Chapter 2, Problem 2.4.16P

A rigid bar of weight W = SOO N hangs from three equally spaced vertical wines( length L = 150 mm, spacing a = 50 mm J: two of steel and one of aluminum. The wires also support a load P acting on the bar. The diameter of the steel wires is d_s= 2 mm, and the diameter of the

aluminum wire is d = A mm. a Assume £,=210 GPa and E_B« 70 GPa.

1. What load P_allowcan be supported at the mitl-point of the bar (x = a) if the allowable stress in the steel wires is 220 MPa and in the aluminum wire is 80 MPa? (See figure part
(b) What is /*, _{I k w}» if the load is positioned at .v = all¹? (See figure part a.)

(c) Repeat part (b) if the second and third wires are switched as shown in the figure part b.

  

  

  

To determine

The maximum load P_allow at the mid-point of the bar.

Answer to Problem 2.4.16P

  $P_S=1504{\rm N}$

Explanation of Solution

Given Information:

A rigid bar of some eight hangs from three equally spaced vertical wires as shown in figure below:

Weight of the bar W = 800 N

Length of wire L = 150 mm

  $d_a=2mm$ And $d_s=4mm$

  $E_a=70G{\rm P}a$

  $E_s=210G{\rm P}a$

  ${\sigma }_s=220{\rm M}{\rm P}a$ And ${\sigma }_a=80{\rm M}{\rm P}a$

  ${\displaystyle \sum F_v=0}$

  $2R_S+R_A=P+W$ ........(1)

Using reaction R_A of aluminum bar as redundant reaction

Elongation of each of the steel wire due to load $P+W$ if aluminum wire is cut.

  ${\delta }_1=(\frac{P+W}{2})\frac{L_{}}{A_S{E}_S}$

Upward displacement due to shortening of steel wires and elongation of aluminum wire under redundant R_A

  ${\delta }_2=(R_A)\left(\frac{L_{}}{2A_S{E}_S}+\frac{L_{}}{A_A{E}_A}\right)$

But, ${\delta }_1={\delta }_2$

  $R_A=(P+W)\left(\frac{E_A{A}_A}{2A_S{E}_S+E_A{A}_A}\right)$

Using relation (1),

  $R_s=\frac{(P+W)-(P+W)\left(\frac{E_A{A}_A}{2A_S{E}_S+E_A{A}_A}\right)}{2}$

  $R_s=(P+W)\left(\frac{E_s{A}_s}{2A_S{E}_S+E_A{A}_A}\right)$

To calculate the value of stress, use following expression.

  ${\sigma }_A=\frac{R_A}{A_A}$ and ${\sigma }_s=\frac{R_s}{A_s}$

Hence,

  ${\sigma }_A=(P+W)\left(\frac{E_A}{2A_S{E}_S+E_A{A}_A}\right)$

  ${\sigma }_s=(P+W)\left(\frac{E_s}{2A_S{E}_S+E_A{A}_A}\right)$

From both the expression, we will get value of $P_{allow}$ . Whichever value is the lower most is the allowable value of P.

  $P_A=1713{\rm N}$ and $P_S=1504{\rm N}$

Hence, steel wire controls the value of $P_{allow}$ .

To determine

The maximum load P_allow at position $x=\frac{a}{2}$ .

Answer to Problem 2.4.16P

  $P_S=820{\rm N}$

Explanation of Solution

Given Information:

A rigid bar of some eight hangs from three equally spaced vertical wires as shown in figure below:

Weight of the bar W = 800 N

Length of wire L = 150 mm

  $d_a=2mm$ And $d_s=4mm$

  $E_a=70G{\rm P}a$

  $E_s=210G{\rm P}a$

  ${\sigma }_s=220{\rm M}{\rm P}a$ And ${\sigma }_a=80{\rm M}{\rm P}a$

Elongation of each of the steel wire due to load P +W if aluminum wire is cut. Computing elongation of left and right steel wire.

