Question

The load is supported by the four 304 stainless steel wires (E=28x10^3 Ksi) that are connected to the rigid members, AB and DC. Determine the vertical displacement of the 500-lb load if the members were originally horizontal when the load was applied. Each wire has a cross-sectional area of 0.025 in^2

Answer 1

Concepts and Reason

The external force and couple moment acting on a body can be reduced to an equivalent resultant force and resultant couple moment. When this resultant force and resultant couple moment is both equal to zero then the body is said to be in equilibrium.

The major assumption for applying these equilibrium equations is that the body remains rigid.

To apply these equilibrium equations we need to know the known and unknown forces that act on the body. When all the supports are removed by replacing them with forces that prevents the translation of body in a given direction that diagram is called free body diagram.

Fundamentals

‎Write the equilibrium equations.

$\begin{array}{l}\\{F_R} = \sum {\bf{F}} = 0\\\\{\left( {{M_R}} \right)_O} = \sum {{{\bf{M}}_O}} = 0\\\end{array}$

Here, the resultant force is ${F_R}$ and the resultant moment about any arbitrary point is ${\left( {{M_R}} \right)_O}$ .

Calculate the magnitude of force using the trigonometric relation:

$\left| F \right| = \sqrt {{F_x}^2 + {F_y}^2}$
‎Here, the component of force in x-direction is ${F_x}$ and the component of force in y-direction is ${F_y}$ .

Calculate the weight of an object when mass is given using the following relation:

$W = mg$

Here, the mass of the particle is $m$ and the acceleration due to gravity is $g$ .

Sign Convention for force: Upward and right forces are positive.

Sign convention for moment: Anti clockwise moment is positive and clockwise moment is negative.

Draw the free body diagram of the rigid members.

Take moments about point A.

$\begin{array}{l}\\\sum {{M_A}} = 0\\\\{F_{BG}}\left( {3 + 1} \right) - 500\left( 3 \right) = 0\\\end{array}$

Here, the tension in the wire BG is ${F_{BG}}$ .

$\begin{array}{l}\\{F_{BG}}\left( {3 + 1} \right) - 500\left( 3 \right) = 0\\\\{F_{BG}} = 375{\rm{ lb}}\\\end{array}$

Balance the vertical forces on the rigid body AB.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{F_{AH}} + {F_{BG}} - 500 = 0\\\end{array}$

Here, the tension in the wire AH is ${F_{AH}}$ .

Substitute 375 lb for ${F_{BG}}$ .

$\begin{array}{l}\\{F_{AH}} + 375 - 500 = 0\\\\{F_{AH}} = 125{\rm{ lb}}\\\end{array}$

Take moments about point D.

$\begin{array}{l}\\\sum {{M_D}} = 0\\\\{F_{CF}}\left( {2 + 1} \right) - {F_{AH}}\left( 1 \right) = 0\\\end{array}$

Here, the tension in the wire CF is ${F_{CF}}$ .

Substitute 125 lb for ${F_{AH}}$ .

$\begin{array}{l}\\{F_{CF}}\left( {2 + 1} \right) - 125\left( 1 \right) = 0\\\\{F_{CF}} = 41.67{\rm{ lb}}\\\end{array}$

Balance the vertical forces on the rigid body DC.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{F_{DE}} + {F_{CF}} - {F_{AH}} = 0\\\end{array}$

Here, the tension in the wire DE is ${F_{DE}}$ .

Substitute 125 lb for ${F_{AH}}$ and 41.67 lb for ${F_{CF}}$ .

$\begin{array}{l}\\{F_{DE}} + 41.67 - 125 = 0\\\\{F_{DE}} = 83.33{\rm{ lb}}\\\end{array}$

Calculate the displacement at point D.

${\delta _D} = \frac{{{F_{DE}}{L_{DE}}}}{{AE}}$

Here, the length of wire DE is ${L_{DE}}$ , the area of cross section of each wire is A, and the modulus of elasticity is E.

Substitute 83.33 lb for ${F_{DE}}$ , 36 in for ${L_{DE}}$ , $0.025{\rm{ i}}{{\rm{n}}^2}$ for A, and $28 \times {10^6}{\rm{ psi}}$ for E.

$\begin{array}{c}\\{\delta _D} = \frac{{83.33 \times 36}}{{0.025 \times 28 \times {{10}^6}}}\\\\ = 0.0042857{\rm{ in}}\\\end{array}$

Calculate the displacement at point C.

${\delta _C} = \frac{{{F_{CF}}{L_{CF}}}}{{AE}}$

Here, the length of wire CF is ${L_{CF}}$ .

Substitute 41.67 lb for ${F_{CF}}$ , 36 in for ${L_{CF}}$ , $0.025{\rm{ i}}{{\rm{n}}^2}$ for A, and $28 \times {10^6}{\rm{ psi}}$ for E.

$\begin{array}{c}\\{\delta _C} = \frac{{41.67 \times 36}}{{0.025 \times 28 \times {{10}^6}}}\\\\ = 0.0021429{\rm{ in}}\\\end{array}$

Draw the displacement diagram for member AB.

From the displacement diagram, use the similarity of triangles,

Here, the displacement at point H is ${\delta _H}$ .

Substitute 0.0042857 in for ${\delta _D}$ and 0.0021429 in for ${\delta _C}$ .

$\begin{array}{c}\\{\delta _H} = \frac{1}{3}\left[ {2\left( {0.0042857} \right) + 0.0021429} \right]\\\\ = 0.0035713{\rm{ in}}\\\end{array}$

Calculate the displacement at point A.

Here, the length of wire AH is ${L_{AH}}$ .

Substitute 125 lb for ${F_{AH}}$ , 21.6 in for ${L_{AH}}$ , $0.025{\rm{ i}}{{\rm{n}}^2}$ for A, and $28 \times {10^6}{\rm{ psi}}$ for E.

$\begin{array}{c}\\{\delta _A} = 0.0035713 + \frac{{125 \times 21.6}}{{0.025 \times 28 \times {{10}^6}}}\\\\ = 0.0035713 + 0.0038571\\\\ = 0.0074284{\rm{ in}}\\\end{array}$

Calculate the displacement of point B,

${\delta _B} = \frac{{{F_{BG}}{L_{BG}}}}{{AE}}$

Here, the length of wire BG is ${L_{BG}}$ .

Substitute 375 lb for ${F_{BG}}$ , 60 in for ${L_{BG}}$ , $0.025{\rm{ i}}{{\rm{n}}^2}$ for A, and $28 \times {10^6}{\rm{ psi}}$ for E.

$\begin{array}{c}\\{\delta _B} = \frac{{375 \times 60}}{{0.025 \times 28 \times {{10}^6}}}\\\\ = 0.032143{\rm{ in}}\\\end{array}$

From the displacement diagram, use the similarity of triangles,

Here, the displacement at point I is ${\delta _I}$ .

Substitute 0.0074284 in for ${\delta _A}$ and 0.032143 for ${\delta _B}$ .

Ans:

The vertical displacement of the load is 0.02596 in.