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## Q. 4.4

Determine the force in each member of the Warren truss shown in $Fig. 4.19(a)$ by the method of joints. ## Verified Solution

Static Determinacy The truss has $13$ members and $8$ joints and is supported by $3$ reactions. Because $m + r = 2j$ and the reactions and the members of the truss are properly arranged, it is statically determinate.

Zero-Force Members It can be seen from Fig. 4.19(a) that at joint $G$, three members, $CG , FG$, and $GH$, are connected, of which $FG$ and $GH$ are collinear and $CG$ is not. Since no external load is applied at joint $G$, member $CG$ is a zero-force member.

$F_{CG} = 0$                                              Ans.

From the dimensions of the truss, we find that all inclined members have slopes of $3:4$, as shown in Fig. 4.19(a). The free-body diagram of the entire truss is shown in Fig. 4.19(b). As a joint with two or fewer unknowns—which should not be collinear—cannot be found, we calculate the support reactions. (Although joint G has only two unknown forces, $F_{FG}$ and $F_{GH}$, acting on it, these forces are collinear, so they cannot be determined from the joint equilibrium equation,$\sum{F_x}=0$.)

Reactions By using proportions,

$A_y=24(\frac{3}{4})+30(\frac{1}{2})+12(\frac{1}{4})=36$

$\sum{F_y}=0$                      $E_y = (24 + 30 + 12) – 36 = 30 k$

$\sum{F_x}=0$                     $A_x=0$

Joint $A$ Focusing our attention on joint $A$ in Fig. 4.19(b), we observe that in order to satisfy $\sum{F_y}=0$, the vertical component of $F_{AF}$ must push downward into the joint with a magnitude of $36 k$ to balance the upward reaction of $36 k$. The slope of member $AF$ is $3:4$, so the magnitude of the horizontal component of $F_{AF}$ is $(4=3)(36)$, or $48 k$. Thus, the force in member $AF$ is compressive, with a magnitude of $F_{AF}=\sqrt{(48)^2+(36)^2}=60k$.

$F_{AF} = 60 k (C)$                                 Ans.

With the horizontal component of $F_{AF}$ now known, we can see from the figure that in order for $\sum{F_x}=0$ to be satisfied, $F_{AB}$ must pull to the right with a magnitude of $48 k$ to balance the horizontal component of $F_{AF}$ of $48 k$ acting to the left. Therefore, member $AB$ is in tension with a force of $48 k$.

$F_{AB} = 48 k (T)$                                   Ans.

Joint $B$ Next, we consider the equilibrium of joint $B$. Applying $\sum{F_x}=0$ , we obtain $F_{BC}$.

$F_{BC} = 48 k (T)$                                       Ans.

From $\sum{F_y}=0$ , we obtain $F_{BF}$.

$F_{BF} = 24 k (T)$                                     Ans.

Joint $F$ This joint now has two unknowns, $F_{CF}$ and $F_{FG}$, so they can be determined by applying the equations of equilibrium as follows. We can see from Fig. 4.19(b) that in order to satisfy $\sum{F_y}=0$ , the vertical component of $F_{CF}$ must pull downward on joint $F$ with a magnitude of $36 – 24 = 12 k$. Using the $3:4$ slope of member $CF$, we obtain the magnitude of the horizontal component as $(4/3)(12) = 16 k$ and the magnitude of $F_{CF}$ itself as $20 k$.

$F_{CF} = 20 k (T)$                           Ans.

Considering the equilibrium of joint $F$ in the horizontal direction $(\sum{F_x} = 0)$, it should be obvious from Fig. 4.19(b) that $F_{FG}$ must push to the left on the joint with a magnitude of $48 + 16 = 64 k$.

$F_{FG} = 64 k (C)$                           Ans.

Joint $G$ Similarly, by applying $\sum{F_x}=0$ , we obtain $F_{GH}$.

$F_{GH} = 64 k (C)$                        Ans.

Note that the second equilibrium equation, $\sum{F_y}=0$, at this joint has already been utilized in the identification of member $CG$ as a zero-force member.

Joint $C$ By considering equilibrium in the vertical direction, $\sum{F_y}=0$ , we observe (from Fig. 4.19(b)) that member $CH$ should be in tension and that the magnitude of the vertical component of its force must be equal to $30 – 12 = 18 k$. Therefore, the magnitudes of the horizontal component of $F_{CH}$ and of $F_{CH}$ itself are $24 k$ and $30 k$, respectively, as shown in Fig. 4.19(b).

$F_{CH} = 30 k (T)$                          Ans.

By considering equilibrium in the horizontal direction, $\sum{F_x}=0$, we observe that member $CD$ must be in tension and that the magnitude of its force should be equal to $48 + 16 – 24 = 40 k$.

$F_{CD} = 40 k (T)$                            Ans.

Joint $D$ By applying $\sum{F_x}=0$, we obtain $F_{DE}$.

$F_{DE} = 40 k (T)$                           Ans.

From $\sum{F_y}=0$ , we determine $F_{DH}$.

$F_{DH} = 12 k (T)$                             Ans.

Joint $E$ Considering the vertical components of all the forces acting at joint $E$, we find that in order to satisfy $\sum{F_y}=0$ , the vertical component of $F_{EH}$ must push downward into joint $E$ with a magnitude of $30 k$ to balance the upward reaction $E_y = 30 k$. The magnitude of the horizontal component of $F_{EH}$ is equal to $(4/3)(30)$, or $40 k$. Thus, $F_{EH}$ is a compressive force with a magnitude of $50 k$.

$F_{EH} = 50 k (C)$                       Ans.

Checking Computations To check our computations, we apply the following remaining joint equilibrium equations (see Fig. 4.19(b)). At joint $E$,

$+\longrightarrow \sum{F_x} = -40 + 40 = 0$               Checks

At joint $H$,

$+\longrightarrow \sum{F_x} = 64 – 24 – 40 = 0$               Checks

$+\uparrow \sum{F_y} = -18 – 12 + 30 = 0$                     Checks 