Chapter 5, Problem 5.1P

through 5.7 Calculate the forces in all members of the trusses shown, using the method of joints.

To determine

The forces in the member from method of joints.

Answer to Problem 5.1P

  $F_{AC}=5kips\left(tension\right)$

  $F_{BC}=F_{AB}=7.07kips\left(compression\right)$

Explanation of Solution

Given:

Two reactions at the pin support is Ax and Ay.

At roller support one reaction is Cy.

Force on the member $=10kips$

Free body diagram of the truss.

Taking moment about A

  $\begin{array}{l}\sum M_A=0\\ C_Y\left(L\right)-10\left(\frac{L}{2}\right)=0\\ C_Y\left(L\right)-5L=0\\ C_Y=5kips\end{array}$

From equilibrium equations

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

  $\begin{array}{l}\sum F_y=0\\ A_y+C_y-10=0\\ A_y=5kips\end{array}$

Considering joint A

From the equilibrium equation

  $\begin{array}{l}\sum F_y=0\\ A_y+F_{AB}\sin {45}^{\circ }=0\\ 5+F_{AB}\sin {45}^{\circ }=0\\ F_{AB}=-7.07kips\left(compression\right)\\ \sum F_x=0\\ A_x+F_{AC}+F_{AB}\cos {45}^{\circ }=0\\ 0+F_{AC}+\left(-7.07\right)\cos {45}^{\circ }=0\\ F_{AC}=5kips\left(tension\right)\end{array}$

Considering joint C

  $\begin{array}{l}\sum F_y=0\\ C_y+F_{BC}\sin {45}^{\circ }=0\\ 5+F_{BC}\sin {45}^{\circ }=0\\ F_{BC}=-7.07kips\\ F_{BC}=7.07kips\left(compression\right)\end{array}$

Conclusion:

Forces in the member $F_{BC}=F_{AB}=7.07kips\left(compression\right)$ and $F_{AC}=5kips\left(tension\right)$ .