Chapter 6, Problem 27P

(a)

To determine

Calculate the reactions and member forces of three-hinged, trussed arch.

Answer to Problem 27P

Case A:

The support reaction at A is $\underset{\_}{A_y=45\text{kN}\uparrow }$ and $\underset{\_}{A_x=67.5\text{kN}\to }$.

The support reaction at G is $\underset{\_}{G_y=45\text{kN}\uparrow }$ and $\underset{\_}{G_x=67.5\text{kN}\leftarrow }$.

Case B:

The support reaction at A is $\underset{\_}{A_y=25\text{kN}\uparrow }$ and $\underset{\_}{A_x=37.5\text{kN}\to }$.

The support reaction at G is $\underset{\_}{G_y=65\text{kN}\uparrow }$ and $\underset{\_}{G_x=97.5\text{kN}\leftarrow }$.

Explanation of Solution

Given information:

The structure of the arch is given in Figure 6P.27.

Case A: $90\text{kN}$ force only acting at joint D.

Assumption:

Apply the sign convention for calculating the equations of equilibrium as below.

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.
• For the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.
• For moment equilibrium condition, take the clockwise moment as positive and counter clockwise moment as negative.

Calculation:

Draw the free body diagram as shown in Figure 1.

Calculate the support reactions $A_y$ and $G_y$ at support A and G.

Due to symmetry:

$\begin{array}{c}{A}_y=G_y=\frac{90\text{kN}}{2}\\ =45\text{kN}\end{array}$

Calculate the horizontal reaction at supports.

Take moment at D.

$\begin{array}{c}\left(45\times 27\right)-\left(A_x\times 18\right)=0\\ A_x=67.5\text{kN}\to \\ G_x=67.5\text{kN}\leftarrow \end{array}$

Therefore, the support reaction at A is $\underset{\_}{A_y=45\text{kN}\uparrow }$ and $\underset{\_}{A_x=67.5\text{kN}\to }$.

Therefore, the support reaction at G is $\underset{\_}{G_y=45\text{kN}\uparrow }$ and $\underset{\_}{G_x=67.5\text{kN}\leftarrow }$.

Find the member forces using method of joints as in Table 1

Free body diagram for joints	Calculation of Member forces
	Joint A. $\begin{array}{c}{\displaystyle \sum F_x=0}\\ F_{AB}\cos 45°=-67.5\text{kN}\\ F_{AB}=-95.45\text{kN}\left(\text{Compression}\right)\end{array}$ $\begin{array}{c}{\displaystyle \sum F_y=0}\\ F_{AM}=-45\text{kN}-F_{AB}\sin 45°\\ F_{AM}=-45\text{kN}-\left(-95.45\right)\sin 45°\\ =22.5\text{kN}\left(\text{Tension}\right)\end{array}$
	Joint M. $\begin{array}{c}{\displaystyle \sum F_y=0}\\ -F_{MB}\sin 45=F_{AM}\\ -F_{MB}\sin 45=22.45\text{kN}\\ F_{MB}=-31.82\text{kN}\left(\text{Compression}\right)\end{array}$ $\begin{array}{c}{\displaystyle \sum F_x=0}\\ F_{ML}+F_{MB}\cos 45=0\\ F_{ML}+\left(-31.82\right)\cos 45=0\\ F_{ML}=22.5\text{kN}\left(\text{Tension}\right)\end{array}$
	Joint B. $\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_{BC}\cos 33.69-F_{MB}\cos 45-F_{AB}\cos 45=0\\ F_{BC}\cos 33.69-\left(-31.82\right)\cos 45-\left(-95.45\right)\cos 45=0\\ F_{BC}=-108.15\text{kN}\left(\text{Compression}\right)\end{array}$ $\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{BL}+F_{BC}\sin 33.69+F_{MB}\sin 45-F_{AB}\sin 45=0\\ \left\{\begin{array}{l}{F}_{BL}+\left(-108.15\right)\sin 33.69+\left(-31.82\right)\sin 45\\ -\left(-95.45\right)\sin 45\end{array}\right\}=0\\ F_{BL}=15.3\text{kN}\\ F_{BL}\approx \text{15}\text{kN}\left(\text{tension}\right)\end{array}$
	Joint L. $\begin{array}{l}{\displaystyle \sum F_y=0}\\ -F_{BL}-F_{LC}\sin 18.43=0\\ -15-F_{LC}\sin 18.43=0\\ F_{LC}=-47.4\text{kN}\left(\text{Compression}\right)\end{array}$ $\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{KL}-F_{ML}+F_{LC}\cos 18.43=0\\ F_{KL}-22.5+\left(-47.4\right)\cos 18.43=0\\ F_{KL}=67.5\text{kN}\left(\text{Tension}\right)\end{array}$
	Joint K. $\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{KC}=0\\ {\displaystyle \sum F_x=0}\\ F_{KD}=F_{KL}\\ =67.5\text{kN}\left(\text{Tension}\right)\end{array}$
	Joint D. $\begin{array}{l}{\displaystyle \sum F_y=0}\\ -F_{KD}-\frac{1}{2}{F}_{CD}\cos 18.43=0\\ -67.5-\frac{1}{2}{F}_{CD}\cos 18.43=0\\ F_{CD}=-142.3\text{kN}\left(\text{Compression}\right)\end{array}$

