f6-1.determine the force in each member of the truss.state if the members are in tension or compression.

Explanation:

Step1: Calculate support reactions

Sum of moments about point $A$: $\sum M_A = 0$. Let the vertical reaction at $C$ be $C_y$ and the horizontal reaction at $A$ be $A_x$ and vertical reaction at $A$ be $A_y$. $C_y\times2 - 2\times1=0$, so $C_y = 1$ kN. Sum of vertical forces $\sum F_y=0$, $A_y + C_y=0$, so $A_y=- 1$ kN. Sum of horizontal forces $\sum F_x = 0$, $A_x - 2=0$, so $A_x = 2$ kN.

Step2: Analyze joint $A$

At joint $A$, assume forces in members $AB$ and $AD$ are $F_{AB}$ and $F_{AD}$ respectively. Sum of vertical forces $\sum F_y = 0$, $A_y+F_{AD}\sin45^{\circ}=0$. $-1 + F_{AD}\sin45^{\circ}=0$, so $F_{AD}=\sqrt{2}$ kN (tension). Sum of horizontal forces $\sum F_x = 0$, $A_x - F_{AB}-F_{AD}\cos45^{\circ}=0$. $2 - F_{AB}-\sqrt{2}\times\frac{\sqrt{2}}{2}=0$, so $F_{AB}=1$ kN (tension).

Step3: Analyze joint $C$

At joint $C$, assume force in member $BC$ is $F_{BC}$ and in member $CD$ is $F_{CD}$. Sum of vertical forces $\sum F_y = 0$, $C_y - F_{CD}\sin45^{\circ}=0$. $1 - F_{CD}\sin45^{\circ}=0$, so $F_{CD}=\sqrt{2}$ kN (tension). Sum of horizontal forces $\sum F_x = 0$, $-F_{BC}-F_{CD}\cos45^{\circ}=0$. $-F_{BC}-\sqrt{2}\times\frac{\sqrt{2}}{2}=0$, so $F_{BC}=- 1$ kN (compression).

Step4: Analyze joint $D$

Sum of vertical forces $\sum F_y = 0$, $F_{AD}\sin45^{\circ}-F_{CD}\sin45^{\circ}=0$ (satisfied). Sum of horizontal forces $\sum F_x = 0$, $F_{AD}\cos45^{\circ}+F_{CD}\cos45^{\circ}-2 - F_{BD}=0$. $\sqrt{2}\times\frac{\sqrt{2}}{2}+\sqrt{2}\times\frac{\sqrt{2}}{2}-2 - F_{BD}=0$, so $F_{BD}=0$.

Answer:

$F_{AB}=1$ kN (tension), $F_{AD}=\sqrt{2}$ kN (tension), $F_{BC}=-1$ kN (compression), $F_{BD}=0$, $F_{CD}=\sqrt{2}$ kN (tension)