

Here, $S_y$ is the yield strength and $P$ is the load applied.

The expression for mass of the specimen is given as,

  $m=\rho \times A\times l$

Here, $m$ is the mass of the specimen, and $\rho$ is the density of the specimen.

Calculation:

Refer to table 3.5 “Properties of various aluminum alloy” value obtained from this table is,

  $S_y=215\text{MPa}$

The expression for the area of the specimen can be calculated as,

  $\begin{array}{c}A=\frac{P}{S_y}\\ A=\frac{4000\text{N}}{\left(290000\times {10}^3\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)}\\ =1.379\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The expression for mass of the specimen can be calculated as,

  $\begin{array}{c}m=\rho \times A\times l\\ m=8500\text{kg}/{\text{m}}^3\times 8.888\times {10}^{-5}{\text{m}}^2\times 1\text{m}\\ =0.7555\text{kg}\end{array}$

Conclusions:

Therefore, the minimum weight of a $1\text{m}$ long tension member AZ31B-F magnesium is $0.7555kg$ .

(f)

To determine

The minimum weight of a $1\text{m}$ long tension member of pure copper.

Answer to Problem 3.54P

The minimum weight of a $1\text{m}$ long tension member of pure copper is $0.5194kg$

Explanation of Solution

Given:

The length of the wire is $l=1\text{m}$ .

The load supported by the member is $P=4\text{kN}$ .

Formula used:

The expression for the area of the specimen is given as,

  $A=\frac{P}{S_y}$

Here, $S_y$ is the yield strength and $P$ is the load applied.

The expression for mass of the specimen is given as,

  $m=\rho \times A\times l$

Here, $m$ is the mass of the specimen, and $\rho$ is the density of the specimen.

Calculation:

The expression for the area of the specimen can be calculated as,

  $\begin{array}{c}A=\frac{P}{S_y}\\ A=\frac{4000\text{N}}{\left(70000\times {10}^3\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)}\\ =5.7142\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The expression for mass of the specimen can be calculated as,

  $\begin{array}{c}m=\rho \times A\times l\\ m=8960\text{kg}/{\text{m}}^3\times 5.797\times {10}^{-5}{\text{m}}^2\times 1\text{m}\\ =0.5194\text{kg}\end{array}$

Conclusions:

Therefore, the minimum weight of a $1\text{m}$ long tension member pure copper is $0.5194kg$ .