Question

Each bar of the truss shown in the figure below has a diameter of 20 mm. Suppose that P = 33kN .

Determine the average normal stress in bar BC of the truss.

Determine the average normal stress in the bar AC of the truss.

Determine the average normal stress in the bar ABof the truss.

Each bar of the truss shown in the figure below has a diameter of 20 mm. Suppose that P = 33kN . Determine the average normal stress in bar BC of the truss. Determine the average normal stress in the bar AC of the truss. Determine the average normal stress in the bar ABof the truss.

Answer 1

Concepts and reason

The reactions forces induced in the joints of the truss can be calculated by using force equilibrium equations along horizontal and vertical direction, and moment equilibrium equation.

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

Before applying these equilibrium equations, distinction should be made between 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.

The concept of method of joints is used to calculate the force in the particular member in a truss. The applications of truss are bridges and cranes.

Fundamentals

Force is a vector quantity. It is represented by a magnitude, direction, and point of application.

The direction of force can be either compressive or tensile.

Truss:

Truss is a two force member. The forces are applied only at joints. The members in trusses cannot exceed certain limit. Depending on number of members and number of joints, the truss is categorized in to various types.

The forces in members are determined by various methods.

• Method of sections

• Method of joints

Method of sections:

Method of the sections is used to determine forces in any member of the truss. It uses the principle of equilibrium which says that when a part of the truss is in equilibrium then the complete truss is said to be in equilibrium. While considering a section of a truss care should be taken so that not more than three unknown members of truss are a part of section cut.

Consider the following equilibrium conditions to determine the force of the member s in the section cut:

Summation of all forces in x direction is zero, $\sum {{F_x}} = 0$

Summation of all forces in y direction is zero, $\sum {{F_y}} = 0$

Moment about arbitrary point is zero, $\sum {{M_O}} = 0$

Methods of joints:

According to this method, a single joint is considered and the forces acting on that joint is also considered. At the joint, one needs to apply the equations of equilibrium.

Consider the equations of equilibrium as follows:

Summation of all forces in x direction is zero, $\sum {{F_x}} = 0$

Summation of all forces in y direction is zero, $\sum {{F_y}} = 0$

Moment about arbitrary point is zero, $\sum {{M_O}} = 0$

There is generalized way of find the forces in the members. These are as follows:

• When a joint has two collinear forces and load acts on it, then the members experience same force but in opposite direction.

• When a joint has three members, with 2 collinear members and no load acts on them, then the third member has zero force in it.

General sign convention for moment: The moment is considered positive in counter-clockwise direction and negative in clockwise direction.

Draw the free body diagram of the truss.

Calculate the angle between rods BC and AB.

$\begin{array}{l}\\\tan \theta = \frac{{1.5}}{2}\\\\\theta = 36.869^\circ \\\end{array}$

Apply force equilibrium equation along the horizontal direction.

$\begin{array}{l}\\\sum {{F_x}} = 0\\\\P + {A_x} = 0\\\\{A_x} = - P\\\end{array}$

Here, P is applied force and ${A_x}$ is the horizontal reaction force at point A.

Substitute ${\rm{33kN}}$ for P.

${A_x} = - {\rm{33kN}}$

Apply moment equilibrium equation about point A.

$\begin{array}{l}\\\sum {{M_A}} = 0\\\\ - P \times 1.5 + {B_y} \times 2 = 0\\\\{B_y} \times 2 = P \times 0.75\\\end{array}$

Here, ${B_y}$ is the vertical reaction force at point B.

Substitute ${\rm{33kN}}$ for P.

$\begin{array}{c}\\{B_y} = 33 \times 0.75\\\\ = 24.75\,{\rm{kN}}\\\end{array}$

Apply force equilibrium equation along the vertical direction.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{B_y} + {A_x} = 0\\\\{A_y} = - {B_y}\\\end{array}$

Here, ${A_y}$ is the vertical reaction force at point A.

${A_y} = {\rm{ - 24}}{\rm{.75}}\;{\rm{kN}}$

Consider joint A as follows:

Apply force equilibrium equation along the horizontal direction.

