[mechanics] Truss analysis: Practice

 

For a truss given below, can we solve the force acting on every member? Surprisingly, yes. Here’s how.

 

 

First, we have to solve the reactions on $H$ and $L$.

 

$$\begin{array}{rl}\mathrm{\Sigma }{M}_L=3Pa-4Ha=0\to & H=\frac{3}{4}P\\ \mathrm{\Sigma }{F}_y=H+L-P=0\to & L=\frac{1}{4}P\end{array}$$

 

 

Second, let’s get rid of zero-force members. For joint $D$ to be in equilibrium, $F_{DI}$ should be zero because there’s nothing to cancel it out on the opposite side. In a similar manner, $EI$, $FK$, $GK$ are also zero-force members.

 

 

Oh, it looks like we can do more. Members $AI$, $BJ$, $CK$ are free to go.

 

 

Removed members are force-free, but they still need to be in the truss to make the whole structure rigid. Remaining members are only the ones under tension or compression, but without the zero-force members they can not make a rigid truss. Interesting, right?

 

Third, I will use the method of joints because we want to get all the forces on members. Equilibrium of joint $H$ gives

 

 

$$\begin{array}{rl}\mathrm{\Sigma }{F}_y=\frac{F_{DH}}{\sqrt{2}}+\frac{3}{4}P=0\to & F_{DH}=F_{AD}=\frac{-3\sqrt{2}}{4}P\\ \mathrm{\Sigma }{F}_x=F_{HI}+\frac{F_{DH}}{\sqrt{2}}=0\to & F_{HI}=F_{IJ}=\frac{3}{4}P.\end{array}$$

 

Equilibrium on joint $A$ gives

 

 

$$\begin{array}{rlrl}\mathrm{\Sigma }{F}_y& =\frac{3\sqrt{2}}{4}P\cdot \frac{1}{\sqrt{2}}-P-\frac{F_{AE}}{\sqrt{2}}=0& & \to F_{AE}=F_{EJ}=-\frac{\sqrt{2}}{4}P\\ \mathrm{\Sigma }{F}_x& =\frac{3\sqrt{2}}{4}P\cdot \frac{1}{\sqrt{2}}+F_{AB}+\frac{F_{AE}}{\sqrt{2}}=0& & \to F_{AB}=F_{BC}=-\frac{1}{2}P.\end{array}$$

 

Equilibrium on joint $J$ gives

 

 

$$\begin{array}{rlrl}\mathrm{\Sigma }{F}_y& =-\frac{\sqrt{2}}{4}P\cdot \frac{1}{\sqrt{2}}+\frac{F_{FJ}}{\sqrt{2}}=0& & \to F_{FJ}=F_{CF}=\frac{\sqrt{2}}{4}P\\ \mathrm{\Sigma }{F}_x& =-\frac{3}{4}P+\frac{\sqrt{2}}{4}P\cdot \frac{1}{\sqrt{2}}+\frac{F_{FJ}}{\sqrt{2}}+F_{JK}=0& & \to F_{JK}=F_{KL}=\frac{1}{4}P.\end{array}$$

 

Equilibrium on joint $L$ gives

 

 

$$\begin{array}{rlrl}\mathrm{\Sigma }{F}_y& =\frac{F_{GL}}{\sqrt{2}}+\frac{1}{4}P=0& & \to F_{GL}=F_{CG}=-\frac{\sqrt{2}}{4}P\end{array}$$

 

which confirms that

 

$$\begin{array}{rl}\mathrm{\Sigma }{F}_x& =-\frac{1}{4}P-\frac{F_{GL}}{\sqrt{2}}=0.\end{array}$$

 

We didn’t check the joint $C$, but let’s make it sure.

 

 

$$\begin{array}{rl}\mathrm{\Sigma }{F}_x& =\frac{1}{2}P-\frac{\sqrt{2}}{4}P\cdot \frac{1}{\sqrt{2}}-\frac{\sqrt{2}}{4}P\cdot \frac{1}{\sqrt{2}}=0\\ \mathrm{\Sigma }{F}_y& =-\frac{\sqrt{2}}{4}P\cdot \frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{4}P\cdot \frac{1}{\sqrt{2}}=0\end{array}$$

 

Phew, we made it.

 

- lazy engineer