  ${\delta }_{1L}=(\frac{3P}{4}+\frac{W}{2})\frac{L_{}}{A_S{E}_S}$

  ${\delta }_{1R}=(\frac{P}{4}+\frac{W}{2})\frac{L_{}}{A_S{E}_S}$

  ${\delta }_1=\frac{{\delta }_{1L}+{\delta }_{1R}}{2}$

  ${\delta }_1=\left(\frac{P+W}{2}\right)\left(\frac{L}{A_S{E}_S}\right)$ ......…..(2)

Upward displacement due to shortening of steel wires and elongation of aluminum wire under redundant R_A.:

  ${\delta }_2=(R_A)\left(\frac{L_{}}{2A_S{E}_S}+\frac{L_{}}{A_A{E}_A}\right)$ ........…...(3)

Equating (2) and (3) and solving for R_{A :}

  $R_A=(P+W)\frac{E_A{A}_A}{E_A{A}_A+2E_S{A}_S}$

  $R_{SL}=\frac{3P}{4}+\frac{W}{2}-\frac{R_a}{2}$

  $R_{SL}=\frac{P{E}_A{A}_A+6P{E}_S{A}_S+4W{E}_S{A}_S}{4E_A{A}_A+6E_S{A}_S}$

  ${\sigma }_S=\frac{P{E}_A{A}_A+6P{E}_S{A}_S+4W{E}_S{A}_S}{4E_A{A}_A+6E_S{A}_S}\left(\frac{1}{A_S}\right)$

Solving for P_allowwe get value based on stress in steel.

  $P_S=820{\rm N}$

From part (a) value of P_allow based on stress in aluminium . $$

  $P_a=1713{\rm N}$

Hence, considering smaller of two values. Thus, aloowable load is $P_S=820{\rm N}$

To determine

The allowable load P_allow if the second and third wires are interchange.

Answer to Problem 2.4.16P

  $P_{Sa}=703N$

Explanation of Solution

Given Information:

A rigid bar of some eight hangs from three equally spaced vertical wires as shown in figure below:

Weight of the bar W = 800 N

Length of wire L = 150 mm

  $d_a=2mm$ And $d_s=4mm$

  $E_a=70G{\rm P}a$

  $E_s=210G{\rm P}a$

  ${\sigma }_s=220{\rm M}{\rm P}a$ And ${\sigma }_a=80{\rm M}{\rm P}a$

Cut aluminum wire and apply P and W , compute forces in left and right steel wire

  $R_{SL}=\frac{P}{2}$ and $R_{SR}=\frac{P}{2}+W$

  ${\delta }_{1L}=(\frac{P}{2})\frac{L_{}}{A_S{E}_S}$

  ${\delta }_{1R}=(\frac{P}{2}+W)\frac{L_{}}{A_S{E}_S}$

By geometry, for aluminum wire at far right

  ${\delta }_1=(\frac{P}{2}+2W)\frac{L_{}}{A_S{E}_S}$

Apply redundant R_A at right wire, compute wire force and displacement at aluminum wire:

  $R_{SL}=-R_A$ , $R_{SR}=2R_A$

  ${\delta }_2=R_A\left(\frac{5L_{}}{A_S{E}_S}+\frac{L_{}}{A_A{E}_A}\right)$

  ${\delta }_1={\delta }_2$

  $R_A=\frac{E_A{A}_{A}P+4E_A{A}_{A}W}{10E_A{A}_A+2E_S{A}_S}$

And

  ${\sigma }_{Aa}=\frac{E_{A}P+4E_{A}W}{10E_A{A}_A+2E_S{A}_S}$

  $P_{Aa}=\frac{{\sigma }_{Aa}(10E_A{A}_A+2E_S{A}_S)-4E_{A}W}{E_A}$

  $P_{Aa}=1713N$

Calculate the forces in steel wire, then P_allow for steel wires using superposition:

  $R_{SL}=\frac{P}{2}+R_A$

  $R_{SL}=\frac{P}{2}+\frac{E_A{A}_{A}P+4E_A{A}_{A}W}{10E_A{A}_A+2E_S{A}_S}$

  $R_{SL}=\frac{6E_A{A}_{A}P+E_S{A}_{S}P+4E_A{A}_{A}W}{10E_A{A}_A+2E_S{A}_S}$

This is larger than R_SR. Hence using R_SL${\sigma }_{Sa}=\frac{R_{SL}}{A_S}$

  $P_{Sa}=\frac{10{\sigma }_{Sa}{E}_A{A}_S{A}_A+2{\sigma }_{Sa}{E}_S{A}_S^2-4E_A{A}_{A}W}{6E_A{A}_A+E_S{A}_S}$

  $P_{Sa}=703N$

Hence, Steel controls the value of P_allow.