$\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_{AB}} + {A_x} = 0\\\\{F_{AB}} = - {A_x}\\\end{array}$

Here, ${F_{AB}}$ is the force in the rod AB.

Substitute $- {\rm{33kN}}$ for ${A_x}$ .

${F_{AB}} = 33\,{\rm{kN}}$

Apply force equilibrium equation along the vertical direction.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{F_{AC}} + {A_y} = 0\\\\{F_{AC}} = - {A_y}\\\end{array}$

Substitute ${\rm{ - 24}}{\rm{.75}}\;{\rm{kN}}$ for ${A_y}$ .

$\begin{array}{c}\\{F_{AC}} = - \left( { - 24.75kN} \right)\\\\{\rm{ = }}24.75{\rm{kN}}\\\end{array}$

Consider joint B as follows:

Apply force equilibrium equation along the vertical direction.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{F_{BC}}\sin \theta + {B_y} = 0\\\\{F_{BC}} = - \frac{{{B_y}}}{{\sin \theta }}\\\end{array}$

Substitute $24.75\,{\rm{kN}}$ for ${B_y}$ .

$\begin{array}{c}\\{F_{BC}} = - \frac{{24.75}}{{\sin 36.869}}\\\\ = - 41.25{\rm{kN}}\\\end{array}$

The diameters of the all bars are same. Therefore, the areas of the all bars are equal.

Calculate the area of bars as follows:

$A = \frac{\pi }{4} \times {d^2}$

Here, d is the diameter of the bar.

Substitute ${\rm{20}}\,{\rm{mm}}$ for d.

$\begin{array}{c}\\\;A = \frac{\pi }{4} \times {20^2}\\\\ = 314.159{\rm{m}}{{\rm{m}}^{\rm{2}}}\\\end{array}$

Calculate the stress in the bar AB as follows:

${\sigma _{AB}} = \frac{{{F_{AB}}}}{A}$

Substitute $33\,{\rm{kN}}$ for ${F_{AB}}$ and $314.159{\rm{m}}{{\rm{m}}^{\rm{2}}}$ for A.

$\begin{array}{l}\\{\sigma _{AB}} = \frac{{33{\rm{kN}}}}{{314.159{\rm{m}}{{\rm{m}}^{\rm{2}}}}} \times \frac{{1\,{\rm{MPa}}}}{{1\,{\rm{N/m}}{{\rm{m}}^{\rm{2}}}}}\\\\{\sigma _{AB}} = 105.14{\rm{MPa}}\left( T \right)\\\end{array}$

Calculate the stress in the bar BC as follows:

${\sigma _{BC}} = \frac{{{F_{BC}}}}{A}$

Substitute $- 41.25{\rm{kN}}$ for ${F_{BC}}$ and $314.159\,{\rm{m}}{{\rm{m}}^{\rm{2}}}$ for A.

$\begin{array}{l}\\{\sigma _{BC}} = \frac{{ - 41.25{\rm{kN}}}}{{314.159{\rm{m}}{{\rm{m}}^{\rm{2}}}}} \times \frac{{1\,{\rm{MPa}}}}{{1\,{\rm{N/m}}{{\rm{m}}^{\rm{2}}}}}\\\\{\sigma _{BC}} = - 131.30{\rm{MPa}}\\\\{\sigma _{BC}} = 131.30{\rm{MPa}}\left( C \right)\\\end{array}$

Calculate the stress in the bar AC as follows:

${\sigma _{AC}} = \frac{{{F_{AC}}}}{A}$

Substitute $24.75\,{\rm{kN}}$ for ${F_{AC}}$ and $314.159\,{\rm{m}}{{\rm{m}}^{\rm{2}}}$ for A.

$\begin{array}{l}\\{\sigma _{AC}} = \frac{{24.75\,{\rm{kN}}}}{{314.159{\rm{m}}{{\rm{m}}^{\rm{2}}}}} \times \frac{{1\,{\rm{MPa}}}}{{1\,{\rm{N/m}}{{\rm{m}}^{\rm{2}}}}}\\\\{\sigma _{AC}} = 78.781{\rm{MPa}}\left( T \right)\\\end{array}$

Ans:

The stress in the member AB is $105.14{\rm{MPa}}\left( T \right)$ .