Chapter 2, Problem 2.4.4P

A horizontal rigid bar ABC is pinned at end A and supported by two cables at points B and C. A vertical load P = 10 kN acts at end C of the bar. The two cables are made of steel with a modulus elasticity E = 200 GPa and have the same cross-sectional area. Calculate the minimum cross-sectional area of each cable if the yield stress of the cable is 400 MPa and the factor of safely is 2.0. Consider load P only; ignore the weight of bar ABC and the cables.

  

To determine

The minimum cross-section area of each cable.

Answer to Problem 2.4.4P

The minimum cross-section area of each cable is $70.3\times {10}^{-6}{\text{m}}^2$ .

Explanation of Solution

The vertical load is $10\text{kN}$ , modulus of elasticity is $200\text{GPa}$ , yield stress of the cable is $400\text{MPa}$and factor of safety is $2.0$ .

The below figure represents the free body diagram of the cable.

  

Figure-(1)

Here, the length of the portion AB is $a$ , length of the portion BC is $b$and length of the AD is $h$ .

Write the expression for the length of the cable 1.

  $L_1=\sqrt{a^2+h^2}$…… (I)

Here, length of the cable 1 is $L_1$ .

Write the expression for the length of the cable 2.

  $L_2=\sqrt{{\left(a+b\right)}^2+h^2}$…… (II)

Here, length of the cable 2 is $L_2$ .

Write the expression for the angle for cable 1 which makes from horizontal.

  ${\theta }_1={\tan }^{-1}\left(\frac{h}{a}\right)$…… (III)

Write the expression for the angle for cable 1 which makes from horizontal.

  ${\theta }_2={\tan }^{-1}\left(\frac{h}{a+b}\right)$…… (IV)

Write the expression for the moment about the point A.

  $T_1\sin {\theta }_1\left(a\right)+T_2\sin {\theta }_2\left(a+b\right)-P\left(a+b\right)=0$…… (V)

Here, the tension in cable 1 is $T_1$and tension in cable 2 is $T_2$ .

The below figure represents the displacement diagram.

  

Figure-(2)

Here, displacement for cable 1 is ${\delta }_1$ , displacement for cable 2 is ${\delta }_2$ , the displacement of B to B’ is ${\Delta }_1$and the displacement of C to C’ is ${\Delta }_2$ .

Write the expression for the displacement assuming this is a small displacement.

  $\frac{a}{{\Delta }_1}=\frac{a+b}{{\Delta }_2}$…… (VI)

Write the expression for the displacement of portion AB.

  ${\Delta }_1=\frac{{\delta }_1}{\sin {\theta }_1}$…… (VII)

Write the expression for the displacement of portion AC.

  ${\Delta }_2=\frac{{\delta }_2}{\sin {\theta }_2}$ …… (VIII)

Write the expression for the deflection in cable 1.

  ${\delta }_1=\frac{T_1{L}_1}{AE}$…… (IX)

Here, area of the cross-section is $A$ , modulus of elasticity is $E$and the deflection in cable 1is ${\delta }_1$ .

Write the expression for the deflection in cable 2.

  ${\delta }_2=\frac{T_2{L}_2}{AE}$…… (X)

Here, the deflection in cable 2 is ${\delta }_2$ .

Write the expression for the stress equation considering factor of safety for cable 1.

  $\frac{{\sigma }_y}{fs}=\frac{T_1}{A}$ …… (XI)

Here, the factor of safety is $fs$ , yield stress is ${\sigma }_y$and area is $A$ .

Write the expression for the stress equation considering factor of safety for cable 2.

  $\frac{{\sigma }_y}{fs}=\frac{T_2}{A}$…… (XII)

Calculation:

Substitute $1.5\text{m}$for $a$and $1.5\text{m}$for $h$in Equation (I).

  $\begin{array}{c}{L}_1=\sqrt{{\left(1.5\text{m}\right)}^2+{\left(1.5\text{m}\right)}^2}\\ =\sqrt{2.25{\text{m}}^2+2.25{\text{m}}^2}\\ =\sqrt{4.50{\text{m}}^2}\\ =2.12\text{m}\end{array}$

Substitute $1.5\text{m}$for $h$and $3\text{m}$for $\left(a+b\right)$in Equation (II).

  $\begin{array}{c}{L}_2=\sqrt{{\left(3\text{m}\right)}^2+{\left(1.5\text{m}\right)}^2}\\ =\sqrt{11.25{\text{m}}^2}\\ =3.35\text{m}\end{array}$

Substitute $1.5\text{m}$for $a$and $1.5\text{m}$for $h$in Equation (III).

  $\begin{array}{c}{\theta }_1={\tan }^{-1}\left(\frac{1.5\text{m}}{1.5\text{m}}\right)\\ ={\tan }^{-1}\left(1\right)\\ =45°\end{array}$

Substitute $1.5\text{m}$for $h$and $3\text{m}$for $\left(a+b\right)$in Equation (IV).

