Question

As shown, a load of magnitude is applied to a structure. Assuming that all members are weightless and that ABEF and BCD are rigid bodies, find the magnitudes ofthe reactions at D and G needed to maintain the structure in equilibrium.

Assume the following values: a = 3.30m , b = 2.20m , and P = 104N.

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 ${F_y}$ .

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 structure.

Draw the free body diagram of portion FG.

Consider the portion FG and apply moment balance about F.

$\begin{array}{l}\\\sum {{M_F}} = 0\\\\{G_y} \times b = 0\\\\{G_y} = 0\\\end{array}$

Consider the entire truss and apply moment balance about B.

$\begin{array}{l}\\\sum {{M_B}} = 0\\\\{G_y}\left( {a + b + b + } \right) - {D_x}\left( {a + b} \right) + P\left( b \right) = 0\\\\0\left( {a + b + b + } \right) - {D_x}\left( {3.3 + 2.2} \right) + 104\left( {2.2} \right) = 0\\\\{D_x} = 41.6\,{\rm{N}}\\\end{array}$

Consider horizontal force balance for the entire truss.

$\begin{array}{l}\\\sum {{F_x}} = 0\\\\{D_x} + {G_x} = 0\\\\{G_x} = - {D_x}\\\\{G_x} = - 41.6\,{\rm{N}}\\\end{array}$

Consider vertical force balance for the entire truss.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{D_y} + {G_y} - P = 0\\\\{D_y} + 0 - 104 = 0\\\\{D_y} = 104\,{\rm{N}}\\\end{array}$

Calculate the magnitude of reaction at D.

$\begin{array}{c}\\\left| D \right| = \sqrt {{D_x}^2 + {D_y}^2} \\\\ = \sqrt {{{\left( {41.6} \right)}^2} + {{\left( {104} \right)}^2}} \\\\ = 112.01\,{\rm{N}}\\\end{array}$

Calculate the magnitude of reaction at G.

$\begin{array}{c}\\\left| G \right| = \sqrt {{G_x}^2 + {G_y}^2} \\\\ = \sqrt {{{\left( { - 41.6} \right)}^2} + {{\left( 0 \right)}^2}} \\\\ = 41.6\,{\rm{N}}\\\end{array}$
‎

Ans:

Therefore, the magnitude of reaction at D is $112.01\,{\rm{N}}$ .

Therefore, the magnitude of reaction at G is $41.6\,{\rm{N}}$ .