Chapter 5, Problem 70P

To determine

Plot the shear diagram, bending moment diagram, axial force diagram, and the qualitative deflected shape of the frame.

Explanation of Solution

Write the condition for static instability, determinacy and indeterminacy of plane frames as follows:

$3m+r<3j+e_c\text{statically}\text{unstable}\text{frame}$        (1)

$3m+r=3j+e_c\text{statically}\text{determinate}\text{frame}$        (2)

$3m+r>3j+e_c\text{statically}\text{indeterminate}\text{frame}$        (3)

Here, number of members is m, number of external reactions is r, the number of joints is j, and the number of elastic hinges is $e_c$.

Find the degree of static indeterminacy (i) using the equation;

$i=\left(3m+r\right)-\left(3j+e_c\right)$        (4)

Refer to the Figure in the question;

The number of members (m) is 3.

The number of external reactions (r) is 4.

The number of joints (j) is 4.

The number of elastic hinges $e_c$ is 1.

Substitute the values in Equation (2);

$\begin{array}{l}3\left(3\right)+4=3\left(4\right)+1\\ 13=13\\ \text{statically}\text{determinate}\text{frame}\end{array}$

Show the free-body diagram of the entire frame as in Figure 1.

Refer Figure 1,

Consider entire frame.

Take moment about point A.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_A=0}\\ B_y\left(30\right)-2\left(30\right)\left(\frac{30}{2}\right)-30\left(20\right)=0\\ 30B_y-900-600=0\\ B_y=50\text{k}\uparrow \end{array}$

Find the vertical reaction at point A by resolving the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ A_y-2\left(30\right)+B_y=0\\ A_y-60+50=0\\ A_y=10\text{k}\uparrow \end{array}$

Find the horizontal reaction at point A by resolving the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -A_x+30-B_x=0\end{array}$        (1)

Consider the section DEB.

Find the horizontal reaction at point B by taking moment about the hinge at D.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D^{DEB}=0}\\ B_y\left(15\right)-2\left(15\right)\left(\frac{15}{2}\right)-B_x\left(20\right)=0\\ 50\left(15\right)-225-B_x\left(20\right)=0\\ B_x=26.25\text{k}\leftarrow \end{array}$

Substitute 26.25 k for $B_x$ in Equation (1).

$\begin{array}{l}-A_x+30-26.25=0\\ A_x=3.75\text{k}\leftarrow \end{array}$

Show the free-body diagram of the members and joints of the entire frame as in Figure 2.

Consider point A:

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_Y=0}\\ 10-A_Y^{AC}=0\\ A_Y^{AC}=10\text{k}\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_X=0}\\ A_X^{AC}-3.75=0\\ A_X^{AC}=3.75\text{k}\end{array}$

Consider the member AC:

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_Y=0}\\ A_Y^{AC}-C_Y^{AC}=0\\ 10-C_Y^{AC}=0\\ C_Y^{AC}=10\text{k}\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_X=0}\\ -A_X^{AC}+C_X^{AC}=0\\ -3.75+C_X^{AC}=0\\ C_X^{AC}=3.75\text{k}\end{array}$

Take moment about the point C.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C^{AC}=0}\\ M_C^{AC}-A_X^{AC}\left(20\right)=0\\ M_C^{AC}-3.75\left(20\right)=0\\ M_C^{AC}=75\text{k-ft}\end{array}$

Consider the point C:

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_Y=0}\\ C_Y^{AC}-C_Y^{CE}=0\\ 10-C_Y^{CE}=0\\ C_Y^{CE}=10\text{k}\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_X=0}\\ 30-C_X^{AC}-C_X^{CE}=0\\ 30-3.75-C_X^{CE}=0\\ C_X^{CE}=26.25\text{k}\end{array}$

Take moment about the point C.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C=0}\\ -M_C^{AC}+M_C^{CE}=0\\ -75+M_C^{CE}=0\\ M_C^{CE}=75\text{k-ft}\end{array}$

Consider the member CDE:

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_Y=0}\\ C_Y^{CE}-2\left(30\right)+E_Y^{CE}=0\\ 10-60+E_Y^{CE}=0\\ E_Y^{CE}=50\text{k}\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_X=0}\\ C_X^{CE}-E_X^{CE}=0\\ 26.25-E_X^{CE}=0\\ E_X^{CE}=26.25\text{k}\end{array}$

Take moment about the point E.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_E^{EC}=0}\\ -M_C^{CE}-C_Y^{CE}\left(30\right)+2\left(30\right)\left(\frac{30}{2}\right)-M_E^{CE}=0\\ -75-10\left(30\right)+900-M_E^{CE}=0\\ M_E^{CE}=525\text{k-ft}\end{array}$

Consider the point E:

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_Y=0}\\ -E_Y^{CE}+E_Y^{BE}=0\\ -50+E_Y^{BE}=0\\ E_Y^{BE}=50\text{k}\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_X=0}\\ E_X^{CE}-E_X^{BE}=0\\ 26.25-E_X^{BE}=0\\ E_X^{BE}=26.25\text{k}\end{array}$

Take moment about the point E.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_E=0}\\ M_E^{CE}-M_E^{BE}=0\\ 525-M_E^{BE}=0\\ M_E^{BE}=525\text{k-ft}\end{array}$

Consider the point B:

Resolve the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_Y=0}\\ -E_Y^{BE}+B_Y^{BE}=0\\ -50+B_Y^{BE}=0\\ B_Y^{BE}=50\text{k}\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_X=0}\\ E_X^{BE}-B_X^{BE}=0\\ 26.25-B_X^{BE}=0\\ B_X^{BE}=26.25\text{k}\end{array}$

Plot the moment end forces of the frame as in Figure 3.

Refer to the moment end force diagram plot the shear diagram, bending moment diagram, and the axial force diagrams.

Plot the shear force diagram as in Figure 4.

Refer to the shear force diagram, the maximum bending moment occurs at point F where the shear force changes sign.

Use similar triangle concept for the region CE:

$\begin{array}{l}\frac{10}{x}=\frac{50}{30-x}\\ 300-10x=50x\\ 60x=300\\ x=5\text{ft}\text{from}\text{point}C\end{array}$

Plot the bending moment diagram as in Figure 5.

Plot the axial force diagram as in Figure 6.

Plot the qualitative deflected shape as in Figure 7.