  $\begin{array}{c}{\theta }_2={\tan }^{-1}\left(\frac{1\text{m}}{3\text{m}}\right)\\ ={\tan }^{-1}\left(0.333\text{m}\right)\\ =26.56°\end{array}$

Substitute $10\text{kN}$for $P$ , $1.5\text{m}$for $h$ , $3\text{m}$for $\left(a+b\right)$ , $45°$for ${\theta }_1$and $26.56°$for ${\theta }_2$in Equation (V).

  $\begin{array}{l}{T}_1\left(\sin 45°\right)\left(1.5\text{m}\right)+T_2\sin \left(26.56°\right)\left(3\text{m}\right)-\left(10\text{kN}\left(3\text{m}\right)\right)=0\\ T_1\left(\sin 45°\right)\left(1.5\text{m}\right)+T_2\sin \left(26.56°\right)\left(3\text{m}\right)-\left(10\text{kN}\left(\frac{1000\text{N}}{1\text{kN}}\right)\left(3\text{m}\right)\right)=0\end{array}$

  $\left(1.0607\text{m}\right)T_1+\left(1.3416\text{m}\right)T_2=30000\text{N}\cdot \text{m}$ …… (XIII)

Substitute $1.5\text{m}$for $h$and $3\text{m}$for $\left(a+b\right)$in Equation (VI).

  $\frac{1.5\text{m}}{{\Delta }_1}=\frac{3\text{m}}{{\Delta }_2}$

  ${\Delta }_2=2{\Delta }_1$…… (XIV)

Substitute $\frac{{\delta }_1}{\sin {\theta }_1}$for ${\Delta }_1$and $\frac{{\delta }_2}{\sin {\theta }_2}$for ${\Delta }_2$in Equation (XIV).

  $\frac{{\delta }_2}{\sin {\theta }_2}=2\frac{{\delta }_1}{\sin {\theta }_1}$…… (XV)

Substitute $\frac{T_1{L}_1}{AE}$for ${\delta }_1$and $\frac{T_2{L}_2}{AE}$for ${\delta }_2$in Equation (XV).

  $\frac{\frac{T_2{L}_2}{AE}}{\sin {\theta }_2}=2\frac{\frac{T_1{L}_1}{AE}}{\sin {\theta }_1}$

  $\frac{T_2{L}_2}{\sin {\theta }_2}=2\frac{T_1{L}_1}{\sin {\theta }_1}$…… (XVI)

Substitute $2.12\text{m}$for $L_1$ , $3.35\text{m}$for $L_2$ , $45°$for ${\theta }_1$and $26.56°$for ${\theta }_2$in Equation (XVI).

  $\begin{array}{l}\frac{T_2\left(3.35\text{m}\right)}{\sin \left(26.56°\right)}=2\frac{T_1\left(2.12\text{m}\right)}{\sin \left(45°\right)}\\ \left(7.4921\text{m}\right)T_2=\left(5.996\text{m}\right)T_1\end{array}$

  $T_2=0.80035T_1$…… (XVII)

Substitute $0.80035T_1$for $T_2$in Equation (XIII).

  $\begin{array}{l}\left(1.0607\text{m}\right)T_1+\left(1.3416\text{m}\right)\left(0.80035T_1\right)=30000\text{N}\cdot \text{m}\\ \left(1.0607\text{m}\right)T_1+\left(1.073749\text{m}\left(T_1\right)\right)=30000\text{N}\cdot \text{m}\\ T_1=14055.23\text{N}\end{array}$

Substitute $14055.23\text{N}$for $T_1$in Equation (XVII).

  $\begin{array}{c}{T}_2=0.80035\left(14055.23\text{N}\right)\\ =11248.96\text{N}\end{array}$

Substitute $14055.23\text{N}$for $T_1$ , $400\text{MPa}$for ${\sigma }_y$and $2$for$fs$in Equation (XI).

  $\begin{array}{l}\frac{400\text{MPa}}{2}=\frac{14055.23\text{N}}{A}\\ A=\frac{14055.23\text{N}}{200\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)}\\ A=\left(70.3\times {10}^{-6}\text{N}/\text{Pa}\right)\left(\frac{1}{\left(\frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)}\right)\\ A=70.3\times {10}^{-6}{\text{m}}^2\end{array}$

Substitute $11248.96\text{N}$for $T_2$ , $400\text{MPa}$for ${\sigma }_y$and $2$for $fs$in Equation (XII).

  $\begin{array}{l}\frac{400\text{MPa}}{2}=\frac{11248.96\text{N}}{A}\\ A=\frac{11248.96\text{N}}{200\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)}\\ A=\left(56.2\times {10}^{-6}\text{N}/\text{Pa}\right)\frac{1}{\left(\frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)}\\ A=56.2\times {10}^{-6}{\text{m}}^2.\end{array}$

Conclusion:

The minimum cross- section area is calculated by equating modulus of stress, elasticity, cable and factor of safety,height and length of cable.