Question 4.12: Determine the reactions at the supports and the force in eac...
Determine the reactions at the supports and the force in each member of the space truss shown in Fig. 4.32(a).
Static Determinacy The truss contains 9 members and 5 joints and is supported by 6 reactions. Because m +r = 3j and the reactions and the members of the truss are properly arranged, it is statically determinate.
Member Projections The projections of the truss members in the x, y, and z directions, as obtained from Fig. 4.32(a), as well as their lengths computed from these projections, are tabulated in Table 4.1.
Zero-Force Members It can be seen from Fig. 4.32(a) that at joint D, three members, AD,CD, and DE, are connected. Of these members, AD and CD lie in the same (xz) plane, whereas DE does not. Since no external loads or reactions are applied at the joint, member DE is a zero-force member.
F_{DE} = 0Having identified DE as a zero-force member, we can see that since the two remaining members AD and CD are not collinear, they must also be zero-force members.
F_{AD} = 0F_{CD} = 0
Reactions See Fig. 4.32(a).
+\swarrow\sum{F_{z} } =0B_z + 60 = 0
B_z = -60 kN
B_z = 60 kN\nearrow
+\circlearrowleft \sum{M_{y} } =0
B_x(2) + 60(4) – 60(2) = 0
B_x = -60 kN
B_x = 60 kN \longleftarrow
| TABLE 4.1 | ||||
| Projection | ||||
| Member | x (m) | y (m) | z (m) | Length (m) |
| AB | 4 | 0 | 0 | 4.0 |
| BC | 0 | 0 | 2 | 2.0 |
| CD | 4 | 0 | 0 | 4.0 |
| AD | 0 | 0 | 2 | 2.0 |
| AC | 4 | 0 | 2 | 4.47 |
| AE | 2 | 4 | 1 | 4.58 |
| BE | 2 | 4 | 1 | 4.58 |
| CE | 2 | 4 | 1 | 4.58 |
| DE | 2 | 4 | 1 | 4.58 |
-60 + C_x =0
C_x = 60 kN\longrightarrow
+\circlearrowleft \sum{M_{x} } =0
-A_y(2) – B_y(2) + 100(1) + 60(4) = 0
A_y + B_y = 170 (1)
+\uparrow \sum{F_y}=0A_y + B_y + C_y – 100 = 0
By substituting Eq. (1) into Eq. (2), we obtain
C_y = -70 kNC_y = 70 kN \downarrow
+\circlearrowleft \sum{M_{z} } =0
B_y(4) – 70(4) – 100(2) = 0
B_y = 120 kN \uparrow
By substituting B_y = 120 kN into Eq. (1), we obtain A_y.
A_y = 50 kN \uparrowJoint A See Fig. 4.32(b).
+\uparrow \sum{F_y}=0 50 + (\frac{y_{AE}}{L_{AE}}) F_{AE}=0
in which the second term on the left-hand side represents the y component of F_{AE}. Substituting the values of y and L for member AE from Table 4.1, we write
50 + (\frac{4}{4.58}) F_{AE}=0F_{AE} = -57.25 kN
F_{AE} = 57.25 kN (C)
Similarly, we apply the remaining equilibrium equations:
+\swarrow \sum{F_z}=0 -(\frac{2}{4.47})F_{AC} +(\frac{1}{4.58})(57.25) = 0
F_{AC} = 28.0 kN (T)
+\longrightarrow \sum{F_x}=0 F_{AB}+(\frac{4}{4.47})(28) – (\frac{2}{4.58})(57.25) = 0
F_{AB} =0
Joint B (See Fig. 4.32(c).)
+\longrightarrow \sum{F_x}=0 -(\frac{2}{4.58})F_{BE} – 60 = 0
F_{BE} = -137.4 kN
F_{BE} = 137.4 kN (C)
+\swarrow \sum{F_z}=0 -60 – F_{BC}+(\frac{1}{4.58})(137.4) = 0
F_{BC} = -30 kN
F_{BC} = 30 kN (C)
As all the unknown forces at joint B have been determined, we will use the remaining equilibrium equation to check our computations:
+\uparrow \sum{F_y}=120 – (\frac{4}{4.58})(137.4) = 0Joint C See Fig. 4.32(d).
+\uparrow \sum{F_y}=0 -70 + (\frac{4}{4.58})F_{CE} = 0
F_{CE} = 80.15 kN (T)
Checking Computations At joint C (Fig. 4.32(d)),
+\longrightarrow \sum{F_x}=60 -(\frac{2}{4.58})(80.15) – (\frac{4}{4.47})(28) =0+\swarrow \sum{F_z}= -30 +(\frac{2}{4.47})(28) – (\frac{1}{4.58})(80.15) =0
At joint E (Fig. 4.32(e)),
+\longrightarrow \sum{F_x}= \frac{2}{4.58}(57.32 – 137.4 + 80.15) = 0+\uparrow \sum{F_y}= -100 + (\frac{4}{4.58})(57.32 + 137.4 – 80.15) = 0
+\swarrow \sum{F_z}= 60 -(\frac{1}{4.58})(57.32 + 137.4 + 80.15) = 0