Table 1

Show the calculated member forces as in Table 2

Member	Magnitude of Force
AB, GF	$95.45\text{kN}\left(\text{Compression}\right)$
BC, FE	$108.15\text{kN}\left(\text{Compression}\right)$
CD, ED	$142.3\text{kN}\left(\text{Compression}\right)$
KD, DJ	$67.5\text{kN}\left(\text{Tension}\right)$
KL, JI	$67.5\text{kN}\left(\text{Tension}\right)$
LM, IH	$22.5\text{kN}\left(\text{Tension}\right)$
AM, GH	$22.5\text{kN}\left(\text{Tension}\right)$
BM, FH	$31.82\text{kN}\left(\text{Compression}\right)$
BL, FI	$\text{15}\text{kN}\left(\text{tension}\right)$
CL, EI	$47.4\text{kN}\left(\text{Compression}\right)$
KC, JE	0

Table 2

Consider case B.

Case B: Both $90\text{kN}$ and $60\text{kN}$ at joint D and M.

Draw the free body diagram as shown in Figure 7

Support reactions:

Take moment at G.

$\begin{array}{l}\left(A_y\times 54\right)-\left(90\times 27\right)+\left(60\times 18\right)=0\\ A_y=25\text{kN}\end{array}$

Apply ${\displaystyle \sum F_y}=0$.

$\begin{array}{l}{A}_y+G_y=90\\ 25+G_y=90\\ G_y=65\text{kN}\end{array}$

Calculate the horizontal reaction at supports.

Take moment at D.

$\begin{array}{c}\left(25\times 27\right)-\left(A_x\times 18\right)=0\\ A_x=37.5\text{kN}\to \end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x+60=G_x\\ 37.5\text{kN}+60=G_x\\ G_x=97.5\text{kN}\leftarrow \end{array}$

$G_x=67.5\text{kN}\to$

Therefore, the support reaction at A is $\underset{\_}{A_y=25\text{kN}\uparrow }$ and $\underset{\_}{A_x=37.5\text{kN}\to }$.

Therefore, the support reaction at G is $\underset{\_}{G_y=65\text{kN}\uparrow }$ and $\underset{\_}{G_x=97.5\text{kN}\leftarrow }$.

Find the member forces using method of joints similarly as case A.

Tabulate the member force values as shown in Table 3.

Member	Magnitude of Force
AB	$53\text{kN}\left(\text{Compression}\right)$
BC	$60\text{kN}\left(\text{Compression}\right)$
CD	$79\text{kN}\left(\text{Compression}\right)$
KD	$22.5\text{kN}\left(\text{compression}\right)$
KL	$22.5\text{kN}\left(\text{compression}\right)$
LM	$47.5\text{kN}\left(\text{compression}\right)$
AM	$12.5\text{kN}\left(\text{Tension}\right)$
BM	$17.7\text{kN}\left(\text{Compression}\right)$
BL	$\text{8}\text{.33}\text{kN}\left(\text{tension}\right)$
CL	$26.4\text{kN}\left(\text{Compression}\right)$
KC	0
GF	$137.88\text{kN}\left(\text{Compression}\right)$
FE	$156.2\text{kN}\left(\text{Compression}\right)$
ED	$205.55\text{kN}\left(\text{Compression}\right)$
DJ	$97.5\text{kN}\left(\text{tension}\right)$
JI	$97.5\text{kN}\left(\text{tension}\right)$
IH	$32.5\text{kN}\left(\text{tension}\right)$
GH	$32.5\text{kN}\left(\text{tension}\right)$
FH	$45.96\text{kN}\left(\text{Compression}\right)$
FI	$\text{21}\text{.7}\text{kN}\left(\text{tension}\right)$
EI	$68.5\text{kN}\left(\text{Compression}\right)$
JE	0

Table 3

Show the member forces as in Figure 8.

(b)

To determine

Find the maximum axial force in the arch.

Answer to Problem 27P

The maximum axial force in the arch is $\underset{\_}{F_{DE}=205.55\text{kN}\left(\text{compression}\right)}$, $\underset{\_}{F_{EF}=156.21\text{kN}\left(\text{compression}\right)}$, and $\underset{\_}{F_{FG}=137.88\text{kN}\left(\text{compression}\right)}$.

Explanation of Solution

Refer Table 3 and Figure 8.

The maximum axial force in the arch is,

$F_{DE}=205.55\text{kN}\left(\text{compression}\right)$

$F_{EF}=156.21\text{kN}\left(\text{compression}\right)$

$F_{FG}=137.88\text{kN}\left(\text{compression}\right)$

Therefore, the maximum axial force in the arch is $\underset{\_}{F_{DE}=205.55\text{kN}\left(\text{compression}\right)}$, $\underset{\_}{F_{EF}=156.21\text{kN}\left(\text{compression}\right)}$, and $\underset{\_}{F_{FG}=137.88\text{kN}\left(\text{compression}\right)}$.